Introduction
The invention of the bipolar transistor
in 1948 ushered in a revolution in electronics. Technical feats
previously requiring relatively large, mechanically fragile,
power-hungry vacuum tubes were suddenly achievable with tiny,
mechanically rugged, power-thrifty specks of crystalline silicon. This
revolution made possible the design and manufacture of lightweight,
inexpensive electronic devices that we now take for granted.
Understanding how transistors function is of paramount importance to
anyone interested in understanding modern electronics.
My intent here is to focus as exclusively
as possible on the practical function and application of bipolar
transistors, rather than to explore the quantum world of semiconductor
theory. Discussions of holes and electrons are better left to another
chapter in my opinion. Here I want to explore how to use these
components, not analyze their intimate internal details. I don't mean to
downplay the importance of understanding semiconductor physics, but
sometimes an intense focus on solid-state physics detracts from
understanding these devices' functions on a component level. In taking
this approach, however, I assume that the reader possesses a certain
minimum knowledge of semiconductors: the difference between "P" and "N"
doped semiconductors, the functional characteristics of a PN (diode)
junction, and the meanings of the terms "reverse biased" and "forward
biased." If these concepts are unclear to you, it is best to refer to
earlier chapters in this book before proceeding with this one.
A bipolar transistor consists of a
three-layer "sandwich" of doped (extrinsic) semiconductor materials,
either P-N-P or N-P-N. Each layer forming the transistor has a specific
name, and each layer is provided with a wire contact for connection to a
circuit. Shown here are schematic symbols and physical diagrams of these
two transistor types:
The only functional difference between a
PNP transistor and an NPN transistor is the proper biasing (polarity) of
the junctions when operating. For any given state of operation, the
current directions and voltage polarities for each type of transistor
are exactly opposite each other.
Bipolar transistors work as
current-controlled current regulators. In other words, they
restrict the amount of current that can go through them according to a
smaller, controlling current. The main current that is controlled
goes from collector to emitter, or from emitter to collector, depending
on the type of transistor it is (PNP or NPN, respectively). The small
current that controls the main current goes from base to emitter,
or from emitter to base, once again depending on the type of transistor
it is (PNP or NPN, respectively). According to the confusing standards
of semiconductor symbology, the arrow always points against the
direction of electron flow:
Bipolar transistors are called bipolar
because the main flow of electrons through them takes place in two
types of semiconductor material: P and N, as the main current goes from
emitter to collector (or visa-versa). In other words, two types of
charge carriers -- electrons and holes -- comprise this main current
through the transistor.
As you can see, the controlling
current and the controlled current always mesh together through
the emitter wire, and their electrons always flow against the
direction of the transistor's arrow. This is the first and foremost rule
in the use of transistors: all currents must be going in the proper
directions for the device to work as a current regulator. The small,
controlling current is usually referred to simply as the base current
because it is the only current that goes through the base wire of the
transistor. Conversely, the large, controlled current is referred to as
the collector current because it is the only current that goes
through the collector wire. The emitter current is the sum of the base
and collector currents, in compliance with Kirchhoff's Current Law.
If there is no current through the base
of the transistor, it shuts off like an open switch and prevents current
through the collector. If there is a base current, then the transistor
turns on like a closed switch and allows a proportional amount of
current through the collector. Collector current is primarily limited by
the base current, regardless of the amount of voltage available to push
it. The next section will explore in more detail the use of bipolar
transistors as switching elements.
- REVIEW:
- Bipolar transistors are so named
because the controlled current must go through two types of
semiconductor material: P and N. The current consists of both electron
and hole flow, in different parts of the transistor.
- Bipolar transistors consist of either
a P-N-P or an N-P-N semiconductor "sandwich" structure.
- The three leads of a bipolar
transistor are called the Emitter, Base, and
Collector.
- Transistors function as current
regulators by allowing a small current to control a larger
current. The amount of current allowed between collector and emitter
is primarily determined by the amount of current moving between base
and emitter.
- In order for a transistor to properly
function as a current regulator, the controlling (base) current and
the controlled (collector) currents must be going in the proper
directions: meshing additively at the emitter and going against
the emitter arrow symbol.
The transistor as a switch
Because a transistor's collector current
is proportionally limited by its base current, it can be used as a sort
of current-controlled switch. A relatively small flow of electrons sent
through the base of the transistor has the ability to exert control over
a much larger flow of electrons through the collector.
Suppose we had a lamp that we wanted to
turn on and off by means of a switch. Such a circuit would be extremely
simple:
For the sake of illustration, let's
insert a transistor in place of the switch to show how it can control
the flow of electrons through the lamp. Remember that the controlled
current through a transistor must go between collector and emitter.
Since it's the current through the lamp that we want to control, we must
position the collector and emitter of our transistor where the two
contacts of the switch are now. We must also make sure that the lamp's
current will move against the direction of the emitter arrow
symbol to ensure that the transistor's junction bias will be correct:
In this example I happened to choose an
NPN transistor. A PNP transistor could also have been chosen for the
job, and its application would look like this:
The choice between NPN and PNP is really
arbitrary. All that matters is that the proper current directions are
maintained for the sake of correct junction biasing (electron flow going
against the transistor symbol's arrow).
Going back to the NPN transistor in our
example circuit, we are faced with the need to add something more so
that we can have base current. Without a connection to the base wire of
the transistor, base current will be zero, and the transistor cannot
turn on, resulting in a lamp that is always off. Remember that for an
NPN transistor, base current must consist of electrons flowing from
emitter to base (against the emitter arrow symbol, just like the lamp
current). Perhaps the simplest thing to do would be to connect a switch
between the base and collector wires of the transistor like this:
If the switch is open, the base wire of
the transistor will be left "floating" (not connected to anything) and
there will be no current through it. In this state, the transistor is
said to be cutoff. If the switch is closed, however, electrons
will be able to flow from the emitter through to the base of the
transistor, through the switch and up to the left side of the lamp, back
to the positive side of the battery. This base current will enable a
much larger flow of electrons from the emitter through to the collector,
thus lighting up the lamp. In this state of maximum circuit current, the
transistor is said to be saturated.
Of course, it may seem pointless to use a
transistor in this capacity to control the lamp. After all, we're still
using a switch in the circuit, aren't we? If we're still using a switch
to control the lamp -- if only indirectly -- then what's the point of
having a transistor to control the current? Why not just go back to our
original circuit and use the switch directly to control the lamp
current?
There are a couple of points to be made
here, actually. First is the fact that when used in this manner, the
switch contacts need only handle what little base current is necessary
to turn the transistor on, while the transistor itself handles the
majority of the lamp's current. This may be an important advantage if
the switch has a low current rating: a small switch may be used to
control a relatively high-current load. Perhaps more importantly,
though, is the fact that the current-controlling behavior of the
transistor enables us to use something completely different to turn the
lamp on or off. Consider this example, where a solar cell is used to
control the transistor, which in turn controls the lamp:
Or, we could use a thermocouple to
provide the necessary base current to turn the transistor on:
Even a microphone of sufficient voltage
and current output could be used to turn the transistor on, provided its
output is rectified from AC to DC so that the emitter-base PN junction
within the transistor will always be forward-biased:
The point should be quite apparent by
now: any sufficient source of DC current may be used to turn the
transistor on, and that source of current need only be a fraction of the
amount of current needed to energize the lamp. Here we see the
transistor functioning not only as a switch, but as a true amplifier:
using a relatively low-power signal to control a relatively large
amount of power. Please note that the actual power for lighting up the
lamp comes from the battery to the right of the schematic. It is not as
though the small signal current from the solar cell, thermocouple, or
microphone is being magically transformed into a greater amount of
power. Rather, those small power sources are simply controlling
the battery's power to light up the lamp.
- REVIEW:
- Transistors may be used as switching
elements to control DC power to a load. The switched (controlled)
current goes between emitter and collector, while the controlling
current goes between emitter and base.
- When a transistor has zero current
through it, it is said to be in a state of cutoff (fully
nonconducting).
- When a transistor has maximum current
through it, it is said to be in a state of saturation (fully
conducting).
Meter check of a transistor
Bipolar transistors are constructed of a
three-layer semiconductor "sandwich," either PNP or NPN. As such, they
register as two diodes connected back-to-back when tested with a
multimeter's "resistance" or "diode check" functions:
Here I'm assuming the use of a multimeter
with only a single continuity range (resistance) function to check the
PN junctions. Some multimeters are equipped with two separate continuity
check functions: resistance and "diode check," each with its own
purpose. If your meter has a designated "diode check" function, use that
rather than the "resistance" range, and the meter will display the
actual forward voltage of the PN junction and not just whether or not it
conducts current.
Meter readings will be exactly opposite,
of course, for an NPN transistor, with both PN junctions facing the
other way. If a multimeter with a "diode check" function is used in this
test, it will be found that the emitter-base junction possesses a
slightly greater forward voltage drop than the collector-base junction.
This forward voltage difference is due to the disparity in doping
concentration between the emitter and collector regions of the
transistor: the emitter is a much more heavily doped piece of
semiconductor material than the collector, causing its junction with the
base to produce a higher forward voltage drop.
Knowing this, it becomes possible to
determine which wire is which on an unmarked transistor. This is
important because transistor packaging, unfortunately, is not
standardized. All bipolar transistors have three wires, of course, but
the positions of the three wires on the actual physical package are not
arranged in any universal, standardized order.
Suppose a technician finds a bipolar
transistor and proceeds to measure continuity with a multimeter set in
the "diode check" mode. Measuring between pairs of wires and recording
the values displayed by the meter, the technician obtains the following
data:
- Meter touching wire 1 (+) and 2 (-): "OL"
- Meter touching wire 1 (-) and 2 (+): "OL"
- Meter touching wire 1 (+) and 3 (-):
0.655 volts
- Meter touching wire 1 (-) and 3 (+): "OL"
- Meter touching wire 2 (+) and 3 (-):
0.621 volts
- Meter touching wire 2 (-) and 3 (+): "OL"
The only combinations of test points
giving conducting meter readings are wires 1 and 3 (red test lead on 1
and black test lead on 3), and wires 2 and 3 (red test lead on 2 and
black test lead on 3). These two readings must indicate forward
biasing of the emitter-to-base junction (0.655 volts) and the
collector-to-base junction (0.621 volts).
Now we look for the one wire common to
both sets of conductive readings. It must be the base connection of the
transistor, because the base is the only layer of the three-layer device
common to both sets of PN junctions (emitter-base and collector-base).
In this example, that wire is number 3, being common to both the 1-3 and
the 2-3 test point combinations. In both those sets of meter readings,
the black (-) meter test lead was touching wire 3, which tells us
that the base of this transistor is made of N-type semiconductor
material (black = negative). Thus, the transistor is an PNP type with
base on wire 3, emitter on wire 1 and collector on wire 2:
Please note that the base wire in this
example is not the middle lead of the transistor, as one might
expect from the three-layer "sandwich" model of a bipolar transistor.
This is quite often the case, and tends to confuse new students of
electronics. The only way to be sure which lead is which is by a meter
check, or by referencing the manufacturer's "data sheet" documentation
on that particular part number of transistor.
Knowing that a bipolar transistor behaves
as two back-to-back diodes when tested with a conductivity meter is
helpful for identifying an unknown transistor purely by meter readings.
It is also helpful for a quick functional check of the transistor. If
the technician were to measure continuity in any more than two or any
less than two of the six test lead combinations, he or she would
immediately know that the transistor was defective (or else that it
wasn't a bipolar transistor but rather something else -- a distinct
possibility if no part numbers can be referenced for sure
identification!). However, the "two diode" model of the transistor fails
to explain how or why it acts as an amplifying device.
To better illustrate this paradox, let's
examine one of the transistor switch circuits using the physical diagram
rather than the schematic symbol to represent the transistor. This way
the two PN junctions will be easier to see:
A grey-colored diagonal arrow shows the
direction of electron flow through the emitter-base junction. This part
makes sense, since the electrons are flowing from the N-type emitter to
the P-type base: the junction is obviously forward-biased. However, the
base-collector junction is another matter entirely. Notice how the
grey-colored thick arrow is pointing in the direction of electron flow
(upwards) from base to collector. With the base made of P-type material
and the collector of N-type material, this direction of electron flow is
clearly backwards to the direction normally associated with a PN
junction! A normal PN junction wouldn't permit this "backward" direction
of flow, at least not without offering significant opposition. However,
when the transistor is saturated, there is very little opposition to
electrons all the way from emitter to collector, as evidenced by the
lamp's illumination!
Clearly then, something is going on here
that defies the simple "two-diode" explanatory model of the bipolar
transistor. When I was first learning about transistor operation, I
tried to construct my own transistor from two back-to-back diodes, like
this:
My circuit didn't work, and I was
mystified. However useful the "two diode" description of a transistor
might be for testing purposes, it doesn't explain how a transistor can
behave as a controlled switch.
What happens in a transistor is this: the
reverse bias of the base-collector junction prevents collector current
when the transistor is in cutoff mode (that is, when there is no base
current). However, when the base-emitter junction is forward biased by
the controlling signal, the normally-blocking action of the
base-collector junction is overridden and current is permitted through
the collector, despite the fact that electrons are going the "wrong way"
through that PN junction. This action is dependent on the quantum
physics of semiconductor junctions, and can only take place when the two
junctions are properly spaced and the doping concentrations of the three
layers are properly proportioned. Two diodes wired in series fail to
meet these criteria, and so the top diode can never "turn on" when it is
reversed biased, no matter how much current goes through the bottom
diode in the base wire loop.
That doping concentrations play a crucial
part in the special abilities of the transistor is further evidenced by
the fact that collector and emitter are not interchangeable. If the
transistor is merely viewed as two back-to-back PN junctions, or merely
as a plain N-P-N or P-N-P sandwich of materials, it may seem as though
either end of the transistor could serve as collector or emitter. This,
however, is not true. If connected "backwards" in a circuit, a
base-collector current will fail to control current between collector
and emitter. Despite the fact that both the emitter and collector layers
of a bipolar transistor are of the same doping type (either N or
P), they are definitely not identical!
So, current through the emitter-base
junction allows current through the reverse-biased base-collector
junction. The action of base current can be thought of as "opening a
gate" for current through the collector. More specifically, any given
amount of emitter-to-base current permits a limited amount of
base-to-collector current. For every electron that passes through the
emitter-base junction and on through the base wire, there is allowed a
certain, restricted number of electrons to pass through the
base-collector junction and no more.
In the next section, this
current-limiting behavior of the transistor will be investigated in more
detail.
- REVIEW:
- Tested with a multimeter in the
"resistance" or "diode check" modes, a transistor behaves like two
back-to-back PN (diode) junctions.
- The emitter-base PN junction has a
slightly greater forward voltage drop than the collector-base PN
junction, due to more concentrated doping of the emitter semiconductor
layer.
- The reverse-biased base-collector
junction normally blocks any current from going through the transistor
between emitter and collector. However, that junction begins to
conduct if current is drawn through the base wire. Base current can be
thought of as "opening a gate" for a certain, limited amount of
current through the collector.
Active mode operation
When a transistor is in the fully-off
state (like an open switch), it is said to be cutoff. Conversely,
when it is fully conductive between emitter and collector (passing as
much current through the collector as the collector power supply and
load will allow), it is said to be saturated. These are the two
modes of operation explored thus far in using the transistor as a
switch.
However, bipolar transistors don't have
to be restricted to these two extreme modes of operation. As we learned
in the previous section, base current "opens a gate" for a limited
amount of current through the collector. If this limit for the
controlled current is greater than zero but less than the maximum
allowed by the power supply and load circuit, the transistor will
"throttle" the collector current in a mode somewhere between cutoff and
saturation. This mode of operation is called the active mode.
