Discrete Semiconductor Circuit Experiments - Basic Electronic Tutorials

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Introduction

A semiconductor device is one made of silicon or any number of other specially prepared materials designed to exploit the unique properties of electrons in a crystal lattice, where electrons are not as free to move as in a conductor, but are far more mobile than in an insulator. A discrete device is one contained in its own package, not built on a common semiconductor substrate with other components, as is the case with ICs, or integrated circuits. Thus, "discrete semiconductor circuits" are circuits built out of individual semiconductor components, connected together on some kind of circuit board or terminal strip. These circuits employ all the components and concepts explored in the previous chapters, so a firm comprehension of DC and AC electricity is essential before embarking on these experiments.

Just for fun, one circuit is included in this section using a vacuum tube for amplification instead of a semiconductor transistor. Before the advent of transistors, "vacuum tubes" were the workhorses of the electronics industry: used to make rectifiers, amplifiers, oscillators, and many other circuits. Though now considered obsolete for most purposes, there are still some applications for vacuum tubes, and it can be fun building and operating circuits using these devices.

Commutating diode

PARTS AND MATERIALS

6 volt battery
Power transformer, 120VAC step-down to 12VAC (Radio Shack catalog # 273-1365, 273-1352, or 273-1511).
One 1N4001 rectifying diode (Radio Shack catalog # 276-1101)
One neon lamp (Radio Shack catalog # 272-1102)
Two toggle switches, SPST ("Single-Pole, Single-Throw")

A power transformer is specified, but any iron-core inductor will suffice, even the home-made inductor or transformer from the AC experiments chapter!

The diode need not be an exact model 1N4001. Any of the "1N400X" series of rectifying diodes are suitable for the task, and they are quite easy to obtain.

I recommend household light switches for their low cost and durability.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 1, chapter 16: "RC and L/R Time Constants"

Lessons In Electric Circuits, Volume 3, chapter 3: "Diodes and Rectifiers"

LEARNING OBJECTIVES

Review inductive "kickback"
Learn how to suppress "kickback" using a diode

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

When assembling the circuit, be very careful of the diode's orientation. The cathode end of the diode (the end marked with a single band) must face the positive (+) side of the battery. The diode should be reverse-biased and nonconducting with switch #1 in the "on" position. Use the high-voltage (120 V) winding of the transformer for the inductor coil. The primary winding of a step-down transformer has more inductance than the secondary winding, and will give a greater lamp-flashing effect.

Set switch #2 to the "off" position. This disconnects the diode from the circuit so that it has no effect. Quickly close and open (turn "on" and then "off") switch #1. When that switch is opened, the neon bulb will flash from the effect of inductive "kickback." Rapid current decrease caused by the switch's opening causes the inductor to create a large voltage drop as it attempts to keep current at the same magnitude and going in the same direction.

Inductive kickback is detrimental to switch contacts, as it causes excessive arcing whenever they are opened. In this circuit, the neon lamp actually diminishes the effect by providing an alternate current path for the inductor's current when the switch opens, dissipating the inductor's stored energy harmlessly in the form of light and heat. However, there is still a fairly high voltage dropped across the opening contacts of switch #1, causing undue arcing and shortened switch life.

If switch #2 is closed (turned "on"), the diode will now be a part of the circuit. Quickly close and open switch #1 again, noting the difference in circuit behavior. This time, the neon lamp does not flash. Connect a voltmeter across the inductor to verify that the inductor is still receiving full battery voltage with switch #1 closed. If the voltmeter registers only a small voltage with switch #1 "on," the diode is probably connected backward, creating a short-circuit.

Half-wave rectifier

PARTS AND MATERIALS

Low-voltage AC power supply (6 volt output)
6 volt battery
One 1N4001 rectifying diode (Radio Shack catalog # 276-1101)
Small "hobby" motor, permanent-magnet type (Radio Shack catalog # 273-223 or equivalent)
Audio detector with headphones
0.1 µF capacitor (Radio Shack catalog # 272-135 or equivalent)

The diode need not be an exact model 1N4001. Any of the "1N400X" series of rectifying diodes are suitable for the task, and they are quite easy to obtain.

See the AC experiments chapter for detailed instructions on building the "audio detector" listed here. If you haven't built one already, you're missing a simple and valuable tool for experimentation.

A 0.1 µF capacitor is specified for "coupling" the audio detector to the circuit, so that only AC reaches the detector circuit. This capacitor's value is not critical. I've used capacitors ranging from 0.27 µF to 0.015 µF with success. Lower capacitor values attenuate low-frequency signals to a greater degree, resulting in less sound intensity from the headphones, so use a greater-value capacitor value if you experience difficulty hearing the tone(s).

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 3: "Diodes and Rectifiers"

LEARNING OBJECTIVES

Function of a diode as a rectifier
Permanent-magnet motor operation on AC versus DC power
Measuring "ripple" voltage with a voltmeter

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

Connect the motor to the low-voltage AC power supply through the rectifying diode as shown. The diode only allows current to pass through during one half-cycle of a full positive-and-negative cycle of power supply voltage, eliminating one half-cycle from ever reaching the motor. As a result, the motor only "sees" current in one direction, albeit a pulsating current, allowing it to spin in one direction.

Take a jumper wire and short past the diode momentarily, noting the effect on the motor's operation:

As you can see, permanent-magnet "DC" motors do not function well on alternating current. Remove the temporary jumper wire and reverse the diode's orientation in the circuit. Note the effect on the motor.

Measure DC voltage across the motor like this:

Then, measure AC voltage across the motor as well:

Most digital multimeters do a good job of discriminating AC from DC voltage, and these two measurements show the DC average and AC "ripple" voltages, respectively of the power "seen" by the motor. Ripple voltage is the varying portion of the voltage, interpreted as an AC quantity by measurement equipment although the voltage waveform never actually reverses polarity. Ripple may be envisioned as an AC signal superimposed on a steady DC "bias" or "offset" signal. Compare these measurements of DC and AC with voltage measurements taken across the motor while powered by a battery:

Batteries give very "pure" DC power, and as a result there should be very little AC voltage measured across the motor in this circuit. Whatever AC voltage is measured across the motor is due to the motor's pulsating current draw as the brushes make and break contact with the rotating commutator bars. This pulsating current causes pulsating voltages to be dropped across any stray resistances in the circuit, resulting in pulsating voltage "dips" at the motor terminals.

A qualitative assessment of ripple voltage may be obtained by using the sensitive audio detector described in the AC experiments chapter (the same device described as a "sensitive voltage detector" in the DC experiments chapter). Turn the detector's sensitivity down for low volume, and connect it across the motor terminals through a small (0.1 µF) capacitor, like this:

The capacitor acts as a high-pass filter, blocking DC voltage from reaching the detector and allowing easier "listening" of the remaining AC voltage. This is the exact same technique used in oscilloscope circuitry for "AC coupling," where DC signals are blocked from viewing by a series-connected capacitor. With a battery powering the motor, the ripple should sound like a high-pitched "buzz" or "whine." Try replacing the battery with the AC power supply and rectifying diode, "listening" with the detector to the low-pitched "buzz" of the half-wave rectified power:

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

Halfwave rectifier
v1 1 0 sin(0 8.485 60 0 0)
rload 2 0 10k
d1 1 2 mod1
.model mod1 d
.tran .5m 25m
.plot tran v(1,0) v(2,0) 
.end

This simulation plots the input voltage as a sine wave and the output voltage as a series of "humps" corresponding to the positive half-cycles of the AC source voltage. The dynamics of a DC motor are far too complex to be simulated using SPICE, unfortunately.

