Chapter 5   Applying Feedback To Amplifier Circuits.

A major confusion factor has developed in teaching negative feedback because of a split among electrical engineers. One group, the audio engineers, sets out to derive the feedback equation by assuming a generalized amplifier. The equation they derive has a minus sign in the denominator. The other group, control systems engineers, assumes a net 180 degree phase shift in the feedback loop. In other words they assume that the feedback is always negative. The equation they derive has a plus sign in the denominator.

That would be bad enough by itself but the operational amplifiers (op amps) which are so useful have grown out of the control systems branch of electrical engineering. What this means to a teacher teaching a survey course in electronics is that when teaching feedback applied to audio amplifiers the equation with the minus sign must be used and when teaching op amps the equation with the plus sign must be used. That has proven to be so confusing to students that there is doubt as to whether they ever learned anything about feedback. In this book we will define away the problem by adopting one convention and sticking to it. If you look up the subject of feedback in another textbook you may find more confusion than help.

The output of a differential amplifier is given by equation 4.27 which is repeated here.

For the noninverting amplifier beta is given by

Let us be sure we understand all of the terms associated with feedback. The open-loop gain (A) is the gain of the amplifier without any feedback circuits connected. The closed-loop gain (A') is the gain of the amplifier after the feedback has been connected. Beta, (it has no other name,) is that fraction of the output which is fed back to the input.

Example 5.1.

An amplifier has an open-loop gain of 250 and a beta of 0.02. What is its closed-loop gain?

Solution:

The closed-loop gain is given by

A' = A / (1 + A beta) = 250 / (1 + 250 x 0.02) = 41.67.

If we write the loop equations around the loops labeled "equation 5.5" and "equation 5.6" we have

You are challenged to find equation 5.13 in any other text on this level. If all starting assumptions are correct and all of the algebra has been correctly done, the solution must be correct. Careful laboratory measurements using an amplifier with a low open-loop gain have shown that the equation does accurately predict the gain of an inverting amplifier. Equation 5.13 is the correct equation for the gain of an inverting amplifier.

A Difference Which Makes No Difference Is No Difference.

Example 5.2.

An inverting amplifier has an open-loop gain of -10. The values of the feedback resistors are R₁ = R₂ = 10 k ohms. (a) What gain is predicted by equation 5.4? (b) What gain is predicted by equation 5.13? (c) What is the percent error of equation 5.4?

Solution:

For an inverting amplifier beta = R₁ / R₂ and if R₁ = R₂, beta = 1.00. (a) The gain predicted by equation 5.4 is.

A' = A / (1 + A beta) = 10 / (1 + 10 x 1) = -0.9091.

(b) The gain predicted by equation 5.13 is.

A' = A / (1 + beta ( A + 1)) = 10 / (1 + 1 x 11) = -0.8333.

(c) The result of equation 5.13 is the standard; therefore, the percent error is + 9.10%

Example 5.3.

An inverting amplifier has an open-loop gain of -500. R₁ = 10 k ohms and R₂ = 200 k ohms. What is the closed-loop gain of the amplifier?

Solution:<

The value of beta is beta = R₁ / R₂ = 10 k ohms / 200 k ohms = 0.0500. The open-loop gain is so much greater than unity that we can use equation 5.4.

A' = A / (1 + A beta) = 500 / (1 + 500 x 0.05) = -19.23.

Example 5.4.

A noninverting amplifier has the same values as in example 5.3, A = 500, R₁ = 10 k ohms and R₂ = 200 k ohms. What is the closed-loop gain?

Solution:

For a noninverting amplifier the value of beta is

beta = R₁ / (R₁ + R₂) = 10 k ohms / (10 k ohms + 200 k ohms) = 0.047619

The closed-loop gain is.

A' = A / (1 + A beta) = 500 / (1 + 500 x 0.047619) = +20.15.

The gain of an amplifier will change with frequency if there are RC networks inside the amplifier. These circuits may have been introduced intentionally or they may just be there. An example of an intentionally introduced RC circuit is the coupling capacitor between the collector of one BJT and the base of the next BJT. The resistance in the circuit and the capacitor form a high-pass filter. A semiconductor amplifier may not have any, or maybe one, of these circuits but a vacuum tube amplifier may have several.