An automotive analogy for transistor
operation is as follows: cutoff is the condition where there is
no motive force generated by the mechanical parts of the car to make it
move. In cutoff mode, the brake is engaged (zero base current),
preventing motion (collector current). Active mode is when the
automobile is cruising at a constant, controlled speed (constant,
controlled collector current) as dictated by the driver. Saturation
is when the automobile is driving up a steep hill that prevents it from
going as fast as the driver would wish. In other words, a "saturated"
automobile is one where the accelerator pedal is pushed all the way down
(base current calling for more collector current than can be provided by
the power supply/load circuit).
I'll set up a circuit for SPICE
simulation to demonstrate what happens when a transistor is in its
active mode of operation:
"Q" is the standard letter designation
for a transistor in a schematic diagram, just as "R" is for resistor and
"C" is for capacitor. In this circuit, we have an NPN transistor powered
by a battery (V1) and controlled by current through a
current source (I1). A current source is a device that
outputs a specific amount of current, generating as much or as little
voltage as necessary across its terminals to ensure that exact amount of
current through it. Current sources are notoriously difficult to find in
nature (unlike voltage sources, which by contrast attempt to maintain a
constant voltage, outputting as much or as little current in the
fulfillment of that task), but can be simulated with a small collection
of electronic components. As we are about to see, transistors themselves
tend to mimic the constant-current behavior of a current source in their
ability to regulate current at a fixed value.
In the SPICE simulation, I'll set the
current source at a constant value of 20 µA, then vary the voltage
source (V1) over a range of 0 to 2 volts and monitor how much
current goes through it. The "dummy" battery (Vammeter) with
its output of 0 volts serves merely to provide SPICE with a circuit
element for current measurement.
bipolar transistor simulation
i1 0 1 dc 20u
q1 2 1 0 mod1
vammeter 3 2 dc 0
v1 3 0 dc
.model mod1 npn
.dc v1 0 2 0.05
.plot dc i(vammeter)
.end
type npn
is 1.00E-16
bf 100.000
nf 1.000
br 1.000
nr 1.000
v1 i(ammeter) -1.000E-03 0.000E+00 1.000E-03 2.000E-03
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0.000E+00 -1.980E-05 . * . .
5.000E-02 9.188E-05 . .* . .
1.000E-01 6.195E-04 . . * . .
1.500E-01 1.526E-03 . . . * .
2.000E-01 1.914E-03 . . . *.
2.500E-01 1.987E-03 . . . *
3.000E-01 1.998E-03 . . . *
3.500E-01 2.000E-03 . . . *
4.000E-01 2.000E-03 . . . *
4.500E-01 2.000E-03 . . . *
5.000E-01 2.000E-03 . . . *
5.500E-01 2.000E-03 . . . *
6.000E-01 2.000E-03 . . . *
6.500E-01 2.000E-03 . . . *
7.000E-01 2.000E-03 . . . *
7.500E-01 2.000E-03 . . . *
8.000E-01 2.000E-03 . . . *
8.500E-01 2.000E-03 . . . *
9.000E-01 2.000E-03 . . . *
9.500E-01 2.000E-03 . . . *
1.000E+00 2.000E-03 . . . *
1.050E+00 2.000E-03 . . . *
1.100E+00 2.000E-03 . . . *
1.150E+00 2.000E-03 . . . *
1.200E+00 2.000E-03 . . . *
1.250E+00 2.000E-03 . . . *
1.300E+00 2.000E-03 . . . *
1.350E+00 2.000E-03 . . . *
1.400E+00 2.000E-03 . . . *
1.450E+00 2.000E-03 . . . *
1.500E+00 2.000E-03 . . . *
1.550E+00 2.000E-03 . . . *
1.600E+00 2.000E-03 . . . *
1.650E+00 2.000E-03 . . . *
1.700E+00 2.000E-03 . . . *
1.750E+00 2.000E-03 . . . *
1.800E+00 2.000E-03 . . . *
1.850E+00 2.000E-03 . . . *
1.900E+00 2.000E-03 . . . *
1.950E+00 2.000E-03 . . . *
2.000E+00 2.000E-03 . . . *
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
The constant base current of 20 µA sets a
collector current limit of 2 mA, exactly 100 times as much. Notice how
flat the curve is for collector current over the range of battery
voltage from 0 to 2 volts. The only exception to this featureless plot
is at the very beginning, where the battery increases from 0 volts to
0.25 volts. There, the collector current increases rapidly from 0 amps
to its limit of 2 mA.
Let's see what happens if we vary the
battery voltage over a wider range, this time from 0 to 50 volts. We'll
keep the base current steady at 20 µA:
bipolar transistor simulation
i1 0 1 dc 20u
q1 2 1 0 mod1
vammeter 3 2 dc 0
v1 3 0 dc
.model mod1 npn
.dc v1 0 50 2
.plot dc i(vammeter)
.end
type npn
is 1.00E-16
bf 100.000
nf 1.000
br 1.000
nr 1.000
v1 i(ammeter) -1.000E-03 0.000E+00 1.000E-03 2.000E-03
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0.000E+00 -1.980E-05 . * . .
2.000E+00 2.000E-03 . . . *
4.000E+00 2.000E-03 . . . *
6.000E+00 2.000E-03 . . . *
8.000E+00 2.000E-03 . . . *
1.000E+01 2.000E-03 . . . *
1.200E+01 2.000E-03 . . . *
1.400E+01 2.000E-03 . . . *
1.600E+01 2.000E-03 . . . *
1.800E+01 2.000E-03 . . . *
2.000E+01 2.000E-03 . . . *
2.200E+01 2.000E-03 . . . *
2.400E+01 2.000E-03 . . . *
2.600E+01 2.000E-03 . . . *
2.800E+01 2.000E-03 . . . *
3.000E+01 2.000E-03 . . . *
3.200E+01 2.000E-03 . . . *
3.400E+01 2.000E-03 . . . *
3.600E+01 2.000E-03 . . . *
3.800E+01 2.000E-03 . . . *
4.000E+01 2.000E-03 . . . *
4.200E+01 2.000E-03 . . . *
4.400E+01 2.000E-03 . . . *
4.600E+01 2.000E-03 . . . *
4.800E+01 2.000E-03 . . . *
5.000E+01 2.000E-03 . . . *
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Same result! The collector current holds
absolutely steady at 2 mA despite the fact that the battery (v1) voltage
varies all the way from 0 to 50 volts. It would appear from our
simulation that collector-to-emitter voltage has little effect over
collector current, except at very low levels (just above 0 volts). The
transistor is acting as a current regulator, allowing exactly 2 mA
through the collector and no more.
Now let's see what happens if we increase
the controlling (I1) current from 20 µA to 75 µA, once again
sweeping the battery (V1) voltage from 0 to 50 volts and
graphing the collector current:
bipolar transistor simulation
i1 0 1 dc 75u
q1 2 1 0 mod1
vammeter 3 2 dc 0
v1 3 0 dc
.model mod1 npn
.dc v1 0 50 2
.plot dc i(vammeter)
.end
type npn
is 1.00E-16
bf 100.000
nf 1.000
br 1.000
nr 1.000
v1 i(ammeter) -5.000E-03 0.000E+00 5.000E-03 1.000E-02
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0.000E+00 -7.426E-05 . * . .
2.000E+00 7.500E-03 . . . * .
4.000E+00 7.500E-03 . . . * .
6.000E+00 7.500E-03 . . . * .
8.000E+00 7.500E-03 . . . * .
1.000E+01 7.500E-03 . . . * .
1.200E+01 7.500E-03 . . . * .
1.400E+01 7.500E-03 . . . * .
1.600E+01 7.500E-03 . . . * .
1.800E+01 7.500E-03 . . . * .
2.000E+01 7.500E-03 . . . * .
2.200E+01 7.500E-03 . . . * .
2.400E+01 7.500E-03 . . . * .
2.600E+01 7.500E-03 . . . * .
2.800E+01 7.500E-03 . . . * .
3.000E+01 7.500E-03 . . . * .
3.200E+01 7.500E-03 . . . * .
3.400E+01 7.500E-03 . . . * .
3.600E+01 7.500E-03 . . . * .
3.800E+01 7.500E-03 . . . * .
4.000E+01 7.500E-03 . . . * .
4.200E+01 7.500E-03 . . . * .
4.400E+01 7.500E-03 . . . * .
4.600E+01 7.500E-03 . . . * .
4.800E+01 7.500E-03 . . . * .
5.000E+01 7.500E-03 . . . * .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Not surprisingly, SPICE gives us a
similar plot: a flat line, holding steady this time at 7.5 mA -- exactly
100 times the base current -- over the range of battery voltages from
just above 0 volts to 50 volts. It appears that the base current is the
deciding factor for collector current, the V1 battery voltage
being irrelevant so long as it's above a certain minimum level.
This voltage/current relationship is
entirely different from what we're used to seeing across a resistor.
With a resistor, current increases linearly as the voltage across it
increases. Here, with a transistor, current from emitter to collector
stays limited at a fixed, maximum value no matter how high the voltage
across emitter and collector increases.
Often it is useful to superimpose several
collector current/voltage graphs for different base currents on the same
graph. A collection of curves like this -- one curve plotted for each
distinct level of base current -- for a particular transistor is called
the transistor's characteristic curves:
Each curve on the graph reflects the
collector current of the transistor, plotted over a range of
collector-to-emitter voltages, for a given amount of base current. Since
a transistor tends to act as a current regulator, limiting collector
current to a proportion set by the base current, it is useful to express
this proportion as a standard transistor performance measure.
Specifically, the ratio of collector current to base current is known as
the Beta ratio (symbolized by the Greek letter β):
Sometimes the β ratio is designated as "hfe,"
a label used in a branch of mathematical semiconductor analysis known as
"hybrid parameters" which strives to achieve very precise predictions of
transistor performance with detailed equations. Hybrid parameter
variables are many, but they are all labeled with the general letter "h"
and a specific subscript. The variable "hfe" is just another
(standardized) way of expressing the ratio of collector current to base
current, and is interchangeable with "β." Like all ratios, β is unitless.
β for any transistor is determined by its
design: it cannot be altered after manufacture. However, there are so
many physical variables impacting β that it is rare to have two
transistors of the same design exactly match. If a circuit design relies
on equal β ratios between multiple transistors, "matched sets" of
transistors may be purchased at extra cost. However, it is generally
considered bad design practice to engineer circuits with such
dependencies.
It would be nice if the β of a transistor
remained stable for all operating conditions, but this is not true in
real life. For an actual transistor, the β ratio may vary by a factor of
over 3 within its operating current limits. For example, a transistor
with advertised β of 50 may actually test with Ic/Ib
ratios as low as 30 and as high as 100, depending on the amount of
collector current, the transistor's temperature, and frequency of
amplified signal, among other factors. For tutorial purposes it is
adequate to assume a constant β for any given transistor (which is what
SPICE tends to do in a simulation), but just realize that real life is
not that simple!
Sometimes it is helpful for comprehension
to "model" complex electronic components with a collection of simpler,
better-understood components. The following is a popular model shown in
many introductory electronics texts:
This model casts the transistor as a
combination of diode and rheostat (variable resistor). Current through
the base-emitter diode controls the resistance of the collector-emitter
rheostat (as implied by the dashed line connecting the two components),
thus controlling collector current. An NPN transistor is modeled in the
figure shown, but a PNP transistor would be only slightly different
(only the base-emitter diode would be reversed). This model succeeds in
illustrating the basic concept of transistor amplification: how the base
current signal can exert control over the collector current. However, I
personally don't like this model because it tends to miscommunicate the
notion of a set amount of collector-emitter resistance for a given
amount of base current. If this were true, the transistor wouldn't
regulate collector current at all like the characteristic curves
show. Instead of the collector current curves flattening out after their
brief rise as the collector-emitter voltage increases, the collector
current would be directly proportional to collector-emitter voltage,
rising steadily in a straight line on the graph.
A better transistor model, often seen in
more advanced textbooks, is this:
It casts the transistor as a combination
of diode and current source, the output of the current source being set
at a multiple (β ratio) of the base current. This model is far more
accurate in depicting the true input/output characteristics of a
transistor: base current establishes a certain amount of collector
current, rather than a certain amount of collector-emitter
resistance as the first model implies. Also, this model is favored
when performing network analysis on transistor circuits, the current
source being a well-understood theoretical component. Unfortunately,
using a current source to model the transistor's current-controlling
behavior can be misleading: in no way will the transistor ever act as a
source of electrical energy, which the current source symbol
implies is a possibility.
My own personal suggestion for a
transistor model substitutes a constant-current diode for the current
source:
Since no diode ever acts as a source
of electrical energy, this analogy escapes the false implication of the
current source model as a source of power, while depicting the
transistor's constant-current behavior better than the rheostat model.
Another way to describe the constant-current diode's action would be to
refer to it as a current regulator, so this transistor
illustration of mine might also be described as a diode-current
regulator model. The greatest disadvantage I see to this model is
the relative obscurity of constant-current diodes. Many people may be
unfamiliar with their symbology or even of their existence, unlike
either rheostats or current sources, which are commonly known.
- REVIEW:
- A transistor is said to be in its
active mode if it is operating somewhere between fully on
(saturated) and fully off (cutoff).
- Base current tends to regulate
collector current. By regulate, we mean that no more collector
current may exist than what is allowed by the base current.
- The ratio between collector current
and base current is called "Beta" (β) or "hfe".
- β ratios are different for every
transistor, and they tend to change for different operating
conditions.
The common-emitter amplifier
At the beginning of this chapter we saw
how transistors could be used as switches, operating in either their
"saturation" or "cutoff" modes. In the last section we saw how
transistors behave within their "active" modes, between the far limits
of saturation and cutoff. Because transistors are able to control
current in an analog (infinitely divisible) fashion, they find use as
amplifiers for analog signals.
One of the simpler transistor amplifier
circuits to study is the one used previously for illustrating the
transistor's switching ability:
It is called the common-emitter
configuration because (ignoring the power supply battery) both the
signal source and the load share the emitter lead as a common connection
point. This is not the only way in which a transistor may be used as an
amplifier, as we will see in later sections of this chapter:
Before, this circuit was shown to
illustrate how a relatively small current from a solar cell could be
used to saturate a transistor, resulting in the illumination of a lamp.
Knowing now that transistors are able to "throttle" their collector
currents according to the amount of base current supplied by an input
signal source, we should be able to see that the brightness of the lamp
in this circuit is controllable by the solar cell's light exposure. When
there is just a little light shone on the solar cell, the lamp will glow
dimly. The lamp's brightness will steadily increase as more light falls
on the solar cell.
Suppose that we were interested in using
the solar cell as a light intensity instrument. We want to be able to
measure the intensity of incident light with the solar cell by using its
output current to drive a meter movement. It is possible to directly
connect a meter movement to a solar cell for this purpose. In fact, the
simplest light-exposure meters for photography work are designed like
this:
While this approach might work for
moderate light intensity measurements, it would not work as well for low
light intensity measurements. Because the solar cell has to supply the
meter movement's power needs, the system is necessarily limited in its
sensitivity. Supposing that our need here is to measure very low-level
light intensities, we are pressed to find another solution.
Perhaps the most direct solution to this
measurement problem is to use a transistor to amplify the solar
cell's current so that more meter movement needle deflection may be
obtained for less incident light. Consider this approach:
Current through the meter movement in
this circuit will be β times the solar cell current. With a transistor β
of 100, this represents a substantial increase in measurement
sensitivity. It is prudent to point out that the additional power to
move the meter needle comes from the battery on the far right of the
circuit, not the solar cell itself. All the solar cell's current does is
control battery current to the meter to provide a greater meter
reading than the solar cell could provide unaided.