AC source voltage is specified as 8.485 instead of 6 volts because SPICE understands AC voltage in terms of peak value only. A 6 volt RMS sine-wave voltage is actually 8.485 volts peak. In simulations where the distinction between RMS and peak value isn't relevant, I will not bother with an RMS-to-peak conversion like this. To be truthful, the distinction is not terribly important in this simulation, but I discuss it here for your edification.

Full-wave center-tap rectifier

PARTS AND MATERIALS

Low-voltage AC power supply (6 volt output)
Two 1N4001 rectifying diodes (Radio Shack catalog # 276-1101)
Small "hobby" motor, permanent-magnet type (Radio Shack catalog # 273-223 or equivalent)
Audio detector with headphones
0.1 µF capacitor
One toggle switch, SPST ("Single-Pole, Single-Throw")

It is essential for this experiment that the low-voltage AC power supply be equipped with a center tap. A transformer with a non-tapped secondary winding simply will not work for this circuit.

The diodes need not be exact model 1N4001 units. Any of the "1N400X" series of rectifying diodes are suitable for the task, and they are quite easy to obtain.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 3: "Diodes and Rectifiers"

LEARNING OBJECTIVES

Design of a center-tap rectifier circuit
Measuring "ripple" voltage with a voltmeter

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

This rectifier circuit is called full-wave because it makes use of the entire waveform, both positive and negative half-cycles, of the AC source voltage in powering the DC load. As a result, there is less "ripple" voltage seen at the load. The RMS (Root-Mean-Square) value of the rectifier's output is also greater for this circuit than for the half-wave rectifier.

Use a voltmeter to measure both the DC and AC voltage delivered to the motor. You should notice the advantages of the full-wave rectifier immediately by the greater DC and lower AC indications as compared to the last experiment.

An experimental advantage of this circuit is the ease of which it may be "de-converted" to a half-wave rectifier: simply disconnect the short jumper wire connecting the two diodes' cathode ends together on the terminal strip. Better yet, for quick comparison between half and full-wave rectification, you may add a switch in the circuit to open and close this connection at will:

With the ability to quickly switch between half- and full-wave rectification, you may easily perform qualitative comparisons between the two different operating modes. Use the audio signal detector to "listen" to the ripple voltage present between the motor terminals for half-wave and full-wave rectification modes, noting both the intensity and the quality of the tone. Remember to use a coupling capacitor in series with the detector so that it only receives the AC "ripple" voltage and not DC voltage:

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

Fullwave center-tap rectifier
v1 1 0 sin(0 8.485 60 0 0)
v2 0 3 sin(0 8.485 60 0 0)
rload 2 0 10k
d1 1 2 mod1
d2 3 2 mod1
.model mod1 d
.tran .5m 25m
.plot tran v(1,0) v(2,0) 
.end

Full-wave bridge rectifier

PARTS AND MATERIALS

Low-voltage AC power supply (6 volt output)
Four 1N4001 rectifying diodes (Radio Shack catalog # 276-1101)
Small "hobby" motor, permanent-magnet type (Radio Shack catalog # 273-223 or equivalent)

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 3: "Diodes and Rectifiers"

LEARNING OBJECTIVES

Design of a bridge rectifier circuit
Advantages and disadvantages of the bridge rectifier circuit, compared to the center-tap circuit

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

This circuit provides full-wave rectification without the necessity of a center-tapped transformer. In applications where a center-tapped, or split-phase, source is unavailable, this is the only practical method of full-wave rectification.

In addition to requiring more diodes than the center-tap circuit, the full-wave bridge suffers a slight performance disadvantage as well: the additional voltage drop caused by current having to go through two diodes in each half-cycle rather than through only one. With a low-voltage source such as the one you're using (6 volts RMS), this disadvantage is easily measured. Compare the DC voltage reading across the motor terminals with the reading obtained from the last experiment, given the same AC power supply and the same motor.

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

Fullwave bridge rectifier
v1 1 0 sin(0 8.485 60 0 0)
rload 2 3 10k
d1 3 1 mod1
d2 1 2 mod1
d3 3 0 mod1
d4 0 2 mod1
.model mod1 d
.tran .5m 25m
.plot tran v(1,0) v(2,3) 
.end

Rectifier/filter circuit

PARTS AND MATERIALS

Low-voltage AC power supply
Bridge rectifier pack (Radio Shack catalog # 276-1185 or equivalent)
Electrolytic capacitor, 1000 µF, at least 25 WVDC (Radio Shack catalog # 272-1047 or equivalent)
Four "banana" jack style binding posts, or other terminal hardware, for connection to potentiometer circuit (Radio Shack catalog # 274-662 or equivalent)
Metal box
12-volt light bulb, 25 watt
Lamp socket

A bridge rectifier "pack" is highly recommended over constructing a bridge rectifier circuit from individual diodes, because such "packs" are made to bolt onto a metal heat sink. A metal box is recommended over a plastic box for its ability to function as a heat sink for the rectifier.

A larger capacitor value is fine to use in this experiment, so long as its working voltage is high enough. To be safe, choose a capacitor with a working voltage rating at least twice the RMS AC voltage output of the low-voltage AC power supply.

High-wattage 12-volt lamps may be purchased from recreational vehicle (RV) and boating supply stores. Common sizes are 25 watt and 50 watt. This lamp will be used as a "heavy" load for the power supply.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 2, chapter 8: "Filters"

LEARNING OBJECTIVES

Capacitive filter function in an AC/DC power supply
Importance of heat sinks for power semiconductors

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

This experiment involves constructing a rectifier and filter circuit for attachment to the low-voltage AC power supply constructed earlier. With this device, you will have a source of low-voltage, DC power suitable as a replacement for a battery in battery-powered experiments. If you would like to make this device its own, self-contained 120VAC/DC power supply, you may add all the componentry of the low-voltage AC supply to the "AC in" side of this circuit: a transformer, power cord, and plug. Even if you don't choose to do this, I recommend using a metal box larger than necessary to provide room for additional voltage regulation circuitry you might choose to add to this project later.

The bridge rectifier unit should be rated for a current at least as high as the transformer's secondary winding is rated for, and for a voltage at least twice as high as the RMS voltage of the transformer's output (this allows for peak voltage, plus an additional safety margin). The Radio Shack rectifier specified in the parts list is rated for 25 amps and 50 volts, more than enough for the output of the low-voltage AC power supply specified in the AC experiments chapter.

Rectifier units of this size are often equipped with "quick-disconnect" terminals. Complementary "quick-disconnect" lugs are sold that crimp onto the bare ends of wire. This is the preferred method of terminal connection. You may solder wires directly to the lugs of the rectifier, but I recommend against direct soldering to any semiconductor component for two reasons: possible heat damage during soldering, and difficulty of replacing the component in the event of failure.

Semiconductor devices are more prone to failure than most of the components covered in these experiments thus far, and so if you have any intent of making a circuit permanent, you should build it to be maintained. "Maintainable construction" involves, among other things, making all delicate components replaceable. It also means making "test points" accessible to meter probes throughout the circuit, so that troubleshooting may be executed with a minimum of inconvenience. Terminal strips inherently provide test points for taking voltage measurements, and they also allow for easy disconnection of wires without sacrificing connection durability.

