CHAPTER 7

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Amplifiers and Feedback

We will cover some basic concepts of negative feedback first before showing some of the amplifier circuits employing negative feedback. For most amplifiers, negative feedback is used for a variety of purposes that include setting the gain of the amplifier with a couple of resistors, reducing distortion, and increasing frequency response, just to name a few. Voltage-controlled amplifiers are often used with negative-feedback systems such as automatic gain control circuits, and an automatic level control amplifier for audio signals will be presented. Finally, we will present a couple of examples where positive feedback in an amplifier is not desirable.


What Are Negative-Feedback Systems?

Generally, a negative-feedback system is where a reference signal is at the input of the system, and the output of the system is following the reference signal in some manner. If the output strays too low or too high, the output is compared with the input reference signal, and a correction signal in the opposite direction is given to maintain the desired output. For example, probably one of the oldest negative-feedback systems is the biological systems in mammals by which a body temperature is maintained. For example, for humans, the body temperature is regulated to about 98.6 degrees Fahrenheit.

In terms of a mechanical system, a motor with weights (aka a governor) attached to arms that will fly open when the speed is too high and collapse if the speed is too slow is a negative-feedback system. (Think of an ice skater who uses the extension of the arms to control spin speed.) In electric motors for tape recorders built in the mid-1960s, two weights were attached to the contacts of two switches that shorted out two resistors in series with the motor’s winding. When the speed was too high, the weights flew out to open the switch contacts. The motor was reduced in current because the two series resistors were now enabled to reduce the current into the motor, which then reduced the speed of the motor. When the motor slowed down sufficiently, the weights collapsed such that the switches were turned back on to short out the series resistors to deliver more current to the motor and speed it up. There is then a constant back and forth of the motor’s speed going slow and fast due to the governor’s arms extending and collapsing, which results in a long-term constant speed. If the motor with its governor is designed correctly and is coupled with a flywheel to average out the short-term speed variations of the motor, the tape speed is constant during the record and playback modes. A model of a negative-feedback system is shown in Figure 7-1.

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FIGURE 7-1 A negative-feedback system.

A reference voltage is fed to the input, and the output of the system is feedback and is scaled or attenuated or reduced in strength and then subtracted from the reference input signal. A difference signal then provides an error signal to the voltage gain amplifier.

At first, the block diagram in Figure 7-1 may seem abstract. For instance, what does it mean to have an attenuated output subtracted from the reference input? So let’s take a look at a concrete example, a heating system with a thermostat and furnace (see Figure 7-2).

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FIGURE 7-2 A furnace heating system.

A thermostat sets the desired room temperature (e.g., 70 degrees Fahrenheit). The furnace turns on, and some of its heat is transferred to the sensor. When the room is heated to 70 degrees or slightly higher, the furnace turns off. As the room cools below 70 degrees, the sensor develops a voltage to turn the furnace back on.

Now let’s take a look at whether the sensor is insulated or put in the colder corner of the room. What will happen is that the furnace will stay on until that part of the corner of the room is heated to 70 degrees. But this will cause the rest of the room to be hotter than 70 degrees. Alternatively, do not move the sensor but wrap some heating insulation around it. This will cause the sensor to receive attenuated heat, and thus the room will be heated to beyond 70 degrees because the sensor has to reach 70 degrees after the heat goes through the wrapped insulation.

In both of these cases, if the output heat is attenuated by distance to the sensor or by insulating heat from the sensor, the result will be a higher output from the heater, which leads to higher than 70 degrees in the room. The negative-feedback system is merely compensating to have the sensor reach 70 degrees whether it has been put in a normal area of the room, a remote location, or the sensor has been blocked from the heat by having insulation around it.

A generalized electronic feedback system consists of a differencing circuit that has the input reference signal subtracted by a return, or attenuated, output signal. For a specific feedback amplifier, see the right block diagram of Figure 7-3 that contains a differential input amplifier that mimics the subtractor circuit with (+) and (–) inputs.

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FIGURE 7-3 (From left to right) A generalized feedback system and a specific feedback amplifier system.

Negative-feedback systems include a feedback network. In the furnace heater system, the sensor acts as the feedback network. Positioning the sensor determines the received heat and thus determines whether the room is hotter than the set temperature at the input via the thermostat. In an amplifier, the feedback network usually attenuates the signal via a resistor voltage divider such that the output signal is of higher amplitude than the input signal. In a voltage follower configuration, the feedback network does not attenuate the output signal to the inverting input of the amplifier. The gain of this specific negative-feedback system is unity or one.


Another Look at the Non-Inverting-Gain Operation Amplifier (Op Amp)

Chapter 6 introduced the non-inverting-gain amplifier. We will look at this configuration with a slightly different view for this chapter. See Figure 7-4. Intuitively, we observe that the input voltage is applied to the (+) input terminal, so we should expect that the output signal in in phase. That is, a positive voltage into the input yields a positive output voltage. Note: For the input notations of amplifiers, the “(+)” input is equivalent to the “+” input, and the “(–)” input is equivalent to the “–” input.

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FIGURE 7-4 A non-inverting-gain amplifier using negative feedback with a voltage divider circuit, RF and R1.

For an example, first look at the open-loop gain of the system by removing RF and setting R1 = 0 Ω. The open loop gain A0 refers to the raw gain of the amplifier prior to adding a feedback network around it.

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where V(+) is the voltage at the (+) input terminal, and V(–) is the voltage at the (–) input terminal. This says that the output voltage of the op amp is equal to the voltage across the (+) and (–) inputs multiplied by the open-loop gain A0. This relationship holds whether the op amp has no feedback or has negative feedback.