Because the transistor is a
current-regulating device, and because meter movement indications are
based on the amount of current through their movement coils, meter
indication in this circuit should depend only on the amount of current
from the solar cell, not on the amount of voltage provided by the
battery. This means the accuracy of the circuit will be independent of
battery condition, a significant feature! All that is required of the
battery is a certain minimum voltage and current output ability to be
able to drive the meter full-scale if needed.
Another way in which the common-emitter
configuration may be used is to produce an output voltage derived
from the input signal, rather than a specific output current.
Let's replace the meter movement with a plain resistor and measure
voltage between collector and emitter:
With the solar cell darkened (no
current), the transistor will be in cutoff mode and behave as an open
switch between collector and emitter. This will produce maximum voltage
drop between collector and emitter for maximum Voutput, equal
to the full voltage of the battery.
At full power (maximum light exposure),
the solar cell will drive the transistor into saturation mode, making it
behave like a closed switch between collector and emitter. The result
will be minimum voltage drop between collector and emitter, or almost
zero output voltage. In actuality, a saturated transistor can never
achieve zero voltage drop between collector and emitter due to the two
PN junctions through which collector current must travel. However, this
"collector-emitter saturation voltage" will be fairly low, around
several tenths of a volt, depending on the specific transistor used.
For light exposure levels somewhere
between zero and maximum solar cell output, the transistor will be in
its active mode, and the output voltage will be somewhere between zero
and full battery voltage. An important quality to note here about the
common-emitter configuration is that the output voltage is inversely
proportional to the input signal strength. That is, the output
voltage decreases as the input signal increases. For this reason, the
common-emitter amplifier configuration is referred to as an inverting
amplifier.
A quick SPICE simulation will verify our
qualitative conclusions about this amplifier circuit:
common-emitter amplifier
i1 0 1 dc
q1 2 1 0 mod1
r 3 2 5000
v1 3 0 dc 15
.model mod1 npn
.dc i1 0 50u 2u
.plot dc v(2,0)
.end
type npn
is 1.00E-16
bf 100.000
nf 1.000
br 1.000
nr 1.000
i1 v(2) 0.000E+00 5.000E+00 1.000E+01 1.500E+01
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0.000E+00 1.500E+01 . . . *
2.000E-06 1.400E+01 . . . * .
4.000E-06 1.300E+01 . . . * .
6.000E-06 1.200E+01 . . . * .
8.000E-06 1.100E+01 . . . * .
1.000E-05 1.000E+01 . . * .
1.200E-05 9.000E+00 . . * . .
1.400E-05 8.000E+00 . . * . .
1.600E-05 7.000E+00 . . * . .
1.800E-05 6.000E+00 . . * . .
2.000E-05 5.000E+00 . * . .
2.200E-05 4.000E+00 . * . . .
2.400E-05 3.000E+00 . * . . .
2.600E-05 2.000E+00 . * . . .
2.800E-05 1.000E+00 . * . . .
3.000E-05 2.261E-01 .* . . .
3.200E-05 1.850E-01 .* . . .
3.400E-05 1.694E-01 * . . .
3.600E-05 1.597E-01 * . . .
3.800E-05 1.527E-01 * . . .
4.000E-05 1.472E-01 * . . .
4.200E-05 1.427E-01 * . . .
4.400E-05 1.388E-01 * . . .
4.600E-05 1.355E-01 * . . .
4.800E-05 1.325E-01 * . . .
5.000E-05 1.299E-01 * . . .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
At the beginning of the simulation where
the current source (solar cell) is outputting zero current, the
transistor is in cutoff mode and the full 15 volts from the battery is
shown at the amplifier output (between nodes 2 and 0). As the solar
cell's current begins to increase, the output voltage proportionally
decreases, until the transistor reaches saturation at 30 µA of base
current (3 mA of collector current). Notice how the output voltage trace
on the graph is perfectly linear (1 volt steps from 15 volts to 1 volt)
until the point of saturation, where it never quite reaches zero. This
is the effect mentioned earlier, where a saturated transistor can never
achieve exactly zero voltage drop between collector and emitter due to
internal junction effects. What we do see is a sharp output voltage
decrease from 1 volt to 0.2261 volts as the input current increases from
28 µA to 30 µA, and then a continuing decrease in output voltage from
then on (albeit in progressively smaller steps). The lowest the output
voltage ever gets in this simulation is 0.1299 volts, asymptotically
approaching zero.
So far, we've seen the transistor used as
an amplifier for DC signals. In the solar cell light meter example, we
were interested in amplifying the DC output of the solar cell to drive a
DC meter movement, or to produce a DC output voltage. However, this is
not the only way in which a transistor may be employed as an amplifier.
In many cases, what is desired is an AC amplifier for amplifying
alternating current and voltage signals. One common application
of this is in audio electronics (radios, televisions, and public-address
systems). Earlier, we saw an example where the audio output of a tuning
fork could be used to activate a transistor as a switch. Let's see if we
can modify that circuit to send power to a speaker rather than to a
lamp:
In the original circuit, a full-wave
bridge rectifier was used to convert the microphone's AC output signal
into a DC voltage to drive the input of the transistor. All we cared
about here was turning the lamp on with a sound signal from the
microphone, and this arrangement sufficed for that purpose. But now we
want to actually reproduce the AC signal and drive a speaker. This means
we cannot rectify the microphone's output anymore, because we need
undistorted AC signal to drive the transistor! Let's remove the bridge
rectifier and replace the lamp with a speaker:
Since the microphone may produce voltages
exceeding the forward voltage drop of the base-emitter PN (diode)
junction, I've placed a resistor in series with the microphone. Let's
simulate this circuit now in SPICE and see what happens:
common-emitter amplifier
vinput 1 0 sin (0 1.5 2000 0 0)
r1 1 2 1k
q1 3 2 0 mod1
rspkr 3 4 8
v1 4 0 dc 15
.model mod1 npn
.tran 0.02m 0.74m
.plot tran v(1,0) i(v1)
.end
legend:
*: v(1)
+: i(v1)
v(1)
(*)--- -2.000E+00 -1.000E+00 0.000E+00 1.000E+00 2.000E+00
(+)--- -8.000E-02 -6.000E-02 -4.000E-02 -2.000E-02 0.000E+00
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0.000E+00 . . * . +
3.725E-01 . . . * . +
7.195E-01 . . . * . +.
1.024E+00 . . . *+ .
1.264E+00 . . + . * .
1.420E+00 . . + . . * .
1.493E+00 . +. . . * .
1.470E+00 . .+ . . * .
1.351E+00 . . + . . * .
1.154E+00 . . . + . * .
8.791E-01 . . . * . + .
5.498E-01 . . . * . +
1.877E-01 . . . * . +
-1.872E-01 . . * . . +
-5.501E-01 . . * . . +
-8.815E-01 . . * . . +
-1.151E+00 . * . . . +
-1.352E+00 . * . . . +
-1.472E+00 . * . . . +
-1.491E+00 . * . . . +
-1.422E+00 . * . . . +
-1.265E+00 . * . . . +
-1.022E+00 . * . . +
-7.205E-01 . . * . . +
-3.723E-01 . . * . . +
3.040E-06 . . * . +
3.724E-01 . . . * . +
7.205E-01 . . . * . +.
1.022E+00 . . . * + .
1.265E+00 . . + . * .
1.422E+00 . . + . . * .
1.491E+00 . +. . . * .
1.473E+00 . .+ . . * .
1.352E+00 . . + . . * .
1.151E+00 . . . + . * .
8.814E-01 . . . * . + .
5.501E-01 . . . * . +
1.880E-01 . . . * . +
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
The simulation plots both the input
voltage (an AC signal of 1.5 volt peak amplitude and 2000 Hz frequency)
and the current through the 15 volt battery, which is the same as the
current through the speaker. What we see here is a full AC sine wave
alternating in both positive and negative directions, and a half-wave
output current waveform that only pulses in one direction. If we were
actually driving a speaker with this waveform, the sound produced would
be horribly distorted.
What's wrong with the circuit? Why won't
it faithfully reproduce the entire AC waveform from the microphone? The
answer to this question is found by close inspection of the transistor
diode-regulating diode model:
Collector current is controlled, or
regulated, through the constant-current mechanism according to the pace
set by the current through the base-emitter diode. Note that both
current paths through the transistor are monodirectional: one way
only! Despite our intent to use the transistor to amplify an AC
signal, it is essentially a DC device, capable of handling
currents in a single direction only. We may apply an AC voltage input
signal between the base and emitter, but electrons cannot flow in that
circuit during the part of the cycle that reverse-biases the
base-emitter diode junction. Therefore, the transistor will remain in
cutoff mode throughout that portion of the cycle. It will "turn on" in
its active mode only when the input voltage is of the correct polarity
to forward-bias the base-emitter diode, and only when that voltage is
sufficiently high to overcome the diode's forward voltage drop. Remember
that bipolar transistors are current-controlled devices: they
regulate collector current based on the existence of base-to-emitter
current, not base-to-emitter voltage.
The only way we can get the transistor to
reproduce the entire waveform as current through the speaker is to keep
the transistor in its active mode the entire time. This means we must
maintain current through the base during the entire input waveform
cycle. Consequently, the base-emitter diode junction must be kept
forward-biased at all times. Fortunately, this can be accomplished with
the aid of a DC bias voltage added to the input signal. By
connecting a sufficient DC voltage in series with the AC signal source,
forward-bias can be maintained at all points throughout the wave cycle:
common-emitter amplifier
vinput 1 5 sin (0 1.5 2000 0 0)
vbias 5 0 dc 2.3
r1 1 2 1k
q1 3 2 0 mod1
rspkr 3 4 8
v1 4 0 dc 15
.model mod1 npn
.tran 0.02m 0.78m
.plot tran v(1,0) i(v1)
.end
legend:
*: v(1)
+: i(v1)
v(1)
(*)--- 0.000E+00 1.000E+00 2.000E+00 3.000E+00 4.000E+00
(+)--- -3.000E-01 -2.000E-01 -1.000E-01 0.000E+00 1.000E-01
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
2.300E+00 . . + . * . .
2.673E+00 . . + . * . .
3.020E+00 . +. . * .
3.322E+00 . + . . . * .
3.563E+00 . + . . . * .
3.723E+00 . + . . . * .
3.790E+00 . + . . . *.
3.767E+00 . + . . . *.
3.657E+00 . + . . . * .
3.452E+00 . + . . . * .
3.177E+00 . + . . . * .
2.850E+00 . .+ . * . .
2.488E+00 . . + . * . .
2.113E+00 . . + . * . .
1.750E+00 . . * . + . .
1.419E+00 . . * . + . .
1.148E+00 . . * . + . .
9.493E-01 . *. . + . .
8.311E-01 . * . . + .
8.050E-01 . * . . + .
8.797E-01 . * . . +. .
1.039E+00 . .* . + . .
1.275E+00 . . * . + . .
1.579E+00 . . * . + . .
1.929E+00 . . *+ . .
2.300E+00 . . + . * . .
2.673E+00 . . + . * . .
3.019E+00 . +. . * .
3.322E+00 . + . . . * .
3.564E+00 . + . . . * .
3.722E+00 . + . . . * .
3.790E+00 . + . . . *.
3.768E+00 . + . . . *.
3.657E+00 . + . . . * .
3.451E+00 . + . . . * .
3.178E+00 . + . . . * .
2.851E+00 . .+ . * . .
2.488E+00 . . + . * . .
2.113E+00 . . + . * . .
1.748E+00 . . * . + . .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
With the bias voltage source of 2.3 volts
in place, the transistor remains in its active mode throughout the
entire cycle of the wave, faithfully reproducing the waveform at the
speaker. Notice that the input voltage (measured between nodes 1 and 0)
fluctuates between about 0.8 volts and 3.8 volts, a peak-to-peak voltage
of 3 volts just as expected (source voltage = 1.5 volts peak). The
output (speaker) current varies between zero and almost 300 mA, 180o
out of phase with the input (microphone) signal.
The following illustration is another
view of the same circuit, this time with a few oscilloscopes ("scopemeters")
connected at crucial points to display all the pertinent signals:
The need for biasing a transistor
amplifier circuit to obtain full waveform reproduction is an important
consideration. A separate section of this chapter will be devoted
entirely to the subject biasing and biasing techniques. For now, it is
enough to understand that biasing may be necessary for proper voltage
and current output from the amplifier.
Now that we have a functioning amplifier
circuit, we can investigate its voltage, current, and power gains. The
generic transistor used in these SPICE analyses has a β of 100, as
indicated by the short transistor statistics printout included in the
text output (these statistics were cut from the last two analyses for
brevity's sake):
type npn
is 1.00E-16
bf 100.000
nf 1.000
br 1.000
nr 1.000
β is listed under the abbreviation
"bf," which actually stands for "beta, forward". If we
wanted to insert our own β ratio for an analysis, we could have done so
on the .model
line of the SPICE netlist.
Since β is the ratio of collector current
to base current, and we have our load connected in series with the
collector terminal of the transistor and our source connected in series
with the base, the ratio of output current to input current is equal to
beta. Thus, our current gain for this example amplifier is 100, or 40
dB.
Voltage gain is a little more complicated
to figure than current gain for this circuit. As always, voltage gain is
defined as the ratio of output voltage divided by input voltage. In
order to experimentally determine this, we need to modify our last SPICE
analysis to plot output voltage rather than output current so we have
two voltage plots to compare:
common-emitter amplifier
vinput 1 5 sin (0 1.5 2000 0 0)
vbias 5 0 dc 2.3
r1 1 2 1k
q1 3 2 0 mod1
rspkr 3 4 8
v1 4 0 dc 15
.model mod1 npn
.tran 0.02m 0.78m
.plot tran v(1,0) v(4,3)
.end
legend:
*: v(1)
+: v(4,3)
v(1)
(*+)- 0.000E+00 1.000E+00 2.000E+00 3.000E+00 4.000E+00
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
2.300E+00 . . + . * . .
2.673E+00 . . + . * . .
3.020E+00 . . + . * .
3.322E+00 . . +. . * .
3.563E+00 . . . + . * .
3.723E+00 . . . + . * .
3.790E+00 . . . + . * .
3.767E+00 . . . + . * .
3.657E+00 . . . + . * .
3.452E+00 . . + . * .
3.177E+00 . . + . . * .
2.850E+00 . . + . * . .
2.488E+00 . . + . * . .
2.113E+00 . + . * . .
1.750E+00 . + . * . . .
1.419E+00 . + . * . . .
1.148E+00 . + . * . . .
9.493E-01 .+ *. . . .
8.311E-01 + * . . . .
8.050E-01 + * . . . .
8.797E-01 .+ * . . . .
1.039E+00 . + .* . . .
1.275E+00 . + . * . . .
1.579E+00 . + . * . . .
1.929E+00 . + . *. . .
2.300E+00 . . + . * . .
2.673E+00 . . + . * . .
3.019E+00 . . + . * .
3.322E+00 . . +. . * .
3.564E+00 . . . + . * .
3.722E+00 . . . + . * .
3.790E+00 . . . + . * .
3.768E+00 . . . + . * .
3.657E+00 . . . + . * .
3.451E+00 . . + . * .
3.178E+00 . . + . . * .
2.851E+00 . . + . * . .
2.488E+00 . . + . * . .
2.113E+00 . + . * . .
1.748E+00 . + . * . . .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Plotted on the same scale (from 0 to 4
volts), we see that the output waveform ("+") has a smaller peak-to-peak
amplitude than the input waveform ("*"), in addition to being at a lower
bias voltage, not elevated up from 0 volts like the input. Since voltage
gain for an AC amplifier is defined by the ratio of AC amplitudes, we
can ignore any DC bias separating the two waveforms. Even so, the input
waveform is still larger than the output, which tells us that the
voltage gain is less than 1 (a negative dB figure).