Bolt the rectifier unit to the inside of the metal box. The box's surface area will act as a radiator, keeping the rectifier unit cool as it passes high currents. Any metal radiator surface designed to lower the operating temperature of an electronic component is called a heat sink. Semiconductor devices in general are prone to damage from overheating, so providing a path for heat transfer from the device(s) to the ambient air is very important when the circuit in question may handle large amounts of power.

A capacitor is included in the circuit to act as a filter to reduce ripple voltage. Make sure that you connect the capacitor properly across the DC output terminals of the rectifier, so that the polarities match. Being an electrolytic capacitor, it is sensitive to damage by polarity reversal. In this circuit especially, where the internal resistance of the transformer and rectifier are low and the short-circuit current consequently is high, the potential for damage is great. Warning: a failed capacitor in this circuit will likely explode with alarming force!

After the rectifier/filter circuit is built, connect it to the low-voltage AC power supply like this:

Measure the AC voltage output by the low-voltage power supply. Your meter should indicate approximately 6 volts if the circuit is connected as shown. This voltage measurement is the RMS voltage of the AC power supply.

Now, switch your multimeter to the DC voltage function and measure the DC voltage output by the rectifier/filter circuit. It should read substantially higher than the RMS voltage of the AC input measured before. The filtering action of the capacitor provides a DC output voltage equal to the peak AC voltage, hence the greater voltage indication:

Measure the AC ripple voltage magnitude with a digital voltmeter set to AC volts (or AC millivolts). You should notice a much smaller ripple voltage in this circuit than what was measured in any of the unfiltered rectifier circuits previously built. Feel free to use your audio detector to "listen" to the AC ripple voltage output by the rectifier/filter unit. As usual, connect a small "coupling" capacitor in series with the detector so that it does not respond to the DC voltage, but only the AC ripple. Very little sound should be heard.

After taking unloaded AC ripple voltage measurements, connect the 25 watt light bulb to the output of the rectifier/filter circuit like this:

Re-measure the ripple voltage present between the rectifier/filter unit's "DC out" terminals. With a heavy load, the filter capacitor becomes discharged between rectified voltage peaks, resulting in greater ripple than before:

If less ripple is desired under heavy-load conditions, a larger capacitor may be used, or a more complex filter circuit may be built using two capacitors and an inductor:

If you choose to build such a filter circuit, be sure to use an iron-core inductor for maximum inductance, and one with thick enough wire to safely handle the full rated current of power supply. Inductors used for the purpose of filtering are sometimes referred to as chokes, because they "choke" AC ripple voltage from getting to the load. If a suitable choke cannot be obtained, the secondary winding of a step-down power transformer like the type used to step 120 volts AC down to 12 or 6 volts AC in the low-voltage power supply may be used. Leave the primary (120 volt) winding open:

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

Fullwave bridge rectifier
v1 1 0 sin(0 8.485 60 0 0)
rload 2 3 10k
c1 2 3 1000u ic=0
d1 3 1 mod1
d2 1 2 mod1
d3 3 0 mod1
d4 0 2 mod1
.model mod1 d
.tran .5m 25m
.plot tran v(1,0) v(2,3) 
.end

You may decrease the value of R_load in the simulation from 10 kΩ to some lower value to explore the effects of loading on ripple voltage. As it is with a 10 kΩ load resistor, the ripple is undetectable on the waveform plotted by SPICE.

Voltage regulator

PARTS AND MATERIALS

Four 6 volt batteries
Zener diode, 12 volt -- type 1N4742 (Radio Shack catalog # 276-563 or equivalent)
One 10 kΩ resistor

Any low-voltage zener diode is appropriate for this experiment. The 1N4742 model listed here (zener voltage = 12 volts) is but one suggestion. Whatever diode model you choose, I highly recommend one with a zener voltage rating greater than the voltage of a single battery, for maximum learning experience. It is important that you see how a zener diode functions when exposed to a voltage less than its breakdown rating.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 3: "Diodes and Rectifiers"

LEARNING OBJECTIVES

Zener diode function

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

Build this simple circuit, being sure to connect the diode in "reverse-bias" fashion (cathode positive and anode negative), and measure the voltage across the diode with one battery as a power source. Record this voltage drop for future reference. Also, measure and record the voltage drop across the 10 kΩ resistor.

Modify the circuit by connecting two 6-volt batteries in series, for 12 volts total power source voltage. Re-measure the diode's voltage drop, as well as the resistor's voltage drop, with a voltmeter:

Connect three, then four 6-volt batteries together in series, forming an 18 volt and 24 volt power source, respectively. Measure and record the diode's and resistor's voltage drops for each new power supply voltage. What do you notice about the diode's voltage drop for these four different source voltages? Do you see how the diode voltage never exceeds a level of 12 volts? What do you notice about the resistor's voltage drop for these four different source voltage levels?

Zener diodes are frequently used as voltage regulating devices, because they act to clamp the voltage drop across themselves at a predetermined level. Whatever excess voltage is supplied by the power source becomes dropped across the series resistor. However, it is important to note that a zener diode cannot make up for a deficiency in source voltage. For instance, this 12-volt zener diode does not drop 12 volts when the power source is only 6 volts strong. It is helpful to think of a zener diode as a voltage limiter: establishing a maximum voltage drop, but not a minimum voltage drop.

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

Zener diode
v1 1 0
r1 1 2 10k
d1 0 2 mod1
.model mod1 d bv=12
.dc v1 18 18 1
.print dc v(2,0)
.end

A zener diode may be simulated in SPICE with a normal diode, the reverse breakdown parameter (bv=12) set to the desired zener breakdown voltage.

Transistor as a switch

PARTS AND MATERIALS

Two 6-volt batteries
One NPN transistor -- models 2N2222 or 2N3403 recommended (Radio Shack catalog # 276-1617 is a package of fifteen NPN transistors ideal for this and other experiments)
One 100 kΩ resistor
One 560 Ω resistor
One light-emitting diode (Radio Shack catalog # 276-026 or equivalent)

Resistor values are not critical for this experiment. Neither is the particular light emitting diode (LED) selected.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 4: "Bipolar Junction Transistors"

LEARNING OBJECTIVES

Current amplification of a bipolar junction transistor

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

The red wire shown in the diagram (the one terminating in an arrowhead, connected to one end of the 100 kΩ resistor) is intended to remain loose, so that you may touch it momentarily to other points in the circuit.

If you touch the end of the loose wire to any point in the circuit more positive than it, such as the positive side of the DC power source, the LED should light up. It takes 20 mA to fully illuminate a standard LED, so this behavior should strike you as interesting, because the 100 kΩ resistor to which the loose wire is attached restricts current through it to a far lesser value than 20 mA. At most, a total voltage of 12 volts across a 100 kΩ resistance yields a current of only 0.12 mA, or 120 µA! The connection made by your touching the wire to a positive point in the circuit conducts far less current than 1 mA, yet through the amplifying action of the transistor, is able to control a much greater current through the LED.

Try using an ammeter to connect the loose wire to the positive side of the power source, like this:

You may have to select the most sensitive current range on the meter to measure this small flow. After measuring this controlling current, try measuring the LED's current (the controlled current) and compare magnitudes. Don't be surprised if you find a ratio in excess of 200 (the controlled current 200 times as great as the controlling current)!