Something never shown in schematic diagrams but always present in real circuits is stray capacitance. Remember that a capacitor consists of two conductors separated by an insulator. Any time there are two wires close together there is capacitance between them. That means that there are invisible capacitors connected from the collector of each transistor to ground and from the base of every transistor to ground. These capacitors combine with the Thevenin's resistance of the circuits to form low-pass filters. Each tiny low-pass filter causes the frequency response of the amplifier to roll off (gain decreases as frequency increases). Moreover, each transistor (BJT or FET) has its own built-in low-pass filter. Each device has its own frequency limits which are imposed by such things as collector-to-base capacitance and charge mobility within the semiconductor material.

The equations for phase shift (chapter 2) tell us that at the cutoff frequency the phase shift is 45 degrees. As the frequency gets higher than the cutoff frequency of a low- pass filter (or lower than the cutoff frequency of a high- pass filter) the phase shift quickly approaches 90 degrees. At the same time the gain is falling at a rate of a factor of 10 for each factor of 10 change in frequency.

Because the individual RC filters are isolated from one another by transistor stages, each one contributes to the total without affecting each other. If there are two RC filters with similar cutoff frequencies, the gain beyond cutoff will change by a factor of 100 for every factor of 10 change in frequency. The phase shift will approach 180 degrees. If there are three RC filters, the gain will change by a factor of 1000 for every factor of 10 change in frequency and the phase shift will approach 270 degrees. Gain change and phase shift go together like many other pairs of things you may have heard of. "You can't have one without the other." Thus, it is possible to know what the phase shift is even if you have no way of measuring it directly.

In the common operational amplifier, (op amp), the collector of the preceding transistor stage is directly coupled, not capacitively coupled, to the base of the next stage. This makes the frequency response go down to zero frequency which is DC. This is one of the few cases in electronics where it is possible to get something all the way to zero. Without any high pass filters in the amplifier the low pass filters that make the response roll off at the high frequencies are our only concern.

In some transistor amplifiers and all vacuum tube amplifiers there are high pass filters between each stage. These are capacitors in series with the signal path which are necessary to block the DC on the plate or collector of one stage from altering the bias on the grid or base of the following stage. The capacitors in conjunction with the resistors that connect to the control element of the amplifying device form high pass filters. These High-pass filters introduce phase shift and can cause low frequency oscillation if not designed properly.

The amount of phase shift depends upon the rate at which the gain is changing with frequency, which is determined by how many RC low-pass (or high-pass) circuits there are inside the amplifier. The rate of gain change is geometric. This means that a changing gain will be a straight line on a log- log graph. Figure 5.3 shows a log-log graph with various slopes (rates of change) plotted on it.

A rate of change of one decade decrease in gain for each decade increase in frequency is referred to as a slope of -1. If the gain decreases two decades for one decade increase in frequency, the slope is -2. If the gain should be increasing as frequency increases, the slope would be positive.

In the following discussion the phase shifts will be given with respect to the noninverting input. The phase shifts for the inverting input will be the algebraic sum of the stated phase shift and 180 degrees. If the slope of the frequency response is a negative 1, the phase shift will be -90 degrees. If the slope of the frequency response is -2, the phase shift will be -180 degrees. If the gain is falling with a slope of -2, the noninverting and the inverting inputs will be effectively interchanged. A connection which we thought would give negative feedback instead gives positive feedback. Positive feedback is the situation wherein the input signal is reinforced by the signal which is fed back instead of being partially canceled. When the input signal is reinforced, the output signal increases in amplitude. This causes the fed back signal to be increased which, in turn, causes the output signal to increase. This is a run- away condition and the input signal is no longer necessary; the amplifier is supplying its own input. An example is the earsplitting squeal of a sound system. This process is called oscillation and a device which oscillates is called an oscillator. When negative feedback becomes positive feedback we have an oscillator instead of an amplifier.

It would do no good to reverse the connections to the inverting and noninverting inputs. There would always be a part of the frequency spectrum where the gain is flat and there would be no extra phase shift. The circuit would oscillate because the feedback would be positive for those frequencies where the frequency response is flat.

