14 June 2010

Solar Panel using Diodes!

Intro:Make a Solar Panel using Diodes!

Make a Solar Panel using Diodes!
So, solar panels are made of silicon...diodes are made of silicon.  Have you ever wondered if diodes conduct current induced by light?  Sure you have, and so have I, so let's experiment and make a solar panel using 1N4148 (run-of-the-mill) diodes.   This is just a brief instructable, as the full construction is left up to you, intrepid reader, as a fun project.

Check out the video below to see how much voltage you can get from just four diodes.
 

Step 1: Silicon Diode

Silicon Diode
A silicon diode is a two-lead semiconductor that gates current flow in one direction.  The symbol below shows how a diode is lined up with the schematic symbol.  The image was taken from http://www.gadgetjq.com/single_fire_tach_adapter_diode.jpg for copyright purposes.  Diodes are used in circuits that convert AC voltages to DC voltages, and also as voltage regulators, clamps, and multipliers.

Current flows in the direction of the arrow.  A few other terms of technical use are:


Forward-biased
is when the anode is more positive than the cathode, and reverse-bias is the opposite:  the anode is made more negative in voltage than the cathode.

There are different kinds of diodes, too.  Pin diodes, germanium diodes, schottky diodes, rectifier diodes (p-n junction diodes) name most of them.  In this instructable we're going to work with silicon-based pin diodes, although if you're really curious I'd encourage you trying the different kinds of diodes to see how it all shakes out.


Diodes are pretty cheap.  You can pick up a
pack of 50 from Radio Shack for around $3.

Step:2 Test it for yourself

Test it for yourself
Test it for yourselfGo ahead and dig out a few diodes from your tacklebox, toolbox, bead drawer, or whatever you keep all your electronic goodies in and put them in parallel.  Connect your voltmeter to either side and take a reading in ambient light.  I get about 4-5mV in ambient light.

Next, grab your maglight and while still taking a reading shine a focused beam on the diodes and see what your voltmeter says.  In my configuration using the diodes I had at hand, I was able to get more than 100mV from four diodes.  That's not too shabby, especially if you're shuffling that voltage off to a capacitor to either save for later or to build up a larger charge to do something more useful, like light an LED (yeah, like that's more useful) or run your garbage disposal.


You don't have to worry too much about which direction the strip is facing as long as you face the strips all in the same direction.

Step: 3 Get Funky

Get Funky
So you're generating voltage with your flashlight and a handful of diodes.....wow.  So what's next?  That's up to you.  You could cover a PCB with them and see how surface area corresponds to voltage, when you get diminishing returns, what the optimum diode to voltage ratio is, etc.   You could add in capacitors....throw in an op amp or even a charge pump and turn mV into V.

I dropped $5 at R.S. and got two packs of 50 diodes and etched a PCB.  Your designs are endless.
 

Step:4 Final Thoughts

Final Thoughts
I hope I was able to share something new with you and that you now have a new-found love for green energy.  Don't get to crazy though; a rooftop of diodes will get you in trouble with your significant other.
BEST OF LUCK TRY IT!
 


 
 

13 June 2010

DIGITAL INTEGRATED CIRCUITS (Full Chapter)

Introduction:

Digital circuits are circuits dealing with signals restricted to the extreme limits of zero and some full amount. This stands in contrast to analog circuits, in which signals are free to vary continuously between the limits imposed by power supply voltage and circuit resistances. These circuits find use in "true/false" logical operations and digital computation.
The circuits in this chapter make use of IC, or integrated circuit, components. Such components are actually networks of interconnected components manufactured on a single wafer of semiconducting material. Integrated circuits providing a multitude of pre-engineered functions are available at very low cost, benefitting students, hobbyists and professional circuit designers alike. Most integrated circuits provide the same functionality as "discrete" semiconductor circuits at higher levels of reliability and at a fraction of the cost.
Circuits in this chapter will primarily use CMOS technology, as this form of IC design allows for a broad range of power supply voltage while maintaining generally low power consumption levels. Though CMOS circuitry is susceptible to damage from static electricity (high voltages will puncture the insulating barriers in the MOSFET transistors), modern CMOS ICs are far more tolerant of electrostatic discharge than the CMOS ICs of the past, reducing the risk of chip failure by mishandling. Proper handling of CMOS involves the use of anti-static foam for storage and transport of IC's, and measures to prevent static charge from building up on your body (use of a grounding wrist strap, or frequently touching a grounded object).
Circuits using TTL technology require a regulated power supply voltage of 5 volts, and will not tolerate any substantial deviation from this voltage level. Any TTL circuits in this chapter will be adequately labeled as such, and it will be expected that you realize its unique power supply requirements.
When building digital circuits using integrated circuit "chips," it is highly recommended that you use a breadboard with power supply "rail" connections along the length. These are sets of holes in the breadboard that are electrically common along the entire length of the board. Connect one to the positive terminal of a battery, and the other to the negative terminal, and DC power will be available to any area of the breadboard via connection through short jumper wires:

With so many of these integrated circuits having "reset," "enable," and "disable" terminals needing to be maintained in a "high" or "low" state, not to mention the VDD (or VCC) and ground power terminals which require connection to the power supply, having both terminals of the power supply readily available for connection at any point along the board's length is very useful.
Most breadboards that I have seen have these power supply "rail" holes, but some do not. Up until this point, I've been illustrating circuits using a breadboard lacking this feature, just to show how it isn't absolutely necessary. However, digital circuits seem to require more connections to the power supply than other types of breadboard circuits, making this feature more than just a convenience.

Basic gate function

PARTS AND MATERIALS
  • 4011 quad NAND gate (Radio Shack catalog # 276-2411)
  • Eight-position DIP switch (Radio Shack catalog # 275-1301)
  • Ten-segment bargraph LED (Radio Shack catalog # 276-081)
  • One 6 volt battery
  • Two 10 kΩ resistors
  • Three 470 Ω resistors
Caution! The 4011 IC is CMOS, and therefore sensitive to static electricity!

CROSS-REFERENCES
Lessons In Electric Circuits, Volume 4, chapter 3: "Logic Gates"

LEARNING OBJECTIVES
  • Purpose of a "pulldown" resistor
  • How to experimentally determine the truth table of a gate
  • How to connect logic gates together
  • How to create different logical functions by using NAND gates

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
To begin, connect a single NAND gate to two input switches and one LED, as shown. At first, the use of an 8-position switch and a 10-segment LED bargraph may seem excessive, since only two switches and one LED are needed to show the operation of a single NAND gate. However, the presence of those extra switches and LEDs make it very convenient to expand the circuit, and help make the circuit layout both clean and compact.
It is highly recommended that you have a datasheet for the 4011 chip available when you build your circuit. Don't just follow the illustration shown above! It is important that you develop the skill of reading datasheets, especially "pinout" diagrams, when connecting IC terminals to other circuit elements. The datasheet's connection diagram is an essential piece of information to have. Shown here is my own rendition of what any 4011 datasheet shows:

In the breadboard illustration, I've shown the circuit built using the lower-left NAND gate: pin #'s 1 and 2 are the inputs, and pin #3 is the output. Pin #'s 14 and 7 conduct DC power to all four gate circuits inside the IC chip, "VDD" representing the positive side of the power supply (+V), and "Gnd" representing the negative side of the power supply (-V), or ground. Sometimes the negative power supply terminal will be labeled "VSS" instead of "Gnd" on a datasheet, but it means the same thing.
Digital logic circuitry does not make use of split power supplies as op-amps do. Like op-amp circuits, though, ground is still the implicit point of reference for all voltage measurements. If I were to speak of a "high" signal being present on a certain pin of the chip, I would mean that there was full voltage between that pin and the negative side of the power supply (ground).
Note how all inputs of the unused gates inside the 4011 chip are connected either to VDD or ground. This is not a mistake, but an act of intentional design. Since the 4011 is a CMOS integrated circuit, and CMOS circuit inputs left unconnected (floating) can assume any voltage level merely from intercepting a static electric charge from a nearby object, leaving inputs floating means that those unused gates may receive any random combinations of "high" and "low" signals.
Why is this undesirable, if we aren't using those gates? Who cares what signals they receive, if we are not doing anything with their outputs? The problem is, if static voltage signals appear at the gate inputs that are not fully "high" or fully "low," the gates' internal transistors may begin to turn on in such a way as to draw excessive current. At worst, this could lead to damage of the chip. At best it means excessive power consumption. It matters little if we choose to connect these unused gate inputs "high" (VDD) or "low" (ground), so long as we connect them to one of those two places. In the breadboard illustration, I show all the top inputs connected to VDD, and all the bottom inputs (of the unused gates) connected to ground. This was done merely because those power supply rail holes were closer and did not require long jumper wires!
Please note that none of the unused gate outputs have been connected to VDD or ground, and for good reason! If I were to do that, I may be forcing a gate to assume the opposite output state that its trying to achieve, which is a complicated way of saying that I would have created a short-circuit. Imagine a gate that is supposed to output a "high" logic level (for a NAND gate, this would be true if any of its inputs were "low"). If such a gate were to have its output terminal directly connected to ground, it could never reach a "high" state (being made electrically common to ground through the jumper wire connection). Instead, its upper (P-channel) output transistor would be turned on in vain, sourcing maximum current to a nonexistent load. This would very likely damage the gate! Gate output terminals, by their very nature, generate their own logic levels and never "float" in the same way that CMOS gate inputs do.
The two 10 kΩ resistors are placed in the circuit to avoid floating input conditions on the used gate. With a switch closed, the respective input will be directly connected to VDD and therefore be "high." With a switch open, the 10 kΩ "pulldown" resistor provides a resistive connection to ground, ensuring a secure "low" state at the gate's input terminal. This way, the input will not be susceptible to stray static voltages.
With the NAND gate connected to the two switches and one LED as shown, you are ready to develop a "truth table" for the NAND gate. Even if you already know what a NAND gate truth table looks like, this is a good exercise in experimentation: discovering a circuit's behavioral principles by induction. Draw a truth table on a piece of paper like this:

The "A" and "B" columns represent the two input switches, respectively. When the switch is on, its state is "high" or 1. When the switch is off, its state is "low," or 0, as ensured by its pulldown resistor. The gate's output, of course, is represented by the LED: whether it is lit (1) or unlit (0). After placing the switches in every possible combination of states and recording the LED's status, compare the resulting truth table with what a NAND gate's truth table should be.
As you can imagine, this breadboard circuit is not limited to testing NAND gates. Any gate type may be tested with two switches, two pulldown resistors, and an LED to indicate output status. Just be sure to double-check the chip's "pinout" diagram before substituting it pin-for-pin in place of the 4011. Not all "quad" gate chips have the same pin assignments!
An improvement you might want to make to this circuit is to assign a couple of LEDs to indicate input status, in addition to the one LED assigned to indicate the output. This makes operation a little more interesting to observe, and has the further benefit of indicating if a switch fails to close (or open) by showing the true input signal to the gate, rather than forcing you to infer input status from switch position:


NOR gate S-R latch

PARTS AND MATERIALS
  • 4001 quad NOR gate (Radio Shack catalog # 276-2401)
  • Eight-position DIP switch (Radio Shack catalog # 275-1301)
  • Ten-segment bargraph LED (Radio Shack catalog # 276-081)
  • One 6 volt battery
  • Two 10 kΩ resistors
  • Two 470 Ω resistors
  • Two 100 Ω resistors
Caution! The 4001 IC is CMOS, and therefore sensitive to static electricity!