As you can see, the transistor is acting as a kind of electrically-controlled switch, switching current on and off to the LED at the command of a much smaller current signal conducted through its base terminal.

To further illustrate just how miniscule the controlling current is, remove the loose wire from the circuit and try "bridging" the unconnected end of the 100 kΩ resistor to the power source's positive pole with two fingers of one hand. You may need to wet the ends of those fingers to maximize conductivity:

Try varying the contact pressure of your fingers with these two points in the circuit to vary the amount of resistance in the controlling current's path. Can you vary the brightness of the LED by doing so? What does this indicate about the transistor's ability to act as more than just a switch; i.e. as a variable

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

Transistor as a switch
v1 1 0
r1 1 2 100k
r2 1 3 560
d1 3 4 mod2
q1 4 2 0 mod1
.model mod1 npn bf=200
.model mod2 d is=1e-28
.dc v1 12 12 1
.print dc v(2,0) v(4,0) v(1,2) v(1,3) v(3,4)
.end

In this simulation, the voltage drop across the 560 Ω resistor v(1,3) turns out to be 10.26 volts, indicating a LED current of 18.32 mA by Ohm's Law (I=E/R). R₁'s voltage drop (voltage between nodes 1 and 2) ends up being 11.15 volts, which across 100 kΩ gives a current of only 111.5 µA. Obviously, a very small current is exerting control over a much larger current in this circuit.

In case you were wondering, the is=1e-28 parameter in the diode's .model line is there to make the diode act more like an LED with a higher forward voltage drop.

Static electricity sensor

PARTS AND MATERIALS

One N-channel junction field-effect transistor, models 2N3819 or J309 recommended (Radio Shack catalog # 276-2035 is the model 2N3819)
One 6 volt battery
One 100 kΩ resistor
One light-emitting diode (Radio Shack catalog # 276-026 or equivalent)
Plastic comb

The particular junction field-effect transistor, or JFET, model used in this experiment is not critical. P-channel JFETs are also okay to use, but are not as popular as N-channel transistors.

Beware that not all transistors share the same terminal designations, or pinouts, even if they share the same physical appearance. This will dictate how you connect the transistors together and to other components, so be sure to check the manufacturer's specifications (component datasheet), easily obtained from the manufacturer's website. Beware that it is possible for the transistor's package and even the manufacturer's datasheet to show incorrect terminal identification diagrams! Double-checking pin identities with your multimeter's "diode check" function is highly recommended. For details on how to identify junction field-effect transistor terminals using a multimeter, consult chapter 5 of the Semiconductor volume (volume III) of this book series.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 5: "Junction Field-Effect Transistors"

LEARNING OBJECTIVES

How the JFET is used as an on/off switch
How JFET current gain differs from a bipolar transistor

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

This experiment is very similar to the previous experiment using a bipolar junction transistor (BJT) as a switching device to control current through an LED. In this experiment, a junction field-effect transistor is used instead, giving dramatically improved sensitivity.

Build this circuit and touch the loose wire end (the wire shown in red on the schematic diagram and in the illustration, connected to the 100 kΩ resistor) with your hand. Simply touching this wire will likely have an effect on the LED's status. This circuit makes a fine sensor of static electricity! Try scuffing your feet on a carpet and then touching the wire end if no effect on the light is seen yet.

For a more controlled test, touch the wire with one hand and alternately touch the positive (+) and negative (-) terminals of the battery with one finger of your other hand. Your body acts as a conductor (albeit a poor one), connecting the gate terminal of the JFET to either terminal of the battery as you touch them. Make note which terminal makes the LED turn on and which makes the LED turn off. Try to relate this behavior with what you've read about JFETs in chapter 5 of the Semiconductor volume.

The fact that a JFET is turned on and off so easily (requiring so little control current), as evidenced by full on-and-off control simply by conduction of a control current through your body, demonstrates how great of a current gain it has. With the BJT "switch" experiment, a much more "solid" connection between the transistor's gate terminal and a source of voltage was needed to turn it on. Not so with the JFET. In fact, the mere presence of static electricity can turn it on and off at a distance.

To further experiment with the effects of static electricity on this circuit, brush your hair with the plastic comb and then wave the comb near the transistor, watching the effect on the LED. The action of combing your hair with a plastic object creates a high static voltage between the comb and your body. The strong electric field produced between these two objects should be detectable by this circuit from a significant distance!

In case you're wondering why there is no 560 Ω "dropping" resistor to limit current through the LED, many small-signal JFETs tend to self-limit their controlled current to a level acceptable by LEDs. The model 2N3819, for example, has a typical saturated drain current (I_DSS) of 10 mA and a maximum of 20 mA. Since most LEDs are rated at a forward current of 20 mA, there is no need for a dropping resistor to limit circuit current: the JFET does it intrinsically.

Pulsed-light sensor

PARTS AND MATERIALS

Two 6-volt batteries
One NPN transistor -- models 2N2222 or 2N3403 recommended (Radio Shack catalog # 276-1617 is a package of fifteen NPN transistors ideal for this and other experiments)
One light-emitting diode (Radio Shack catalog # 276-026 or equivalent)
Audio detector with headphones

If you don't have an audio detector already constructed, you can use a nice set of audio headphones (closed-cup style, that completely covers your ears) and a 120V/6V step-down transformer to build a sensitive audio detector without volume control or overvoltage protection, just for this experiment.

Connect these portions of the headphone stereo plug to the transformer's secondary (6 volt) winding:

Try both the series and the parallel connection schemes for the loudest sound.

If you haven't made an audio detector as outlined in both the DC and AC experiments chapters, you really should -- it is a valuable piece of test equipment for your collection.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 4: "Bipolar Junction Transistors"

LEARNING OBJECTIVES

How to use a transistor as a crude common-emitter amplifier
How to use an LED as a light sensor

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

This circuit detects pulses of light striking the LED and converts them into relatively strong audio signals to be heard through the headphones. LEDs have the little-known ability to produce voltage when exposed to light, in a manner not unlike a semiconductor solar cell. By itself, the LED does not produce enough electrical power to drive the audio detector circuit, so a transistor is used to amplify the LED's signals. If the LED is exposed to a pulsing source of light, a tone will be heard in the headphones.

Sources of light suitable for this experiment include fluorescent and neon lamps, which blink rapidly with the 60 Hz AC power energizing them. You may also try using bright sunlight for a steady light source, then waving your fingers in front of the LED. The rapidly passing shadows will cause the LED to generate pulses of voltage, creating a brief "buzzing" sound in the headphones.

With a little imagination, it is not difficult to grasp the concept of transmitting audio information -- such as music or speech -- over a beam of pulsing light. Given a suitable "transmitter" circuit to pulse an LED on and off with the positive and negative crests of an audio waveform from a microphone, the "receiver" circuit shown here would convert those light pulses back into audio signals.

