Figure 5.1 shows an amplifier with negative feedback applied. The amplifier is a differential amplifier which was discussed in the closing section of chapter 4. The signal generator is applied to the noninverting input of the amplifier. A fraction, beta, of the output signal is fed back to the inverting input. The difference between the two signals is amplified by the factor A and appears at the output.The output of a differential amplifier is given by equation 4.27 which is repeated here.
VO = A (V1 - V2) (5.1) As shown in Figure 5.1 V1 is the input signal and V2 is the fraction of the output signal beta VO. If we make these substitutions into equation 5.1 we haveVO = A (VIN - beta VO). (5.2) Solving this equation for VO / VIN we have the feedback equation.VO / Vin = A / (1 + A beta) (5.3) The term VO / VIN is called the closed loop gain and is given the symbol A', thus,A' = A / (1 + A beta) (5.4) where A' is the closed-loop gain, A is the open-loop gain and beta is the fraction of the output which is fed back to the input. This equation is to amplifiers as Ohm's law is to electric circuits. Memorize it.For the noninverting amplifier beta is given by
beta = R1 / (R1 + R2) (5.5) This beta is not to be confused with the BJT beta. They are not the same thing nor are they related in any way.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 byNote that we indicate this gain as a positive number. The circuit of Figure 5.1 is a noninverting amplifier which has no phase inversion. Such an amplifier has the signal applied to the noninverting input.A' = A / (1 + A beta) = 250 / (1 + 250 x 0.02) = 41.67.
Back to Fun with Transistors.
Back to Fun with Tubes.
Back to Table of Contents.
Back to top.
5.3 The Special Case of the Inverting Amplifier.
When the signal is applied to the inverting input, the circuit is called an inverting amplifier. An AC signal applied to the amplifier will be inverted by 180 degrees. When the value of the gain is stated it must be given as a negative quantity. The circuit of an inverting amplifier is shown in Figure 5.2. If you look in other texts similar to this one you are not likely to find the derivation of the gain of an inverting amplifier. The reason is that the equation which results from an exact derivation is slightly different from the one for a noninverting amplifier. Let us derive the equation and then we will discuss the difference.
Figure 5.2 Inverting Amplifier with Equation Loops Indicated.
For a verbal description click here.
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' = 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%
A' = A / (1 + A beta) = 500 / (1 + 500 x 0.05) = -19.23.
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.
Back to Fun with Transistors.
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.
Figure 5.3 Bode Diagram Showing Compensation of an Amplifier.
For a verbal description click here.
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.
Figure 5.4 Schematic of Compensation Network.
For a verbal description click here.
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.
Figure 5.5 Bode Diagram with Low Frequency Role Off.
For a verbal description click here.
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.
Figure 5.6 Vacuum Tube Amplifier With Feedback.
For a verbal description click here.
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.
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 AMax.
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 fc1 and fc2 usually are easy to measure. For op amps and some very high gain band pass amplifiers the values of fcx 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.
fc2 = fhu / AMax = 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.
Figure 5.7 Bode Diagram for Example 5.6.
For a verbal description click here.
The first step is to find fc1 and fc2.
fc1 = flu AMax = 0.1 x 1000 = 100 Hz.
fc2 = fhu / AMax = 108 / 1000 = 105 Hz.
In order to decide which equation to use we must split the frequency response curve at the center of the flat portion. The center is the geometric mean of 100 Hz and 100,000 Hz.
fcc is given by,
fcc = sqrt(100 x 100,000) = 3.16 kHz
(a), (b), and (c), are all below the center frequency and (d), (e), and (f), are all above it. Therefore we will use equation 5.15.2 for (a), (b), and (c), and 5.15.1 for (d), (e), and (f).
From equation 5.15.2 we have,
(a) A = 49.94,
(b) A = 196.1,
(c) A = 995.0
Now changing to equation 5.15.1,
(d) A = 980.6.
(e) A = 447.2,
(f) A = 24.99.
Figure 5.7.1 Curves showing open-loop gain and
For a verbal description click here.
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.
(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
Back to Fun with Transistors.
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.
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.
A' = E' A / E = 2.5% x 5000 / 35% = 357.
Back to Fun with Transistors.
Figure 5.8 Bode Diagram of Closed and Open-loop Gains.
For a verbal description click here.
Back to Fun with Transistors.
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.
Figure 5.9 Laboratory Tested Circuit. Gain = 47, Frequency
For a verbal description click here.
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.
Figure 5.10 Laboratory Tested Circuit. Gain = 85,
For a verbal description click here.
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 VCC. In the case
of Figure 5.10 the base of Q2 is at the same potential as the
collector of Q1, and so VCE + VRE 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 VCC and the output voltage can almost swing from ground to
VCC.
Figure 5.11 Laboratory Tested Circuit. Gain = 24,
For a verbal description click here.
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.
Figure 5.11.1 Vacuum Tube Gain Block.
For a verbal description click here.