When the open-loop gain is very large, the voltage difference across V(+) and V(–) is very small. Put another way, via algebraic manipulation of dividing A0 on both sides of Equation (7-1), the output voltage divided by the open-loop gain is just the voltage across the (+) and (–) inputs, which says

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For example, if the open-loop gain A0 = 100,000, a common figure for many op amps such as the LM741, TL082, or NE5532, and the output is +10 volts DC, the voltage across the (+) and (–) inputs is +10 volts/100,000, or 0.10 mV, which is one-tenth of a thousandth of a volt. See Equation (7-3).

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Thus, for all practical purposes, we say that the voltage at the (+) and (–) inputs is essentially the same, within a few millivolts give or take. See Equation (7-4). A consequence of having a very high open-loop gain results in that the voltages at the input terminals are essentially the same potential.

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Now let’s look at a couple of examples. Let RF = R1. We know with equal-valued resistors, a divide-by-two voltage divider is formed from the output of the op amp to the (–) input. Therefore, if we connect +1 volt at the (+) input, we should also get +1 volt at the (–) input by virtue of the fact that the two input voltages have the same potential when negative feedback is applied and the negative-feedback system is working properly with a large open-loop gain A0; that is, it is not oscillating, or the output is not latched or stuck to one of the power-supply voltages.

We then ask the question, what voltage divided by two is +1 volt? The answer is +2 volts at the output of the op amp—at least, intuitively, one might guess that.

So what is really happening here? The op amp via negative feedback is “servoing” the output to provide a voltage when attenuated by a factor of two via the voltage divider to drive the voltage at the (–) input to match the voltage at the (+) input. Again, some books will just state that there is a virtual short circuit between the (+) and (–) inputs of an op amp when negative feedback is applied. But this virtual short circuit is really not a short circuit at all. It is the feedback system trying to do its best to change the output of the amplifier to match the (–) input voltage to the (+) input voltage after going through the feedback network. If the output voltage, for example, overshoots and exceeds +2 volts, there will be a bit more than +1 volt at the negative input terminal of the op amp, which is larger in magnitude that the input voltage at the (+) terminal of exactly +1 volt.

By nature of the (–) input voltage being larger than the (+) input voltage, the output voltage then has to move down from slightly greater than +2 volts to something lower. Similarly, if the output voltage is a little less than +2 volts, yielding slightly less than +1 volt at the (–) input, the input voltage at the (+) input terminal at exactly +1 volt would be greater than the voltage at the (–) input and thus would cause the output voltage to increase. Thus, there is a self-correcting nature about negative-feedback systems or amplifiers. We then get an equilibrium state such that the two input voltages are the same.

What if we change RF to 9 kΩ and keep R1 = 1 kΩ? We end up with a divide by 10 voltage divider circuit where the voltage divider formula from Vout to the (–) input terminal is [R1/(R1 + RF)] = voltage divider formula. This leads to V(–) = Vout[R1/(R1 + RF)] = Vout[1 kΩ/(1 kΩ + 9 kΩ)] = Vout[1 kΩ/(1 + 9) kΩ] = Vout(1 kΩ/10 kΩ) = Vout (1/10). Therefore, if we have +1 volt at the (+) input, what voltage at the output when divided by 10 gives +1 volt at the (–) input? Again, just by guessing, the answer would be 10 volts.

Therefore, the gain of the two examples is defined as the ratio of the output voltage to the input voltage. For RF = R1 = 1 kΩ, the gain Vout/Vin = +2 and for RF = 9 kΩ and R1 = 1 kΩ, and thus, Vout/Vin = +10. If we look more closely, we find that the gain is just the reciprocal of the voltage divider formula [R1/(R1 + RF)] → Vout/Vin = (R1 + RF)/R1, which sometimes is expressed as Vout/Vin = (RF + R1)/R1 and sometimes an alternate expression, is Vout/Vin = [1 + (RF/R1)].

Let’s look at another way of looking at the equation V(–) = Vout[R1/(R1 + RF)]. We know via negative feedback that V(+) = V(–) and that Vin is connected to the (+) input of the op amp in Figure 7-4. So Vin = V(+) = V(–) via negative feedback. Therefore,

V(–) = Vin = Vout[R1/(R1 + RF)] or Vin = Vout[R1/(R1 + RF)]

which leads to when we divide by Vout on both sides:

Vin/Vout = [R1/(R1 + RF)]

and if we take the reciprocal of both sides, we get the gain as:

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Inverting-Gain Configuration

The inverting op amp configuration is not used as often as the noninverting amplifier circuit. Noninverting op amp circuits allow for high input resistance so as not to load the input signal and also lend themselves to low-noise amplifier configurations such as microphone and phono-cartridge preamps or even moving-coil pre-preamps.

When one uses the inverting-gain configuration, there is a higher likelihood of loading down the input signal source and generating more noise. However, the inverting-gain amplifier is often used as photodiode or transresistance amplifier, where the input signal is a current and the output signal is a voltage. The gain of a transresistance amplifier is Vout/Iin, and if we recall Ohm’s law, we know that something that has voltage divided by current will yield a resistance. Thus, Vout/Iin is measured in resistance. We will discuss the transresistance amplifier later. For now, Figure 7-5 shows an inverting-gain voltage amplifier.

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FIGURE 7-5 An inverting-gain op amp circuit.

