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In Reply to: Amplifier Modifications posted by Hawk28 on February 29, 2012 at 06:34:14:
The 'load' of a preamp driving a poweramp has to do with CURRENT flow. A higher load-resistor value DECREASES the load of the preamp because the preamp drives less current at the same level-control setting. Increasing the value of the poweramp's load resistor (which is wired to neutral/common/ground) decreases the preamp's load.
Hawk, if you have some skills and experience at soldering and changing small components such as resistors on circuitboards--does you poweramp have a printed-circuit board or is it hardwired?--changing the load resistor probably will be easy or at least doable. If you don't, I suggest you have someone else do it.
And yes, I have done this, so to speak. 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. Most recently, I added 220K in series with the signal to several 47K-input-impedance Marantz MA-24s, which increased their input impedances and decreased their Voltage gain from around 30dB to around 10dB.
What are the output and input impedances of your pre- and poweramps now?
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Follow Ups
- 'Loading' a preamp with a poweramp, input impedance, and the load resistor - jeffreybehr 10:33:05 02/29/12 (5)
- Goldpoint shows how to accomplish this... - wheezer 12:12:10 03/02/12 (0)
- 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)