CROSS-REFERENCES
Lessons In Electric Circuits, Volume 4, chapter 3: "Logic Gates"
Lessons In Electric Circuits, Volume 4, chapter 10: "Multivibrators"

LEARNING OBJECTIVES
  • The effects of positive feedback in a digital circuit
  • What is meant by the "invalid" state of a latch circuit
  • What a race condition is in a digital circuit
  • The importance of valid "high" CMOS signal voltage levels

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
The 4001 integrated circuit is a CMOS quad NOR gate, identical in input, output, and power supply pin assignments to the 4011 quad NAND gate. Its "pinout," or "connection," diagram is as such:

When two NOR gates are cross-connected as shown in the schematic diagram, there will be positive feedback from output to input. That is, the output signal tends to maintain the gate in its last output state. Just as in op-amp circuits, positive feedback creates hysteresis. This tendency for the circuit to remain in its last output state gives it a sort of "memory." In fact, there are solid-state computer memory technologies based on circuitry like this!
If we designate the left switch as the "Set" input and the right switch as the "Reset," the left LED will be the "Q" output and the right LED the "Q-not" output. With the Set input "high" (switch on) and the Reset input "low," Q will go "high" and Q-not will go "low." This is known as the set state of the circuit. Making the Reset input "high" and the Set input "low" reverses the latch circuit's output state: Q "low" and Q-not "high." This is known as the reset state of the circuit. If both inputs are placed into the "low" state, the circuit's Q and Q-not outputs will remain in their last states, "remembering" their prior settings. This is known as the latched state of the circuit.
Because the outputs have been designated "Q" and "Q-not," it is implied that their states will always be complementary (opposite). Thus, if something were to happen that forced both outputs to the same state, we would be inclined to call that mode of the circuit "invalid." This is exactly what will happen if we make both Set and Reset inputs "high:" both Q and Q-not outputs will be forced to the same "low" logic state. This is known as the invalid or illegal state of the circuit, not because something has gone wrong, but because the outputs have failed to meet the expectations established by their labels.
Since the "latched" state is a hysteretic condition whereby the last output states are "remembered," one might wonder what will happen if the circuit powers up this way, with no previous state to hold. To experiment, place both switches in their off positions, making both Set and Reset inputs low, then disconnect one of the battery wires from the breadboard. Then, quickly make and break contact between that battery wire and its proper connection point on the breadboard, noting the status of the two LEDs as the circuit is powered up again and again:

When a latch circuit such as this is powered up into its "latched" state, the gates race against each other for control. Given the "low" inputs, both gates try to output "high" signals. If one of the gates reaches its "high" output state before the other, that "high" state will be fed back to the other gate's input to force its output "low," and the race is won by the faster gate.
Invariably, one gate wins the race, due to internal variations between gates in the chip, and/or external resistances and capacitances that act to delay one gate more than the other. What this usually means is that the circuit tends to power up in the same mode, over and over again. However, if you are persistent in your powering/unpowering cycles, you should see at least a few times where the latch circuit powers up latched in the opposite state from normal.
Race conditions are generally undesirable in any kind of system, as they lead to unpredictable operation. They can be particularly troublesome to locate, as this experiment shows, because of the unpredictability they create. Imagine a scenario, for instance, where one of the two NOR gates was exceptionally slow-acting, due to a defect in the chip. This handicap would cause the other gate to win the power-up race every time. In other words, the circuit will be very predictable on power-up with both inputs "low." However, suppose that the unusual chip were to be replaced by one with more evenly matched gates, or by a chip where the other NOR gate were consistently slower. Normal circuit behavior is not supposed to change when a component is replaced, but if race conditions are present, a change of components may very well do just that.
Due to the inherent race tendency of an S-R latch, one should not design a circuit with the expectation of a consistent power-up state, but rather use external means to "force" the race so that the desired gate always "wins."
An interesting modification to try in this circuit is to replace one of the 470 Ω LED "dropping" resistors with a lower-value unit, such as 100 Ω. The obvious effect of this alteration will be increased LED brightness, as more current is allowed through. A not-so-obvious effect will also result, and it is this effect which holds great learning value. Try replacing one of the 470 Ω resistors with a 100 Ω resistor, and operate the input signal switches through all four possible setting combinations, noting the behavior of the circuit.
You should note that the circuit refuses to latch in one of its states (either Set or Reset), but only in the other state, when the input switches are both set "low" (the "latch" mode). Why is this? Take a voltmeter and measure the output voltage of the gate whose output is "high" when both inputs are "low." Note this voltage indication, then set the input switches in such a way that the other state (either Reset or Set) is forced, and measure the output voltage of the other gate when its output is "high." Note the difference between the two gate output voltage levels, one gate loaded by an LED with a 470 Ω resistor, and the other loaded by an LED with a 100 Ω resistor. The one loaded down by the "heavier" load (100 Ω resistor) will be much less: so much less that this voltage will not be interpreted by the other NOR gate's input as a "high" signal at all as it is fed back! All logic gates have permissible "high" and "low" input signal voltage ranges, and if the voltage of a digital signal falls outside this permissible range, it might not be properly interpreted by the receiving gate. In a latch circuit such as this, which depends on a solid "high" signal fed back from the output of one gate to the input of the other, a "weak" signal will not be able to maintain the positive feedback necessary to keep the circuit latched in one of its states.
This is one reason I favor the use of a voltmeter as a logic "probe" for determining digital signal levels, rather than an actual logic probe with "high" and "low" lights. A logic probe may not indicate the presence of a "weak" signal, whereas a voltmeter definitely will by means of its quantitative indication. This type of problem, common in circuits where different "families" of integrated circuits are mixed (TTL and CMOS, for example), can only be found with test equipment providing quantitative measurements of signal level.

NAND gate S-R enabled latch

PARTS AND MATERIALS
  • 4011 quad NAND gate (Radio Shack catalog # 276-2411)
  • Eight-position DIP switch (Radio Shack catalog # 275-1301)
  • Ten-segment bargraph LED (Radio Shack catalog # 276-081)
  • One 6 volt battery
  • Three 10 kΩ resistors
  • Two 470 Ω resistors
Caution! The 4011 IC is CMOS, and therefore sensitive to static electricity!

CROSS-REFERENCES
Lessons In Electric Circuits, Volume 4, chapter 3: "Logic Gates"
Lessons In Electric Circuits, Volume 4, chapter 10: "Multivibrators"

LEARNING OBJECTIVES
  • Principle and function of an enabled latch circuit

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
Although this circuit uses NAND gates instead of NOR gates, its behavior is identical to that of the NOR gate S-R latch (a "high" Set input drives Q "high," and a "high" Reset input drives Q-not "high"), except for the presence of a third input: the Enable. The purpose of the Enable input is to enable or disable the Set and Reset inputs from having effect over the circuit's output status. When the Enable input is "high," the circuit acts just like the NOR gate S-R latch. When the Enable input is "low," the Set and Reset inputs are disabled and have no effect whatsoever on the outputs, leaving the circuit in its latched state.
This kind of latch circuit (also called a gated S-R latch), may be constructed from two NOR gates and two AND gates, but the NAND gate design is easier to build since it makes use of all four gates in a single integrated circuit.


NAND gate S-R flip-flop

PARTS AND MATERIALS
  • 4011 quad NAND gate (Radio Shack catalog # 276-2411)
  • 4001 quad NOR gate (Radio Shack catalog # 276-2401)
  • Eight-position DIP switch (Radio Shack catalog # 275-1301)
  • Ten-segment bargraph LED (Radio Shack catalog # 276-081)
  • One 6 volt battery
  • Three 10 kΩ resistors
  • Two 470 Ω resistors
Caution! The 4011 IC is CMOS, and therefore sensitive to static electricity!
Although the parts list calls for a ten-segment LED unit, the illustration shows two individual LEDs being used instead. This is due to lack of room on my breadboard to mount the switch assembly, two integrated circuits, and the bargraph. If you have room on your breadboard, feel free to use the bargraph as called for in the parts list, and as shown in prior latch circuits.

CROSS-REFERENCES
Lessons In Electric Circuits, Volume 4, chapter 3: "Logic Gates"
Lessons In Electric Circuits, Volume 4, chapter 10: "Multivibrators"

LEARNING OBJECTIVES
  • The difference between a gated latch and a flip-flop
  • How to build a "pulse detector" circuit
  • Learn the effects of switch contact "bounce" on digital circuits

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
The only difference between a gated (or enabled) latch and a flip-flop is that a flip-flop is enabled only on the rising or falling edge of a "clock" signal, rather than for the entire duration of a "high" enable signal. Converting an enabled latch into a flip-flop simply requires that a "pulse detector" circuit be added to the Enable input, so that the edge of a clock pulse generates a brief "high" Enable pulse:

The single NOR gate and three inverter gates create this effect by exploiting the propagation delay time of multiple, cascaded gates. In this experiment, I use three NOR gates with paralleled inputs to create three inverters, thus using all four NOR gates of a 4001 integrated circuit:

Normally, when using a NOR gate as an inverter, one input would be grounded while the other acts as the inverter input, to minimize input capacitance and increase speed. Here, however, slow response is desired, and so I parallel the NOR inputs to make inverters rather than use the more conventional method.
Please note that this particular pulse detector circuit produces a "high" output pulse at every falling edge of the clock (input) signal. This means that the flip-flop circuit should be responsive to the Set and Reset input states only when the middle switch is moved from "on" to "off," not from "off" to "on."
When you build this circuit, though, you may discover that the outputs respond to Set and Reset input signals during both transitions of the Clock input, not just when it is switched from a "high" state to a "low" state. The reason for this is contact bounce: the effect of a mechanical switch rapidly making-and-breaking when its contacts are first closed, due to the elastic collision of the metal contact pads. Instead of the Clock switch producing a single, clean low-to-high signal transition when closed, there will most likely be several low-high-low "cycles" as the contact pads "bounce" upon off-to-on actuation. The first high-to-low transition caused by bouncing will trigger the pulse detector circuit, enabling the S-R latch for that moment in time, making it responsive to the Set and Reset inputs.
Ideally, of course, switches are perfect and bounce-free. In the real world, though, contact bounce is a very common problem for digital gate circuits operated by switch inputs, and must be understood well if it is to be overcome.