To be honest, this low voltage gain is
not characteristic to all common-emitter amplifiers. In this case
it is a consequence of the great disparity between the input and load
resistances. Our input resistance (R1) here is 1000 Ω, while
the load (speaker) is only 8 Ω. Because the current gain of this
amplifier is determined solely by the β of the transistor, and because
that β figure is fixed, the current gain for this amplifier won't change
with variations in either of these resistances. However, voltage gain
is dependent on these resistances. If we alter the load resistance,
making it a larger value, it will drop a proportionately greater voltage
for its range of load currents, resulting in a larger output waveform.
Let's try another simulation, only this time with a 30 Ω load instead of
an 8 Ω load:
common-emitter amplifier
vinput 1 5 sin (0 1.5 2000 0 0)
vbias 5 0 dc 2.3
r1 1 2 1k
q1 3 2 0 mod1
rspkr 3 4 30
v1 4 0 dc 15
.model mod1 npn
.tran 0.02m 0.78m
.plot tran v(1,0) v(4,3)
.end
legend:
*: v(1)
+: v(4,3)
v(1)
(*)-- 0.000E+00 1.000E+00 2.000E+00 3.000E+00 4.000E+00
(+)-- -5.000E+00 0.000E+00 5.000E+00 1.000E+01 1.500E+01
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
2.300E+00 . . + . * . .
2.673E+00 . . .+ * . .
3.020E+00 . . . + * .
3.322E+00 . . . + . * .
3.563E+00 . . . + . * .
3.723E+00 . . . + . * .
3.790E+00 . . . + . * .
3.767E+00 . . . + . * .
3.657E+00 . . . + . * .
3.452E+00 . . . + . * .
3.177E+00 . . . + . * .
2.850E+00 . . . + * . .
2.488E+00 . . +. * . .
2.113E+00 . . + . * . .
1.750E+00 . . + * . . .
1.419E+00 . . +* . . .
1.148E+00 . . x . . .
9.493E-01 . *.+ . . .
8.311E-01 . * + . . .
8.050E-01 . * + . . .
8.797E-01 . * .+ . . .
1.039E+00 . .*+ . . .
1.275E+00 . . +* . . .
1.579E+00 . . + * . . .
1.929E+00 . . + *. . .
2.300E+00 . . + . * . .
2.673E+00 . . .+ * . .
3.019E+00 . . . + * .
3.322E+00 . . . + . * .
3.564E+00 . . . + . * .
3.722E+00 . . . + . * .
3.790E+00 . . . + . * .
3.768E+00 . . . + . * .
3.657E+00 . . . + . * .
3.451E+00 . . . + . * .
3.178E+00 . . . + . * .
2.851E+00 . . . + * . .
2.488E+00 . . +. * . .
2.113E+00 . . + . * . .
1.748E+00 . . + * . . .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
This time the output voltage waveform is
significantly greater in amplitude than the input waveform. This may not
be obvious at first, since the two waveforms are plotted on different
scales: the input on a scale of 0 to 4 volts and the output on a scale
of -5 to 15 volts. Looking closely, we can see that the output waveform
("+") crests between 0 and about 9 volts: approximately 3 times the
amplitude of the input voltage.
We can perform another computer analysis
of this circuit, only this time instructing SPICE to analyze it from an
AC point of view, giving us peak voltage figures for input and output
instead of a time-based plot of the waveforms:
common-emitter amplifier
vinput 1 5 ac 1.5
vbias 5 0 dc 2.3
r1 1 2 1k
q1 3 2 0 mod1
rspkr 3 4 30
v1 4 0 dc 15
.model mod1 npn
.ac lin 1 2000 2000
.print ac v(1,0) v(4,3)
.end
freq v(1) v(4,3)
2.000E+03 1.500E+00 4.418E+00
Peak voltage measurements of input and
output show an input of 1.5 volts and an output of 4.418 volts. This
gives us a voltage gain ratio of 2.9453 (4.418 V / 1.5 V), or 9.3827 dB.
Because the current gain of the
common-emitter amplifier is fixed by β, and since the input and output
voltages will be equal to the input and output currents multiplied by
their respective resistors, we can derive an equation for approximate
voltage gain:
As you can see, the predicted results for
voltage gain are quite close to the simulated results. With perfectly
linear transistor behavior, the two sets of figures would exactly match.
SPICE does a reasonable job of accounting for the many "quirks" of
bipolar transistor function in its analysis, hence the slight mismatch
in voltage gain based on SPICE's output.
These voltage gains remain the same
regardless of where we measure output voltage in the circuit: across
collector and emitter, or across the series load resistor as we did in
the last analysis. The amount of output voltage change for any
given amount of input voltage will remain the same. Consider the two
following SPICE analyses as proof of this. The first simulation is
time-based, to provide a plot of input and output voltages. You will
notice that the two signals are 180o out of phase with each
other. The second simulation is an AC analysis, to provide simple, peak
voltage readings for input and output:
common-emitter amplifier
vinput 1 5 sin (0 1.5 2000 0 0)
vbias 5 0 dc 2.3
r1 1 2 1k
q1 3 2 0 mod1
rspkr 3 4 30
v1 4 0 dc 15
.model mod1 npn
.tran 0.02m 0.74m
.plot tran v(1,0) v(3,0)
.end
legend:
*: v(1)
+: v(3)
v(1)
(*)-- 0.000E+00 1.000E+00 2.000E+00 3.000E+00 4.000E+00
(+)-- 0.000E+00 5.000E+00 1.000E+01 1.500E+01 2.000E+01
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
2.300E+00 . . . + * . .
2.673E+00 . . +. * . .
3.020E+00 . . + . * .
3.324E+00 . . + . . * .
3.564E+00 . . + . . * .
3.720E+00 . . + . . * .
3.793E+00 . . + . . * .
3.770E+00 . . + . . * .
3.651E+00 . . + . . * .
3.454E+00 . . + . . * .
3.179E+00 . . + . . * .
2.850E+00 . . + . * . .
2.488E+00 . . .+ * . .
2.113E+00 . . . * + . .
1.750E+00 . . * . + . .
1.418E+00 . . * . + . .
1.149E+00 . . * . + . .
9.477E-01 . *. . +. .
8.277E-01 . * . . + .
8.091E-01 . * . . + .
8.781E-01 . * . . +. .
1.035E+00 . * . + . .
1.278E+00 . . * . + . .
1.579E+00 . . * . + . .
1.928E+00 . . *. + . .
2.300E+00 . . . + * . .
2.672E+00 . . +. * . .
3.020E+00 . . + . * .
3.322E+00 . . + . . * .
3.565E+00 . . + . . * .
3.722E+00 . . + . . * .
3.791E+00 . . + . . * .
3.773E+00 . . + . . * .
3.652E+00 . . + . . * .
3.451E+00 . . + . . * .
3.181E+00 . . + . . * .
2.850E+00 . . + . * . .
2.488E+00 . . .+ * . .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
common-emitter amplifier
vinput 1 5 ac 1.5
vbias 5 0 dc 2.3
r1 1 2 1k
q1 3 2 0 mod1
rspkr 3 4 30
v1 4 0 dc 15
.model mod1 npn
.ac lin 1 2000 2000
.print ac v(1,0) v(3,0)
.end
freq v(1) v(3)
2.000E+03 1.500E+00 4.418E+00
We still have a peak output voltage of
4.418 volts with a peak input voltage of 1.5 volts. The only difference
from the last set of simulations is the phase of the output
voltage.
So far, the example circuits shown in
this section have all used NPN transistors. PNP transistors are just as
valid to use as NPN in any amplifier configuration, so long as
the proper polarity and current directions are maintained, and the
common-emitter amplifier is no exception. The inverting behavior and
gain properties of a PNP transistor amplifier are the same as its NPN
counterpart, just the polarities are different:
- REVIEW:
- Common-emitter
transistor amplifiers are so-called because the input and output
voltage points share the emitter lead of the transistor in common with
each other, not considering any power supplies.
- Transistors are essentially DC
devices: they cannot directly handle voltages or currents that reverse
direction. In order to make them work for amplifying AC signals, the
input signal must be offset with a DC voltage to keep the transistor
in its active mode throughout the entire cycle of the wave. This is
called biasing.
- If the output voltage is measured
between emitter and collector on a common-emitter amplifier, it will
be 180o out of phase with the input voltage waveform. For
this reason, the common-emitter amplifier is called an inverting
amplifier circuit.
- The current gain of a common-emitter
transistor amplifier with the load connected in series with the
collector is equal to β. The voltage gain of a common-emitter
transistor amplifier is approximately given here:
-
- Where "Rout" is the
resistor connected in series with the collector and "Rin"
is the resistor connected in series with the base.
The common-collector amplifier
Our next transistor configuration to
study is a bit simpler in terms of gain calculations. Called the
common-collector configuration, its schematic diagram looks like
this:
It is called the common-collector
configuration because (ignoring the power supply battery) both the
signal source and the load share the collector lead as a common
connection point:
It should be apparent that the load
resistor in the common-collector amplifier circuit receives both the
base and collector currents, being placed in series with the emitter.
Since the emitter lead of a transistor is the one handling the most
current (the sum of base and collector currents, since base and
collector currents always mesh together to form the emitter current), it
would be reasonable to presume that this amplifier will have a very
large current gain (maximum output current for minimum input current).
This presumption is indeed correct: the current gain for a
common-collector amplifier is quite large, larger than any other
transistor amplifier configuration. However, this is not necessarily
what sets it apart from other amplifier designs.
Let's proceed immediately to a SPICE
analysis of this amplifier circuit, and you will be able to immediately
see what is unique about this amplifier:
common-collector amplifier
vin 1 0
q1 2 1 3 mod1
v1 2 0 dc 15
rload 3 0 5k
.model mod1 npn
.dc vin 0 5 0.2
.plot dc v(3,0)
.end
type npn
is 1.00E-16
bf 100.000
nf 1.000
br 1.000
nr 1.000
vin v(3) 0.000E+00 2.000E+00 4.000E+00 6.000E+00
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0.000E+00 7.500E-08 * . . .
2.000E-01 7.501E-08 * . . .
4.000E-01 2.704E-06 * . . .
6.000E-01 4.954E-03 * . . .
8.000E-01 1.221E-01 .* . . .
1.000E+00 2.989E-01 . * . . .
1.200E+00 4.863E-01 . * . . .
1.400E+00 6.777E-01 . * . . .
1.600E+00 8.712E-01 . * . . .
1.800E+00 1.066E+00 . * . . .
2.000E+00 1.262E+00 . * . . .
2.200E+00 1.458E+00 . * . . .
2.400E+00 1.655E+00 . * . . .
2.600E+00 1.852E+00 . *. . .
2.800E+00 2.049E+00 . * . .
3.000E+00 2.247E+00 . . * . .
3.200E+00 2.445E+00 . . * . .
3.400E+00 2.643E+00 . . * . .
3.600E+00 2.841E+00 . . * . .
3.800E+00 3.039E+00 . . * . .
4.000E+00 3.237E+00 . . * . .
4.200E+00 3.436E+00 . . * . .
4.400E+00 3.634E+00 . . * . .
4.600E+00 3.833E+00 . . *. .
4.800E+00 4.032E+00 . . * .
5.000E+00 4.230E+00 . . . * .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Unlike the common-emitter amplifier from
the previous section, the common-collector produces an output voltage in
direct rather than inverse proportion to the rising input
voltage. As the input voltage increases, so does the output voltage.
More than that, a close examination reveals that the output voltage is
nearly identical to the input voltage, lagging behind only about
0.77 volts.
This is the unique quality of the
common-collector amplifier: an output voltage that is nearly equal to
the input voltage. Examined from the perspective of output voltage
change for a given amount of input voltage change, this
amplifier has a voltage gain of almost exactly unity (1), or 0 dB. This
holds true for transistors of any β value, and for load resistors of any
resistance value.
It is simple to understand why the output
voltage of a common-collector amplifier is always nearly equal to the
input voltage. Referring back to the diode-regulating diode transistor
model, we see that the base current must go through the base-emitter PN
junction, which is equivalent to a normal rectifying diode. So long as
this junction is forward-biased (the transistor conducting current in
either its active or saturated modes), it will have a voltage drop of
approximately 0.7 volts, assuming silicon construction. This 0.7 volt
drop is largely irrespective of the actual magnitude of base current, so
we can regard it as being constant:
Given the voltage polarities across the
base-emitter PN junction and the load resistor, we see that they must
add together to equal the input voltage, in accordance with Kirchhoff's
Voltage Law. In other words, the load voltage will always be about 0.7
volts less than the input voltage for all conditions where the
transistor is conducting. Cutoff occurs at input voltages below 0.7
volts, and saturation at input voltages in excess of battery (supply)
voltage plus 0.7 volts.
Because of this behavior, the
common-collector amplifier circuit is also known as the
voltage-follower or emitter-follower amplifier, in reference
to the fact that the input and load voltages follow each other so
closely.
Applying the common-collector circuit to
the amplification of AC signals requires the same input "biasing" used
in the common-emitter circuit: a DC voltage must be added to the AC
input signal to keep the transistor in its active mode during the entire
cycle. When this is done, the result is a non-inverting amplifier:
common-collector amplifier
vin 1 4 sin(0 1.5 2000 0 0)
vbias 4 0 dc 2.3
q1 2 1 3 mod1
v1 2 0 dc 15
rload 3 0 5k
.model mod1 npn
.tran .02m .78m
.plot tran v(1,0) v(3,0)
.end
legend:
*: v(1)
+: v(3)
v(1) 0.000E+00 1.000E+00 2.000E+00 3.000E+00 4.000E+00
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
2.300E+00 . . + . * . .
2.673E+00 . . +. * . .
3.020E+00 . . . + * .
3.322E+00 . . . + . * .
3.563E+00 . . . + . * .
3.723E+00 . . . +. * .
3.790E+00 . . . + * .
3.767E+00 . . . + * .
3.657E+00 . . . +. * .
3.452E+00 . . . + . * .
3.177E+00 . . . + . * .
2.850E+00 . . .+ * . .
2.488E+00 . . + . * . .
2.113E+00 . . + . * . .
1.750E+00 . + * . . .
1.419E+00 . + . * . . .
1.148E+00 . + . * . . .
9.493E-01 . + *. . . .
8.311E-01 .+ * . . . .
8.050E-01 .+ * . . . .
8.797E-01 . + * . . . .
1.039E+00 . + .* . . .
1.275E+00 . + . * . . .
1.579E+00 . + . * . . .
1.929E+00 . . + *. . .
2.300E+00 . . + . * . .
2.673E+00 . . +. * . .
3.019E+00 . . . + * .
3.322E+00 . . . + . * .
3.564E+00 . . . + . * .
3.722E+00 . . . +. * .
3.790E+00 . . . + * .
3.768E+00 . . . + * .
3.657E+00 . . . +. * .
3.451E+00 . . . + . * .
3.178E+00 . . . + . * .
2.851E+00 . . .+ * . .
2.488E+00 . . + . * . .
2.113E+00 . . + . * . .