In Figure 5.2 the loops and polarity of all voltage
drops are clearly marked. Consider these markings as a
"snapshot" taken at the peak of the sine wave. Current flows
from left to right through R1 and R2 thus causing the voltage
drops to be as indicated on R1 and R2. Because of the nature
of the amplifier the output voltage (VO) must have the
opposite polarity as the input voltage (Vsig).
A Difference Which Makes No Difference Is No Difference.
If the open-loop gain A is much greater than unity,
equation 5.13 may be approximated by equation 5.4 with little
detectable error. In actual practice open-loop gains are
almost never less than 200 and may be as high as 107. Adding
one to these large numbers will not make any practical
difference. Unless the open-loop gain is less than 50, you
need not bother with equation 5.13. Use equation 5.4.
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.
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.
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
Back to Fun with Tubes.
Back to Table of Contents.
Back to top.
5.4 Frequency Response and Stability.
Until now we have treated amplifiers as having a phase
shift of 0 degrees (noninverting) or 180 degrees (inverting).
As long as the frequency response is flat (gain does not
change with frequency), these rules of phase shift apply.
If the gain is rising or falling with frequency, there will
be an extra phase shift which may cause trouble.
It should be pointed out that the slope of real
frequency response curves does not break suddenly from one
slope to another. The curves make gradual and smooth
transitions from one slope to the other as we saw in chapter
two. What has been drawn in Figure 5.3 are the
tangents to the curves. The tangents from the flat portion
and the -1 slope portion intersect directly over the -3 dB
point. That is why the -3 dB frequency is sometimes called
the corner frequency.
The frequency where the red curve breaks from flat to a slope of -1, 10 Hz in this case, is given by,
The black line plot in figure 5.5 is of the amplifier in figure 5.6. Capacitors C1 and C5, if they were numbered, are outside of the feedback loop. C3 is the cathode bypass of the second amplifier and common practice is to set such capacitors so the corner frequency for the RC circuit is very much less than one Hz. Experience has shown that the cathode RC has little effect on the phase shift of the amplifier even if the corner frequency is about one Hz. The resistor values have been fudged so the corner frequencies will fall on decade marks on the Bode plot. These values are not necessarily realistic.
The value of beta is 1/10 so the amplifier does not have to be compensated to unity gain. A beta must be compensated to unity which means the gain needs only to be compensated to 10 as shown by the green line. To do this one of the break points must be moved down to 1 Hz and the other one up to 100 Hz.
Gain at Any Frequency, and Unity Gain Bandwidth (Gain-bandwidth Product).
An amplifier which has been properly compensated will
have its gain change by one decade for every decade change in
frequency, for all values of A beta greater than unity. A
compensated amplifier may have a very large gain at middle
frequencies but its gain can become quite low at the extremes
of frequency near the unity gain frequencies.
Where f is the frequency at which the gain is desired, A is the desired gain, fc1 is the lower cutoff or corner frequency, fc2 is the upper cutoff or corner frequency, and AMax is the maximum gain. For an AC coupled amplifier, one that roles off at both ends of the frequency spectrum, AMax is the maximum gain that the amplifier reaches at mid band. For a DC coupled amplifier such as an op amp the maximum gain and the DC gain are the same thing. As long as you don't interchange fc1 and fc2 you shouldn't have any trouble knowing which equation to use.
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 fc2.
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 Bode diagram for this amplifier is shown in Figure
5.7. Making a Bode diagram is a good way to avoid
mistakes in problems of this kind.
The Feedback Equation with Sloping Gains.
The examples above show how the magnitude of the amplifier gain varies with frequency. In order to make absolutely correct calculations of closed loop gain, taking phase shift into account, we would have to use the j operator. There is no mathematically rigorous way around it. You might suppose that we could use the Pythagorean theorem as we did in our study of RC circuits but the phase shift is not ninety degrees in the region around the corner frequency. The equation that produced the green line in figure 5.7.1 looks like this.
That equation seems to work but it trades off accuracy in one part of the frequency spectrum for inaccuracy in another. A graph will show what's going on.
three calculations of closed-loop gain.
The graph shows frequency on a log scale versus dB. The numbers along the horizontal axis are the power to which you must raise 10, to obtain the frequency. The scale is from 0.1 to 100,000 Hz. Remember that ten to the 0 power is one. The red curve shows the open loop gain of a DC coupled amplifier with a DC gain of 10,000 and a unity gain bandwidth of 100 kHz. Notice that the gain is down by 3 dB at 10 Hz as it is supposed to be.
These equations were not so much derived as they were cloodged together to get the right answer. The light blue line behaves the way the real frequency response does. However, equations 5.4.1 and 5.4.2 will not be required in homework or test problems. They are presented for reference only.
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:
Back to Fun with Tubes.
Back to Table of Contents.
Back to top.
5.5 Gain Accuracy and Precision.
Amplifiers without feedback have rather unpredictable
gains. Transistor parameters vary with temperature and from
one transistor to the next. Even if a one-of-a-kind
amplifier were designed and constructed, its gain would
change with temperature. If a transistor ever had to be
replaced the gain would most likely change drastically.