LED sequencer

PARTS AND MATERIALS
  • 4017 decade counter/divider (Radio Shack catalog # 276-2417)
  • 555 timer IC (Radio Shack catalog # 276-1723)
  • Ten-segment bargraph LED (Radio Shack catalog # 276-081)
  • One SPST switch
  • One 6 volt battery
  • 10 kΩ resistor
  • 1 MΩ resistor
  • 0.1 µF capacitor (Radio Shack catalog # 272-135 or equivalent)
  • Coupling capacitor, 0.047 to 0.001 µF
  • Ten 470 Ω resistors
  • Audio detector with headphones
Caution! The 4017 IC is CMOS, and therefore sensitive to static electricity!
Any single-pole, single-throw switch is adequate. A household light switch will work fine, and is readily available at any hardware store.
The audio detector will be used to assess signal frequency. If you have access to an oscilloscope, the audio detector is unnecessary.

CROSS-REFERENCES
Lessons In Electric Circuits, Volume 4, chapter 3: "Logic Gates"
Lessons In Electric Circuits, Volume 4, chapter 4: "Switches"
Lessons In Electric Circuits, Volume 4, chapter 11: "Counters"

LEARNING OBJECTIVES
  • Use of a 555 timer circuit to produce "clock" pulses (astable multivibrator)
  • Use of a 4017 decade counter/divider circuit to produce a sequence of pulses
  • Use of a 4017 decade counter/divider circuit for frequency division
  • Using a frequency divider and timepiece (watch) to measure frequency
  • Purpose of a "pulldown" resistor
  • Learn the effects of switch contact "bounce" on digital circuits
  • Use of a 555 timer circuit to "debounce" a mechanical switch (monostable multivibrator)

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
The model 4017 integrated circuit is a CMOS counter with ten output terminals. One of these ten terminals will be in a "high" state at any given time, with all others being "low," giving a "one-of-ten" output sequence. If low-to-high voltage pulses are applied to the "clock" (Clk) terminal of the 4017, it will increment its count, forcing the next output into a "high" state.
With a 555 timer connected as an astable multivibrator (oscillator) of low frequency, the 4017 will cycle through its ten-count sequence, lighting up each LED, one at a time, and "recycling" back to the first LED. The result is a visually pleasing sequence of flashing lights. Feel free to experiment with resistor and capacitor values on the 555 timer to create different flash rates.
Try disconnecting the jumper wire leading from the 4017's "Clock" terminal (pin #14) to the 555's "Output" terminal (pin #3) where it connects to the 555 timer chip, and hold its end in your hand. If there is sufficient 60 Hz power-line "noise" around you, the 4017 will detect it as a fast clock signal, causing the LEDs to blink very rapidly.
Two terminals on the 4017 chip, "Reset" and "Clock Enable," are maintained in a "low" state by means of a connection to the negative side of the battery (ground). This is necessary if the chip is to count freely. If the "Reset" terminal is made "high," the 4017's output will be reset back to 0 (pin #3 "high," all other output pins "low"). If the "Clock Enable" is made "high," the chip will stop responding to the clock signal and pause in its counting sequence.
If the 4017's "Reset" terminal is connected to one of its ten output terminals, its counting sequence will be cut short, or truncated. You may experiment with this by disconnecting the "Reset" terminal from ground, then connecting a long jumper wire to the "Reset" terminal for easy connection to the outputs at the ten-segment LED bargraph. Notice how many (or how few) LEDs light up with the "Reset" connected to any one of the outputs:

Counters such as the 4017 may be used as digital frequency dividers, to take a clock signal and produce a pulse occurring at some integer factor of the clock frequency. For example, if the clock signal from the 555 timer is 200 Hz, and the 4017 is configured for a full-count sequence (the "Reset" terminal connected to ground, giving a full, ten-step count), a signal with a period ten times as long (20 Hz) will be present at any of the 4017's output terminals. In other words, each output terminal will cycle once for every ten cycles of the clock signal: a frequency ten times as slow.
To experiment with this principle, connect your audio detector between output 0 (pin #3) of the 4017 and ground, through a very small capacitor (0.047 µF to 0.001 µF). The capacitor is used for "coupling" AC signals only, to that you may audibly detect pulses without placing a DC (resistive) load on the counter chip output. With the 4017 "Reset" terminal grounded, you will have a full-count sequence, and you will hear a "click" in the headphones every time the "0" LED lights up, corresponding to 1/10 of the 555's actual output frequency:

In fact, knowing this mathematical relationship between clicks heard in the headphone and the clock frequency allows us to measure the clock frequency to a fair degree of precision. Using a stopwatch or other timepiece, count the number of clicks heard in one full minute while connected to the 4017's "0" output. Using a 1 MΩ resistor and 0.1 µF capacitor in the 555 timing circuit, and a power supply voltage of 13 volts (instead of 6), I counted 79 clicks in one minute from my circuit. Your circuit may produce slightly different results. Multiply the number of pulses counted at the "0" output by 10 to obtain the number of cycles produced by the 555 timer during that same time (my circuit: 79 x 10 = 790 cycles). Divide this number by 60 to obtain the number of timer cycles elapsed in each second (my circuit: 790/60 = 13.17). This final figure is the clock frequency in Hz.
Now, leaving one test probe of the audio detector connected to ground, take the other test probe (the one with the coupling capacitor connected in series) and connect it to pin #3 of the 555 timer. The buzzing you hear is the undivided clock frequency:

By connecting the 4017's "Reset" terminal to one of the output terminals, a truncated sequence will result. If we are using the 4017 as a frequency divider, this means the output frequency will be a different factor of the clock frequency: 1/9, 1/8, 1/7, 1/6, 1/5, 1/4, 1/3, or 1/2, depending on which output terminal we connect the "Reset" jumper wire to. Re-connect the audio detector test probe to output "0" of the 4017 (pin #3), and connect the "Reset" terminal jumper to the sixth LED from the left on the bargraph. This should produce a 1/5 frequency division ratio:

Counting the number of clicks heard in one minute again, you should obtain a number approximately twice as large as what was counted with the 4017 configured for a 1/10 ratio, because 1/5 is twice as large a ratio as 1/10. If you do not obtain a count that is exactly twice what you obtained before, it is because of error inherent to the method of counting cycles: coordinating your sense of hearing with the display of a stopwatch or other time-keeping device.
Try replacing the 1 MΩ timing resistor in the 555 circuit with one of greatly lesser value, such as 10 kΩ. This will increase the clock frequency driving the 4017 chip. Use the audio detector to listen to the divided frequency at pin #3 of the 4017, noting the different tones produced as you move the "Reset" jumper wire to different outputs, creating different frequency division ratios. See if you can produce octaves by dividing the original frequency by 2, then by 4, and then by 8 (each descending octave represents one-half the previous frequency). Octaves are readily distinguished from other divided frequencies by their similar pitches to the original tone.
A final lesson that may be learned from this circuit is that of switch contact "bounce." For this, you will need a switch to provide clock signals to the 4017 chip, instead of the 555 timer. Re-connect the "Reset" jumper wire to ground to enable a full ten-step count sequence, and disconnect the 555's output from the 4017's "Clock" input terminal. Connect a switch in series with a 10 kΩ pulldown resistor, and connect this assembly to the 4017 "Clock" input as shown:



The purpose of a "pulldown" resistor is to provide a definite "low" logic state when the switch contact opens. Without this resistor in place, the 4017's "Clock" input wire would be floating whenever the switch contact was opened, leaving it susceptible to interference from stray static voltages or electrical "noise," either one capable of making the 4017 count randomly. With the pulldown resistor in place, the 4017's "Clock" input will have a definite, albeit resistive, connection to ground, providing a stable "low" logic state that precludes any interference from static electricity or "noise" coupled from nearby AC circuit wiring.
Actuate the switch on and off, noting the action of the LEDs. With each off-to-on switch transition, the 4017 should increment once in its count. However, you may notice some strange behavior: sometimes, the LED sequence will "skip" one or even several steps with a single switch closure. Why is this? It is due to very rapid, mechanical "bouncing" of the switch contacts. When two metallic contacts are brought together rapidly as does happen inside most switches, there will be an elastic collision. This collision results in the contacts making and breaking very rapidly as they "bounce" off one another. Normally, this "bouncing" is much to rapid for you to see its effects, but in a digital circuit such as this where the counter chip is able to respond to very quick clock pulses, these "bounces" are interpreted as distinct clock signals, and the count incremented accordingly.
One way to combat this problem is to use a timing circuit to produce a single pulse for any number of input pulse signals received within a short amount of time. The circuit is called a monostable multivibrator, and any technique eliminating the false pulses caused by switch contact "bounce" is called debouncing.
The 555 timer circuit is capable of functioning as a debouncer, if the "Trigger" input is connected to the switch as such:



Please note that since we are using the 555 once again to provide a clock signal to the 4017, we must re-connect pin #3 of the 555 chip to pin #14 of the 4017 chip! Also, if you have altered the values of the resistor or capacitor in the 555 timer circuit, you should return to the original 1 MΩ and 0.1 µF components.
Actuate the switch again and note the counting behavior of the 4017. There should be no more "skipped" counts as there were before, because the 555 timer outputs a single, crisp pulse for every on-to-off actuation (notice the inversion of operation here!) of the switch. It is important that the timing of the 555 circuit be appropriate: the time to charge the capacitor should be longer than the "settling" period of the switch (the time required for the contacts to stop bouncing), but not so long that the timer would "miss" a rapid sequence of switch actuations, if they were to occur.

Simple combination lock

PARTS AND MATERIALS
  • 4001 quad NOR gate (Radio Shack catalog # 276-2401)
  • 4070 quad XOR gate (Radio Shack catalog # 900-6906)
  • Two, eight-position DIP switches (Radio Shack catalog # 275-1301)
  • Two light-emitting diodes (Radio Shack catalog # 276-026 or equivalent)
  • Four 1N914 "switching" diodes (Radio Shack catalog # 276-1122)
  • Ten 10 kΩ resistors
  • Two 470 Ω resistors
  • Pushbutton switch, normally open (Radio Shack catalog # 275-1556)
  • Two 6 volt batteries
Caution! Both the 4001 and 4070 ICs are CMOS, and therefore sensitive to static electricity!
This experiment may be built using only one 8-position DIP switch, but the concept is easier to understand if two switch assemblies are used. The idea is, one switch acts to hold the correct code for unlocking the lock, while the other switch serves as a data entry point for the person trying to open the lock. In real life, of course, the switch assembly with the "key" code set on it must be hidden from the sight of the person opening the lock, which means it must be physically located elsewhere from where the data entry switch assembly is. This requires two switch assemblies. However, if you understand this concept clearly, you may build a working circuit with only one 8-position switch, using the left four switches for data entry and the right four switches to hold the "key" code.
For extra effect, choose different colors of LED: green for "Go" and red for "No go."