Intuitively, this configuration is somewhat more mysterious or difficult to understand. We can draw one good guess, that if the voltage is positive at Vin, the output Vout has to be a negative voltage because R1 is connected to the (–) input. Thus, we can also surmise that the output is out of phase with the input, so if the input voltage is negative, the output has to be positive.

However, we can still take the one feature we know about an op amp when negative feedback is applied, and that is that the voltages at the (+) and (–) inputs are essentially the same. This means that the voltage difference between the (+) and (–) inputs is zero. Given just this information, let’s see what happens when RF = R1 = 1 kΩ and Vin at R1 = +1 volt. We see that the (–) input terminal is grounded or is at 0 volt. Therefore, the op amp will generate an output to “servo” the voltage at the (–) input terminal to 0 volt. The input and output voltages can be represented by two voltage sources, Vin and Vout, as seen in Figure 7-6, along with RF and R1.

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FIGURE 7-6 Voltage sources Vin and Vout with resistors RF and R1.

Given that Vin = +1 volt DC and that the two resistors are equal in resistance, what voltage must Vout be in order to have 0 volt where the two equal resistors are connected? Our first guess would be –1 volt DC = Vout because we can redraw the circuit from Figure 7-6 as a summing circuit, as shown in Figure 7-7.

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FIGURE 7-7 Vin and Vout redrawn as a summing circuit.

If we have equal resistors and a summing circuit, the signal at the output of the resistor network Vsum_out = (½)(Vin + Vout). Specifically, if Vin = +1 volt and Vout = –1 volt, then

Vsum_out = V(–) = (½)(+1 volt + –1 volt) = 0 volt

Now let’s see what happens when we change one of the resistors RF to 10 kΩ and keep R1 = 1 kΩ. We no longer have a simple summer because the resistor values are unequal. However, we do know this: if there is a 1 volt drop across R1, we will get 0 volt at the junction of the two resistors. Put another way, because R1 = 1 kΩ, we need 1 mA flowing through it to develop 1 volt across R1. But the total resistance of the divider circuit is R1 + RF = 1 kΩ + 10 kΩ = 11 kΩ.

We have +1 volt at Vin, and we need a total of 11 volts across R1 and RF via voltage sources Vin and Vout. The voltage across the series resistances of R1 and RF, which total 11 kΩ, is just the potential difference between Vin and Vout, which is (Vin – Vout). We know that Vout should be a negative voltage such that Vin – Vout = +11 volts. One of the concepts we learned about subtracting a negative number is that a positive number results, at least in this case. So we have the following: +1 volt minus a negative voltage Vout should equal +11 volts. At this point, if you guessed –10 volts, you are correct. Therefore, Vout = –10 volts.

What has happened here, then? The op amp via negative feedback “servoed” the output voltage to provide a negative voltage such that at the (–) input terminal there is 0 volt via RF and R1, and Vin = +1 volt. In this feedback amplifier, as shown in Figure 7-5, there are actually two reference voltages going on here. The first one is the (+) input being connected to 0 volt or ground. This anchors what the voltage at the (–) input terminal will be. The second reference voltage is Vin, which determines as a whole what the output signal is going to be. There is much more complicated math that can be shown, but for now, with the examples shown, we so far know that for an inverting amplifier as shown in Figure 7-5:

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With Vin = +1 volt, when RF = R1 = 1 kΩ, Vout is –1 volt, and when RF= 10 kΩ with R1 = 1 kΩ, Vout = –10 volts. The gain Vout/Vin is then deduced as:

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Alternatively, divide Equation (7-6) by Vin on both sides to get Equation (7-7). Note that if RF < R1 or, put another way, R1 > RF, the magnitude of inverting gain is less than 1. For example, make R1 = 100 kΩ and RF = 1 kΩ, and we get a gain of –1/100 = –0.01. This sets up a unique feature. With a gain of less than 1, the input voltage can exceed the power-supply voltage (e.g., typically ±12 volts) of the op amp. For example, with a gain of –0.01, an input signal of +200 volts DC as Vin into R1 will only yield an output voltage of –2 volts DC.

NOTE The maximum input signal voltage into the (+) input of a noninverting-gain op amp configuration (e.g., Figure 7-4) is generally less than the power-supply voltage.

We know that volts divided by amps equals a resistance. So we can also rewrite Equation (7-6) in terms of an input current; that is, Iin = Vin/R1. Then we can reinterpret Equation (7-6) as the following:

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Note that Iin is the “positive” current flowing into the (–) input terminal via R1. In a photodiode circuit with the anode grounded and the cathode connected to the (–) terminal, the photocurrent Iphoto = –Iin because the photocurrent is flowing down and away in the opposite direction from the (–) input terminal. See Figure 7-8.

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FIGURE 7-8 A photodiode amplifier using an inverting op amp configuration.

In Figure 7-8, the photodiode preamp is also known as a transresistance amplifier because the input signal is a current and the output signal is a voltage. The gain or transfer function that characterizes this amplifier is Vout/Iin, which has a unit of resistance. But because the input is a signal current, we call Vout/Iin a transresistance transfer function and we describe the gain magnitude in terms more of volts per amp or milliamp. For example, with RF = 1 kΩ, the transfer function for the transresistance amplifier is 1 volt per 1 mA. That is, if 1 mA is inputted, the output will have a voltage magnitude of 1 mA × 1 kΩ = 1 volt. Therefore, we get 1 volt (output) per 1 mA of input. Also, note that (1 volt per mA) = 1 volt/mA = 1 kΩ = RF.