Voltage follower

PARTS AND MATERIALS

One NPN transistor -- models 2N2222 or 2N3403 recommended (Radio Shack catalog # 276-1617 is a package of fifteen NPN transistors ideal for this and other experiments)
Two 6-volt batteries
Two 1 kΩ resistors
One 10 kΩ potentiometer, single-turn, linear taper (Radio Shack catalog # 271-1715)

Beware that not all transistors share the same terminal designations, or pinouts, even if they share the same physical appearance. This will dictate how you connect the transistors together and to other components, so be sure to check the manufacturer's specifications (component datasheet), easily obtained from the manufacturer's website. Beware that it is possible for the transistor's package and even the manufacturer's datasheet to show incorrect terminal identification diagrams! Double-checking pin identities with your multimeter's "diode check" function is highly recommended. For details on how to identify bipolar transistor terminals using a multimeter, consult chapter 4 of the Semiconductor volume (volume III) of this book series.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 4: "Bipolar Junction Transistors"

LEARNING OBJECTIVES

Purpose of circuit "ground" when there is no actual connection to earth ground
Using a shunt resistor to measure current with a voltmeter
Measure amplifier voltage gain
Measure amplifier current gain
Amplifier impedance transformation

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

Again, beware that the transistor you select for this experiment may not have the same terminal designations shown here, and so the breadboard layout shown in the illustration may not be correct for you. In my illustrations, I show all TO-92 package transistors with terminals labeled "CBE": Collector, Base, and Emitter, from left to right. This is correct for the model 2N2222 transistor and some others, but not for all; not even for all NPN-type transistors! As usual, check with the manufacturer for details on the particular component(s) you choose for a project. With bipolar junction transistors, it is easy enough to verify terminal assignments with a multimeter.

The voltage follower is the safest and easiest transistor amplifier circuit to build. Its purpose is to provide approximately the same voltage to a load as what is input to the amplifier, but at a much greater current. In other words, it has no voltage gain, but it does have current gain.

Note that the negative (-) side of the power supply is shown in the schematic diagram to be connected to ground, as indicated by the symbol in the lower-left corner of the diagram. This does not necessarily represent a connection to the actual earth. What it means is that this point in the circuit -- and all points electrically common to it -- constitute the default reference point for all voltage measurements in the circuit. Since voltage is by necessity a quantity relative between two points, a "common" point of reference designated in a circuit gives us the ability to speak meaningfully of voltage at particular, single points in that circuit.

For example, if I were to speak of voltage at the base of the transistor (V_B), I would mean the voltage measured between the transistor's base terminal and the negative side of the power supply (ground), with the red probe touching the base terminal and the black probe touching ground. Normally, it is nonsense to speak of voltage at a single point, but having an implicit reference point for voltage measurements makes such statements meaningful:

Build this circuit, and measure output voltage versus input voltage for several different potentiometer settings. Input voltage is the voltage at the potentiometer's wiper (voltage between the wiper and circuit ground), while output voltage is the load resistor voltage (voltage across the load resistor, or emitter voltage: between emitter and circuit ground). You should see a close correlation between these two voltages: one is just a little bit greater than the other (about 0.6 volts or so?), but a change in the input voltage gives almost equal change in the output voltage. Because the relationship between input change and output change is almost 1:1, we say that the AC voltage gain of this amplifier is nearly 1.

Not very impressive, is it? Now measure current through the base of the transistor (input current) versus current through the load resistor (output current). Before you break the circuit and insert your ammeter to take these measurements, consider an alternative method: measure voltage across the base and load resistors, whose resistance values are known. Using Ohm's Law, current through each resistor may be easily calculated: divide the measured voltage by the known resistance (I=E/R). This calculation is particularly easy with resistors of 1 kΩ value: there will be 1 milliamp of current for every volt of drop across them. For best precision, you may measure the resistance of each resistor rather than assume an exact value of 1 kΩ, but it really doesn't matter much for the purposes of this experiment. When resistors are used to take current measurements by "translating" a current into a corresponding voltage, they are often referred to as shunt resistors.

You should expect to find huge differences between input and output currents for this amplifier circuit. In fact, it is not uncommon to experience current gains well in excess of 200 for a small-signal transistor operating at low current levels. This is the primary purpose of a voltage follower circuit: to boost the current capacity of a "weak" signal without altering its voltage.

Another way of thinking of this circuit's function is in terms of impedance. The input side of this amplifier accepts a voltage signal without drawing much current. The output side of this amplifier delivers the same voltage, but at a current limited only by load resistance and the current-handling ability of the transistor. Cast in terms of impedance, we could say that this amplifier has a high input impedance (voltage dropped with very little current drawn) and a low output impedance (voltage dropped with almost unlimited current-sourcing capacity).

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

Voltage follower
v1 1 0
rpot1 1 2 5k
rpot2 2 0 5k
rbase 2 3 1k
rload 4 0 1k
q1 1 3 4 mod1
.model mod1 npn bf=200
.dc v1 12 12 1
.print dc v(2,0) v(4,0) v(2,3)
.end

When this simulation is run through the SPICE program, it shows an input voltage of 5.937 volts and an output voltage of 5.095 volts, with an input current of 25.35 µA (2.535E-02 volts dropped across the 1 kΩ R_base resistor). Output current is, of course, 5.095 mA, inferred from the output voltage of 5.095 volts dropped across a load resistance of exactly 1 kΩ. You may change the "potentiometer" setting in this circuit by adjusting the values of R_pot1 and R_pot2, always keeping their sum at 10 kΩ.

Common-emitter amplifier

PARTS AND MATERIALS

One NPN transistor -- model 2N2222 or 2N3403 recommended (Radio Shack catalog # 276-1617 is a package of fifteen NPN transistors ideal for this and other experiments)
Two 6-volt batteries
One 10 kΩ potentiometer, single-turn, linear taper (Radio Shack catalog # 271-1715)
One 1 MΩ resistor
One 100 kΩ resistor
One 10 kΩ resistor
One 1.5 kΩ resistor

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 4: "Bipolar Junction Transistors"

LEARNING OBJECTIVES

Design of a simple common-emitter amplifier circuit
How to measure amplifier voltage gain
The difference between an inverting and a noninverting amplifier
Ways to introduce negative feedback in an amplifier circuit

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

Build this circuit and measure output voltage (voltage measured between the transistor's collector terminal and ground) and input voltage (voltage measured between the potentiometer's wiper terminal and ground) for several position settings of the potentiometer. I recommend determining the output voltage range as the potentiometer is adjusted through its entire range of motion, then choosing several voltages spanning that output range to take measurements at. For example, if full rotation on the potentiometer drives the amplifier circuit's output voltage from 0.1 volts (low) to 11.7 volts (high), choose several voltage levels between those limits (1 volt, 3 volts, 5 volts, 7 volts, 9 volts, and 11 volts). Measuring the output voltage with a meter, adjust the potentiometer to obtain each of these predetermined voltages at the output, noting the exact figure for later reference. Then, measure the exact input voltage producing that output voltage, and record that voltage figure as well.

In the end, you should have a table of numbers representing several different output voltages along with their corresponding input voltages. Take any two pairs of voltage figures and calculate voltage gain by dividing the difference in output voltages by the difference in input voltages. For example, if an input voltage of 1.5 volts gives me an output voltage of 7.0 volts and an input voltage of 1.66 volts gives me an output voltage of 1.0 volt, the amplifier's voltage gain is (7.0 - 1.0)/(1.66 - 1.5), or 6 divided by 0.16: a gain ratio of 37.50.