In order to support oscillation two conditions must be met: 1) the net phase shift around the feedback loop must be zero and 2) the gain must be greater than unity. In order to prevent a feedback amplifier from oscillating, we must see to it that both conditions are not met at the same frequency. If at any given frequency only one of the two conditions is met, oscillation will not occur. The act of making internal changes to the amplifier so that both conditions will not be met at the same frequency is called amplifier compensation.

An amplifier which has not been compensated may well have a frequency response like that of the black line in Figure 5.3. A properly compensated amplifier could have a frequency response like the green line in the figure.

Examine the black curve in the above figure. It goes out flat to 100 kHz. Actually the frequency response is down 3 dB at 100 kHz but the Bode plot is drawn this way. At 100 kHz the plot breaks to a slope of -1. That means that one of the transistors started rolling off the frequency response. Another transistor in the amplifier begins to roll off at 1 MHz. This causes the Bode plot to break to a slope of -2. The plot falls by 40 dB, a factor of 100, between 1 and 10 MHz. At 10 MHz another transistor kicks in and the slope breaks to a slope of -3. After that the curve falls by 3 decades, 60 dB between 10 and 100 MHz.

What makes a feedback circuit oscillate is the loop gain not the gain of the amplifier. If this amplifier were to be used with a beta of 1/10000, it would not oscillate. 10000 is the value of gain where the slope breaks from -1 to -2. If a gain of 10000 is combined with a beta of 1/10000 the value of A beta is unity where the Bode plot breaks from -1 to -2 and the amplifier will be stable. Such a small value of beta would not be very useful but in audio amplifiers where the value of beta is fixed by the designer and does not change the Bode plot for A beta is used to compensate the amplifier instead of the plot of A alone.

On the other hand an op amp is frequently used with 100% feedback which is a beta of unity. To be useful an op amp must be compensated to unity gain. The way to convert the black curve in figure 5.3 into the green curve is to insert a passive circuit like that of figure 5.4 into the amplifier. This circuit produces the Bode plot of the red curve in the figure.

The amplifier whose frequency response is plotted in Figure 5.3 is DC coupled. Its frequency response goes flat all the way down to DC (zero frequency). An AC coupled amplifier might have a frequency response like that of Figure 5.5. In this example the upper end of the frequency response falls with a slope of -1 because it has already been compensated. The amplifier will not oscillate in this frequency range. At the low end, however, shown by the black line, the gain has a slope of +2 and is greater than unity. If feedback were applied to this amplifier, it would oscillate somewhere in the 0.3 to 10 hertz frequency range. Some internal changes will have to be made to this amplifier to make the positive portion of the gain slope become +1 from unity gain up to maximum.

The capacitor values have been set the way a real, but amateur, designer would set them given the resistor values shown. Both break points have been set at 10 Hz which seems to be the reasonable thing to do. However the slope of the plot is plus 2 and the designer will find to his sorrow and possibly amazement that the circuit oscillates at a very low frequency. This kind of oscillation is often known as motorboating because it reminded someone of the sound of an old fashioned inboard motorboat.

It would not be good to move the C₄ corner frequency up to 100 Hz because the output would rise below 100 Hz. So that one will be moved down to 1 Hz and the C₂ corner frequency will be moved up to 100 Hz. This will give the green curve in figure 5.5. To make this amplifier stable C₂ would have to be decreased to 0.001 microfarads and C₄ increased to 1.0 microfarads.

Gain at Any Frequency, and Unity Gain Bandwidth (Gain-bandwidth Product).

Let us take as an example a properly compensated AC coupled amplifier whose Bode plot is shown in figure 5.7. Its lower unity gain frequency is 0.1 hertz and its upper unity gain frequency is 100 MHz. The tangents to the frequency response curves (often called the Bode plot) are shown in Figure 5.7. As you can see from the figure, the gain is 10 at 1 Hz and 10 MHz.

To calculate the closed-loop gain of a feedback amplifier at any frequency we must know the open-loop gain at that particular frequency. We will recycle equations 2.24.4 and 2.24.5 from chapter two. You don't have to go back to look at them unless you want to see how they were derived. Those equations have a maximum value of unity. To adapt them for use here they must be multiplied by the maximum gain of the amplifier A_Max.

