Hi George,
Yes, I’m pretty certain that what I said is correct.
The example you cited, resulting in the cap seeing 50K, applies if the pot is set at max. But since the amplifier with the 100K input impedance is connected to the wiper of the pot consider what would happen if the pot were set, for example, such that 10% of the pot’s resistance is between the wiper and ground, and 90% of the pot’s resistance is between the wiper and the point where the pot and the cap are connected.
With that setting the resistance between the point where the cap and the pot are connected together and ground (i.e., what the cap sees), assuming the same 100K input impedance of the amplifier, would be:
Cap sees: 90K + (10K in parallel with 100K) = 99.1K, not 50K
The rolloff of the resulting high pass filter in that example will be determined by the combination of 0.1 uf and 99.1K. The rolloff (and the corresponding phase shift) will occur at the point where the cap and the pot are connected together, and it will also occur at the wiper of the pot which in turn is wired to the output connector. The signal at the output connector and the signal at the point where the cap and pot are connected together will differ in terms of amplitude, of course, but not in terms of frequency response (putting aside the possible effects of cable capacitance and amplifier input capacitance on high frequencies).
Finally, to be sure we are looking at the same schematic, I am looking at the one shown on the last page of the manual.
Best regards,
-- Al
Yes, I’m pretty certain that what I said is correct.
The example you cited, resulting in the cap seeing 50K, applies if the pot is set at max. But since the amplifier with the 100K input impedance is connected to the wiper of the pot consider what would happen if the pot were set, for example, such that 10% of the pot’s resistance is between the wiper and ground, and 90% of the pot’s resistance is between the wiper and the point where the pot and the cap are connected.
With that setting the resistance between the point where the cap and the pot are connected together and ground (i.e., what the cap sees), assuming the same 100K input impedance of the amplifier, would be:
Cap sees: 90K + (10K in parallel with 100K) = 99.1K, not 50K
The rolloff of the resulting high pass filter in that example will be determined by the combination of 0.1 uf and 99.1K. The rolloff (and the corresponding phase shift) will occur at the point where the cap and the pot are connected together, and it will also occur at the wiper of the pot which in turn is wired to the output connector. The signal at the output connector and the signal at the point where the cap and pot are connected together will differ in terms of amplitude, of course, but not in terms of frequency response (putting aside the possible effects of cable capacitance and amplifier input capacitance on high frequencies).
Finally, to be sure we are looking at the same schematic, I am looking at the one shown on the last page of the manual.
Best regards,
-- Al