Two Projects: A Photodiode Sensor and an Automatic Level Control Amplifier

Photodiode Sensor

Speaking of photodiodes, let’s build a photodiode sensor using an op amp and comparator to sense whether a white or light piece of paper is present. The circuit consists of a white light-emitting diode (LED), a photodiode (a real one, not using an LED this time), and a dual op amp. The first section of the dual op amp will amplify the photocurrent reflected from LED1, and the second section of the op amp will be used as a comparator. The comparator is set to turn on another LED, LED2, to indicate the presence of the white paper (see Figures 7-9 and 7-10).

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FIGURE 7-9 Schematic diagram of a white paper sensor.

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FIGURE 7-10 Prototype circuit of the white paper sensor.

LED1 is positioned below photodiode PD1 so that stray light will not leak into the photodiode. Thus, the photodiode will only pick up reflected light from the white paper that is illuminated by LED1.

Parts List

         • One LM358 or TLC272 op amp for U1 with an 8 pin IC socket

         • One 22 pF capacitor for C1 (C1 is optional)

         • One 1μF capacitor for C2

         • One 9 volt battery and connector for BT1

         • Two 1N914 or 1N4148 diodes for CR1 and CR2

         • One SFH229 or other photodiode for PD1

         • One 24,000 mcd (millicandela) white LED for LED1

         • One green or red LED for LED2

         • Two 560 Ω resistors for R1 and R4 (other values between 330 Ω and 1 kΩ may be used)

         • One 10 kΩ resistor for R2

         • One 1 MΩ resistor for resistor R3

     • One makeshift barrier made of a generally opaque material such as cardboard (optional)

Here are a few notes to add on the photodiode sensor circuit. First, the cathode of photodiode PD1 is biased up to about +0.6 volt via CR2 so that if a phototransistor is substituted, there is sufficient DC bias on the phototransistor. Also, biasing the cathode of the photodiode reduces its inherent reverse-bias capacitance across the cathode and anode. A +1.2 volt reference voltage source is established by the series connection of CR1 and CR2 with R2, providing DC forward bias for these two diodes. At the anode of CR1, the 1.2 volt reference voltage is connected to the inverting input of U1B, the second section of the dual op amp; U1B is used as a comparator instead of a negative-feedback amplifier. The output of transresistance amplifier U1A provides a voltage output that is +0.6 volt via CR2 and the product of the photocurrent from PD1 and R3 = 1 MΩ. So, when there is no photocurrent, pin 1, the output of U1A, is sitting at +0.6 volt. Any photocurrent via R3 then adds a voltage on top of the +0.6 volt. The output of the transresistance amplifier is then connected to the (+) input of the comparator at pin 5 of U1B. When the photocurrent exceeds about 0.6 μA via the light reflected from the paper from LED1 to PD1, the transresistance amplifier will output a voltage of more than +1.2 volts, which then causes comparator U1B to output a high-level output of nearly the power-supply voltage (~+7.5 volts DC) at pin 7, which then supplies current to turn on LED2.

NOTE The output of the transresistance amplifier at pin 1 = (photocurrent x 1MΩ) + 0.6 volt.

Controlling Audio Signal Levels

The next project is an audio signal automatic level control amplifier. This amplifier combines circuits such as a DC restoration, voltage doubler, op amps, and a bipolar transistor voltage-controlled attenuator.

An automatic level control (ALC) amplifier will allow normal operation of amplifying until a certain amplitude level is exceeded. Then the gain of the amplifier is decreased to limit the amplitude of the AC signal in a nondistorting manner. ALC amplifiers are very useful when recording or amplifying a live event signal from a microphone so that the signal is limited in the dynamic range so as to prevent distortion in the recording or amplifying device (see Figures 7-11 and 7-12).

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FIGURE 7-11 Block diagram and schematic of the ALC amplifier.

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FIGURE 7-12 Prototype circuit of the audio ALC (automatic level control) amplifier.

This amplifier includes an inverting op amp input amplifier U1A, which has a gain of –R2/R1 = –10 kΩ/47 kΩ = –0.212. The output of U1A is fed to a 30 kΩ series resistor that forms a voltage divider circuit via the collector-to-ground variable resistance of Q1. This variable resistance is varied by a control voltage from R9. The output of the variable voltage divider is AC coupled via C5 to a non-inverting-gain amplifier U1B, whose gain is [1 + (R7/R6)] = [1 + (220 kΩ/3.3 kΩ)] ≈ 67. The AC signal output is coupled via output resistor R11 to Vout to be connected to an amplifier or recorder. However, the AC signal output of U1B is also converted to a DC voltage by a DC restoration circuit, C8 and CR1. At the cathode of CR1, the audio or AC signal has the negative peak clamped or DC restored to nearly 0 volt via germanium diode CR1. The peak of the clamped AC signal at the cathode of CR1 is transferred by rectifier CR2 to a hold capacitor C9. The voltage from C9 is drained slowly via R12, a 1 MΩ resistor. Thus, R12 controls the release time of this peak limiter amplifier circuit.

