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In Reply to: 'Loading' a preamp with a poweramp, input impedance, and the load resistor posted by jeffreybehr on February 29, 2012 at 10:33:05:
"I achieved your goal and my goal of decreasing the Voltage gains of several poweramps by adding a resistor in series. That does indeed increase the resistance, decrease the preamp's load, and decrease the Voltage gain of the poweramp."
This is completely incorrect or stated in such a way as to make no sense.
The power amp's gain is fixed by the internal circuit topology. Nothing you do to the signal BEFORE the input to the amp will change the gain of the amp IN ANY WAY. It is already fixed. If you reduce the amplitude of the signal BEFORE the input of the amp you will get less signal out of the amp because the gain of the amp is constant. A smaller signal in will result in a smaller signal out.
This, of course, has nothing to do with the input impedance, this is simply attenuating a signal using a resistor in SERIES forming a voltage divider with the load resistor. This, naturally, adds noise to the signal and degrades the sound; so is not recommended unless you can find a "perfect" resistor.
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Follow Ups
- RE: 'Loading' a preamp with a poweramp, input impedance, and the load resistor - Palustris 11:11:58 02/29/12 (3)
- Perhaps I should have written 'poweramp system' instead of just 'poweramp'. - jeffreybehr 21:46:17 02/29/12 (0)
- no,,,, that makes perfect sense - bwb 14:29:39 02/29/12 (1)
- RE: no,,,, that makes perfect sense - Hawk28 16:24:28 02/29/12 (0)
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