CROSS-REFERENCES
Lessons In Electric Circuits, Volume 4, chapter 3: "Logic Gates"

LEARNING OBJECTIVES
  • Using XOR gates as bit comparators
  • How to build simple gate functions with diodes and a pullup/down resistor
  • Using NOR gates as controlled inverters

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
This circuit illustrates the use of XOR (Exclusive-OR) gates as bit comparators. Four of these XOR gates compare the respective bits of two 4-bit binary numbers, each number "entered" into the circuit via a set of switches. If the two numbers match, bit for bit, the green "Go" LED will light up when the "Enter" pushbutton switch is pressed. If the two numbers do not exactly match, the red "No go" LED will light up when the "Enter" pushbutton is pressed.
Because four bits provides a mere sixteen possible combinations, this lock circuit is not very sophisticated. If it were used in a real application such as a home security system, the "No go" output would have to be connected to some kind of siren or other alarming device, so that the entry of an incorrect code would deter an unauthorized person from attempting another code entry. Otherwise, it would not take much time to try all combinations (0000 through 1111) until the correct one was found! In this experiment, I do not describe how to work this circuit into a real security system or lock mechanism, but only how to make it recognize a pre-entered code.
The "key" code that must be matched at the data entry switch array should be hidden from view, of course. If this were part of a real security system, the data entry switch assembly would be located outside the door, and the key code switch assembly behind the door with the rest of the circuitry. In this experiment, you will likely locate the two switch assemblies on two different breadboards, but it is entirely possible to build the circuit using just a single (8-position) DIP switch assembly. Again, the purpose of the experiment is not to make a real security system, but merely to introduce you to the principle of XOR gate code comparison.
It is the nature of an XOR gate to output a "high" (1) signal if the input signals are not the same logic state. The four XOR gates' output terminals are connected through a diode network which functions as a four-input OR gate: if any of the four XOR gates outputs a "high" signal -- indicating that the entered code and the key code are not identical -- then a "high" signal will be passed on to the NOR gate logic. If the two 4-bit codes are identical, then none of the XOR gate outputs will be "high," and the pull-down resistor connected to the common sides of the diodes will provide a "low" signal state to the NOR logic.
The NOR gate logic performs a simple task: prevent either of the LEDs from turning on if the "Enter" pushbutton is not pressed. Only when this pushbutton is pressed can either of the LEDs energize. If the Enter switch is pressed and the XOR outputs are all "low," the "Go" LED will light up, indicating that the correct code has been entered. If the Enter switch is pressed and any of the XOR outputs are "high," the "No go" LED will light up, indicating that an incorrect code has been entered. Again, if this were a real security system, it would be wise to have the "No go" output do something that deters an unauthorized person from discovering the correct code by trial-and-error. In other words, there should be some sort of penalty for entering an incorrect code. Let your imagination guide your design of this detail!

3-bit binary counter

PARTS AND MATERIALS
  • 555 timer IC (Radio Shack catalog # 276-1723)
  • One 1N914 "switching" diode (Radio Shack catalog # 276-1122)
  • Two 10 kΩ resistors
  • One 100 µF capacitor (Radio Shack catalog # 272-1028)
  • 4027 dual J-K flip-flop (Radio Shack catalog # 900-4394)
  • Ten-segment bargraph LED (Radio Shack catalog # 276-081)
  • Three 470 Ω resistors
  • One 6 volt battery
Caution! The 4027 IC is CMOS, and therefore sensitive to static electricity!

CROSS-REFERENCES
Lessons In Electric Circuits, Volume 4, chapter 10: "Multivibrators"
Lessons In Electric Circuits, Volume 4, chapter 11: "Counters"

LEARNING OBJECTIVES
  • Using the 555 timer as a square-wave oscillator
  • How to make an asynchronous counter using J-K flip-flops

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
In a sense, this circuit "cheats" by using only two J-K flip-flops to make a three-bit binary counter. Ordinarily, three flip-flops would be used -- one for each binary bit -- but in this case we can use the clock pulse (555 timer output) as a bit of its own. When you build this circuit, you will find that it is a "down" counter. That is, its count sequence goes from 111 to 110 to 101 to 100 to 011 to 010 to 001 to 000 and then back to 111. While it is possible to construct an "up" counter using J-K flip-flops, this would require additional components and introduce more complexity into the circuit.
The 555 timer operates as a slow, square-wave oscillator with a duty cycle of approximately 50 percent. This duty cycle is made possible by the use of a diode to "bypass" the lower resistor during the capacitor's charging cycle, so that the charging time constant is only RC and not 2RC as it would be without the diode in place.
It is highly recommended, in this experiment as in all experiments, to build the circuit in stages: identify portions of the circuit with specific functions, and build those portions one at a time, testing each one and verifying its performance before building the next. A very common mistake of new electronics students is to build an entire circuit at once without testing sections of it during the construction process, and then be faced with the possibility of several problems simultaneously when it comes time to finally apply power to it. Remember that a small amount of extra attention paid to detail near the beginning of a project is worth an enormous amount of troubleshooting work near the end! Students who make the mistake of not testing circuit portions before attempting to operate the entire circuit often (falsely) think that the time it would take to test those sections is not worth it, and then spend days trying to figure out what the problem(s) might be with their experiment.
Following this philosophy, build the 555 timer circuit first, before even plugging the 4027 IC into the breadboard. Connect the 555's output (pin #3) to the "Least Significant Bit" (LSB) LED, so that you have visual indication of its status. Make sure that the output oscillates in a slow, square-wave pattern (LED is "lit" for about as long as it is "off" in a cycle), and that it is a reliable signal (no erratic behavior, no unexplained pauses). If the 555 timer is not working properly, neither will the rest of the counter circuit! Once the timer circuit has been proven good, proceed to plug the 4027 IC into the breadboard and complete the rest of the necessary connections between it, the 555 timer circuit, and the LED assembly.

7-segment display

PARTS AND MATERIALS
  • 4511 BCD-to-7seg latch/decoder/driver (Radio Shack catalog # 900-4437)
  • Common-cathode 7-segment LED display (Radio Shack catalog # 276-075)
  • Eight-position DIP switch (Radio Shack catalog # 275-1301)
  • Four 10 kΩ resistors
  • Seven 470 Ω resistors
  • One 6 volt battery
Caution! The 4511 IC is CMOS, and therefore sensitive to static electricity!

CROSS-REFERENCES
Lessons In Electric Circuits, Volume 4, chapter 9: "Combinational Logic Functions"

LEARNING OBJECTIVES
  • How to use the 4511 7-segment decoder/display driver IC
  • Gain familiarity with the BCD code
  • How to use 7-segment LED assemblies to create decimal digit displays
  • How to identify and use both "active-low" and "active-high" logic inputs

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
This experiment is more of an introduction to the 4511 decoder/display driver IC than it is a lesson in how to "build up" a digital function from lower-level components. Since 7-segment displays are very common components of digital devices, it is good to be familiar with the "driving" circuits behind them, and the 4511 is a good example of a typical driver IC.
Its operating principle is to input a four-bit BCD (Binary-Coded Decimal) value, and energize the proper output lines to form the corresponding decimal digit on the 7-segment LED display. The BCD inputs are designated A, B, C, and D in order from least-significant to most-significant. Outputs are labeled a, b, c, d, e, f, and g, each letter corresponding to a standardized segment designation for 7-segment displays. Of course, since each LED segment requires its own dropping resistor, we must use seven 470 Ω resistors placed in series between the 4511's output terminals and the corresponding terminals of the display unit.
Most 7-segment displays also provide for a decimal point (sometimes two!), a separate LED and terminal designated for its operation. All LEDs inside the display unit are made common to each other on one side, either cathode or anode. The 4511 display driver IC requires a common-cathode 7-segment display unit, and so that is what is used here.
After building the circuit and applying power, operate the four switches in a binary counting sequence (0000 to 1111), noting the 7-segment display. A 0000 input should result in a decimal "0" display, a 0001 input should result in a decimal "1" display, and so on through 1001 (decimal "9"). What happens for the binary numbers 1010 (10) through 1111 (15)? Read the datasheet on the 4511 IC and see what the manufacturer specifies for operation above an input value of 9. In the BCD code, there is no real meaning for 1010, 1011, 1100, 1101, 1110, or 1111. These are binary values beyond the range of a single decimal digit, and so have no function in a BCD system. The 4511 IC is built to recognize this, and output (or not output!) accordingly.
Three inputs on the 4511 chip have been permanently connected to either Vdd or ground: the "Lamp Test," "Blanking Input," and "Latch Enable." To learn what these inputs do, remove the short jumpers connecting them to either power supply rail (one at a time!), and replace the short jumper with a longer one that can reach the other power supply rail. For example, remove the short jumper connecting the "Latch Enable" input (pin #5) to ground, and replace it with a long jumper wire that can reach all the way to the Vdd power supply rail. Experiment with making this input "high" and "low," observing the results on the 7-segment display as you alter the BCD code with the four input switches. After you've learned what the input's function is, connect it to the power supply rail enabling normal operation, and proceed to experiment with the next input (either "Lamp Test" or "Blanking Input").
Once again, the manufacturer's datasheet will be informative as to the purpose of each of these three inputs. Note that the "Lamp Test" (LT) and "Blanking Input" (BI) input labels are written with boolean complementation bars over the abbreviations. Bar symbols designate these inputs as active-low, meaning that you must make each one "low" in order to invoke its particular function. Making an active-low input "high" places that particular input into a "passive" state where its function will not be invoked. Conversely, the "Latch Enable" (LE) input has no complementation bar written over its abbreviation, and correspondingly it is shown connected to ground ("low") in the schematic so as to not invoke that function. The "Latch Enable" input is an active-high input, which means it must be made "high" (connected to Vdd) in order to invoke its function.

ANALOG INTEGRATED CIRCUITS (Full Chapter)

Introduction:

Analog circuits are circuits dealing with signals free to vary from zero to full power supply voltage. This stands in contrast to digital circuits, which almost exclusively employ "all or nothing" signals: voltages restricted to values of zero and full supply voltage, with no valid state in between those extreme limits. Analog circuits are often referred to as linear circuits to emphasize the valid continuity of signal range forbidden in digital circuits, but this label is unfortunately misleading. Just because a voltage or current signal is allowed to vary smoothly between the extremes of zero and full power supply limits does not necessarily mean that all mathematical relationships between these signals are linear in the "straight-line" or "proportional" sense of the word. As you will see in this chapter, many so-called "linear" circuits are quite nonlinear in their behavior, either by necessity of physics or by design.
The circuits in this chapter make use of IC, or integrated circuit, components. Such components are actually networks of interconnected components manufactured on a single wafer of semiconducting material. Integrated circuits providing a multitude of pre-engineered functions are available at very low cost, benefitting students, hobbyists and professional circuit designers alike. Most integrated circuits provide the same functionality as "discrete" semiconductor circuits at higher levels of reliability and at a fraction of the cost. Usually, discrete-component circuit construction is favored only when power dissipation levels are too high for integrated circuits to handle.
Perhaps the most versatile and important analog integrated circuit for the student to master is the operational amplifier, or op-amp. Essentially nothing more than a differential amplifier with very high voltage gain, op-amps are the workhorse of the analog design world. By cleverly applying feedback from the output of an op-amp to one or more of its inputs, a wide variety of behaviors may be obtained from this single device. Many different models of op-amp are available at low cost, but circuits described in this chapter will incorporate only commonly available op-amp models.