The frequency at which to change from using 5.15.1 to 5.15.2 is at the center frequency. This may not be what you assume. You must use the geometric center frequency which is given by,

In a device such as an audio amplifier the values of f_c1 and f_c2 usually are easy to measure. For op amps and some very high gain band pass amplifiers the values of f_cx are likely to be very difficult to measure so the unity gain frequency is measured and stated in specifications. To calculate the corner frequency from the unity gain frequency we use the following equations.

Example 5.5.

A DC coupled (op amp) amplifier has a DC gain of 100,000 and a gain-bandwidth product (unity gain frequency) of 3 MHz. What is its gain at (a) 100 kHz, (b) 20 kHz, (c) 1000 Hz and (d) 5 Hz?

Solution:

The Bode diagram for this amplifier would be similar in appearance to the green curve in Figure 5.3 but the values would be different. There are no positive slopes in the diagram and so we apply equation 5.15.5 to find f_c2.

f_c2 = f_hu / A_Max = 3 x 10⁶ / 10⁵ = 30 Hz.

(a) Using equation 5.15.1 we have

A = 10⁵ / sqrt((10⁵ / 30)² + 1) = 30

(b) A = 150

(c) A = 2,999

(d) A = 98,639.

Example 5.6.

An AC coupled amplifier has a lower unity gain frequency of 0.1 Hz and an upper unity gain frequency of 100 MHz. Its mid-band (maximum) gain is 1,000. What is its gain at (a) 5 Hz, (b) 20 Hz, (c) 1 kHz, (d) 20 kHz, (e) 200 kHz and (f) 4 MHz?

Solution:

The Feedback Equation with Sloping Gains.

The dark blue curve is the gain calculated from equation 5.4. Notice that it is down 6 dB at 100 Hz but it should be 3 dB down.

The green curve is calculated from the equation above. It is correct in this area of the spectrum, being 3 dB down at 100 Hz but the low frequency (DC) part of the curve is off by 1 dB, approximately 10%. As beta is made smaller this error gets worse. For a beta of 0.0001 the DC gain should be 5,000 which is 74 dB. The above equation, which has no number, gives 7,071 which is 77 dB. Such large errors in DC gain calculation cannot be tolerated so the no-number equation will be considered invalid and will not be used.

The equation which produced the light blue line is given below, along with its low frequency companion.

Example 5.7.

A beta value of .01 is applied to the amplifier of example 5.5. What is the closed-loop gain at frequencies of (a) 100 kHz, (b) 20 kHz, (c) 1 kHz and (d) 5 Hz?

Solution:

(a) A' = A / (1 + A beta) = 30 / (1 + 30 x 0.01) = 23.08

(b) A' = 60.00

(c) A' = 96.77

(d) A' = 99.90

Amplifiers which are used in physics laboratories are involved with measurement of an electrical quantity. A changing gain in a measurement circuit will render the measurement useless. Negative feedback will stabilize the gain of an amplifier. How sensitive the closed-loop gain is to changes in open-loop gain can be determined through the use of calculus.

We will begin with the feedback equation

This equation gives us one of the more important benefits of negative feedback, which is to stabilize the gain of an amplifier. Let us illustrate with an example.

Example 5.8.

An amplifier has an open-loop gain of 10⁵ which may vary by as much as plus or minus 30%. The value of beta is 0.010. What is (a) the closed-loop gain, (b) how much is this gain in error referenced to 1 / beta, and (c) what is the percent variation in gain due to changes in open-loop gain?

Solution:

(a) The closed loop gain is

A' = 10⁵ / (1 + 10⁵ x 0.01) = 99.90

(b) Error referenced to 100 is -0.1%

(c) By equation 5.24

E' = E A' / A = 30 x 99.90 / 10⁵ = 0.0300%.

It is often necessary to calculate how much open-loop gain is required or how much closed-loop gain can be obtained for a certain amount of gain error.

Example 5.9.

An amplifier has an open-loop gain of 5,000 which can vary by as much as 35 %. What is the maximum closed- loop gain available if the closed-loop gain variation is not to exceed 2.5 %?

Solution:

Solving equation 5.24 for A' gives,

A' = E' A / E = 2.5% x 5000 / 35% = 357.

Frequency Response.

Distortion and Noise.

Input and Output Impedance.

It is not intended that you should memorize these circuits; they are given for possible future reference. Some day if you are in graduate school or a job and you need a circuit to accomplish some task, you may just find one here. Where practical, hints will be given on how changes may be made to the circuit to help you to adapt it to your particular situation.