You should immediately notice two characteristics while taking these voltage measurements: first, that the input-to-output effect is "reversed;" that is, an increasing input voltage results in a decreasing output voltage. This effect is known as signal inversion, and this kind of amplifier as an inverting amplifier. Secondly, this amplifier exhibits a very strong voltage gain: a small change in input voltage results in a large change in output voltage. This should stand in stark contrast to the "voltage follower" amplifier circuit discussed earlier, which had a voltage gain of about 1.

Common-emitter amplifiers are widely used due to their high voltage gain, but they are rarely used in as crude a form as this. Although this amplifier circuit works to demonstrate the basic concept, it is very susceptible to changes in temperature. Try leaving the potentiometer in one position and heating the transistor by grasping it firmly with your hand or heating it with some other source of heat such as an electric hair dryer (WARNING: be careful not to get it so hot that your plastic breadboard melts!). You may also explore temperature effects by cooling the transistor: touch an ice cube to its surface and note the change in output voltage.

When the transistor's temperature changes, its base-emitter diode characteristics change, resulting in different amounts of base current for the same input voltage. This in turn alters the controlled current through the collector terminal, thus affecting output voltage. Such changes may be minimized through the use of signal feedback, whereby a portion of the output voltage is "fed back" to the amplifier's input so as to have a negative, or canceling, effect on voltage gain. Stability is improved at the expense of voltage gain, a compromise solution, but practical nonetheless.

Perhaps the simplest way to add negative feedback to a common-emitter amplifier is to add some resistance between the emitter terminal and ground, so that the input voltage becomes divided between the base-emitter PN junction and the voltage drop across the new resistance:

Repeat the same voltage measurement and recording exercise with the 1.5 kΩ resistor installed, calculating the new (reduced) voltage gain. Try altering the transistor's temperature again and noting the output voltage for a steady input voltage. Does it change more or less than without the 1.5 kΩ resistor?

Another method of introducing negative feedback to this amplifier circuit is to "couple" the output to the input through a high-value resistor. Connecting a 1 MΩ resistor between the transistor's collector and base terminals works well:

Although this different method of feedback accomplishes the same goal of increased stability by diminishing gain, the two feedback circuits will not behave identically. Note the range of possible output voltages with each feedback scheme (the low and high voltage values obtained with a full sweep of the input voltage potentiometer), and how this differs between the two circuits.

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

Common-emitter amplifier
vsupply 1 0 dc 12
vin 3 0
rc 1 2 10k
rb 3 4 100k
q1 2 4 0 mod1
.model mod1 npn bf=200
.dc vin 0 2 0.05
.plot dc v(2,0) v(3,0)
.end

This SPICE simulation sets up a circuit with a variable DC voltage source (vin) as the input signal, and measures the corresponding output voltage between nodes 2 and 0. The input voltage is varied, or "swept," from 0 to 2 volts in 0.05 volt increments. Results are shown on a plot, with the input voltage appearing as a straight line and the output voltage as a "step" figure where the voltage begins and ends level, with a steep change in the middle where the transistor is in its active mode of operation.

Multi-stage amplifier

PARTS AND MATERIALS

Three NPN transistors -- model 2N2222 or 2N3403 recommended (Radio Shack catalog # 276-1617 is a package of fifteen NPN transistors ideal for this and other experiments)
Two 6-volt batteries
One 10 kΩ potentiometer, single-turn, linear taper (Radio Shack catalog # 271-1715)
One 1 MΩ resistor
Three 100 kΩ resistors
Three 10 kΩ resistors

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 4: "Bipolar Junction Transistors"

LEARNING OBJECTIVES

Design of a multi-stage, direct-coupled common-emitter amplifier circuit
Effect of negative feedback in an amplifier circuit

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

By connecting three common-emitter amplifier circuit together -- the collector terminal of the previous transistor to the base (resistor) of the next transistor -- the voltage gains of each stage compound to give a very high overall voltage gain. I recommend building this circuit without the 1 MΩ feedback resistor to begin with, to see for yourself just how high the unrestricted voltage gain is. You may find it impossible to adjust the potentiometer for a stable output voltage (that isn't saturated at full supply voltage or zero), the gain being so high.

Even if you can't adjust the input voltage fine enough to stabilize the output voltage in the active range of the last transistor, you should be able to tell that the output-to-input relationship is inverting; that is, the output tends to drive to a high voltage when the input goes low, and visa-versa. Since any one of the common-emitter "stages" is inverting in itself, an even number of staged common-emitter amplifiers gives noninverting response, while an odd number of stages gives inverting. You may experience these relationships by measuring the collector-to-ground voltage at each transistor while adjusting the input voltage potentiometer, noting whether or not the output voltage increases or decreases with an increase in input voltage.

Connect the 1 MΩ feedback resistor into the circuit, coupling the collector of the last transistor to the base of the first. Since the overall response of this three-stage amplifier is inverting, the feedback signal provided through the 1 MΩ resistor from the output of the last transistor to the input of the first should be negative in nature. As such, it will act to stabilize the amplifier's response and minimize the voltage gain. You should notice the reduction in gain immediately by the decreased sensitivity of the output signal on input signal changes (changes in potentiometer position). Simply put, the amplifier isn't nearly as "touchy" as it was without the feedback resistor in place.

As with the simple common-emitter amplifier discussed in an earlier experiment, it is a good idea here to make a table of input versus output voltage figures with which you may calculate voltage gain.

Experiment with different values of feedback resistance. What effect do you think a decrease in feedback resistance have on voltage gain? What about an increase in feedback resistance? Try it and find out!

An advantage of using negative feedback to "tame" a high-gain amplifier circuit is that the resulting voltage gain becomes more dependent upon the resistor values and less dependent upon the characteristics of the constituent transistors. This is good, because it is far easier to manufacture consistent resistors than consistent transistors. Thus, it is easier to design an amplifier with predictable gain by building a staged network of transistors with an arbitrarily high voltage gain, then mitigate that gain precisely through negative feedback. It is this same principle that is used to make operational amplifier circuits behave so predictably.

This amplifier circuit is a bit simplified from what you will normally encounter in practical multi-stage circuits. Rarely is a pure common-emitter configuration (i.e. with no emitter-to-ground resistor) used, and if the amplifier's service is for AC signals, the inter-stage coupling is often capacitive with voltage divider networks connected to each transistor base for proper biasing of each stage. Radio-frequency amplifier circuits are often transformer-coupled, with capacitors connected in parallel with the transformer windings for resonant tuning.

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

Multi-stage amplifier
vsupply 1 0 dc 12
vin 2 0
r1 2 3 100k
r2 1 4 10k
q1 4 3 0 mod1
r3 4 7 100k
r4 1 5 10k
q2 5 7 0 mod1
r5 5 8 100k
r6 1 6 10k
q3 6 8 0 mod1
rf 3 6 1meg
.model mod1 npn bf=200
.dc vin 0 2.5 0.1
.plot dc v(6,0) v(2,0)
.end

This simulation plots output voltage against input voltage, and allows comparison between those variables in numerical form: a list of voltage figures printed to the left of the plot. You may calculate voltage gain by taking any two analysis points and dividing the difference in output voltages by the difference in input voltages, just like you do for the real circuit.

Experiment with different feedback resistance values (rf) and see the impact on overall voltage gain. Do you notice a pattern? Here's a hint: the overall voltage gain may be closely approximated by using the resistance figures of r1 and rf, without reference to any other circuit component!