Voltage comparator

PARTS AND MATERIALS
  • Operational amplifier, model 1458 or 353 recommended (Radio Shack catalog # 276-038 and 900-6298, respectively)
  • Three 6 volt batteries
  • Two 10 kΩ potentiometers, linear taper (Radio Shack catalog # 271-1715)
  • One light-emitting diode (Radio Shack catalog # 276-026 or equivalent)
  • One 330 Ω resistor
  • One 470 Ω resistor
This experiment only requires a single operational amplifier. The model 1458 and 353 are both "dual" op-amp units, with two complete amplifier circuits housed in the same 8-pin DIP package. I recommend that you purchase and use "dual" op-amps over "single" op-amps even if a project only requires one, because they are more versatile (the same op-amp unit can function in projects requiring only one amplifier as well as in projects requiring two). In the interest of purchasing and stocking the least number of components for your home laboratory, this makes sense.

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

LEARNING OBJECTIVES
  • How to use an op-amp as a comparator

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
A comparator circuit compares two voltage signals and determines which one is greater. The result of this comparison is indicated by the output voltage: if the op-amp's output is saturated in the positive direction, the noninverting input (+) is a greater, or more positive, voltage than the inverting input (-), all voltages measured with respect to ground. If the op-amp's voltage is near the negative supply voltage (in this case, 0 volts, or ground potential), it means the inverting input (-) has a greater voltage applied to it than the noninverting input (+).
This behavior is much easier understood by experimenting with a comparator circuit than it is by reading someone's verbal description of it. In this experiment, two potentiometers supply variable voltages to be compared by the op-amp. The output status of the op-amp is indicated visually by the LED. By adjusting the two potentiometers and observing the LED, one can easily comprehend the function of a comparator circuit.
For greater insight into this circuit's operation, you might want to connect a pair of voltmeters to the op-amp input terminals (both voltmeters referenced to ground) so that both input voltages may be numerically compared with each other, these meter indications compared to the LED status:

Comparator circuits are widely used to compare physical measurements, provided those physical variables can be translated into voltage signals. For instance, if a small generator were attached to an anemometer wheel to produce a voltage proportional to wind speed, that wind speed signal could be compared with a "set-point" voltage and compared by an op-amp to drive a high wind speed alarm:


Precision voltage follower

PARTS AND MATERIALS
  • Operational amplifier, model 1458 or 353 recommended (Radio Shack catalog # 276-038 and 900-6298, respectively)
  • Three 6 volt batteries
  • One 10 kΩ potentiometer, linear taper (Radio Shack catalog # 271-1715)

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

LEARNING OBJECTIVES
  • How to use an op-amp as a voltage follower
  • Purpose of negative feedback
  • Troubleshooting strategy

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
In the previous op-amp experiment, the amplifier was used in "open-loop" mode; that is, without any feedback from output to input. As such, the full voltage gain of the operational amplifier was available, resulting in the output voltage saturating for virtually any amount of differential voltage applied between the two input terminals. This is good if we desire comparator operation, but if we want the op-amp to behave as a true amplifier, we need it to exhibit a manageable voltage gain.
Since we do not have the luxury of disassembling the integrated circuitry of the op-amp and changing resistor values to give a lesser voltage gain, we are limited to external connections and componentry. Actually, this is not a disadvantage as one might think, because the combination of extremely high open-loop voltage gain coupled with feedback allows us to use the op-amp for a much wider variety of purposes, much easier than if we were to exercise the option of modifying its internal circuitry.
If we connect the output of an op-amp to its inverting (-) input, the output voltage will seek whatever level is necessary to balance the inverting input's voltage with that applied to the noninverting (+) input. If this feedback connection is direct, as in a straight piece of wire, the output voltage will precisely "follow" the noninverting input's voltage. Unlike the voltage follower circuit made from a single transistor (see chapter 5: Discrete Semiconductor Circuits), which approximated the input voltage to within several tenths of a volt, this voltage follower circuit will output a voltage accurate to within mere microvolts of the input voltage!
Measure the input voltage of this circuit with a voltmeter connected between the op-amp's noninverting (+) input terminal and circuit ground (the negative side of the power supply), and the output voltage between the op-amp's output terminal and circuit ground. Watch the op-amp's output voltage follow the input voltage as you adjust the potentiometer through its range.
You may directly measure the difference, or error, between output and input voltages by connecting the voltmeter between the op-amp's two input terminals. Throughout most of the potentiometer's range, this error voltage should be almost zero.
Try moving the potentiometer to one of its extreme positions, far clockwise or far counterclockwise. Measure error voltage, or compare output voltage against input voltage. Do you notice anything unusual? If you are using the model 1458 or model 353 op-amp for this experiment, you should measure a substantial error voltage, or difference between output and input. Many op-amps, the specified models included, cannot "swing" their output voltage exactly to full power supply ("rail") voltage levels. In this case, the "rail" voltages are +18 volts and 0 volts, respectively. Due to limitations in the 1458's internal circuitry, its output voltage is unable to exactly reach these high and low limits. You may find that it can only go within a volt or two of the power supply "rails." This is a very important limitation to understand when designing circuits using operational amplifiers. If full "rail-to-rail" output voltage swing is required in a circuit design, other op-amp models may be selected which offer this capability. The model 3130 is one such op-amp.
Precision voltage follower circuits are useful if the voltage signal to be amplified cannot tolerate "loading;" that is, if it has a high source impedance. Since a voltage follower by definition has a voltage gain of 1, its purpose has nothing to do with amplifying voltage, but rather with amplifying a signal's capacity to deliver current to a load.
Voltage follower circuits have another important use for circuit builders: they allow for simple linear testing of an op-amp. One of the troubleshooting techniques I recommend is to simplify and rebuild. Suppose that you are building a circuit using one or more op-amps to perform some advanced function. If one of those op-amps seems to be causing a problem and you suspect it may be faulty, try re-connecting it as a simple voltage follower and see if it functions in that capacity. An op-amp that fails to work as a voltage follower certainly won't work as anything more complex!

COMPUTER SIMULATION
Schematic with SPICE node numbers:


Netlist (make a text file containing the following text, verbatim):
Voltage follower
vinput 1 0
rbogus 1 0 1meg
e1 2 0 1 2 999meg
rload 2 0 10k
.dc vinput 5 5 1
.print dc v(1,0) v(2,0) v(1,2)
.end

An ideal operational amplifier may be simulated in SPICE using a dependent voltage source (e1 in the netlist). The output nodes are specified first (2 0), then the two input nodes, non-inverting input first (1 2). Open-loop gain is specified last (999meg) in the dependent voltage source line.
Because SPICE views the input impedance of a dependent source as infinite, some finite amount of resistance must be included to avoid an analysis error. This is the purpose of Rbogus: to provide DC path to ground for the Vinput voltage source. Such "bogus" resistances should be arbitrarily large. In this simulation I chose 1 MΩ for an Rbogus value.
A load resistor is included in the circuit for much the same reason: to provide a DC path for current at the output of the dependent voltage source. As you can see, SPICE doesn't like open circuits!

Noninverting amplifier

PARTS AND MATERIALS
  • Operational amplifier, model 1458 or 353 recommended (Radio Shack catalog # 276-038 and 900-6298, respectively)
  • Three 6 volt batteries
  • Two 10 kΩ potentiometers, linear taper (Radio Shack catalog # 271-1715)

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

LEARNING OBJECTIVES
  • How to use an op-amp as a single-ended amplifier
  • Using divided, negative feedback

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
This circuit differs from the voltage follower in only one respect: output voltage is "fed back" to the inverting (-) input through a voltage-dividing potentiometer rather than being directly connected. With only a fractionmultiple of the voltage sensed at the noninverting (+) input in keeping the input differential voltage near zero. In other words, the op-amp will now function as an amplifier with a controllable voltage gain, that gain being established by the position of the feedback potentiometer (R2). of the output voltage fed back to the inverting input, the op-amp will output a corresponding
Set R2 to approximately mid-position. This should give a voltage gain of about 2. Measure both input and output voltage for several positions of the input potentiometer R1. Move R2 to a different position and re-take voltage measurements for several positions of R1. For any given R2 position, the ratio between output and input voltage should be the same.
You will also notice that the input and output voltages are always positive with respect to ground. Because the output voltage increases in a positive direction for a positive increase of the input voltage, this amplifier is referred to as noninverting. If the output and input voltages were related to one another in an inverse fashion (i.e. positive increasing input voltage results in positive decreasing or negative increasing output), then the amplifier would be known as an inverting type.
The ability to leverage an op-amp in this fashion to create an amplifier with controllable voltage gain makes this circuit an extremely useful one. It would take quite a bit more design and troubleshooting effort to produce a similar circuit using discrete transistors.
Try adjusting R2 for maximum and minimum voltage gain. What is the lowest voltage gain attainable with this amplifier configuration? Why do you think this is?

COMPUTER SIMULATION
Schematic with SPICE node numbers:


Netlist (make a text file containing the following text, verbatim):
Noninverting amplifier
vinput 1 0
r2 3 2 5k
r1 2 0 5k
rbogus 1 0 1meg
e1 3 0 1 2 999meg
rload 3 0 10k
.dc vinput 5 5 1
.print dc v(1,0) v(3,0)
.end

With R1 and R2 set equally to 5 kΩ in the simulation, it mimics the feedback potentiometer of the real circuit at mid-position (50%). To simulate the potentiometer at the 75% position, set R2 to 7.5 kΩ and R1 to 2.5 kΩ.