1.748E+00 . + * . . .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Here's another view of the circuit, this
time with oscilloscopes connected to several points of interest:
Since this amplifier configuration
doesn't provide any voltage gain (in fact, in practice it actually has a
voltage gain of slightly less than 1), its only amplifying factor
is current. The common-emitter amplifier configuration examined in the
previous section had a current gain equal to the β of the transistor,
being that the input current went through the base and the output (load)
current went through the collector, and β by definition is the ratio
between the collector and emitter currents. In the common-collector
configuration, though, the load is situated in series with the emitter,
and thus its current is equal to the emitter current. With the emitter
carrying collector current and base current, the load in this
type of amplifier has all the current of the collector running through
it plus the input current of the base. This yields a current gain
of β plus 1:
Once again, PNP transistors are just as
valid to use in the common-collector configuration as NPN transistors.
The gain calculations are all the same, as is the non-inverting behavior
of the amplifier. The only difference is in voltage polarities and
current directions:
A popular application of the
common-collector amplifier is for regulated DC power supplies, where an
unregulated (varying) source of DC voltage is clipped at a specified
level to supply regulated (steady) voltage to a load. Of course, zener
diodes already provide this function of voltage regulation:
However, when used in this direct
fashion, the amount of current that may be supplied to the load is
usually quite limited. In essence, this circuit regulates voltage across
the load by keeping current through the series resistor at a high enough
level to drop all the excess power source voltage across it, the zener
diode drawing more or less current as necessary to keep the voltage
across itself steady. For high-current loads, an plain zener diode
voltage regulator would have to be capable of shunting a lot of current
through the diode in order to be effective at regulating load voltage in
the event of large load resistance or voltage source changes.
One popular way to increase the
current-handling ability of a regulator circuit like this is to use a
common-collector transistor to amplify current to the load, so that the
zener diode circuit only has to handle the amount of current necessary
to drive the base of the transistor:
There's really only one caveat to this
approach: the load voltage will be approximately 0.7 volts less than the
zener diode voltage, due to the transistor's 0.7 volt base-emitter drop.
However, since this 0.7 volt difference is fairly constant over a wide
range of load currents, a zener diode with a 0.7 volt higher rating can
be chosen for the application.
Sometimes the high current gain of a
single-transistor, common-collector configuration isn't enough for a
particular application. If this is the case, multiple transistors may be
staged together in a popular configuration known as a Darlington pair,
just an extension of the common-collector concept:
Darlington pairs essentially place one
transistor as the common-collector load for another transistor, thus
multiplying their individual current gains. Base current through the
upper-left transistor is amplified through that transistor's emitter,
which is directly connected to the base of the lower-right transistor,
where the current is again amplified. The overall current gain is as
follows:
Voltage gain is still nearly equal to 1
if the entire assembly is connected to a load in common-collector
fashion, although the load voltage will be a full 1.4 volts less than
the input voltage:
Darlington pairs may be purchased as
discrete units (two transistors in the same package), or may be built up
from a pair of individual transistors. Of course, if even more current
gain is desired than what may be obtained with a pair, Darlington
triplet or quadruplet assemblies may be constructed.
- REVIEW:
- Common-collector
transistor amplifiers are so-called because the input and output
voltage points share the collector lead of the transistor in common
with each other, not considering any power supplies.
- The output voltage on a
common-collector amplifier will be in phase with the input voltage,
making the common-collector a non-inverting amplifier circuit.
- The current gain of a common-collector
amplifier is equal to β plus 1. The voltage gain is approximately
equal to 1 (in practice, just a little bit less).
- A Darlington pair is a pair of
transistors "piggybacked" on one another so that the emitter of one
feeds current to the base of the other in common-collector form. The
result is an overall current gain equal to the product
(multiplication) of their individual common-collector current gains (β
plus 1).
The common-base amplifier
The final transistor amplifier
configuration we need to study is the common-base. This
configuration is more complex than the other two, and is less common due
to its strange operating characteristics.
It is called the common-base
configuration because (DC power source aside), the signal source and the
load share the base of the transistor as a common connection point:
Perhaps the most striking characteristic
of this configuration is that the input signal source must carry the
full emitter current of the transistor, as indicated by the heavy arrows
in the first illustration. As we know, the emitter current is greater
than any other current in the transistor, being the sum of base and
collector currents. In the last two amplifier configurations, the signal
source was connected to the base lead of the transistor, thus handling
the least current possible.
Because the input current exceeds all
other currents in the circuit, including the output current, the current
gain of this amplifier is actually less than 1 (notice how Rload
is connected to the collector, thus carrying slightly less current than
the signal source). In other words, it attenuates current rather
than amplifying it. With common-emitter and common-collector
amplifier configurations, the transistor parameter most closely
associated with gain was β. In the common-base circuit, we follow
another basic transistor parameter: the ratio between collector current
and emitter current, which is a fraction always less than 1. This
fractional value for any transistor is called the alpha ratio, or
α ratio.
Since it obviously can't boost signal
current, it only seems reasonable to expect it to boost signal voltage.
A SPICE simulation will vindicate that assumption:
common-base amplifier
vin 0 1
r1 1 2 100
q1 4 0 2 mod1
v1 3 0 dc 15
rload 3 4 5k
.model mod1 npn
.dc vin 0.6 1.2 .02
.plot dc v(3,4)
.end
v(3,4) 0.000E+00 5.000E+00 1.000E+01 1.500E+01 2.000E+01
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
5.913E-03 * . . . .
1.274E-02 * . . . .
2.730E-02 * . . . .
5.776E-02 * . . . .
1.193E-01 * . . . .
2.358E-01 .* . . . .
4.370E-01 .* . . . .
7.447E-01 . * . . . .
1.163E+00 . * . . . .
1.682E+00 . * . . . .
2.281E+00 . * . . . .
2.945E+00 . * . . . .
3.657E+00 . * . . . .
4.408E+00 . * . . . .
5.189E+00 . .* . . .
5.995E+00 . . * . . .
6.820E+00 . . * . . .
7.661E+00 . . * . . .
8.516E+00 . . * . . .
9.382E+00 . . * . . .
1.026E+01 . . .* . .
1.114E+01 . . . * . .
1.203E+01 . . . * . .
1.293E+01 . . . * . .
1.384E+01 . . . * . .
1.474E+01 . . . *. .
1.563E+01 . . . . * .
1.573E+01 . . . . * .
1.575E+01 . . . . * .
1.576E+01 . . . . * .
1.576E+01 . . . . * .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Notice how in this simulation the output
voltage goes from practically nothing (cutoff) to 15.75 volts
(saturation) with the input voltage being swept over a range of 0.6
volts to 1.2 volts. In fact, the output voltage plot doesn't show a rise
until about 0.7 volts at the input, and cuts off (flattens) at about
1.12 volts input. This represents a rather large voltage gain with an
output voltage span of 15.75 volts and an input voltage span of only
0.42 volts: a gain ratio of 37.5, or 31.48 dB. Notice also how the
output voltage (measured across Rload) actually exceeds the
power supply (15 volts) at saturation, due to the series-aiding effect
of the the input voltage source.
A second set of SPICE analyses with an AC
signal source (and DC bias voltage) tells the same story: a high voltage
gain.
common-base amplifier
vin 0 1 sin (0 0.12 2000 0 0)
vbias 1 5 dc 0.95
r1 5 2 100
q1 4 0 2 mod1
v1 3 0 dc 15
rload 3 4 5k
.model mod1 npn
.tran 0.02m 0.78m
.plot tran v(1,0) v(4,3)
.end
legend:
*: v(1)
+: v(4,3)
v(1)
(*)-- -2.000E-01 -1.000E-01 0.000E+00 1.000E-01 2.000E-01
(+)-- -1.500E+01 -1.000E+01 -5.000E+00 0.000E+00 5.000E+00
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0.000E+00 . . + * . .
-2.984E-02 . . + * . . .
-5.757E-02 . + . * . . .
-8.176E-02 . + . * . . .
-1.011E-01 . + * . . .
-1.139E-01 . + * . . . .
-1.192E-01 . + * . . . .
-1.174E-01 . + * . . . .
-1.085E-01 . + *. . . .
-9.213E-02 . + .* . . .
-7.020E-02 . + . * . . .
-4.404E-02 . + * . . .
-1.502E-02 . . + * . . .
1.496E-02 . . + . * . .
4.400E-02 . . + . * . .
7.048E-02 . . + * . .
9.214E-02 . . . + *. .
1.081E-01 . . . + .* .
1.175E-01 . . . + . * .
1.196E-01 . . . + . * .
1.136E-01 . . . + . * .
1.009E-01 . . . + * .
8.203E-02 . . .+ * . .
5.764E-02 . . + . * . .
2.970E-02 . . + . * . .
-1.440E-05 . . + * . .
-2.981E-02 . . + * . . .
-5.755E-02 . + . * . . .
-8.178E-02 . + . * . . .
-1.011E-01 . + * . . .
-1.138E-01 . + * . . . .
-1.192E-01 . + * . . . .
-1.174E-01 . + * . . . .
-1.085E-01 . + *. . . .
-9.209E-02 . + .* . . .
-7.020E-02 . + . * . . .
-4.407E-02 . + * . . .
-1.502E-02 . . + * . . .
1.496E-02 . . + . * . .
4.417E-02 . . + . * . .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
As you can see, the input and output
waveforms are in phase with each other. This tells us that the
common-base amplifier is non-inverting.
common-base amplifier
vin 0 1 ac 0.12
vbias 1 5 dc 0.95
r1 5 2 100
q1 4 0 2 mod1
v1 3 0 dc 15
rload 3 4 5k
.model mod1 npn
.ac lin 1 2000 2000
.print ac v(1,0) v(3,4)
.end
freq v(1) v(3,4)
2.000E+03 1.200E-01 5.129E+00
Voltage figures from the second analysis
(AC mode) show a voltage gain of 42.742 (5.129 V / 0.12 V), or 32.617
dB:
Here's another view of the circuit,
showing the phase relations and DC offsets of various signals in the
circuit just simulated:
. . . and for a PNP transistor:
Predicting voltage gain for the
common-base amplifier configuration is quite difficult, and involves
approximations of transistor behavior that are difficult to measure
directly. Unlike the other amplifier configurations, where voltage gain
was either set by the ratio of two resistors (common-emitter), or fixed
at an unchangeable value (common-collector), the voltage gain of the
common-base amplifier depends largely on the amount of DC bias on the
input signal. As it turns out, the internal transistor resistance
between emitter and base plays a major role in determining voltage gain,
and this resistance changes with different levels of current through the
emitter.
While this phenomenon is difficult to
explain, it is rather easy to demonstrate through the use of computer
simulation. What I'm going to do here is run several SPICE simulations
on a common-base amplifier circuit, changing the DC bias voltage
slightly while keeping the AC signal amplitude and all other circuit
parameters constant. As the voltage gain changes from one simulation to
another, different output voltage amplitudes will be noticed as a
result.
Although these analyses will all be
conducted in the AC mode, they were first "proofed" in the transient
analysis mode (voltage plotted over time) to ensure that the entire wave
was being faithfully reproduced and not "clipped" due to improper
biasing. No meaningful calculations of gain can be based on waveforms
that are distorted:
common-base amplifier DC bias = 0.85 volts
vin 0 1 ac 0.08
vbias 1 5 dc 0.85
r1 5 2 100
q1 4 0 2 mod1
v1 3 0 dc 15
rload 3 4 5k
.model mod1 npn
.ac lin 1 2000 2000
.print ac v(1,0) v(3,4)
.end
freq v(1) v(3,4)
2.000E+03 8.000E-02 3.005E+00
common-base amplifier dc bias = 0.9 volts
vin 0 1 ac 0.08
vbias 1 5 dc 0.90
r1 5 2 100
q1 4 0 2 mod1
v1 3 0 dc 15
rload 3 4 5k
.model mod1 npn
.ac lin 1 2000 2000
.print ac v(1,0) v(3,4)
.end
freq v(1) v(3,4)
2.000E+03 8.000E-02 3.264E+00
common-base amplifier dc bias = 0.95 volts
vin 0 1 ac 0.08
vbias 1 5 dc 0.95
r1 5 2 100
q1 4 0 2 mod1
v1 3 0 dc 15
rload 3 4 5k
.model mod1 npn
.ac lin 1 2000 2000
.print ac v(1,0) v(3,4)
.end
freq v(1) v(3,4)
2.000E+03 8.000E-02 3.419E+00
A trend should be evident here: with
increases in DC bias voltage, voltage gain increases as well. We can see
that the voltage gain is increasing because each subsequent simulation
produces greater output voltage for the exact same input signal voltage
(0.08 volts). As you can see, the changes are quite large, and they are
caused by miniscule variations in bias voltage!
The combination of very low current gain
(always less than 1) and somewhat unpredictable voltage gain conspire
against the common-base design, relegating it to few practical
applications.
- REVIEW:
- Common-base
transistor amplifiers are so-called because the input and output
voltage points share the base lead of the transistor in common with
each other, not considering any power supplies.
- The current gain of a common-base
amplifier is always less than 1. The voltage gain is a function of
input and output resistances, and also the internal resistance of the
emitter-base junction, which is subject to change with variations in
DC bias voltage. Suffice to say that the voltage gain of a common-base
amplifier can be very high.
- The ratio of a transistor's collector
current to emitter current is called α. The α value for any transistor
is always less than unity, or in other words, less than 1.
Biasing techniques
In the common-emitter section of this
chapter, we saw a SPICE analysis where the output waveform resembled a
half-wave rectified shape: only half of the input waveform was
reproduced, with the other half being completely cut off. Since our
purpose at that time was to reproduce the entire waveshape, this
constituted a problem. The solution to this problem was to add a small
bias voltage to the amplifier input so that the transistor stayed in
active mode throughout the entire wave cycle. This addition was called a
bias voltage.
There are applications, though, where a
half-wave output is not problematic. In fact, some applications may
necessitate this very type of amplification. Because it is possible
to operate an amplifier in modes other than full-wave reproduction, and
because there are specific applications requiring different ranges of
reproduction, it is useful to describe the degree to which an amplifier
reproduces the input waveform by designating it according to class.
Amplifier class operation is categorized by means of alphabetical
letters: A, B, C, and AB.
Class A
operation is where the entire input waveform is faithfully reproduced.
Although I didn't introduce this concept back in the common-emitter
section, this is what we were hoping to attain in our simulations. Class
A operation can only be obtained when the transistor spends its entire
time in the active mode, never reaching either cutoff or saturation. To
achieve this, sufficient DC bias voltage is usually set at the level
necessary to drive the transistor exactly halfway between cutoff and
saturation. This way, the AC input signal will be perfectly "centered"
between the amplifier's high and low signal limit levels.
Class B
operation is what we had the first time an AC signal was applied to the
common-emitter amplifier with no DC bias voltage. The transistor spent
half its time in active mode and the other half in cutoff with the input
voltage too low (or even of the wrong polarity!) to forward-bias its
base-emitter junction.
By itself, an amplifier operating in
class B mode is not very useful. In most circumstances, the severe
distortion introduced into the waveshape by eliminating half of it would
be unacceptable. However, class B operation is a useful mode of biasing
if two amplifiers are operated as a push-pull pair, each
amplifier handling only half of the waveform at a time:
Transistor Q1 "pushes" (drives
the output voltage in a positive direction with respect to ground),
while transistor Q2 "pulls" the output voltage (in a negative
direction, toward 0 volts with respect to ground). Individually, each of
these transistors is operating in class B mode, active only for one-half
of the input waveform cycle. Together, however, they function as a team
to produce an output waveform identical in shape to the input waveform.
A decided advantage of the class B
(push-pull) amplifier design over the class A design is greater output
power capability. With a class A design, the transistor dissipates a lot
of energy in the form of heat because it never stops conducting current.
