Current Feedback

Author: N.H. Crowhurst

Using current feedback

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 short-circuited, the internal resistance will be the only thing to limit the current - 12/5 = 2.4 amperes. Assume that this 2.4-ampere 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 2.4-ampere output on short-circuit. Now assume that the 15-ohm 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 15-ohm 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 short-circuited, the feedback amplifier with a 10-millivolt input gives a 2.4-ampere output. With the load of 15 ohms on the output, the same 10 millivolts input produces a 1,5-ampere output.

No load results in no feedback

Current feedback raises source resistance

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 10-millivolt input will produce (10/2 X 12) or a 60-volt output. With the current feedback applied and a 10-millivolt input, the effective open-circuit output voltage is 60 volts, whereas the short-circuited 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 15-ohm load is connected, the total effective resistance is (25 + 15) or 40 ohms. This resistance connected across an effective 60-volt 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 short-circuited.

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 short-circuited.