High-impedance voltmeter

PARTS AND MATERIALS
  • Operational amplifier, model TL082 recommended (Radio Shack catalog # 276-1715)
  • Operational amplifier, model LM1458 recommended (Radio Shack catalog # 276-038)
  • Four 6 volt batteries
  • One meter movement, 1 mA full-scale deflection (Radio Shack catalog #22-410)
  • 15 kΩ precision resistor
  • Four 1 MΩ resistors
The 1 mA meter movement sold by Radio Shack is advertised as a 0-15 VDC meter, but is actually a 1 mA movement sold with a 15 kΩ +/- 1% tolerance multiplier resistor. If you get this Radio Shack meter movement, you can use the included 15 kΩ resistor for the resistor specified in the parts list.
This meter experiment is based on a JFET-input op-amp such as the TL082. The other op-amp (model 1458) is used in this experiment to demonstrate the absence of latch-up: a problem inherent to the TL082.
You don't need 1 MΩ resistors, exactly. Any very high resistance resistors will suffice.

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

LEARNING OBJECTIVES
  • Voltmeter loading: its causes and its solution
  • How to make a high-impedance voltmeter using an op-amp
  • What op-amp "latch-up" is and how to avoid it

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
An ideal voltmeter has infinite input impedance, meaning that it draws zero current from the circuit under test. This way, there will be no "impact" on the circuit as the voltage is being measured. The more current a voltmeter draws from the circuit under test, the more the measured voltage will "sag" under the loading effect of the meter, like a tire-pressure gauge releasing air out of the tire being measured: the more air released from the tire, the more the tire's pressure will be impacted in the act of measurement. This loading is more pronounced on circuits of high resistance, like the voltage divider made of 1 MΩ resistors, shown in the schematic diagram.
If you were to build a simple 0-15 volt range voltmeter by connecting the 1 mA meter movement in series with the 15 kΩ precision resistor, and try to use this voltmeter to measure the voltages at TP1, TP2, or TP3 (with respect to ground), you'd encounter severe measurement errors induced by meter "impact:"

Try using the meter movement and 15 kΩ resistor as shown to measure these three voltages. Does the meter read falsely high or falsely low? Why do you think this is?
If we were to increase the meter's input impedance, we would diminish its current draw or "load" on the circuit under test and consequently improve its measurement accuracy. An op-amp with high-impedance inputs (using a JFET transistor input stage rather than a BJT input stage) works well for this application.
Note that the meter movement is part of the op-amp's feedback loop from output to inverting input. This circuit drives the meter movement with a current proportional to the voltage impressed at the noninverting (+) input, the requisite current supplied directly from the batteries through the op-amp's power supply pins, not from the circuit under test through the test probe. The meter's range is set by the resistor connecting the inverting (-) input to ground.
Build the op-amp meter circuit as shown and re-take voltage measurements at TP1, TP2, and TP3. You should enjoy far better success this time, with the meter movement accurately measuring these voltages (approximately 3, 6, and 9 volts, respectively).
You may witness the extreme sensitivity of this voltmeter by touching the test probe with one hand and the most positive battery terminal with the other. Notice how you can drive the needle upward on the scale simply by measuring battery voltage through your body resistance: an impossible feat with the original, unamplified voltmeter circuit. If you touch the test probe to ground, the meter should read exactly 0 volts.
After you've proven this circuit to work, modify it by changing the power supply from dual to split. This entails removing the center-tap ground connection between the 2nd and 3rd batteries, and grounding the far negative battery terminal instead:

This alteration in the power supply increases the voltages at TP1, TP2, and TP3 to 6, 12, and 18 volts, respectively. With a 15 kΩ range resistor and a 1 mA meter movement, measuring 18 volts will gently "peg" the meter, but you should be able to measure the 6 and 12 volt test points just fine.
Try touching the meter's test probe to ground. This should drive the meter needle to exactly 0 volts as before, but it will not! What is happening here is an op-amp phenomenon called latch-up: where the op-amp output drives to a positive voltage when the input common-mode voltage exceeds the allowable limit. In this case, as with many JFET-input op-amps, neither input should be allowed to come close to either power supply rail voltage. With a single supply, the op-amp's negative power rail is at ground potential (0 volts), so grounding the test probe brings the noninverting (+) input exactly to that rail voltage. This is bad for a JFET op-amp, and drives the output strongly positive, even though it doesn't seem like it should, based on how op-amps are supposed to function.
When the op-amp ran on a "dual" supply (+12/-12 volts, rather than a "single" +24 volt supply), the negative power supply rail was 12 volts away from ground (0 volts), so grounding the test probe didn't violate the op-amp's common-mode voltage limit. However, with the "single" +24 volt supply, we have a problem. Note that some op-amps do not "latch-up" the way the model TL082 does. You may replace the TL082 with an LM1458 op-amp, which is pin-for-pin compatible (no breadboard wiring changes needed). The model 1458 will not "latch-up" when the test probe is grounded, although you may still get incorrect meter readings with the measured voltage exactly equal to the negative power supply rail. As a general rule, you should always be sure the op-amp's power supply rail voltages exceed the expected input voltages.

Integrator

PARTS AND MATERIALS
  • Four 6 volt batteries
  • Operational amplifier, model 1458 recommended (Radio Shack catalog # 276-038)
  • One 10 kΩ potentiometer, linear taper (Radio Shack catalog # 271-1715)
  • Two capacitors, 0.1 µF each, non-polarized (Radio Shack catalog # 272-135)
  • Two 100 kΩ resistors
  • Three 1 MΩ resistors
Just about any operational amplifier model will work fine for this integrator experiment, but I'm specifying the model 1458 over the 353 because the 1458 has much higher input bias currents. Normally, high input bias current is a bad characteristic for an op-amp to have in a precision DC amplifier circuit (and especially an integrator circuit!). However, I want the bias current to be high in order that its bad effects may be exaggerated, and so that you will learn one method of counteracting its effects.

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

LEARNING OBJECTIVES
  • Method for limiting the span of a potentiometer
  • Purpose of an integrator circuit
  • How to compensate for op-amp bias current

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
As you can see from the schematic diagram, the potentiometer is connected to the "rails" of the power source through 100 kΩ resistors, one on each end. This is to limit the span of the potentiometer, so that full movement produces a fairly small range of input voltages for the op-amp to operate on. At one extreme of the potentiometer's motion, a voltage of about 0.5 volt (with respect the the ground point in the middle of the series battery string) will be produced at the potentiometer wiper. At the other extreme of motion, a voltage of about -0.5 volt will be produced. When the potentiometer is positioned dead-center, the wiper voltage should measure zero volts.
Connect a voltmeter between the op-amp's output terminal and the circuit ground point. Slowly move the potentiometer control while monitoring the output voltage. The output voltage should be changing at a rate established by the potentiometer's deviation from zero (center) position. To use calculus terms, we would say that the output voltage represents the integral (with respect to time) of the input voltage function. That is, the input voltage level establishes the output voltage rate of change over time. This is precisely the opposite of differentiation, where the derivative of a signal or function is its instantaneous rate of change.
If you have two voltmeters, you may readily see this relationship between input voltage and output voltage rate of change by measuring the wiper voltage (between the potentiometer wiper and ground) with one meter and the output voltage (between the op-amp output terminal and ground) with the other. Adjusting the potentiometer to give zero volts should result in the slowest output voltage rate-of-change. Conversely, the more voltage input to this circuit, the faster its output voltage will change, or "ramp."
Try connecting the second 0.1 µF capacitor in parallel with the first. This will double the amount of capacitance in the op-amp's feedback loop. What affect does this have on the circuit's integration rate for any given potentiometer position?
Try connecting another 1 MΩ resistor in parallel with the input resistor (the resistor connecting the potentiometer wiper to the inverting terminal of the op-amp). This will halve the integrator's input resistance. What affect does this have on the circuit's integration rate?
Integrator circuits are one of the fundamental "building-block" functions of an analog computer. By connecting integrator circuits with amplifiers, summers, and potentiometers (dividers), almost any differential equation could be modeled, and solutions obtained by measuring voltages produced at various points in the network of circuits. Because differential equations describe so many physical processes, analog computers are useful as simulators. Before the advent of modern digital computers, engineers used analog computers to simulate such processes as machinery vibration, rocket trajectory, and control system response. Even though analog computers are considered obsolete by modern standards, their constituent components still work well as learning tools for calculus concepts.
Move the potentiometer until the op-amp's output voltage is as close to zero as you can get it, and moving as slowly as you can make it. Disconnect the integrator input from the potentiometer wiper terminal and connect it instead to ground, like this:




Applying exactly zero voltage to the input of an integrator circuit should, ideally, cause the output voltage rate-of-change to be zero. When you make this change to the circuit, you should notice the output voltage remaining at a constant level or changing very slowly.
With the integrator input still shorted to ground, short past the 1 MΩ resistor connecting the op-amp's noninverting (+) input to ground. There should be no need for this resistor in an ideal op-amp circuit, so by shorting past it we will see what function it provides in this very real op-amp circuit:



As soon as the "grounding" resistor is shorted with a jumper wire, the op-amp's output voltage will start to change, or drift. Ideally, this should not happen, because the integrator circuit still has an input signal of zero volts. However, real operational amplifiers have a very small amount of current entering each input terminal called the bias current. These bias currents will drop voltage across any resistance in their path. Since the 1 MΩ input resistor conducts some amount of bias current regardless of input signal magnitude, it will drop voltage across its terminals due to bias current, thus "offsetting" the amount of signal voltage seen at the inverting terminal of the op-amp. If the other (noninverting) input is connected directly to ground as we have done here, this "offset" voltage incurred by voltage drop generated by bias current will cause the integrator circuit to slowly "integrate" as though it were receiving a very small input signal.
The "grounding" resistor is better known as a compensating resistor, because it acts to compensate for voltage errors created by bias current. Since the bias currents through each op-amp input terminal are approximately equal to each other, an equal amount of resistance placed in the path of each bias current will produce approximately the same voltage drop. Equal voltage drops seen at the complementary inputs of an op-amp cancel each other out, thus nulling the error otherwise induced by bias current.
Remove the jumper wire shorting past the compensating resistor and notice how the op-amp output returns to a relatively stable state. It may still drift some, most likely due to bias voltage error in the op-amp itself, but that is another subject altogether!

COMPUTER SIMULATION
Schematic with SPICE node numbers:


Netlist (make a text file containing the following text, verbatim):
DC integrator
vinput 1 0 dc 0.05
r1 1 2 1meg
c1 2 3 0.1u ic=0
e1 3 0 0 2 999k
.tran 1 30 uic
.plot tran v(1,0) v(3,0)
.end



555 audio oscillator

PARTS AND MATERIALS
  • Two 6 volt batteries
  • One capacitor, 0.1 µF, non-polarized (Radio Shack catalog # 272-135)
  • One 555 timer IC (Radio Shack catalog # 276-1723)
  • Two light-emitting diodes (Radio Shack catalog # 276-026 or equivalent)
  • One 1 MΩ resistor
  • One 100 kΩ resistor
  • Two 510 Ω resistors
  • Audio detector with headphones
  • Oscilloscope (recommended, but not necessary)
A oscilloscope would be useful in analyzing the waveforms produced by this circuit, but it is not essential. An audio detector is a very useful piece of test equipment for this experiment, especially if you don't have an oscilloscope.

