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.
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