The circuit of Figure 5.9 is a two-stage noninverting amplifier with feedback. The feedback is applied from the output, through the voltage divider consisting of R9 and the parallel combination of R₁0 and R₄. For this circuit,

The idea of applying a signal to the emitter of a transistor may seem strange. There is an amplifier configuration (not discussed) which is known as a common base or grounded base amplifier. In such a configuration the signal is applied at the emitter and taken out at the collector. Such an amplifier does not invert the input signal. Applying a positive-going signal to the emitter is the same as applying a negative-going signal to the base. Therefore, if a signal applied to the base is inverted, a signal applied to the emitter is not inverted.

The circuit of Figure 5.9 contains a large number of resistors and capacitors because each transistor is individually biased. The circuit of Figure 5.10 eliminates the DC blocking capacitor between Q₁ and Q₂ and the biasing network in the base of Q₂. A DC voltage exists at the collector of Q₁ and a DC voltage is required at the base of Q₂. The designer's trick is to make the two voltages equal and then eliminate the capacitor and biasing resistors. The base of Q₂ gets its bias from the collector of Q₁.

If the emitter current of Q₂ tries to increase, for whatever reason, the base bias on Q₁ will be increased through the voltage divider consisting of R₁ and R₂. The increased bias on Q₁ will cause its collector current to increase which pulls down its collector voltage. Since the base of Q₂ is tied to the collector of Q₁ the base bias on Q₂ will be decreased, thus mostly canceling the increase in emitter current.

The AC gain is set by the AC feedback path which is from the output, through the voltage divider consisting of R₇ and R₄ to the emitter of Q₁. In this case,

Figure 5.11 shows another variation on the same theme. The advantage of this circuit over Figure 5.10 is that the output voltage swing can be almost equal to V_CC. In the case of Figure 5.10 the base of Q₂ is at the same potential as the collector of Q₁, and so V_CE + V_RE of Q₁ is not available for output voltage swing. The circuit of Figure 5.11 overcomes this by using a PNP and an NPN transistor. The emitter of Q₂ is at V_CC and the output voltage can almost swing from ground to V_CC.

Here is the circuit of a vacuum tube gain block. Gain blocks usually consist of two triodes with feedback from output plate to input triode cathode to diminish distortion to a very low level and tightly control the value of the gain. Here is a circuit I developed that will give a fairly wide range of gain at very low levels of distortion.

The feedback is DC coupled, not through a capacitor, from the plate of the second triode to the cathode of the first.

The 220 k ohm resistor makes up most of the load for the second triode which is where most of the distortion occurs. I don't recommend changing the value of Rf because it effects the operating point of the second triode. If this resistor is changed the cathode resistor in the second triode will have to be readjusted for minimum distortion.

If we reverse the sign of the feedback the feedback equation becomes,

Multivibrator.

If R₂ = R₃ = R and C₁ = C₂ = C then the frequency of oscillation is given by,

Circuits You May Find in Other Texts.

The Phase-shift Oscillator.

There is only one frequency at which the total phase- shift is 180 degrees and it is given by,

Quartz Crystal Oscillator.

Because mechanical waves propagate through the quartz plate at a very high velocity, a thin plate of quartz will have a very high mechanical resonant frequency (roughly from 10 kHz to 100 MHz). The Piezoelectric effect allows this mechanical resonance to be transformed into an electrical resonance.

The quartz crystal behaves as if it were an LC resonant circuit. Crystals have two major advantages over LC circuits. They are 1) the quartz crystal is usually smaller than an LC circuit tuned to the same frequency and 2) the crystal has a much higher Q than an LC circuit tuned to the same frequency. The highest achievable Q values for an LC circuit are 250, although 100 is more typical. Crystal Q values range from 1,000 to 10,000. The only disadvantage of a quartz crystal is that its frequency cannot be adjusted once it leaves the factory.

When a quartz crystal is used as the frequency- determining element of an oscillator (as opposed to an LC circuit), the oscillator frequency is much more accurate and stable. The much higher Q of the crystal is what makes the oscillator frequency so much more stable.

If you apply what you know about switching circuits you will find that the BJT is neither fully on or fully off. That is desirable in this application because the AC signal from the oscillator will drive the switch alternately on and off. If the switch were set fully off, the AC signal from the oscillator might not have enough amplitude to turn it on.