Parts List

         • One NE5532 or LM833 op amp for U1 with 8-pin IC socket

         • Two 1N277, 1N270, 1N34, or 1N60 germanium diodes for CR1 and CR2 (alternatively, two Schottky diodes or rectifiers may be used)

         • One 1N914 or 1N4148 silicon diode for CR3

         • One 2N3904 or 2N4124 transistor for Q1

         • Five 1 μF capacitors for C1, C2, C3, C4, and C6 (C6 can be 2.2 μF)

         • One 0.1 μF capacitor for C5

         • One 0.047 μF capacitor for C7

         • Two 10 μF electrolytic capacitors at 16 volts or more for C8 and C9

         • One 47 kΩ resistor for R1

         • One 10 kΩ resistor for R2

         • One 15 kΩ resistor for R3

         • One 30 kΩ resistor for R4 (33 kΩ can be used instead)

         • One 100 kΩ resistor for R5

         • One 3.3 kΩ resistor for R6

         • One 220 kΩ resistor for R7

         • One 150 Ω resistor for R8

         • One 1.2 kΩ resistor for R9

         • One 200 Ω resistor for R10 (either 180 Ω or 220 Ω can be used instead)

         • One 100 Ω resistor for R11 (any value from 47 Ω to 1 kΩ can be used instead)

     • One 1 MΩ resistor for R12

The ALC circuit was found to limit at around 1.4 volts peak-to-peak output for input signals from about 210 mV to 7,000 mV peak to peak. CR3 is in series with the base of Q1 to allow more negative value voltages at the collector of Q1 without forward biasing its base-collector junction. Also, the series combination of CR3, a silicon diode with a 0.7 volt turn-on voltage, and the 0.7 volt turn-on voltage of Q1 for its base-emitter junction starts the ALC action at about 1.4 volts. The voltage doubler and DC restoration circuit consisting of CR1 and CR2 will take the peak-to-peak AC voltage from the output and transform that AC voltage to a DC voltage. For example, if the AC output voltage is 1 volt peak to peak (e.g., a sine wave with a positive cycle of +0.5 volt and a negative cycle of –0.5 volt), the output of CR2 is then about a +1.0 volt DC signal. Thus, the turn-on voltages from CR3 and Q1, at +0.7 volt and +0.7 volt = 1.4 volts, determines that the peak-to-peak output AC signal is limited to 1.4 volts.


Another Voltage-Controlled Amplifier Circuit

Previously, we saw voltage-controlled attenuators such as the ones shown in Chapters 5 and 6 that used diodes including LEDs, bipolar transistors, and FETs as voltage-controlled resistors. We can also control the DC bias of bipolar transistors, FETs, and even vacuum tubes to vary the amplifier’s gain. A simple voltage-controlled amplifier is a bipolar transistor common emitter or grounded emitter amplifier where the collector current is varied and the base of the transistor is driven with a low-resistance voltage source. For the next experiment, we will use a combination of op amps and a transistor amplifier to show how gain is varied by varying the collector current of the transistor.

By varying the collector current, we will then uncover a characteristic of the bipolar transistor and learn something extra, which is the concept of transconductance. The concept of transresistance was briefly discussed as a current input signal with a voltage output signal. For understanding transconductance, we first need to determine what conductance is. Conductance is that ability to allow electrons to flow through a material such as silver, copper, aluminum, lead, glass, and so on. Intuitively, we know that silver is the best conductor, followed by copper, and one of the worst conductors is glass. Thus, we would say that silver has higher conductivity than copper. Put another way, silver has very high conductivity that has less resistance than copper. Glass, with very poor conductivity (close to zero conductivity), has very high resistance. Therefore, conductance is merely the reciprocal of resistance. That is, conductance, whose symbol is G, has G = 1/R = 1/resistance. From Ohm’s law, we know that V/I = R; thus, G = I/V.

Transconductance, which is the ratio of output current to input voltage, is then measured or expressed in mhos (ohms spelled backwards), or in the more modern era, transconductance is often measured in siemens, whose abbreviation is S. That is, 1 S = 1 mho. Transconductance, whose symbol is gm, is characterized in amps per volts (mhos), or it is more commonly expressed in milliamp(s) per volt (mmhos).

Voltage-Controlled Common Emitter Amplifier

This experiment will measure an AC signal from an AC adapter, which will provide a 60-Hz sine-wave signal. Figure 7-13 shows the schematic. By adjusting the DC voltage to a base resistor R3 and measuring the voltage across an emitter resistor RE1, we will verify that by increasing the DC emitter current, which leads to almost exactly the same increase in DC collector current, the gain of the amplifier will increase proportionally. An autoranging digital volt-ohmmeter (VOM, e.g., Extech MN26 Series or Extech MiniTec 26 Series) or a digital VOM that can measure full-scale 2 volts AC is required. The cheaper digital VOMs start out with 200 volts AC full scale and are not suitable for this experiment since the AC signals from this amplifier will be typically less than 1 volt.

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FIGURE 7-13 Voltage-controlled common emitter amplifier and schematic.

Potentiometer VR1 is adjusted for 500 mV AC (0.500 volt AC) at its slider terminal 2 or at the input of R1 Vin. The AC signal voltage at VB is 1/100 of Vin via resistive divider R1 and R2. To measure transconductance, we measure the AC collector current and then divide by the AC signal voltage at the base of Q1 VB. The resulting voltage across the base-emitter junction of Q1 is then about 5 mV, small enough to ensure accurate readings without gross distortion of the sine wave at the output Vout. This circuit is relatively easy to build, and the reader is encouraged to do so and take measurements. Table 7-1 summarizes various measurements for different DC collector currents.


TABLE 7-1 Measurements on Voltage-Controlled Common Emitter Amplifier

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What we can see from this table is that increasing the DC collector current linearly increases the gain and transconductance gm for this common emitter amplifier.


