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Current FeedbackAuthor: N.H. Crowhurst
The other way in which the output can be sampled is called current feedback. Assume that we have an amplifier that without any load connected gives 12 volts output for an input of 2 millivolts. It has an internal resistance of 5 ohms. When the output is shortcircuited, the internal resistance will be the only thing to limit the current  12/5 = 2.4 amperes. Assume that this 2.4ampere output produces 8 millivolts of feedback and that the current feedback is always in this ratio. This means the amplifier, complete with feedback, requires 10 millivolts input to produce a 2.4ampere output on shortcircuit. Now assume that the 15ohm load is connected to the amplifier, still with 2 millivolts at the input. The original 12 volts at the output will produce a current of (12/20) or 0.6 ampere (which produces 9 volts across the 15ohm load). Because the 2.4 amperes cause 8 millivolts of feedback, the 0.6 ampere will produce 8 X 0.6/2,4) or 2 millivolts feedback. Thus the input required for this condition will be a total of (2 + 2) or 4 millivolts. Assuming that we still apply the original 10 millivolts of input, the output current will then be (0.6 X 10/4) or 1.5 amperes. Now we have a basis for comparison. With the output shortcircuited, the feedback amplifier with a 10millivolt input gives a 2.4ampere output. With the load of 15 ohms on the output, the same 10 millivolts input produces a 1,5ampere output.
We can take the matter one stage further and consider the output with no load at all. If there is no current, there will be no feedback and a 10millivolt input will produce (10/2 X 12) or a 60volt output. With the current feedback applied and a 10millivolt input, the effective opencircuit output voltage is 60 volts, whereas the shortcircuited current is 2.4 amps. This means the effective internal resistance is 60/2.4 or 25 ohms. This checks with the loaded condition as well. When the 15ohm load is connected, the total effective resistance is (25 + 15) or 40 ohms. This resistance connected across an effective 60volt source will allow a current of 1.5 amperes, as calculated earlier. The current feedback has multiplied the effective source resistance by 5. (The actual value is 5 ohms and the effective value is 25 ohms.) The reduction in gain caused by the feedback is 5» only when the amplifier operates shortcircuited. At normal loaded condition, the reduction in gain due to feedback is 2. (Without feedback, 2 millivolts produces the same output as 4 millivolts does with feedback.) Disconnecting the load entirely results in no feedback. Thus, for current feedback, the effective output source resistance is the actual source resistance multiplied by the feedback factor when the amplifier output is shortcircuited.


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