CROSS-REFERENCES
Lessons In Electric Circuits, Volume 4, chapter 10: "Multivibrators"

LEARNING OBJECTIVES
  • How to use the 555 timer as an astable multivibrator
  • Working knowledge of duty cycle

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
The "555" integrated circuit is a general-purpose timer useful for a variety of functions. In this experiment, we explore its use as an astable multivibrator, or oscillator. Connected to a capacitor and two resistors as shown, it will oscillate freely, driving the LEDs on and off with a square-wave output voltage.
This circuit works on the principle of alternately charging and discharging a capacitor. The 555 begins to discharge the capacitor by grounding the Disch terminal when the voltage detected by the Thresh terminal exceeds 2/3 the power supply voltage (Vcc). It stops discharging the capacitor when the voltage detected by the Trig terminal falls below 1/3 the power supply voltage. Thus, when both Thresh and Trig terminals are connected to the capacitor's positive terminal, the capacitor voltage will cycle between 1/3 and 2/3 power supply voltage in a "sawtooth" pattern.
During the charging cycle, the capacitor receives charging current through the series combination of the 1 MΩ and 100 kΩ resistors. As soon as the Disch terminal on the 555 timer goes to ground potential (a transistor inside the 555 connected between that terminal and ground turns on), the capacitor's discharging current only has to go through the 100 kΩ resistor. The result is an RC time constant that is much longer for charging than for discharging, resulting in a charging time greatly exceeding the discharging time.
The 555's Out terminal produces a square-wave voltage signal that is "high" (nearly Vcc) when the capacitor is charging, and "low" (nearly 0 volts) when the capacitor is discharging. This alternating high/low voltage signal drives the two LEDs in opposite modes: when one is on, the other will be off. Because the capacitor's charging and discharging times are unequal, the "high" and "low" times of the output's square-wave waveform will be unequal as well. This can be seen in the relative brightness of the two LEDs: one will be much brighter than the other, because it is on for a longer period of time during each cycle.
The equality or inequality between "high" and "low" times of a square wave is expressed as that wave's duty cycle. A square wave with a 50% duty cycle is perfectly symmetrical: its "high" time is precisely equal to its "low" time. A square wave that is "high" 10% of the time and "low" 90% of the time is said to have a 10% duty cycle. In this circuit, the output waveform has a "high" time exceeding the "low" time, resulting in a duty cycle greater than 50%.
Use the audio detector (or an oscilloscope) to investigate the different voltage waveforms produced by this circuit. Try different resistor values and/or capacitor values to see what effects they have on output frequency or charge/discharge times.

555 ramp generator

PARTS AND MATERIALS
  • Two 6 volt batteries
  • One capacitor, 470 µF electrolytic, 35 WVDC (Radio Shack catalog # 272-1030 or equivalent)
  • One capacitor, 0.1 µF, non-polarized (Radio Shack catalog # 272-135)
  • One 555 timer IC (Radio Shack catalog # 276-1723)
  • Two PNP transistors -- models 2N2907 or 2N3906 recommended (Radio Shack catalog # 276-1604 is a package of fifteen PNP transistors ideal for this and other experiments)
  • Two light-emitting diodes (Radio Shack catalog # 276-026 or equivalent)
  • One 100 kΩ resistor
  • One 47 kΩ resistor
  • Two 510 Ω resistors
  • Audio detector with headphones
The voltage rating on the 470 µF capacitor is not critical, so long as it generously exceeds the maximum power supply voltage. In this particular circuit, that maximum voltage is 12 volts. Be sure you connect this capacitor in the circuit properly, respecting polarity!

CROSS-REFERENCES
Lessons In Electric Circuits, Volume 1, chapter 13: "Capacitors"
Lessons In Electric Circuits, Volume 4, chapter 10: "Multivibrators"

LEARNING OBJECTIVES
  • How to use the 555 timer as an astable multivibrator
  • A practical use for a current mirror circuit
  • Understanding the relationship between capacitor current and capacitor voltage rate-of-change

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
Again, we are using a 555 timer IC as an astable multivibrator, or oscillator. This time, however, we will compare its operation in two different capacitor-charging modes: traditional RC and constant-current.
Connecting test point #1 (TP1) to test point #3 (TP3) using a jumper wire. This allows the capacitor to charge through a 47 kΩ resistor. When the capacitor has reached 2/3 supply voltage, the 555 timer switches to "discharge" mode and discharges the capacitor to a level of 1/3 supply voltage almost immediately. The charging cycle begins again at this point. Measure voltage directly across the capacitor with a voltmeter (a digital voltmeter is preferred), and note the rate of capacitor charging over time. It should rise quickly at first, then taper off as it builds up to 2/3 supply voltage, just as you would expect from an RC charging circuit.
Remove the jumper wire from TP3, and re-connect it to TP2. This allows the capacitor to be charged through the controlled-current leg of a current mirror circuit formed by the two PNP transistors. Measure voltage directly across the capacitor again, noting the difference in charging rate over time as compared to the last circuit configuration.
By connecting TP1 to TP2, the capacitor receives a nearly constant charging current. Constant capacitor charging current yields a voltage curve that is linear, as described by the equation i = C(de/dt). If the capacitor's current is constant, so will be its rate-of-change of voltage over time. The result is a "ramp" waveform rather than a "sawtooth" waveform:

The capacitor's charging current may be directly measured by substituting an ammeter in place of the jumper wire. The ammeter will need to be set to measure a current in the range of hundreds of microamps (tenths of a milliamp). Connected between TP1 and TP3, you should see a current that starts at a relatively high value at the beginning of the charging cycle, and tapers off toward the end. Connected between TP1 and TP2, however, the current will be much more stable.
It is an interesting experiment at this point to change the temperature of either current mirror transistor by touching it with your finger. As the transistor warms, it will conduct more collector current for the same base-emitter voltage. If the controlling transistor (the one connected to the 100 kΩ resistor) is touched, the current decreases. If the controlled transistor is touched, the current increases. For the most stable current mirror operation, the two transistors should be cemented together so that their temperatures never differ by any substantial amount.
This circuit works just as well at high frequencies as it does at low frequencies. Replace the 470 µF capacitor with a 0.1 µF capacitor, and use an audio detector to sense the voltage waveform at the 555's output terminal. The detector should produce an audio tone that is easy to hear. The capacitor's voltage will now be changing much too fast to view with a voltmeter in the DC mode, but we can still measure capacitor current with an ammeter.
With the ammeter connected between TP1 and TP3 (RC mode), measure both DC microamps and AC microamps. Record these current figures on paper. Now, connect the ammeter between TP1 and TP2 (constant-current mode). Measure both DC microamps and AC microamps, noting any differences in current readings between this circuit configuration and the last one. Measuring AC current in addition to DC current is an easy way to determine which circuit configuration gives the most stable charging current. If the current mirror circuit were perfect -- the capacitor charging current absolutely constant -- there would be zero AC current measured by the meter.

PWM power controller

PARTS AND MATERIALS
  • Four 6 volt batteries
  • One capacitor, 100 µF electrolytic, 35 WVDC (Radio Shack catalog # 272-1028 or equivalent)
  • One capacitor, 0.1 µF, non-polarized (Radio Shack catalog # 272-135)
  • One 555 timer IC (Radio Shack catalog # 276-1723)
  • Dual operational amplifier, model 1458 recommended (Radio Shack catalog # 276-038)
  • One NPN power transistor -- (Radio Shack catalog # 276-2041 or equivalent)
  • Three 1N4001 rectifying diodes (Radio Shack catalog # 276-1101)
  • One 10 kΩ potentiometer, linear taper (Radio Shack catalog # 271-1715)
  • One 33 kΩ resistor
  • 12 volt automotive tail-light lamp
  • Audio detector with headphones

CROSS-REFERENCES
Lessons In Electric Circuits, Volume 3, chapter 8: "Operational Amplifiers"
Lessons In Electric Circuits, Volume 2, chapter 7: "Mixed-Frequency AC Signals"

LEARNING OBJECTIVES
  • How to use the 555 timer as an astable multivibrator
  • How to use an op-amp as a comparator
  • How to use diodes to drop unwanted DC voltage
  • How to control power to a load by pulse-width modulation

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
This circuit uses a 555 timer to generate a sawtooth voltage waveform across a capacitor, then compares that signal against a steady voltage provided by a potentiometer, using an op-amp as a comparator. The comparison of these two voltage signals produces a square-wave output from the op-amp, varying in duty cycle according to the potentiometer's position. This variable duty cycle signal then drives the base of a power transistor, switching current on and off through the load. The 555's oscillation frequency is much higher than the lamp filament's ability to thermally cycle (heat and cool), so any variation in duty cycle, or pulse width, has the effect of controlling the total power dissipated by the load over time.

Controlling electrical power through a load by means of quickly switching it on and off, and varying the "on" time, is known as pulse-width modulation, or PWM. It is a very efficient means of controlling electrical power because the controlling element (the power transistor) dissipates comparatively little power in switching on and off, especially if compared to the wasted power dissipated of a rheostat in a similar situation. When the transistor is in cutoff, its power dissipation is zero because there is no current through it. When the transistor is saturated, its dissipation is very low because there is little voltage dropped between collector and emitter while it is conducting current.
PWM is a concept easier understood through experimentation than reading. It would be nice to view the capacitor voltage, potentiometer voltage, and op-amp output waveforms all on one (triple-trace) oscilloscope to see how they relate to one another, and to the load power. However, most of us have no access to a triple-trace oscilloscope, much less any oscilloscope at all, so an alternative method is to slow the 555 oscillator down enough that the three voltages may be compared with a simple DC voltmeter. Replace the 0.1 µF capacitor with one that is 100 µF or larger. This will slow the oscillation frequency down by a factor of at least a thousand, enabling you to measure the capacitor voltage slowly rise over time, and the op-amp output transition from "high" to "low" when the capacitor voltage becomes greater than the potentiometer voltage. With such a slow oscillation frequency, the load power will not be proportioned as before. Rather, the lamp will turn on and off at regular intervals. Feel free to experiment with other capacitor or resistor values to speed up the oscillations enough so the lamp never fully turns on or off, but is "throttled" by quick on-and-off pulsing of the transistor.
When you examine the schematic, you will notice two operational amplifiers connected in parallel. This is done to provide maximum current output to the base terminal of the power transistor. A single op-amp (one-half of a 1458 IC) may not be able to provide sufficient output current to drive the transistor into saturation, so two op-amps are used in tandem. This should only be done if the op-amps in question are overload-protected, which the 1458 series of op-amps are. Otherwise, it is possible (though unlikely) that one op-amp could turn on before the other, and damage result from the two outputs short-circuiting each other (one driving "high" and the other driving "low" simultaneously). The inherent short-circuit protection offered by the 1458 allows for direct driving of the power transistor base without any need for a current-limiting resistor.
The three diodes in series connecting the op-amps' outputs to the transistor's base are there to drop voltage and ensure the transistor falls into cutoff when the op-amp outputs go "low." Because the 1458 op-amp cannot swing its output voltage all the way down to ground potential, but only to within about 2 volts of ground, a direct connection from the op-amp to the transistor would mean the transistor would never fully turn off. Adding three silicon diodes in series drops approximately 2.1 volts (0.7 volts times 3) to ensure there is minimal voltage at the transistor's base when the op-amp outputs go "low."
It is interesting to listen to the op-amp output signal through an audio detector as the potentiometer is adjusted through its full range of motion. Adjusting the potentiometer has no effect on signal frequency, but it greatly affects duty cycle. Note the difference in tone quality, or timbre, as the potentiometer varies the duty cycle from 0% to 50% to 100%. Varying the duty cycle has the effect of changing the harmonic content of the waveform, which makes the tone sound different.
You might notice a particular uniqueness to the sound heard through the detector headphones when the potentiometer is in center position (50% duty cycle -- 50% load power), versus a kind of similarity in sound just above or below 50% duty cycle. This is due to the absence or presence of even-numbered harmonics. Any waveform that is symmetrical above and below its centerline, such as a square wave with a 50% duty cycle, contains no even-numbered harmonics, only odd-numbered. If the duty cycle is below or above 50%, the waveform will not exhibit this symmetry, and there will be even-numbered harmonics. The presence of these even-numbered harmonic frequencies can be detected by the human ear, as some of them correspond to octaves of the fundamental frequency and thus "fit" more naturally into the tone scheme.