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
This example shows that when the open-loop gain is very
high, it has very little effect on the closed-loop gain. The
closed-loop gain is set by beta which in turn is set by the
values of two resistors. Resistors can be obtained as very
high precision units.
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,
Back to Fun with Tubes.
Back to Table of Contents.
Back to top.
5.6 Other Benefits of Negative Feedback.
A fairly obvious question is "If negative feedback
reduces the gain, why use it?" In the previous section we
have seen that applying negative feedback to an amplifier
will control its gain very accurately. That is not the only
benefit of negative feedback. Almost any amplifier parameter
you care to name is improved by applying negative feedback to
it. Any amplifier designer will tell you that gain is the
easiest of all parameters to obtain. The standard practice
is to design an amplifier with much more gain than is needed
and use negative feedback to trade the extra gain for
improvement of all of the other parameters of the amplifier.
Frequency Response.
Frequency response is improved by the application of
negative feedback. The low frequency limits are moved lower
and the high frequency limits are moved higher. For example
look at the Bode diagram in Figure 5.8.
The frequency response of amplifiers is usually given in
terms of the -3 dB points rather than the unity gain points.
The open-loop gain of the amplifier whose Bode plot is given
in Figure 5.8 has a frequency response of 1,000 to 10,000
hertz. Not exactly hi-fi is it? When negative feedback is
applied to reduce the gain to 100, the frequency response
becomes 10 Hz to 1 MHz. Quantitatively the improvement of
frequency response is given by,
Distortion and Noise.
Distortion is defined as any alteration of the original
signal by an electronic circuit. In audio circuits
distortion is expressed as a percentage. When an absolutely
pure sine wave is introduced at the input of an amplifier,
the output consists of the original sine wave plus harmonics
of the original sine wave. The total of the harmonics is
expressed as a percentage of the original sine wave. This is
often called the total harmonic distortion or THD of an
amplifier. To a musician, distortion represents an
alteration of the reproduced sound. To a physicist,
distortion represents nonlinearity and errors in
measurements. Distortion and nonlinearity are reduced by the addition of
negative feedback to an amplifier as follows. Consider THD to be equivalent to nonlinearity.
Input and Output Impedance.
The output impedance is lowered by the application of
negative feedback by,
Back to Fun with Tubes.
Back to Table of Contents.
Back to top.
5.7 Some Practical Circuits.
From time to time in this text, laboratory tested
circuits will be presented. A circuit which actually works
can illustrate a principle as well as one which looks as if it
might work but probably will not. The latter kind of
circuits are what are normally presented in electronics
texts.
Response is from 35 Hz to 600 kHz +0 -3 dB, Input Impedance
is 9 k ohms, Output Impedance is 700 ohms, VCC is 12 volts.
The gain may be altered by changing the value of R9.
The transistors are 2N3904.
The amplifier of Figure 5.9 has two BJT stages. The
signal which is applied to the base of Q1 is inverted by that
stage and applied to the base of Q2 where it is inverted
again. Two inversions make a noninversion and, therefore,
the base of Q1 is a noninverting input. A signal applied to
the emitter of Q1 appears at its collector without being
inverted. When this signal is passed through the Q2 stage it
is inverted. Therefore, the emitter of Q1 is an inverting
input. If negative feedback is to truly be negative, it must
be applied to an inverting input.
frequency Response is from 4 Hz to 230 kHz +0 -3 dB,
input Impedance is 53 k ohms, output Impedance
is 580 ohms, Gain may be adjusted by changing R7.
This circuit is especially interesting because it has
two feedback paths. One is from the emitter of Q2 to the
base of Q1 and the other is from the collector of Q2 (the
output) to the emitter of Q1. Capacitor C2 filters out the
AC signal at the emitter of Q2 and so the feedback path from
the emitter of Q2 to the base of Q1 is DC only. Because Q2
gets its base bias from the collector of Q1, any change in
the operating point of Q1 will be amplified and passed to Q2.
The DC feedback path provides operating point stability for
both stages.
Frequency Response is from 3 Hz to 1 MHz +0 -3 dB,
Input Impedance is 16.5 k ohms, Output Impedance is
332 ohms, VCC is 12 volts. The gain may be altered
by changing the value of R7.
The voltage divider consisting of R1 and R2 provides a
constant bias for the base of Q1. For DC, 1/2 of the output
is fed back to the emitter of Q1 for operating point
stability. R4, R5 and R6 provide the DC feedback to the
emitter of Q1. R7 and C2 combine to reduce the amount of AC
feedback which will raise the AC gain of the amplifier. If
R7 is shorted out, the amplifier will provide its full open-
loop gain for AC. If R7 and C2 are removed from the circuit,
the gain will be 2 for both AC and DC.
This circuit has a lot of open loop gain which is traded off for low distortion. The gain may be changed by changing the cathode resistor, Rk, in the first triode. Changing this resistor has a very small effect on the operating point of either triode. The numbers are given in the table below.