Negative Feedback to Stabilize or Self-Bias Collector Currents

In Figure 7-13, an emitter resistor RE1 and a DC voltage to the base of Q1 via RB1 established the collector current by establishing a voltage across RE1. The question then arises whether we can self-bias a transistor for setting their collector current. The answer is yes, and we have actually seen this previously in Chapter 3, Figure 3-22. See Figure 7-14.

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FIGURE 7-14 One-transistor amplifier from Chapter 3.

In this amplifier, the collector current is set by “defining” the collector voltage. First, let’s make an approximation that the base current of the transistor is low such that the voltage across R2 is approximately 0 volt because the current gain β of transistor Q1 is high such that β > 100. Recall that a high current gain of a transistor results in low base currents. Note that there is a way to find a more exact voltage drop (Vc – Vb) via a set of equations and “lots” of algebra, which we will not show here. For now, and in practice, chasing the exact answer is usually not an efficient use of one’s time. Also, if one is concerned about base currents, just replace Q1 with a high-β transistor such as a 2N5089, where β > 500, and then forget about the base currents that will cause a voltage drop across R2, the 56 kΩ feedback resistor.

We know that the transistor turns on at about +0.6 volt for VBE, and because the emitter is grounded, VB = VBE = +0.6 volt. With about 0 volt across R2, we have VB = VC, so VC = +0.6 volt. Now all we need to know is the battery voltage for BT1, which is 9 volts. Therefore, the voltage across R3 = 10 kΩ, the collector load resistor, is (9 volts – 0.6 volt) = 8.4 volts = VR3 (see Figure 7-14). The collector current then is IC = VR3/R3 = 8.4 volts/10 kΩ = 0.84 mA = IC.

One question is, why did this circuit bias the collector voltage to 0.6 volt in the first place? The answer is that R2 is a feedback resistor, and the transistor amplifier Q1 can be modeled or thought of as an op amp configured as an inverting amplifier with the (+) input terminal not connected to ground but connected to a VBE voltage source of +0.6 volt. See Figure 7-15.

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FIGURE 7-15 Inverting op amp representation of the one-transistor amplifier.

For the simple one-transistor amplifier, the common emitter amplifier is an inverting input amplifier. That is, the more positive the voltage applied to the input at the base, the more collector current is provided, which then forms a larger voltage drop across R3, which is the collector current times the resistance of R3, or IC × R3. But the collector voltage VC is (BT1 – VR3) = (BT1 – IC × R3) = VC. Therefore, VC’s voltage goes down as the base voltage VB goes up, which then characterizes an inverting amplifier. Note that we have negative feedback via R2 from the collector to the base of Q1, and the AC gain ≈ –R2/R1 ≈ –56 kΩ/27 kΩ ≈ –2.

For one more perspective on why this one-transistor amplifier is a negative-feedback system, we note that if the collector voltage VC rises, it will send an increased voltage via R2 to the base of Q1. The increased voltage into the base of Q1 will then increase collector current IC, which will then decrease the collector voltage at VC. Recall that VC = BT1 – IC × R3, so if IC increases, VC has to get smaller because the product IC × R3 or voltage drop across R3 subtracts more from the battery voltage BT1. Thus we have a system that is self-adjusting in terms of collector current and collector voltage.

The next question is, can we adjust VC to some other voltage? The answer is yes, by adding a resistor R4 from the base of Q1 to ground. See Figure 7-16.

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FIGURE 7-16 Modified one-transistor amplifier with R4 to raise the DC collector bias voltage.

But why change the collector voltage in the first place? For the circuit in Figure 7-14, it is actually optimized for approximately symmetrical AC voltage swings of ±0.6 volt, which would imply that BT1 → 1.5 volts, approximately. However, if BT1 is a higher voltage (e.g., 9 volts), the amplifier is limiting itself to going only 0.6 volt in the negative direction and about 8 volts in the positive direction. If we can bias VC to about half of BT1, or (½) × 9 volts = 4.5 volts, a larger symmetrical voltage swing would be provided.

So how do we calculate R4? We know in a voltage divider with a supply voltage VC that the output of the voltage divider to the base of Q1 is

VB = 0.6 volt = VC × R4/(R2 + R4)

In this case, we know approximately what VB is, but we do not know what VC is. If we multiply both sides of the equation by (R2 + R4)/R4, we get

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Let’s try a couple of examples. Suppose that R4 = R2 = 56 kΩ; then VC = VB × (56 kΩ + 56 kΩ)/56 kΩ = 0.6 volt × 2 = 1.2 volts = VC. Suppose that we want to set VC to a specific voltage. How do we find the resistors? We start with Equation (7-9) and divide by VB = 0.6 volt on both sides to get

(R2 + R4)/R4 = VC/VB → (R2/R4 + 1) = VC/0.6 volt

by subtracting 1 from both sides, that is,

R2/R4 = [(VC/0.6 volt) – 1]

and taking the reciprocal of both sides, we get

R4/R2 = 1/[(VC/0.6 volt) – 1]

and multiplying by R2 on both sides, we finally get

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If we have BT1 = 9 volts, how do we set VC to 4.5 volts when R2 = 56 kΩ? Using Equation (7-10), we get

R4 = 56 kΩ/[(4.5 volts/0.6 volt) – 1] = 56 kΩ/(7.5 – 1) = 56 kΩ/6.5= 8.6 kΩ = R4

After grinding out Equation (7-10), I would just try different values for R4. I know that 4.5 volts for VC is about 7 times 0.6 volt. And I know that the ratio R2/R4 is roughly the gain of the system DC-wise. That is, R2/R4 ≈ (R2/R4 +1) if R2/R4 is a lot larger than 1. Thus, R4 will be approximately 56 kΩ/7 ≈ 8 kΩ ≈ R4, which is not too bad of an approximation.