Class B audio amplifier

PARTS AND MATERIALS
  • Four 6 volt batteries
  • Dual operational amplifier, model TL082 recommended (Radio Shack catalog # 276-1715)
  • One NPN power transistor in a TO-220 package -- (Radio Shack catalog # 276-2020 or equivalent)
  • One PNP power transistor in a TO-220 package -- (Radio Shack catalog # 276-2027 or equivalent)
  • One 1N914 switching diode (Radio Shack catalog # 276-1620)
  • One capacitor, 47 µF electrolytic, 35 WVDC (Radio Shack catalog # 272-1015 or equivalent)
  • Two capacitors, 0.22 µF, non-polarized (Radio Shack catalog # 272-1070)
  • One 10 kΩ potentiometer, linear taper (Radio Shack catalog # 271-1715)
Be sure to use an op-amp that has a high slew rate. Avoid the LM741 or LM1458 for this reason.
The closer matched the two transistors are, the better. If possible, try to obtain TIP41 and TIP42 transistors, which are closely matched NPN and PNP power transistors with dissipation ratings of 65 watts each. If you cannot get a TIP41 NPN transistor, the TIP3055 (available from Radio Shack) is a good substitute. Do not use very large (i.e. TO-3 case) power transistors, as the op-amp may have trouble driving enough current to their bases for good operation.

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
  • How to build a "push-pull" class B amplifier using complementary bipolar transistors
  • The effects of "crossover distortion" in a push-pull amplifier circuit
  • Using negative feedback via an op-amp to correct circuit nonlinearities

SCHEMATIC DIAGRAM


ILLUSTRATION


INSTRUCTIONS
This project is an audio amplifier suitable for amplifying the output signal from a small radio, tape player, CD player, or any other source of audio signals. For stereo operation, two identical amplifiers must be built, one for the left channel and other for the right channel. To obtain an input signal for this amplifier to amplify, just connect it to the output of a radio or other audio device like this:

This amplifier circuit also works well in amplifying "line-level" audio signals from high-quality, modular stereo components. It provides a surprising amount of sound power when played through a large speaker, and may be run without heat sinks on the transistors (though you should experiment with it a bit before deciding to forego heat sinks, as the power dissipation varies according to the type of speaker used).
The goal of any amplifier circuit is to reproduce the input waveshape as accurately as possible. Perfect reproduction is impossible, of course, and any differences between the output and input waveshapes is known as distortion. In an audio amplifier, distortion may cause unpleasant tones to be superimposed on the true sound. There are many different configurations of audio amplifier circuitry, each with its own advantages and disadvantages. This particular circuit is called a "class B," push-pull circuit.
Most audio "power" amplifiers use a class B configuration, where one transistor provides power to the load during one-half of the waveform cycle (it pushes) and a second transistor provides power to the load for the other half of the cycle (it pulls). In this scheme, neither transistor remains "on" for the entire cycle, giving each one a time to "rest" and cool during the waveform cycle. This makes for a power-efficient amplifier circuit, but leads to a distinct type of nonlinearity known as "crossover distortion."
Shown here is a sine-wave shape, equivalent to a constant audio tone of constant volume:

In a push-pull amplifier circuit, the two transistors take turns amplifying the alternate half-cycles of the waveform like this:

If the "hand-off" between the two transistors is not precisely synchronized, though, the amplifier's output waveform may look something like this instead of a pure sine wave:

Here, distortion results from the fact that there is a delay between the time one transistor turns off and the other transistor turns on. This type of distortion, where the waveform "flattens" at the crossover point between positive and negative half-cycles, is called crossover distortion. One common method of mitigating crossover distortion is to bias the transistors so that their turn-on/turn-off points actually overlap, so that both transistors are in a state of conduction for a brief moment during the crossover period:

This form of amplification is technically known as class AB rather than class B, because each transistor is "on" for more than 50% of the time during a complete waveform cycle. The disadvantage to doing this, though, is increased power consumption of the amplifier circuit, because during the moments of time where both transistors are conducting, there is current conducted through the transistors that is not going through the load, but is merely being "shorted" from one power supply rail to the other (from -V to +V). Not only is this a waste of energy, but it dissipates more heat energy in the transistors. When transistors increase in temperature, their characteristics change (Vbe forward voltage drop, β, junction resistances, etc.), making proper biasing difficult.
In this experiment, the transistors operate in pure class B mode. That is, they are never conducting at the same time. This saves energy and decreases heat dissipation, but lends itself to crossover distortion. The solution taken in this circuit is to use an op-amp with negative feedback to quickly drive the transistors through the "dead" zone producing crossover distortion and reduce the amount of "flattening" of the waveform during crossover.
The first (leftmost) op-amp shown in the schematic diagram is nothing more than a buffer. A buffer helps to reduce the loading of the input capacitor/resistor network, which has been placed in the circuit to filter out any DC bias voltage out of the input signal, preventing any DC voltage from becoming amplified by the circuit and sent to the speaker where it might cause damage. Without the buffer op-amp, the capacitor/resistor filtering circuit reduces the low-frequency ("bass") response of the amplifier, and accentuates the high-frequency ("treble").
The second op-amp functions as an inverting amplifier whose gain is controlled by the 10 kΩ potentiometer. This does nothing more than provide a volume control for the amplifier. Usually, inverting op-amp circuits have their feedback resistor(s) connected directly from the op-amp output terminal to the inverting input terminal like this:

If we were to use the resulting output signal to drive the base terminals of the push-pull transistor pair, though, we would experience significant crossover distortion, because there would be a "dead" zone in the transistors' operation as the base voltage went from + 0.7 volts to - 0.7 volts:

If you have already constructed the amplifier circuit in its final form, you may simplify it to this form and listen to the difference in sound quality. If you have not yet begun construction of the circuit, the schematic diagram shown above would be a good starting point. It will amplify an audio signal, but it will sound horrible!
The reason for the crossover distortion is that when the op-amp output signal is between + 0.7 volts and - 0.7 volts, neither transistor will be conducting, and the output voltage to the speaker will be 0 volts for the entire 1.4 volts span of base voltage swing. Thus, there is a "zone" in the input signal range where no change in speaker output voltage will occur. Here is where intricate biasing techniques are usually introduced to the circuit to reduce this 1.4 volt "gap" in transistor input signal response. Usually, something like this is done:

The two series-connected diodes will drop approximately 1.4 volts, equivalent to the combined Vbe forward voltage drops of the two transistors, resulting in a scenario where each transistor is just on the verge of turning on when the input signal is zero volts, eliminating the 1.4 volt "dead" signal zone that existed before.
Unfortunately, though, this solution is not perfect: as the transistors heat up from conducting power to the load, their Vbe forward voltage drops will decrease from 0.7 volts to something less, such as 0.6 volts or 0.5 volts. The diodes, which are not subject to the same heating effect because they do not conduct any substantial current, will not experience the same change in forward voltage drop. Thus, the diodes will continue to provide the same 1.4 volt bias voltage even though the transistors require less bias voltage due to heating. The result will be that the circuit drifts into class AB operation, where both transistors will be in a state of conduction part of the time. This, of course, will result in more heat dissipation through the transistors, exacerbating the problem of forward voltage drop change.
A common solution to this problem is the insertion of temperature-compensation "feedback" resistors in the emitter legs of the push-pull transistor circuit:

This solution doesn't prevent simultaneous turn-on of the two transistors, but merely reduces the severity of the problem and prevents thermal runaway. It also has the unfortunate effect of inserting resistance in the load current path, limiting the output current of the amplifier. The solution I opted for in this experiment is one that capitalizes on the principle of op-amp negative feedback to overcome the inherent limitations of the push-pull transistor output circuit. I use one diode to provide a 0.7 volt bias voltage for the push-pull pair. This is not enough to eliminate the "dead" signal zone, but it reduces it by at least 50%:

Since the voltage drop of a single diode will always be less than the combined voltage drops of the two transistors' base-emitter junctions, the transistors can never turn on simultaneously, thereby preventing class AB operation. Next, to help get rid of the remaining crossover distortion, the feedback signal of the op-amp is taken from the output terminal of the amplifier (the transistors' emitter terminals) like this:

The op-amp's function is to output whatever voltage signal it has to in order to keep its two input terminals at the same voltage (0 volts differential). By connecting the feedback wire to the emitter terminals of the push-pull transistors, the op-amp has the ability to sense any "dead" zone where neither transistor is conducting, and output an appropriate voltage signal to the bases of the transistors to quickly drive them into conduction again to "keep up" with the input signal waveform. This requires an op-amp with a high slew ratede/dt) necessary for low-distortion operation. (the ability to produce a fast-rising or fast-falling output voltage), which is why the TL082 op-amp was specified for this circuit. Slower op-amps such as the LM741 or LM1458 may not be able to keep up with the high dv/dt (voltage rate-of-change over time, also known as
Only a couple of capacitors are added to this circuit to bring it into its final form: a 47 µF capacitor connected in parallel with the diode helps to keep the 0.7 volt bias voltage constant despite large voltage swings in the op-amp's output, while a 0.22 µF capacitor connected between the base and emitter of the NPN transistor helps reduce crossover distortion at low volume settings: