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See also: Feedback and Distortion, Positive Feedback, Negative Feedback, Current Feedback, The Cathode Follower and Feedback  
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Voltage FeedbackAuthor: N.H. Crowhurst
There are several ways in which feedback may be distinguished. For example, either the output voltage or the output current can be used as a basis for the output "sampling." The chief difference as far as the operation of the amplifier is concerned is in the effect on the amplifier's effective output impedance. Let us work out a typical example, using voltage feedback. Suppose that an amplifier has to work with a 15ohm resistance as its output load. It has an internal resistance due to the plate resistance of the output tubes of 5 ohms. Without the output load connected, a 10millivolt input produces a 10volt output. Without feedback, a 2millivolt input would produce the 10volt output. This means the gain of the amplifier, A, is 5000, or 74 db. The feedback, (3, is 8 millivolts fed back for 10 volts at the output, 8/10,000 of 4/5000. The loop gain, A(3, thus is 5000 X 4/5000 or 4. The feedback (1 + Ap) is 1 + 4 or 5, which corresponds to 14 db. This means the gain and distortion will be divided by 5. Connecting the 15ohm load causes the same 2 millivolts at the input of the actual amplifier to produce (10 volts X 15 ohms/20 ohms) 7.5 volts output.
Because i is .0008, a 7.5volt output will produce 6 millivolts of feedback. The total input required to give a 7.5volt output with the load as well as the feedback connected is (2 + 6) or 8 millivolts. By simple proportion we can deduce the output with a 10millivolt input. It will be (7.5 X 10/8) or 9.375 volts. Thus the effective drop in the internal resistance (because feedback changes the actual input) is from 10 volts (without the load connected) to 9.375 volts (with the load connected), a difference of 0.625 volt. As it takes a load of 15 ohms to do this, the internal resistance must be 0.625/9J75 X 15 ohms, or 1 ohm. The connecting of feedback, which has reduced the gain by a ratio of 5, has also reduced the effective internal resistance of the output by the same ratio. Without feedback, the internal resistance was 5 ohms; with feedback it is 1 ohm.
We could use the formula instead of the numbers and come out with the same answer  that the internal resistance with voltage feedback is divided by the same factor that the gain is reduced by. An important point to notice (not stated in many textbooks on the subject) is that we must use the gain without the load connected in this formula.


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