At all points in the wave cycle it is in the active (conducting) mode,
conducting substantial current and dropping substantial voltage. This
means there is substantial power dissipated by the transistor throughout
the cycle. In a class B design, each transistor spends half the time in
cutoff mode, where it dissipates zero power (zero current = zero power
dissipation). This gives each transistor a time to "rest" and cool while
the other transistor carries the burden of the load. Class A amplifiers
are simpler in design, but tend to be limited to low-power signal
applications for the simple reason of transistor heat dissipation.
There is another class of amplifier
operation known as class AB, which is somewhere between class A
and class B: the transistor spends more than 50% but less than 100% of
the time conducting current.
If the input signal bias for an amplifier
is slightly negative (opposite of the bias polarity for class A
operation), the output waveform will be further "clipped" than it was
with class B biasing, resulting in an operation where the transistor
spends the majority of the time in cutoff mode:
At first, this scheme may seem utterly
pointless. After all, how useful could an amplifier be if it clips the
waveform as badly as this? If the output is used directly with no
conditioning of any kind, it would indeed be of questionable utility.
However, with the application of a tank circuit (parallel resonant
inductor-capacitor combination) to the output, the occasional output
surge produced by the amplifier can set in motion a higher-frequency
oscillation maintained by the tank circuit. This may be likened to a
machine where a heavy flywheel is given an occasional "kick" to keep it
spinning:
Called class C operation, this
scheme also enjoys high power efficiency due to the fact that the
transistor(s) spend the vast majority of time in the cutoff mode, where
they dissipate zero power. The rate of output waveform decay (decreasing
oscillation amplitude between "kicks" from the amplifier) is exaggerated
here for the benefit of illustration. Because of the tuned tank circuit
on the output, this type of circuit is usable only for amplifying
signals of definite, fixed frequency.
Another type of amplifier operation,
significantly different from Class A, B, AB, or C, is called Class D.
It is not obtained by applying a specific measure of bias voltage as are
the other classes of operation, but requires a radical re-design of the
amplifier circuit itself. It's a little too early in this chapter to
investigate exactly how a class D amplifier is built, but not too early
to discuss its basic principle of operation.
A class D amplifier reproduces the
profile of the input voltage waveform by generating a rapidly-pulsing
squarewave output. The duty cycle of this output waveform (time "on"
versus total cycle time) varies with the instantaneous amplitude of the
input signal. The following plots demonstrate this principle:
The greater the instantaneous voltage of
the input signal, the greater the duty cycle of the output squarewave
pulse. If there can be any goal stated of the class D design, it is to
avoid active-mode transistor operation. Since the output transistor of a
class D amplifier is never in the active mode, only cutoff or saturated,
there will be little heat energy dissipated by it. This results in very
high power efficiency for the amplifier. Of course, the disadvantage of
this strategy is the overwhelming presence of harmonics on the output.
Fortunately, since these harmonic frequencies are typically much greater
than the frequency of the input signal, they can be filtered out by a
low-pass filter with relative ease, resulting in an output more closely
resembling the original input signal waveform. Class D technology is
typically seen where extremely high power levels and relatively low
frequencies are encountered, such as in industrial inverters (devices
converting DC into AC power to run motors and other large devices) and
high-performance audio amplifiers.
A term you will likely come across in
your studies of electronics is something called quiescent, which
is a modifier designating the normal, or zero input signal, condition of
a circuit. Quiescent current, for example, is the amount of current in a
circuit with zero input signal voltage applied. Bias voltage in a
transistor circuit forces the transistor to operate at a different level
of collector current with zero input signal voltage than it would
without that bias voltage. Therefore, the amount of bias in an amplifier
circuit determines its quiescent values.
In a class A amplifier, the quiescent
current should be exactly half of its saturation value (halfway between
saturation and cutoff, cutoff by definition being zero). Class B and
class C amplifiers have quiescent current values of zero, since they are
supposed to be cutoff with no signal applied. Class AB amplifiers have
very low quiescent current values, just above cutoff. To illustrate this
graphically, a "load line" is sometimes plotted over a transistor's
characteristic curves to illustrate its range of operation while
connected to a load resistance of specific value:
A load line is a plot of
collector-to-emitter voltage over a range of base currents. At the
lower-right corner of the load line, voltage is at maximum and current
is at zero, representing a condition of cutoff. At the upper-left corner
of the line, voltage is at zero while current is at a maximum,
representing a condition of saturation. Dots marking where the load line
intersects the various transistor curves represent realistic operating
conditions for those base currents given.
Quiescent operating conditions may be
shown on this type of graph in the form of a single dot along the load
line. For a class A amplifier, the quiescent point will be in the middle
of the load line, like this:
In this illustration, the quiescent point
happens to fall on the curve representing a base current of 40 µA. If we
were to change the load resistance in this circuit to a greater value,
it would affect the slope of the load line, since a greater load
resistance would limit the maximum collector current at saturation, but
would not change the collector-emitter voltage at cutoff. Graphically,
the result is a load line with a different upper-left point and the same
lower-right point:
Note how the new load line doesn't
intercept the 75 µA curve along its flat portion as before. This is very
important to realize because the non-horizontal portion of a
characteristic curve represents a condition of saturation. Having the
load line intercept the 75 µA curve outside of the curve's horizontal
range means that the amplifier will be saturated at that amount of base
current. Increasing the load resistor value is what caused the load line
to intercept the 75 µA curve at this new point, and it indicates that
saturation will occur at a lesser value of base current than before.
With the old, lower-value load resistor
in the circuit, a base current of 75 µA would yield a proportional
collector current (base current multiplied by β). In the first load line
graph, a base current of 75 µA gave a collector current almost twice
what was obtained at 40 µA, as the β ratio would predict. Now, however,
there is only a marginal increase in collector current between base
current values of 75 µA and 40 µA, because the transistor begins to lose
sufficient collector-emitter voltage to continue to regulate collector
current.
In order to maintain linear
(no-distortion) operation, transistor amplifiers shouldn't be operated
at points where the transistor will saturate; that is, in any case where
the load line will not potentially fall on the horizontal portion of a
collector current curve. In this case, we'd have to add a few more
curves to the graph before we could tell just how far we could "push"
this transistor with increased base currents before it saturates.
It appears in this graph that the
highest-current point on the load line falling on the straight portion
of a curve is the point on the 50 µA curve. This new point should be
considered the maximum allowable input signal level for class A
operation. Also for class A operation, the bias should be set so that
the quiescent point is halfway between this new maximum point and
cutoff:
Now that we know a little more about the
consequences of different DC bias voltage levels, it is time to
investigate practical biasing techniques. So far, I've shown a small DC
voltage source (battery) connected in series with the AC input signal to
bias the amplifier for whatever desired class of operation. In real
life, the connection of a precisely-calibrated battery to the input of
an amplifier is simply not practical. Even if it were possible to
customize a battery to produce just the right amount of voltage for any
given bias requirement, that battery would not remain at its
manufactured voltage indefinitely. Once it started to discharge and its
output voltage drooped, the amplifier would begin to drift in the
direction of class B operation.
Take this circuit, illustrated in the
common-emitter section for a SPICE simulation, for instance:
That 2.3 volt "Vbias" battery
would not be practical to include in a real amplifier circuit. A far
more practical method of obtaining bias voltage for this amplifier would
be to develop the necessary 2.3 volts using a voltage divider network
connected across the 15 volt battery. After all, the 15 volt battery is
already there by necessity, and voltage divider circuits are very easy
to design and build. Let's see how this might look:
If we choose a pair of resistor values
for R2 and R3 that will produce 2.3 volts across R3
from a total of 15 volts (such as 8466 Ω for R2 and 1533 Ω
for R3), we should have our desired value of 2.3 volts
between base and emitter for biasing with no signal input. The only
problem is, this circuit configuration places the AC input signal source
directly in parallel with R3 of our voltage divider. This is
not acceptable, as the AC source will tend to overpower any DC voltage
dropped across R3. Parallel components must have the
same voltage, so if an AC voltage source is directly connected across
one resistor of a DC voltage divider, the AC source will "win" and there
will be no DC bias voltage added to the signal.
One way to make this scheme work,
although it may not be obvious why it will work, is to place a
coupling capacitor between the AC voltage source and the voltage
divider like this:
The capacitor forms a high-pass filter
between the AC source and the DC voltage divider, passing almost all of
the AC signal voltage on to the transistor while blocking all DC voltage
from being shorted through the AC signal source. This makes much more
sense if you understand the superposition theorem and how it works.
According to superposition, any linear, bilateral circuit can be
analyzed in a piecemeal fashion by only considering one power source at
a time, then algebraically adding the effects of all power sources to
find the final result. If we were to separate the capacitor and R2--R3
voltage divider circuit from the rest of the amplifier, it might be
easier to understand how this superposition of AC and DC would work.
With only the AC signal source in effect,
and a capacitor with an arbitrarily low impedance at signal frequency,
almost all the AC voltage appears across R3:
With only the DC source in effect, the
capacitor appears to be an open circuit, and thus neither it nor the
shorted AC signal source will have any effect on the operation of the R2--R3
voltage divider:
Combining these two separate analyses, we
get a superposition of (almost) 1.5 volts AC and 2.3 volts DC, ready to
be connected to the base of the transistor:
Enough talk -- it's about time for a
SPICE simulation of the whole amplifier circuit. I'll use a capacitor
value of 100 µF to obtain an arbitrarily low (0.796 Ω) impedance at 2000
Hz:
voltage divider biasing
vinput 1 0 sin (0 1.5 2000 0 0)
c1 1 5 100u
r1 5 2 1k
r2 4 5 8466
r3 5 0 1533
q1 3 2 0 mod1
rspkr 3 4 8
v1 4 0 dc 15
.model mod1 npn
.tran 0.02m 0.78m
.plot tran v(1,0) i(v1)
.end
legend:
*: v(1)
+: i(v1)
v(1)
(*)-- -2.000E+00 -1.000E+00 0.000E+00 1.000E+00 2.000E+00
(+)-- -3.000E-01 -2.000E-01 -1.000E-01 0.000E+00 1.000E-01
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0.000E+00 . . * + . .
3.730E-01 . . + * . .
7.197E-01 . . + . * . .
1.022E+00 . . + . * .
1.263E+00 . . + . . * .
1.423E+00 . + . . * .
1.490E+00 . +. . . * .
1.467E+00 . +. . . * .
1.357E+00 . .+ . . * .
1.152E+00 . . + . . * .
8.774E-01 . . + . * . .
5.505E-01 . . + . * . .
1.878E-01 . . . x . .
-1.870E-01 . . * . + . .
-5.500E-01 . . * . + . .
-8.810E-01 . . * . + .
-1.152E+00 . * . . + .
-1.351E+00 . * . . + .
-1.469E+00 . * . . + .
-1.495E+00 . * . . + .
-1.420E+00 . * . . + .
-1.261E+00 . * . . + .
-1.025E+00 . * . + .
-7.205E-01 . . * . +. .
-3.713E-01 . . * . + . .
1.800E-04 . . * + . .
3.726E-01 . . + * . .
7.194E-01 . . + . * . .
1.022E+00 . . + . * .
1.264E+00 . . + . . * .
1.422E+00 . + . . * .
1.490E+00 . +. . . * .
1.468E+00 . +. . . * .
1.357E+00 . .+ . . * .
1.151E+00 . . + . . * .
8.775E-01 . . + . * . .
5.509E-01 . . + . * . .
1.877E-01 . . . x . .
-1.871E-01 . . * . + . .
-5.522E-01 . . * . + . .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Notice that there is substantial
distortion in the output waveform here: the sine wave is being clipped
during most of the input signal's negative half-cycle. This tells us the
transistor is entering into cutoff mode when it shouldn't (I'm assuming
a goal of class A operation as before). Why is this? This new biasing
technique should give us exactly the same amount of DC bias voltage as
before, right?
With the capacitor and R2--R3
resistor network unloaded, it will provide exactly 2.3 volts worth of DC
bias. However, once we connect this network to the transistor, it is no
longer loaded. Current drawn through the base of the transistor will
load the voltage divider, thus reducing the DC bias voltage available
for the transistor. Using the diode-regulating diode transistor model to
illustrate, the bias problem becomes evident:
A voltage divider's output depends not
only on the size of its constituent resistors, but also on how much
current is being divided away from it through a load. In this case, the
base-emitter PN junction of the transistor is a load that decreases the
DC voltage dropped across R3, due to the fact that the bias
current joins with R3's current to go through R2,
upsetting the divider ratio formerly set by the resistance values of R2
and R3. In order to obtain a DC bias voltage of 2.3 volts,
the values of R2 and/or R3 must be adjusted to
compensate for the effect of base current loading. In this case, we want
to increase the DC voltage dropped across R3, so we
can lower the value of R2, raise the value of R3,
or both.
voltage divider biasing
vinput 1 0 sin (0 1.5 2000 0 0)
c1 1 5 100u
r1 5 2 1k
r2 4 5 6k <--- R2 decreased to 6 k ohms
r3 5 0 4k <--- R3 increased to 4 k ohms
q1 3 2 0 mod1
rspkr 3 4 8
v1 4 0 dc 15
.model mod1 npn
.tran 0.02m 0.78m
.plot tran v(1,0) i(v1)
.end
legend:
*: v(1)
+: i(v1)
v(1)
(*)-- -2.000E+00 -1.000E+00 0.000E+00 1.000E+00 2.000E+00
(+)-- -3.000E-01 -2.000E-01 -1.000E-01 0.000E+00 1.000E-01
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0.000E+00 . . + * . .
3.730E-01 . . + . * . .
7.197E-01 . + . . * . .
1.022E+00 . + . . * .
1.263E+00 . + . . . * .
1.423E+00 .+ . . . * .
1.490E+00 + . . . * .
1.467E+00 + . . . * .
1.357E+00 . + . . . * .
1.152E+00 . + . . . * .
8.774E-01 . + . . * . .
5.505E-01 . +. . * . .
1.878E-01 . . + . * . .
-1.870E-01 . . + * . . .
-5.500E-01 . . * + . .
-8.810E-01 . . * . + . .
-1.152E+00 . * . . + . .
-1.351E+00 . * . . + . .
-1.469E+00 . * . . + . .
-1.495E+00 . * . . +. .
-1.420E+00 . * . . + . .
-1.261E+00 . * . . + . .
-1.025E+00 . * . + . .
-7.205E-01 . . * . + . .
-3.713E-01 . . * + . . .
1.800E-04 . . + * . .
3.726E-01 . . + . * . .
7.194E-01 . + . . * . .
1.022E+00 . + . . * .
1.264E+00 . + . . . * .
1.422E+00 .+ . . . * .
1.490E+00 + . . . * .
1.468E+00 + . . . * .
1.357E+00 . + . . . * .
1.151E+00 . + . . . * .
8.775E-01 . + . . * . .
5.509E-01 . +. . * . .
1.877E-01 . . + . * . .
-1.871E-01 . . + * . . .
-5.522E-01 . . * + . .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
As you can see, the new resistor values
of 6 kΩ and 4 kΩ (R2 and R3, respectively) results
in class A waveform reproduction, just the way we wanted.
- REVIEW:
- Class A
operation is where an amplifier is biased so as to be in the active
mode throughout the entire waveform cycle, thus faithfully reproducing
the whole waveform.
- Class B
operation is where an amplifier is biased so that only half of the
input waveform gets reproduced: either the positive half or the
negative half. The transistor spends half its time in the active mode
and half its time cutoff. Complementary pairs of transistors running
in class B operation are often used to deliver high power
amplification in audio signal systems, each transistor of the pair
handling a separate half of the waveform cycle. Class B operation
delivers better power efficiency than a class A amplifier of similar
output power.
- Class AB
operation is where an amplifier is biased at a point somewhere between
class A and class B.