1. An amplifier has an open-loop gain of 300 and a beta of 0.01. What is its closed-loop gain?

2. A noninverting amplifier has an open-loop gain of 8. R₁ = R₂ = 12 k ohms. What is the closed-loop gain of this amplifier?

3. An inverting amplifier has an open-loop gain of 8. R₁ = R₂ = 12 k ohms. What is the closed-loop gain of this amplifier?

4. A noninverting amplifier has an open-loop gain of 1,000. R₁ = 10 k ohms and R₂ = 40 k ohms. What is the closed-loop gain of this amplifier?

5. An inverting amplifier has an open-loop gain of 1,000. R₁ = 10 k ohms and R₂ = 40 k ohms. What is the closed-loop gain of this amplifier?

6. An amplifier which has a Bode plot similar to that of Figure 5.4 has a DC gain of 75,000 and a gain-bandwidth product of 1.5 MHz. What is its gain at frequencies of (a) 100 kHz, (b) 20 kHz, (c) 1 kHz, (d) 20 Hz, and (e) 5 Hz?

7. If negative feedback is applied to the amplifier of problem 6 and the value of beta is 0.013333, what is the closed-loop gain at frequencies of (a) 100 kHz, (b) 20 kHz, (c) 1 kHz, (d) 20 Hz, and (e) 5 Hz?

8. An amplifier which has a Bode plot similar to that of Figure 5.5 has a lower unity gain frequency of 0.5 Hz, an upper unity gain frequency of 10 MHz and a maximum gain of 300. What is its gain at frequencies of (a) 5 Hz, (b) 20 Hz, (c) 200 Hz, (d) 1 kHz, (e) 20 kHz (f) 50 kHz, and (g) 2 MHz?

9. If negative feedback is applied to the amplifier of problem 8 and the value of beta is 0.025, what is the closed-loop gain at frequencies of (a) 5 Hz, (b) 20 Hz, (c) 200 Hz, (d) 1 kHz, (e) 20 kHz, (f) 50 kHz and (g) 2 MHz?

10. An amplifier is required to have a gain stability of 0.5%. The open-loop amplifier has a gain stability of 20%. If the closed-loop gain is to be 200, what must be the minimum open-loop gain of the amplifier?

11. An amplifier has a closed-loop gain of 1,000 and an open-loop gain of 100,000 plus or minus 50%. What is the closed- loop gain error?

12. An amplifier has an open-loop gain of 300,000 plus or minus 40%. What is the maximum closed-loop gain available if the error in the closed-loop gain is not to exceed plus or minus 1%?

13. What is the calculated value of beta of Figure 5.9?

14. In Figure 5.9 if the open-loop gain is 280, what is the closed-loop gain?

15. In Figure 5.12 if resistor values are as given, what must the values of C₁ and C₂ be to give an oscillation frequency of 60 Hz?

16. In Figure 5.13 if all capacitors marked C are 0.001 uf, what must be the value of the resistors R to give an oscillation frequency of 440 Hz?

17. In Figure 5.14 if there were no AC signal from the oscillator, what would be the value of V_CE of the BJT? Assume beta = 100.

1. 75.

2. 1.6.

3. -0.8.

4. 4.975.

5. -3.98.

6. (a) 15.0, (b) 75.0, (c) 1500, (d) 53,033, (e) 72,761.

7. (a) 12.50, (b) 37.50, (c) 71.43, (d) 74.89, (e) 74.92.
   Note. These calculations are correct not just close.
   If you didn't hit the numbers on the nose,
   you must have done something wrong.
   

8. f_c1 = 150 Hz, f_c2 = 33,333 Hz, f_cc = 2,236 Hz.
   (a) 9.994, (b) 39.65, (c) 240.0, (d) 296.7, (e) 257.2, (f) 166.4, (g) 4.999.

9. (a) 7.996, (b) 19.91, (c) 34.28, (d) 35.25, (e) 34.62, (f) 32.25, (g) 4.444.

10. 8000.

11. 0.5%.

12. 7,500.

13. 0.01799.

14. 46.38.

15. 0.13 microfarad.

16. 148 k ohms.

17. 1.48 volts.