When we look at the DC output at VC, it has a voltage that is the result of multiplying VBE. That is, VC = VBE × (R2 + R4)/R4, where the multiplying factor is (R2 + R4)/R4. A circuit consisting of only Q1, R2, and R4 is sometimes called a VBE (voltage) multiplier circuit, and it has uses in power output stages in amplifiers such as stereo high-fidelity receivers, headphone amplifiers, and many op amps.


Power Output Stages and Using the VBE Multiplier Circuit

In the following schematics, we will show a progression of circuits on how to improve the one-transistor amplifier into a simple power amplifier that can drive a headphone or small loudspeaker. Figure 7-16 as it stands will work as a preamplifier that can be fed to the auxiliary input or CD input of a stereo receiver. However, it does have limited output current drive. One improvement to this circuit is to add an emitter follower amplifier Q2 (see Figure 7-17).

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FIGURE 7-17 One-transistor amplifier with a second transistor added for increased output current capability.

With the addition of Q2 and R5, this amplifier has a lower output resistance at the emitter of Q2, which is useful when driving long cables that have sizable capacitances. The emitter follower amplifier inherently has a voltage gain of around unity or 1 and has an advantage of high input resistance (base terminal of Q2) while having low output resistance and providing improved output current capability at its emitter output terminal. The addition of R6, a series 100 Ω resistor, is to ensure that the emitter follower Q2 does not directly load to the capacitance of an external cable. We can further improve on lowering distortion and frequency response by rerouting R2 so that it applies negative feedback not from the output of Q1 but from the output of Q2. See Figure 7-18.

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FIGURE 7-18 Negative-feedback resistor R2 rerouted to apply feedback at the emitter of Q2.

With R2 “sensing” the output of Q2, at least two things are happening. One is that the output resistance at Q2’s emitter is further lowered from Figure 7-18, and second, there should be a wider frequency response when the signal is taken from the emitter of Q2. A disadvantage of using an emitter follower consisting of Q2 and R5 is that although signals from the positive cycle of an AC waveform have good current capability, the negative cycle or negative flowing current of an AC signal is limited by R5. R5 can be lowered in value (e.g., in Figure 7-18, R5 → 470 Ω) to increase the standing current of Q2. Recall that the emitter current of Q2 is just the DC emitter voltage of Q2 divided by R5. But such lowering of the resistance value of R5 would consume too much power on standby. What would be desirable is to design an output stage that idles low current on standby (e.g., no signal at the input) while having the capability of good current drive for both positive and negative cycles of AC signal. See Figure 7-19.

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FIGURE 7-19 One-transistor amplifier with Q2 and Q3 as a push-pull output stage.

In this circuit, a third transistor Q3 that is PNP allows the capability of increased current in the negative direction. This is needed for the negative cycle of the AC waveform. Q2’s emitter handles the current for the positive cycle of the AC signal, and Q3’s emitter handles the current for the negative cycle. Diodes CR1 and CR2 are used to bias emitter currents of Q2 and Q3. However, because CR1 and CR2 often do not have the same turn-on voltage characteristic as the VBE and VEB turn-on voltages of Q2 and Q3, respectively, there is a good chance that the emitter currents of Q2 and Q3 may be too low, which leads to crossover distortion (i.e., dead-band distortion). If the emitter currents of Q2 and Q3 are too low, one can add a third diode in series, but that may cause the emitter currents of Q2 and Q3 to be too high. And, if CR1 and CR2 initially cause too high emitter currents for Q2 and Q3, one can use one diode instead and, for instance, short out CR2, but that may lead to crossover distortion due to insufficient emitter bias currents in Q2 and Q3. So what can be done to adjust the emitter or collector currents of Q2 and Q3? What would be desirable is a variable diode voltage reference generator. But we have already seen that in Figure 7-14, where VCE, the DC bias collector-to-emitter voltage of Q1, is just VBE(1 + R2/R4) = VCE. We can use another transistor as a VBE voltage multiplier to replace CR1 and CR2. See Figure 7-20.

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FIGURE 7-20 A fourth transistor Q4 replaces CR1 and CR2 for biasing the output transistors.

In Figure 7-20, the collector-to-emitter voltage of Q4 is VBE(1 + R8/R9). R9 is nominally 10 kΩ to set the VCE voltage to about 2 diode voltage drops or 2 × VBE = 2 × 0.6 volt = 1.2 volts = VCE. However, the asterisk in R9 denotes that it can have a selected resistance value, or R9 can be a variable resistor (e.g., 20 kΩ) to have a range from 20 kΩ to about 5 kΩ. The emitter current of Q2 or Q3 is just the voltage across R5 or R7 divided by 2.2 Ω. Setting up the emitter current of Q2 and Q3 requires that the AC input signal be disconnected or turned off.

Note in Figure 7-20 that R3 is lowered from 10 kΩ to 3.3 kΩ to bias Q1 to a higher collector current so that more current is available to the bases of Q2 and Q3 for higher output current drive. Also, R5 and R7 are set to 2.2 Ω, and C2 is increased from 1 μF to 470 μF for better low-frequency bass response when using this amplifier to drive a headphone or small loudspeaker.


Misadventures in Positive Feedback

There is the old saying about amplifiers turning into oscillators and oscillators turning into amplifiers. In this section, we will explore just a couple of examples where positive feedback gives an undesirable result.