- Class C
operation is where an amplifier's bias forces it to amplify only a
small portion of the waveform. A majority of the transistor's time is
spent in cutoff mode. In order for there to be a complete waveform at
the output, a resonant tank circuit is often used as a "flywheel" to
maintain oscillations for a few cycles after each "kick" from the
amplifier. Because the transistor is not conducting most of the time,
power efficiencies are very high for a class C amplifier.
- Class D
operation requires an advanced circuit design, and functions on the
principle of representing instantaneous input signal amplitude by the
duty cycle of a high-frequency squarewave. The output transistor(s)
never operate in active mode, only cutoff and saturation. Thus, there
is very little heat energy dissipated and energy efficiency is high.
- DC bias voltage on the input signal,
necessary for certain classes of operation (especially class A and
class C), may be obtained through the use of a voltage divider and
coupling capacitor rather than a battery connected in series with
the AC signal source.
Input and output coupling
To overcome the challenge of creating
necessary DC bias voltage for an amplifier's input signal without
resorting to the insertion of a battery in series with the AC signal
source, we used a voltage divider connected across the DC power source.
To make this work in conjunction with an AC input signal, we "coupled"
the signal source to the divider through a capacitor, which acted as a
high-pass filter. With that filtering in place, the low impedance of the
AC signal source couldn't "short out" the DC voltage dropped across the
bottom resistor of the voltage divider. A simple solution, but not
without any disadvantages.
Most obvious is the fact that using a
high-pass filter capacitor to couple the signal source to the amplifier
means that the amplifier can only amplify AC signals. A steady, DC
voltage applied to the input would be blocked by the coupling capacitor
just as much as the voltage divider bias voltage is blocked from the
input source. Furthermore, since capacitive reactance is
frequency-dependent, lower-frequency AC signals will not be amplified as
much as higher-frequency signals. Non-sinusoidal signals will tend to be
distorted, as the capacitor responds differently to each of the signal's
constituent harmonics. An extreme example of this would be a
low-frequency square-wave signal:
Incidentally, this same problem occurs
when oscilloscope inputs are set to the "AC coupling" mode. In this
mode, a coupling capacitor is inserted in series with the measured
voltage signal to eliminate any vertical offset of the displayed
waveform due to DC voltage combined with the signal. This works fine
when the AC component of the measured signal is of a fairly high
frequency, and the capacitor offers little impedance to the signal.
However, if the signal is of a low frequency, and/or contains
considerable levels of harmonics over a wide frequency range, the
oscilloscope's display of the waveform will not be accurate.
In applications where the limitations of
capacitive coupling would be intolerable, another solution may be used:
direct coupling. Direct coupling avoids the use of capacitors or
any other frequency-dependent coupling component in favor of resistors.
A direct-coupled amplifier circuit might look something like this:
With no capacitor to filter the input
signal, this form of coupling exhibits no frequency dependence. DC and
AC signals alike will be amplified by the transistor with the same gain
(the transistor itself may tend to amplify some frequencies better than
others, but that is another subject entirely!).
If direct coupling works for DC as well
as for AC signals, then why use capacitive coupling for any
application? One reason might be to avoid any unwanted DC bias
voltage naturally present in the signal to be amplified. Some AC signals
may be superimposed on an uncontrolled DC voltage right from the source,
and an uncontrolled DC voltage would make reliable transistor biasing
impossible. The high-pass filtering offered by a coupling capacitor
would work well here to avoid biasing problems.
Another reason to use capacitive coupling
rather than direct is its relative lack of signal attenuation. Direct
coupling through a resistor has the disadvantage of diminishing, or
attenuating, the input signal so that only a fraction of it reaches the
base of the transistor. In many applications, some attenuation is
necessary anyway to prevent normal signal levels from "overdriving" the
transistor into cutoff and saturation, so any attenuation inherent to
the coupling network is useful anyway. However, some applications
require that there be no signal loss from the input connection to
the transistor's base for maximum voltage gain, and a direct coupling
scheme with a voltage divider for bias simply won't suffice.
So far, we've discussed a couple of
methods for coupling an input signal to an amplifier, but haven't
addressed the issue of coupling an amplifier's output to a load.
The example circuit used to illustrate input coupling will serve well to
illustrate the issues involved with output coupling.
In our example circuit, the load is a
speaker. Most speakers are electromagnetic in design: that is, they use
the force generated by an lightweight electromagnet coil suspended
within a strong permanent-magnet field to move a thin paper or plastic
cone, producing vibrations in the air which our ears interpret as sound.
An applied voltage of one polarity moves the cone outward, while a
voltage of the opposite polarity will move the cone inward. To exploit
cone's full freedom of motion, the speaker must receive true (unbiased)
AC voltage. DC bias applied to the speaker coil tends to offset the cone
from its natural center position, and this tends to limit the amount of
back-and-forth motion it can sustain from the applied AC voltage without
overtraveling. However, our example circuit applies a varying voltage of
only one polarity across the speaker, because the speaker is
connected in series with the transistor which can only conduct current
one way. This situation would be unacceptable in the case of any
high-power audio amplifier.
Somehow we need to isolate the speaker
from the DC bias of the collector current so that it only receives AC
voltage. One way to achieve this goal is to couple the transistor
collector circuit to the speaker through a transformer:
Voltage induced in the secondary
(speaker-side) of the transformer will be strictly due to variations
in collector current, because the mutual inductance of a transformer
only works on changes in winding current. In other words, only
the AC portion of the collector current signal will be coupled to the
secondary side for powering the speaker. The speaker will "see" true
alternating current at its terminals, without any DC bias.
Transformer output coupling works, and
has the added benefit of being able to provide impedance matching
between the transistor circuit and the speaker coil with custom winding
ratios. However, transformers tend to be large and heavy, especially for
high-power applications. Also, it is difficult to engineer a transformer
to handle signals over a wide range of frequencies, which is almost
always required for audio applications. To make matters worse, DC
current through the primary winding adds to the magnetization of the
core in one polarity only, which tends to make the transformer core
saturate more easily in one AC polarity cycle than the other. This
problem is reminiscent of having the speaker directly connected in
series with the transistor: a DC bias current tends to limit how much
output signal amplitude the system can handle without distortion.
Generally, though, a transformer can be designed to handle a lot more DC
bias current than a speaker without running into trouble, so transformer
coupling is still a viable solution in most cases.
Another method to isolate the speaker
from DC bias in the output signal is to alter the circuit a bit and use
a coupling capacitor in a manner similar to coupling the input signal to
the amplifier:
This circuit resembles the more
conventional form of common-emitter amplifier, with the transistor
collector connected to the battery through a resistor. The capacitor
acts as a high-pass filter, passing most of the AC voltage to the
speaker while blocking all DC voltage. Again, the value of this coupling
capacitor is chosen so that its impedance at the expected signal
frequency will be arbitrarily low.
The blocking of DC voltage from an
amplifier's output, be it via a transformer or a capacitor, is useful
not only in coupling an amplifier to a load, but also in coupling one
amplifier to another amplifier. "Staged" amplifiers are often used to
achieve higher power gains than what would be possible using a single
transistor:
While it is possible to directly couple
each stage to the next (via a resistor rather than a capacitor), this
makes the whole amplifier very sensitive to variations in the DC
bias voltage of the first stage, since that DC voltage will be amplified
along with the AC signal until the last stage. In other words, the
biasing of the first stage will affect the biasing of the second stage,
and so on. However, if the stages are capacitively coupled as shown in
the above illustration, the biasing of one stage has no effect on the
biasing of the next, because DC voltage is blocked from passing on to
the next stage.
Transformer coupling between amplifier
stages is also a possibility, but less often seen due to some of the
problems inherent to transformers mentioned previously. One notable
exception to this rule is in the case of radio-frequency amplifiers
where coupling transformers are typically small, have air cores (making
them immune to saturation effects), and can be made part of a resonant
circuit so as to block unwanted harmonic frequencies from passing on to
subsequent stages. The use of resonant circuits assumes that the signal
frequency remains constant, of course, but this is typically the case in
radio circuitry. Also, the "flywheel" effect of LC tank circuits allows
for class C operation for high efficiency:
Having said all this, it must be
mentioned that it is possible to use direct coupling within a
multi-stage transistor amplifier circuit. In cases where the amplifier
is expected to handle DC signals, this is the only alternative.
- REVIEW:
- Capacitive coupling acts like a
high-pass filter on the input of an amplifier. This tends to make the
amplifier's voltage gain decrease at lower signal frequencies.
Capacitive-coupled amplifiers are all but unresponsive to DC input
signals.
- Direct coupling with a series resistor
instead of a series capacitor avoids the problem of
frequency-dependent gain, but has the disadvantage of reducing
amplifier gain for all signal frequencies by attenuating the input
signal.
- Transformers and capacitors may be
used to couple the output of an amplifier to a load, to eliminate DC
voltage from getting to the load.
- Multi-stage amplifiers often make use
of capacitive coupling between stages to eliminate problems with the
bias from one stage affecting the bias of another.
Feedback
If some percentage of an amplifier's
output signal is connected to the input, so that the amplifier amplifies
part of its own output signal, we have what is known as feedback.
Feedback comes in two varieties: positive (also called
regenerative), and negative (also called degenerative).
Positive feedback reinforces the direction of an amplifier's output
voltage change, while negative feedback does just the opposite.
A familiar example of feedback happens in
public-address ("PA") systems where someone holds the microphone too
close to a speaker: a high-pitched "whine" or "howl" ensues, because the
audio amplifier system is detecting and amplifying its own noise.
Specifically, this is an example of positive or regenerative
feedback, as any sound detected by the microphone is amplified and
turned into a louder sound by the speaker, which is then detected by the
microphone again, and so on . . . the result being a noise of steadily
increasing volume until the system becomes "saturated" and cannot
produce any more volume.
One might wonder what possible benefit
feedback is to an amplifier circuit, given such an annoying example as
PA system "howl." If we introduce positive, or regenerative, feedback
into an amplifier circuit, it has the tendency of creating and
sustaining oscillations, the frequency of which determined by the values
of components handling the feedback signal from output to input. This is
one way to make an oscillator circuit to produce AC from a DC
power supply. Oscillators are very useful circuits, and so feedback has
a definite, practical application for us.
Negative feedback, on the other hand, has
a "dampening" effect on an amplifier: if the output signal happens to
increase in magnitude, the feedback signal introduces a decreasing
influence into the input of the amplifier, thus opposing the change in
output signal. While positive feedback drives an amplifier circuit
toward a point of instability (oscillations), negative feedback drives
it the opposite direction: toward a point of stability.
An amplifier circuit equipped with some
amount of negative feedback is not only more stable, but it tends to
distort the input waveform to a lesser degree and is generally capable
of amplifying a wider range of frequencies. The tradeoff for these
advantages (there just has to be a disadvantage to negative
feedback, right?) is decreased gain. If a portion of an amplifier's
output signal is "fed back" to the input in such a way as to oppose any
changes in the output, it will require a greater input signal amplitude
to drive the amplifier's output to the same amplitude as before. This
constitutes a decreased gain. However, the advantages of stability,
lower distortion, and greater bandwidth are worth the tradeoff in
reduced gain for many applications.
Let's examine a simple amplifier circuit
and see how we might introduce negative feedback into it:
The amplifier configuration shown here is
a common-emitter, with a resistor bias network formed by R1
and R2. The capacitor couples Vinput to the
amplifier so that the signal source doesn't have a DC voltage imposed on
it by the R1/R2 divider network. Resistor R3
serves the purpose of controlling voltage gain. We could omit if for
maximum voltage gain, but since base resistors like this are common in
common-emitter amplifier circuits, we'll keep it in this schematic.
Like all common-emitter amplifiers, this
one inverts the input signal as it is amplified. In other words,
a positive-going input voltage causes the output voltage to decrease, or
go in the direction of negative, and visa-versa. If we were to examine
the waveforms with oscilloscopes, it would look something like this:
Because the output is an inverted, or
mirror-image, reproduction of the input signal, any connection between
the output (collector) wire and the input (base) wire of the transistor
will result in negative feedback:
The resistances of R1, R2,
R3, and Rfeedback function together as a
signal-mixing network so that the voltage seen at the base of the
transistor (in reference to ground) is a weighted average of the input
voltage and the feedback voltage, resulting in signal of reduced
amplitude going into the transistor. As a result, the amplifier circuit
will have reduced voltage gain, but improved linearity (reduced
distortion) and increased bandwidth.
A resistor connecting collector to base
is not the only way to introduce negative feedback into this amplifier
circuit, though. Another method, although more difficult to understand
at first, involves the placement of a resistor between the transistor's
emitter terminal and circuit ground, like this:
This new feedback resistor drops voltage
proportional to the emitter current through the transistor, and it does
so in such a way as to oppose the input signal's influence on the
base-emitter junction of the transistor. Let's take a closer look at the
emitter-base junction and see what difference this new resistor makes:
With no feedback resistor connecting the
emitter to ground, whatever level of input signal (Vinput)
makes it through the coupling capacitor and R1/R2/R3
resistor network will be impressed directly across the base-emitter
junction as the transistor's input voltage (VB-E). In other
words, with no feedback resistor, VB-E equals Vinput.
Therefore, if Vinput increases by 100 mV, then VB-E
likewise increases by 100 mV: a change in one is the same as a change in
the other, since the two voltages are equal to each other.
Now let's consider the effects of
inserting a resistor (Rfeedback) between the transistor's
emitter lead and ground:
Note how the voltage dropped across Rfeedback
adds with VB-E to equal Vinput. With Rfeedback
in the Vinput -- VB-E loop, VB-E will
no longer be equal to Vinput. We know that Rfeedback
will drop a voltage proportional to emitter current, which is in turn
controlled by the base current, which is in turn controlled by the
voltage dropped across the base-emitter junction of the transistor (VB-E).
Thus, if Vinput were to increase in a positive direction, it
would increase VB-E, causing more base current, causing more
collector (load) current, causing more emitter current, and causing more
feedback voltage to be dropped across Rfeedback. This
increase of voltage drop across the feedback resistor, though,
subtracts from Vinput to reduce the VB-E, so
that the actual voltage increase for VB-E will be less than
the voltage increase of Vinput. No longer will a 100 mV
increase in Vinput result in a full 100 mV increase for VB-E,
because the two voltages are not equal to each other.
Consequently, the input voltage has less
control over the transistor than before, and the voltage gain for the
amplifier is reduced: just what we expected from negative feedback.
In practical common-emitter circuits,
negative feedback isn't just a luxury; it's a necessity for stable
operation. In a perfect world, we could build and operate a
common-emitter transistor amplifier with no negative feedback, and have
the full amplitude of Vinput impressed across the
transistor's base-emitter junction. This would give us a large voltage
gain. Unfortunately, though, the relationship between base-emitter
voltage and base-emitter current changes with temperature, as predicted
by the "diode equation." As the transistor heats up, there will be less
of a forward voltage drop across the base-emitter junction for any given
current. This causes a problem for us, as the R1/R2
voltage divider network is designed to provide the correct quiescent
current through the base of the transistor so that it will operate in
whatever class of operation we desire (in this example, I've shown the
amplifier working in class-A mode). If the transistor's voltage/current
relationship changes with temperature, the amount of DC bias voltage
necessary for the desired class of operation will change. In this case,
a hot transistor will draw more bias current for the same amount of bias
voltage, making it heat up even more, drawing even more bias current.
The result, if unchecked, is called thermal runaway.