Current mirror

PARTS AND MATERIALS

Two NPN transistors -- models 2N2222 or 2N3403 recommended (Radio Shack catalog # 276-1617 is a package of fifteen NPN transistors ideal for this and other experiments)
Two 6-volt batteries
One 10 kΩ potentiometer, single-turn, linear taper (Radio Shack catalog # 271-1715)
Two 10 kΩ resistors
Four 1.5 kΩ resistors

Small signal transistors are recommended so as to be able to experience "thermal runaway" in the latter portion of the experiment. Larger "power" transistors may not exhibit the same behavior at these low current levels. However, any pair of identical NPN transistors may be used to build a current mirror.

Beware that not all transistors share the same terminal designations, or pinouts, even if they share the same physical appearance. This will dictate how you connect the transistors together and to other components, so be sure to check the manufacturer's specifications (component datasheet), easily obtained from the manufacturer's website. Beware that it is possible for the transistor's package and even the manufacturer's datasheet to show incorrect terminal identification diagrams! Double-checking pin identities with your multimeter's "diode check" function is highly recommended. For details on how to identify bipolar transistor terminals using a multimeter, consult chapter 4 of the Semiconductor volume (volume III) of this book series.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 4: "Bipolar Junction Transistors"

LEARNING OBJECTIVES

How to build a current mirror circuit
Current limitations of a current mirror circuit
Temperature dependence of BJTs
Experience a controlled "thermal runaway" situation

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

A current mirror may be thought of as an adjustable current regulator, the current limit being easily set by a single resistance. It is a rather crude current regulator circuit, but one that finds wide use due to its simplicity. In this experiment, you will get the opportunity to build one of these circuits, explore its current-regulating properties, and also experience some of its practical limitations firsthand.

Build the circuit as shown in the schematic and illustration. You will have one extra 1.5 kΩ fixed-value resistor from the parts specified in the parts list. You will be using it in the last part of this experiment.

The potentiometer sets the amount of current through transistor Q₁. This transistor is connected to act as a simple diode: just a PN junction. Why use a transistor instead of a regular diode? Because it is important to match the junction characteristics of these two transistors when using them in a current mirror circuit. Voltage dropped across the base-emitter junction of Q₁ is impressed across the base-emitter junction of the other transistor, Q₂, causing it to turn "on" and likewise conduct current.

Since voltage across the two transistors' base-emitter junctions is the same -- the two junction pairs being connected in parallel with each other -- so should the current be through their base terminals, assuming identical junction characteristics and identical junction temperatures. Matched transistors should have the same β ratios, as well, so equal base currents means equal collector currents. The practical result of all this is Q₂'s collector current mimicking whatever current magnitude has been established through the collector of Q₁ by the potentiometer. In other words, current through Q₂ mirrors the current through Q₁.

Changes in load resistance (resistance connecting the collector of Q₂ to the positive side of the battery) have no effect on Q₁'s current, and consequently have no effect upon the base-emitter voltage or base current of Q₂. With a constant base current and a nearly constant β ratio, Q₂ will drop as much or as little collector-emitter voltage as necessary to hold its collector (load) current constant. Thus, the current mirror circuit acts to regulate current at a value set by the potentiometer, without regard to load resistance.

Well, that is how it is supposed to work, anyway. Reality isn't quite so simple, as you are about to see. In the circuit diagram shown, the load circuit of Q₂ is completed to the positive side of the battery through an ammeter, for easy current measurement. Rather than solidly connect the ammeter's black probe to a definite point in the circuit, I've marked five test points, TP1 through TP5, for you to touch the black test probe to while measuring current. This allows you to quickly and effortlessly change load resistance: touching the probe to TP1 results in practically no load resistance, while touching it to TP5 results in approximately 14.5 kΩ of load resistance.

To begin the experiment, touch the test probe to TP4 and adjust the potentiometer through its range of travel. You should see a small, changing current indicated by your ammeter as you move the potentiometer mechanism: no more than a few milliamps. Leave the potentiometer set to a position giving a round number of milliamps and move the meter's black test probe to TP3. The current indication should be very nearly the same as before. Move the probe to TP2, then TP1. Again, you should see a nearly unchanged amount of current. Try adjusting the potentiometer to another position, giving a different current indication, and touch the meter's black probe to test points TP1 through TP4, noting the stability of the current indications as you change load resistance. This demonstrates the current regulating behavior of this circuit.

You should note that the current regulation isn't perfect. Despite regulating the current at nearly the value for load resistances between 0 and 4.5 kΩ, there is some variation over this range. The regulation may be much worse if load resistance is allowed to rise too high. Try adjusting the potentiometer so that maximum current is obtained, as indicated with the ammeter test probe connected to TP1. Leaving the potentiometer at that position, move the meter probe to TP2, then TP3, then TP4, and finally TP5, noting the meter's indication at each connection point. The current should be regulated at a nearly constant value until the meter probe is moved to the last test point, TP5. There, the current indication will be substantially lower than at the other test points. Why is this? Because too much load resistance has been inserted into Q₂'s circuit. Simply put, Q₂ cannot "turn on" any more than it already has, to maintain the same amount of current with this great a load resistance as with lesser load resistances.

This phenomenon is common to all current-regulator circuits: there is a limited amount of resistance a current regulator can handle before it saturates. This stands to reason, as any current regulator circuit capable of supplying a constant amount of current through any load resistance imaginable would require an unlimited source of voltage to do it! Ohm's Law (E=IR) dictates the amount of voltage needed to push a given amount of current through a given amount of resistance, and with only 12 volts of power supply voltage at our disposal, a finite limit of load current and load resistance definitely exists for this circuit. For this reason, it may be helpful to think of current regulator circuits as being current limiter circuits, for all they can really do is limit current to some maximum value.

An important caveat for current mirror circuits in general is that of equal temperature between the two transistors. The current "mirroring" taking place between the two transistors' collector circuits depends on the base-emitter junctions of those two transistors having the exact same properties. As the "diode equation" describes, the voltage/current relationship for a PN junction strongly depends on junction temperature. The hotter a PN junction is, the more current it will pass for a given amount of voltage drop. If one transistor should become hotter than the other, it will pass more collector current than the other, and the circuit will no longer "mirror" current as expected. When building a real current mirror circuit using discrete transistors, the two transistors should be epoxy-glued together (back-to-back) so that they remain at approximately the same temperature.

To illustrate this dependence on equal temperature, try grasping one transistor between your fingers to heat it up. What happens to the current through the load resistors as the transistor's temperature increases? Now, let go of the transistor and blow on it to cool it down to ambient temperature. Grasp the other transistor between your fingers to heat it up. What does the load current do now?

In this next phase of the experiment, we will intentionally allow one of the transistors to overheat and note the effects. To avoid damaging a transistor, this procedure should be conducted no longer than is necessary to observe load current begin to "run away." To begin, adjust the potentiometer for minimum current. Next, replace the 10 kΩ R_limit resistor with a 1.5 kΩ resistor. This will allow a higher current to pass through Q₁, and consequently through Q₂ as well.

Place the ammeter's black probe on TP1 and observe the current indication. Move the potentiometer in the direction of increasing current until you read about 10 mA through the ammeter. At that point, stop moving the potentiometer and just observe the current. You will notice current begin to increase all on its own, without further potentiometer motion! Break the circuit by removing the meter probe from TP1 when the current exceeds 30 mA, to avoid damaging transistor Q₂.