One of the misconceptions in using op amps is that it does not matter how the output is connected to one of the inputs. For an op amp to work properly as an amplifying device, there must always be negative feedback. Figure 7-21 provides one example.

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FIGURE 7-21 An erroneous inverting voltage follower that does not work.

I have seen a hobbyist magazine and a ham radio book show a circuit like the one in Figure 7-21. Unfortunately, this circuit will latch to one of the power-supply voltages, or it will oscillate. The problem is that there is no servo system to drive the output voltage going to the (+) input of the op amp to match the input voltage at the (–) input. When a positive voltage is connected to the (–) input of the op amp, the output has to produce a negative-valued voltage. However, we see that the input terminals at pin 2 and pin 3 of the op amp are taking the arithmetic difference of two voltages (input and output voltages) that are opposite in polarity. This results in a large output voltage. This large output voltage builds on itself to eventually saturate the output voltage, or the op amp becomes an oscillator.

In other words, the voltages are in opposite polarities to each other at the (+) and (–) inputs, so it is incorrect to think that the input signal at the (+) terminal can servo the output that is connected to the (–) input to the same voltages. For example in Figure 7-21, if the input = +1 volt, then the output equals some negative voltage, and one cannot reconcile that a positive voltage equals a negative voltage.

In a negative-feedback configuration, the input terminals are taking the arithmetic difference of two like polarity voltages, which results in a smaller output voltage. And that smaller output voltage is under control. The two identical-polarity signals at the input under negative feedback can be driven to the same voltage. Thus, if someone tells you that there is such a thing as a negative or inverting voltage follower, something is wrong. A proper voltage follower is shown in Figure 7-22 with RF = 0 Ω and R1 taken out.

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FIGURE 7-22 A proper voltage follower is implemented when RF = 0 Ω and R1 is removed.

Another problem arises when an op amp is directly connected to a capacitive load. See Figure 7-23. An op amp is a system that consists of at least two phase-shift stages. A third phase-shift stage can cause the op amp circuit in Figure 7-19 to oscillate (see Figure 7-24).

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FIGURE 7-23 Pure capacitive loading can lead to oscillation.

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FIGURE 7-24 An op amp representation with multiple voltage gain stages and low-pass filters.

In Figure 7-24, an op amp is represented by a differential amplifier A1, a voltage-gain amplifier A2, and a unity gain of 1 in the output stage (e.g., emitter follower gain of 1). Equivalent internal low-pass filters are shown via R10, C10, R20, and C20. An internal output resistor of the op amp is shown as Rout. External biasing resistor Rin and external feedback resistor RF and R1 are shown outside the box along with the capacitive load C_Load. Each low-pass filter delivers up to 90 degrees of phase shift. With two stages shown inside the box, less than 180 degrees of phase shift is provided, but the extra phase shift via Rout and C_Load may have just enough added phase shift to push the amplifier into oscillation. The reason is that when there is a net 180 degrees of phase shift eventually going into the inverting input of the amplifier via A1, the total phase shift is 360 degrees, which looks like 0 degree. Or, in other words, the 180 degrees of phase shift from R10, C10, R20, C20, Rout, and C_Load essentially flips the polarity of the (–) input into a (+) input, and positive feedback is provided in a “bad” way that would result in oscillation. Also note that oscillation will occur even if the input signal is 0 (e.g., Vin = 0 in Figure 7-24). A solution to ensure that there is no oscillation is to just add a series resistor as shown in Figure 7-25.

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FIGURE 7-25 An added series resistor isolates the capacitive load to provide freedom from oscillation.

Generally, a series resistor Rseries of 22 Ω or larger will suffice in allowing stability without oscillation when connecting to a capacitive load such as a long cable.


Some Considerations When Choosing Op Amps

So far we have seen an overview of how op amps work. But which op amps do we choose? For easiest construction, I prefer to use dual op amps in the 8 pin dual inline package (DIP). However, single or quad op amp packages can be used as well. For general purpose op amps that have a power supply range of 9 volts to 30 volts across the +V and –V supply terminals, I would use the TL072, TL082, NE5532, LM833, or LM4562 dual op amps. General purpose dual op amps such as the LM1458 or RC4558 can be also used.

For lower voltage projects using between 3 volts and 15 volts across the +V and –V supply terminals, my choices are the TLC272 or TLC27M2 dual op amps for working with frequencies below 20 kHz.

The LM358 dual op amp works from a supply of 3 volts to 30 volts across the +V and –V supply terminals. However, this op amp has a unique output stage that exhibits crossover distortion when loading into less than 10 kΩ. One can “fix” this by adding a pull down resistor of about 3.3 kΩ from the output terminal to the –V supply terminal. Alternatively, a 3.3 kΩ resistor from the output terminal to the +V supply terminal will also work in many situations.

If low voltage capability and “rail to rail” performance is desired from 3 volts to 30 volts across the +V and –V supply terminals, then you may want to consider the surface mount SO-8 (small-outline 8 pin) dual op amps, ISL28208, ISL28217, or ISL28218. However, you will need an adapter board (see Figure 11-26) or very careful soldering techniques to work with these op amps since they are not through-hole devices.

Finally for low noise applications, you may consider the AD797, an 8 pin DIP single op amp.


References

  1. Paul R. Gray and Robert G. Meyer, Analysis and Design of Analog Integrated Circuits, 3rd ed. New York: Wiley, 1993.

  2. Thomas M Frederiksen, “Intuitive IC Op Amps.” Santa Clara, CA: National Semiconductor, 1984.

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