Common-collector amplifiers, however, do
not suffer from thermal runaway. Why is this? The answer has everything
to do with negative feedback:
Note that the common-collector amplifier
has its load resistor placed in exactly the same spot as we had the Rfeedback
resistor in the last circuit: between emitter and ground. This means
that the only voltage impressed across the transistor's base-emitter
junction is the difference between Vinput and Voutput,
resulting in a very low voltage gain (usually close to 1 for a
common-collector amplifier). Thermal runaway is impossible for this
amplifier: if base current happens to increase due to transistor
heating, emitter current will likewise increase, dropping more voltage
across the load, which in turn subtracts from Vinput
to reduce the amount of voltage dropped between base and emitter. In
other words, the negative feedback afforded by placement of the load
resistor makes the problem of thermal runaway self-correcting. In
exchange for a greatly reduced voltage gain, we get superb stability and
immunity from thermal runaway.
By adding a "feedback" resistor between
emitter and ground in a common-emitter amplifier, we make the amplifier
behave a little less like an "ideal" common-emitter and a little more
like a common-collector. The feedback resistor value is typically quite
a bit less than the load, minimizing the amount of negative feedback and
keeping the voltage gain fairly high.
Another benefit of negative feedback,
seen clearly in the common-collector circuit, is that it tends to make
the voltage gain of the amplifier less dependent on the characteristics
of the transistor. Note that in a common-collector amplifier, voltage
gain is nearly equal to unity (1), regardless of the transistor's β.
This means, among other things, that we could replace the transistor in
a common-collector amplifier with one having a different β and not see
any significant changes in voltage gain. In a common-emitter circuit,
the voltage gain is highly dependent on β. If we were to replace the
transistor in a common-emitter circuit with another of differing β, the
voltage gain for the amplifier would change significantly. In a
common-emitter amplifier equipped with negative feedback, the voltage
gain will still be dependent upon transistor β to some degree, but not
as much as before, making the circuit more predictable despite
variations in transistor β.
The fact that we have to introduce
negative feedback into a common-emitter amplifier to avoid thermal
runaway is an unsatisfying solution. It would be nice, after all, to
avoid thermal runaway without having to suppress the amplifier's
inherently high voltage gain. A best-of-both-worlds solution to this
dilemma is available to us if we closely examine the nature of the
problem: the voltage gain that we have to minimize in order to avoid
thermal runaway is the DC voltage gain, not the AC voltage
gain. After all, it isn't the AC input signal that fuels thermal
runaway: it's the DC bias voltage required for a certain class of
operation: that quiescent DC signal that we use to "trick" the
transistor (fundamentally a DC device) into amplifying an AC signal. We
can suppress DC voltage gain in a common-emitter amplifier circuit
without suppressing AC voltage gain if we figure out a way to make the
negative feedback function with DC only. That is, if we only feed back
an inverted DC signal from output to input, but not an inverted AC
signal.
The Rfeedback emitter resistor
provides negative feedback by dropping a voltage proportional to load
current. In other words, negative feedback is accomplished by inserting
an impedance into the emitter current path. If we want to feed back DC
but not AC, we need an impedance that is high for DC but low for AC.
What kind of circuit presents a high impedance to DC but a low impedance
to AC? A high-pass filter, of course!
By connecting a capacitor in parallel
with the feedback resistor, we create the very situation we need: a path
from emitter to ground that is easier for AC than it is for DC:
The new capacitor "bypasses" AC from the
transistor's emitter to ground, so that no appreciable AC voltage will
be dropped from emitter to ground to "feed back" to the input and
suppress voltage gain. Direct current, on the other hand, cannot go
through the bypass capacitor, and so must travel through the feedback
resistor, dropping a DC voltage between emitter and ground which lowers
the DC voltage gain and stabilizes the amplifier's DC response,
preventing thermal runaway. Because we want the reactance of this
capacitor (XC) to be as low as possible, Cbypass
should be sized relatively large. Because the polarity across this
capacitor will never change, it is safe to use a polarized
(electrolytic) capacitor for the task.
Another approach to the problem of
negative feedback reducing voltage gain is to use multi-stage amplifiers
rather than single-transistor amplifiers. If the attenuated gain of a
single transistor is insufficient for the task at hand, we can use more
than one transistor to make up for the reduction caused by feedback.
Here is an example circuit showing negative feedback in a three-stage
common-emitter amplifier:
Note how there is but one "path" for
feedback, from the final output to the input through a single resistor,
Rfeedback. Since each stage is a common-emitter amplifier --
and thus inverting in nature -- and there are an odd number of stages
from input to output, the output signal will be inverted with respect to
the input signal, and the feedback will be negative (degenerative).
Relatively large amounts of feedback may be used without sacrificing
voltage gain, because the three amplifier stages provide so much gain to
begin with.
At first, this design philosophy may seem
inelegant and perhaps even counter-productive. Isn't this a rather crude
way to overcome the loss in gain incurred through the use of negative
feedback, to simply recover gain by adding stage after stage? What is
the point of creating a huge voltage gain using three transistor stages
if we're just going to attenuate all that gain anyway with negative
feedback? The point, though perhaps not apparent at first, is increased
predictability and stability from the circuit as a whole. If the three
transistor stages are designed to provide an arbitrarily high voltage
gain (in the tens of thousands, or greater) with no feedback, it will be
found that the addition of negative feedback causes the overall voltage
gain to become less dependent of the individual stage gains, and
approximately equal to the simple ratio Rfeedback/Rin.
The more voltage gain the circuit has (without feedback), the more
closely the voltage gain will approximate Rfeedback/Rin
once feedback is established. In other words, voltage gain in this
circuit is fixed by the values of two resistors, and nothing more.
This advantage has profound impact on
mass-production of electronic circuitry: if amplifiers of predictable
gain may be constructed using transistors of widely varied β values, it
makes the selection and replacement of components very easy and
inexpensive. It also means the amplifier's gain varies little with
changes in temperature. This principle of stable gain control through a
high-gain amplifier "tamed" by negative feedback is elevated almost to
an art form in electronic circuits called operational amplifiers,
or op-amps. You may read much more about these circuits in a
later chapter of this book!
- REVIEW:
- Feedback
is the coupling of an amplifier's output to its input.
- Positive,
or regenerative feedback has the tendency of making an
amplifier circuit unstable, so that it produces oscillations (AC). The
frequency of these oscillations is largely determined by the
components in the feedback network.
- Negative,
or degenerative feedback has the tendency of making an
amplifier circuit more stable, so that its output changes less
for a given input signal than without feedback. This reduces the gain
of the amplifier, but has the advantage of decreasing distortion and
increasing bandwidth (the range of frequencies the amplifier can
handle).
- Negative feedback may be introduced
into a common-emitter circuit by coupling collector to base, or by
inserting a resistor between emitter and ground.
- An emitter-to-ground "feedback"
resistor is usually found in common-emitter circuits as a preventative
measure against thermal runaway.
- Negative feedback also has the
advantage of making amplifier voltage gain more dependent on resistor
values and less dependent on the transistor's characteristics.
- Common-collector amplifiers have a lot
of negative feedback, due to the placement of the load resistor
between emitter and ground. This feedback accounts for the extremely
stable voltage gain of the amplifier, as well as its immunity against
thermal runaway.
- Voltage gain for a common-emitter
circuit may be re-established without sacrificing immunity to thermal
runaway, by connecting a bypass capacitor in parallel with the
emitter "feedback resistor."
- If the voltage gain of an amplifier is
arbitrarily high (tens of thousands, or greater), and negative
feedback is used to reduce the gain to reasonable levels, it will be
found that the gain will approximately equal Rfeedback/Rin.
Changes in transistor β or other internal component values will have
comparatively little effect on voltage gain with feedback in
operation, resulting in an amplifier that is stable and easy to
design.
Amplifier impedances
*** PENDING ***
Current mirrors
An interesting and often-used circuit
applying the bipolar junction transistor is the so-called current
mirror, which serves as a simple current regulator, supplying nearly
constant current to a load over a wide range of load resistances.
We know that in a transistor operating in
its active mode, collector current is equal to base current multiplied
by the ratio β. We also know that the ratio between collector current
and emitter current is called α. Because collector current is equal to
base current multiplied by β, and emitter current is the sum of the base
and collector currents, α should be mathematically derivable from β. If
you do the algebra, you'll find that α = β/(β+1) for any transistor.
We've seen already how maintaining a
constant base current through an active transistor results in the
regulation of collector current, according to the β ratio. Well, the α
ratio works similarly: if emitter current is held constant, collector
current will remain at a stable, regulated value so long as the
transistor has enough collector-to-emitter voltage drop to maintain it
in its active mode. Therefore, if we have a way of holding emitter
current constant through a transistor, the transistor will work to
regulate collector current at a constant value.
Remember that the base-emitter junction
of a BJT is nothing more than a PN junction, just like a diode, and that
the "diode equation" specifies how much current will go through a PN
junction given forward voltage drop and junction temperature:
If both junction voltage and temperature
are held constant, then the PN junction current will likewise be
constant. Following this rationale, if we were to hold the base-emitter
voltage of a transistor constant, then its emitter current should
likewise be constant, given a constant temperature:
This constant emitter current, multiplied
by a constant α ratio, gives a constant collector current through Rload,
provided that there is enough battery voltage to keep the transistor in
its active mode for any change in Rload's resistance.
Maintaining a constant voltage across the
transistor's base-emitter junction is easy: use a forward-biased diode
to establish a constant voltage of approximately 0.7 volts, and connect
it in parallel with the base-emitter junction:
Now, here's where it gets interesting.
The voltage dropped across the diode probably won't be 0.7 volts
exactly. The exact amount of forward voltage dropped across it depends
on the current through the diode, and the diode's temperature, all in
accordance with the diode equation. If diode current is increased (say,
by reducing the resistance of Rbias), its voltage drop will
increase slightly, increasing the voltage drop across the transistor's
base-emitter junction, which will increase the emitter current by the
same proportion, assuming the diode's PN junction and the transistor's
base-emitter junction are well-matched to each other. In other words,
transistor emitter current will closely equal diode current at any given
time. If you change the diode current by changing the resistance value
of Rbias, then the transistor's emitter current will follow
suit, because the emitter current is described by the same equation as
the diode's, and both PN junctions experience the same voltage drop.
Remember, the transistor's collector
current is almost equal to its emitter current, as the α ratio of a
typical transistor is almost unity (1). If we have control over the
transistor's emitter current by setting diode current with a simple
resistor adjustment, then we likewise have control over the transistor's
collector current. In other words, collector current mimics, or
mirrors, diode current.
Current through resistor Rload
is therefore a function of current set by the bias resistor, the two
being nearly equal. This is the function of the current mirror circuit:
to regulate current through the load resistor by conveniently adjusting
the value of Rbias. It is very easy to create a set amount of
diode current, as current through the diode is described by a simple
equation: power supply voltage minus diode voltage (almost a constant
value), divided by the resistance of Rbias.
To better match the characteristics of
the two PN junctions (the diode junction and the transistor base-emitter
junction), a transistor may be used in place of a regular diode, like
this:
Because temperature is a factor in the
"diode equation," and we want the two PN junctions to behave identically
under all operating conditions, we should maintain the two transistors
at exactly the same temperature. This is easily done using discrete
components by gluing the two transistor cases back-to-back. If the
transistors are manufactured together on a single chip of silicon (as a
so-called integrated circuit, or IC), the designers should
locate the two transistors very close to one another to facilitate heat
transfer between them.
The current mirror circuit shown with two
NPN transistors is sometimes called a current-sinking type,
because the regulating transistor conducts current to the load from
ground ("sinking" current), rather than from the positive side of
the battery ("sourcing" current). If we wish to have a grounded
load, and a current sourcing mirror circuit, we could use PNP
transistors like this:
- REVIEW:
- A current mirror is a
transistor circuit that regulates current through a load resistance,
the regulation point being set by a simple resistor adjustment.
- Transistors in a current mirror
circuit must be maintained at the same temperature for precise
operation. When using discrete transistors, you may glue their cases
together to help accomplish this.
- Current mirror circuits may be found
in two basic varieties: the current sinking configuration,
where the regulating transistor connects the load to ground; and the
current sourcing configuration, where the regulating transistor
connects the load to the positive terminal of the DC power supply.
Transistor ratings and packages
*** INCOMPLETE ***
Like all electrical and electronic
components, transistors are limited in the amounts of voltage and
current they can handle without sustaining damage. Since transistors are
a bit more complex than some of the other components you're used to
seeing at this point, they tend to have more kinds of ratings. What
follows is an itemized description of some typical transistor ratings.
Power dissipation:
When a transistor conducts current between collector and emitter, it
also drops voltage between those two points. At any given time, the
power dissipated by a transistor is equal to the product
(multiplication) of collector current and collector-emitter voltage.
Just like resistors, transistors are rated in terms of how many watts
they can safely dissipate without sustaining damage. High temperature is
the mortal enemy of all semiconductor devices, and bipolar transistors
tend to be more susceptible to thermal damage than most. Power ratings
are always given in reference to the temperature of ambient
(surrounding) air. When transistors are to be used in hotter-than-normal
environments, their power ratings must be derated to avoid a
shortened service life.
Reverse voltages:
As with diodes, bipolar transistors are rated for maximum allowable
reverse-bias voltage across their PN junctions. This includes voltage
ratings for the base-emitter junction, base-collector junction, and also
from collector to emitter. The rating for maximum collector-emitter
voltage can be thought of in terms of the maximum voltage it can
withstand while in full-cutoff mode (no base current). This rating is of
particular importance when using a bipolar transistor as a switch.
Collector current:
A maximum value for collector current will be given by the manufacturer
in amps. Understand that this maximum figure assumes a saturated state
(minimum collector-emitter voltage drop). If the transistor is not
saturated, and in fact is dropping substantial voltage between collector
and emitter, the maximum power dissipation rating will probably be
exceeded before the maximum collector current rating will. Just
something to keep in mind when designing a transistor circuit!
Saturation voltages:
Ideally, a saturated transistor acts as a closed switch contact between
collector and emitter, dropping zero voltage at full collector current.
In reality this is never true. Manufacturers will specify the
maximum voltage drop of a transistor at saturation, both between the
collector and emitter, and also between base and emitter (forward
voltage drop of that PN junction). Collector-emitter voltage drop at
saturation is generally expected to be 0.3 volts or less, but this
figure is of course dependent on the specific type of transistor.
Base-emitter forward voltage drop is very similar to that of an
equivalent diode, which should come as no surprise.
Beta:
The ratio of collector current to base current, β is the fundamental
parameter characterizing the amplifying ability of a bipolar transistor.
β is usually assumed to be a constant figure in circuit calculations,
but unfortunately this is far from true in practice. As such,
manufacturers provide a set of β (or "hfe") figures for a
given transistor over a wide range of operating conditions, usually in
the form of maximum/minimum/typical ratings. It may surprise you to see
just how widely β can be expected to vary within normal operating
limits. One popular small-signal transistor, the 2N3903, is advertised
as having a β ranging from 15 to 150 depending on the amount of
collector current. Generally, β is highest for medium collector
currents, decreasing for very low and very high collector currents.
Alpha:
the ratio of collector current to emitter current, α may be derived from
β, being equal to β/(β+1).
Bipolar transistors come in a wide
variety of physical packages. Package type is primarily dependent upon
the power dissipation of the transistor, much like resistors: the
greater the maximum power dissipation, the larger the device has to be
to stay cool. There are several standardized package types for
three-terminal semiconductor devices, any of which may be used to house
a bipolar transistor. This is an important fact to consider: there are
many other semiconductor devices other than bipolar transistors which
have three connection points. It is impossible to positively
identify a three-terminal semiconductor device without referencing the
part number printed on it, and/or subjecting it to a set of electrical
tests.
BJT quirks
*** PENDING ***
Nonlinearity Temperature drift Thermal
runaway Junction capacitance Noise Mismatch (problem with paralleling
transistors) β cutoff frequency Alpha cutoff frequency
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