If you carefully touch both transistors with a finger, you should notice Q₂ is warm, while Q₁ is cool. Warning: if Q₂'s current has been allowed to "run away" too far or for too long a time, it may become very hot! You can receive a bad burn on your fingertip by touching an overheated semiconductor component, so be careful here!

What just happened to make Q₂ overheat and lose current control? By connecting the ammeter to TP1, all load resistance was removed, so Q₂ had to drop full battery voltage between collector and emitter as it regulated current. Transistor Q₁ at least had the 1.5 kΩ resistance of R_limit in place to drop most of the battery voltage, so its power dissipation was far less than that of Q₂. This gross imbalance of power dissipation caused Q₂ to heat more than Q₁. As the temperature increased, Q₂ began to pass more current for the same amount of base-emitter voltage drop. This caused it to heat up even faster, as it was passing more collector current while still dropping the full 12 volts between collector and emitter. The effect is known as thermal runaway, and it is possible in many bipolar junction transistor circuits, not just current mirrors.

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

Current mirror
v1 1 0
vammeter 1 3 dc 0
rlimit 1 2 10k
rload 3 4 3k
q1 2 2 0 mod1
q2 4 2 0 mod1
.model mod1 npn bf=100
.dc v1 12 12 1
.print dc i(vammeter)
.end

V_ammeter is nothing more than a zero-volt DC battery strategically placed to intercept load current. This is nothing more than a trick to measure current in a SPICE simulation, as no dedicated "ammeter" component exists in the SPICE language.

It is important to remember that SPICE only recognizes the first eight characters of a component's name. The name "vammeter" is okay, but if we were to incorporate more than one current-measuring voltage source in the circuit and name them "vammeter1" and "vammeter2", respectively, SPICE would see them as being two instances of the same component "vammeter" (seeing only the first eight characters) and halt with an error. Something to bear in mind when altering the netlist or programming your own SPICE simulation!

You will have to experiment with different resistance values of R_load in this simulation to appreciate the current-regulating nature of the circuit. With R_limit set to 10 kΩ and a power supply voltage of 12 volts, the regulated current through R_load will be 1.1 mA. SPICE shows the regulation to be perfect (isn't the virtual world of computer simulation so nice?), the load current remaining at 1.1 mA for a wide range of load resistances. However, if the load resistance is increased beyond 10 kΩ, even this simulation shows the load current suffering a decrease as in real life.

JFET current regulator

PARTS AND MATERIALS

One N-channel junction field-effect transistor, models 2N3819 or J309 recommended (Radio Shack catalog # 276-2035 is the model 2N3819)
Two 6-volt batteries
One 10 kΩ potentiometer, single-turn, linear taper (Radio Shack catalog # 271-1715)
One 1 kΩ resistor
One 10 kΩ resistor
Three 1.5 kΩ resistors

For this experiment you will need an N-channel JFET, not a P-channel!

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 5: "Junction Field-Effect Transistors"

Lessons In Electric Circuits, Volume 3, chapter 3: "Diodes and Rectifiers"

LEARNING OBJECTIVES

How to use a JFET as a current regulator
How the JFET is relatively immune to changes in temperature

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

Previously in this chapter, you saw how a pair of bipolar junction transistors (BJTs) could be used to form a current mirror, whereby one transistor would try to maintain the same current through it as through the other, the other's current level being established by a variable resistance. This circuit performs the same task of regulating current, but uses a single junction field-effect transistor (JFET) instead of two BJTs.

The two series resistors R_adjust and R_limit set the current regulation point, while the load resistors and the test points between them serve only to demonstrate constant current despite changes in load resistance.

TP5, at the end of a 10 kΩ resistor, is provided for introducing a large change in load resistance. Connecting the black test probe of your ammeter to that test point gives a combined load resistance of 14.5 kΩ, which will be too much resistance for the transistor to maintain maximum regulated current through. To experience what I'm describing here, touch the black test probe to TP1 and adjust the potentiometer for maximum current. Now, move the black test probe to TP2, then TP3, then TP4. For all these test point positions, the current will remain approximately constant. However, when you touch the black probe to TP5, the current will fall dramatically. Why? Because at this level of load resistance, there is insufficient voltage drop across the transistor to maintain regulation. In other words, the transistor will be saturated as it attempts to provide more current than the circuit resistance will allow.

Move the black test probe back to TP1 and adjust the potentiometer for minimum current. Now, touch the black test probe to TP2, then TP3, then TP4, and finally TP5. What do you notice about the current indication at all these points? When the current regulation point is adjusted to a lesser value, the transistor is able to maintain regulation over a much larger range of load resistance.

An important caveat with the BJT current mirror circuit is that both transistors must be at equal temperature for the two currents to be equal. With this circuit, however, transistor temperature is almost irrelevant. Try grasping the transistor between your fingers to heat it up, noting the load current with your ammeter. Try cooling it down afterward by blowing on it. Not only is the requirement of transistor matching eliminated (due to the use of just one transistor), but the thermal effects are all but eliminated as well due to the relative thermal immunity of the field-effect transistor. This behavior also makes field-effect transistors immune to thermal runaway; a decided advantage over bipolar junction transistors.

An interesting application of this current-regulator circuit is the so-called constant-current diode. Described in the "Diodes and Rectifiers" chapter of volume III, this diode isn't really a PN junction device at all. Instead, it is a JFET with a fixed resistance connected between the gate and source terminals:

A normal PN-junction diode is included in series with the JFET to protect the transistor against damage from reverse-bias voltage, but otherwise the current-regulating facility of this device is entirely provided by the field-effect transistor.

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):

JFET current regulator
vsource 1 0 
rload 1 2 4.5k
j1 2 0 3 mod1
rlimit 3 0 1k
.model mod1 njf
.dc vsource 6 12 0.1
.plot dc i(vsource)
.end

SPICE does not allow for "sweeping" resistance values, so to demonstrate the current regulation of this circuit over a wide range of conditions, I've elected to sweep the source voltage from 6 to 12 volts in 0.1 volt steps. If you wish, you can set rload to different resistance values and verify that the circuit current remains constant. With an rlimit value of 1 kΩ, the regulated current will be 291.8 µA. This current figure will most likely not be the same as your actual circuit current, due to differences in JFET parameters.

Many manufacturers give SPICE model parameters for their transistors, which may be typed in the .model line of the netlist for a more accurate circuit simulation.

Differential amplifier

PARTS AND MATERIALS

Two 6-volt batteries
Two NPN transistors -- models 2N2222 or 2N3403 recommended (Radio Shack catalog # 276-1617 is a package of fifteen NPN transistors ideal for this and other experiments)
Two 10 kΩ potentiometers, single-turn, linear taper (Radio Shack catalog # 271-1715)
Two 22 kΩ resistors
Two 10 kΩ resistors
One 100 kΩ resistor
One 1.5 kΩ resistor

Resistor values are not especially critical in this experiment, but have been chosen to provide high voltage gain for a "comparator-like" differential amplifier behavior.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 3, chapter 4: "Bipolar Junction Transistors"

Lessons In Electric Circuits, Volume 3, chapter 8: "Operational Amplifiers"

LEARNING OBJECTIVES

Basic design of a differential amplifier circuit.
Working definitions of differential and common-mode voltages

SCHEMATIC DIAGRAM