I'm not 100% sure how one determines the input impedance of an amplifier, especially without a schematic. The DC component is easy, but I suspect there may be a significant contribution through the first transistor/tube stage as well. As a first test, I connected my LCR meter to the input terminals of the sub and measured the resistive component at 100Hz and 1KHz, and I got 10 KOhms. I've heard moderm SS amps are this low, so this seems reasonable.
Next I measured the VTA ST70. It measures about 75 KOhms, which is as expected for the combination of the 270 and 10 K fixed internal resistors and a 100 K stepped attenuator.
From what I've read about the PAS, they are expecting a 500 K external load so that the total load seen at the end of the bass pot is 50 K (comprised of 62K || 510K || 500K). It seems like the only consequence of not having a 500 K load is that the tone controls will not function properly. So my first question is is this truly the only consequence, or are there other problems with a PAS driving a low impedance load?
Of course with the 3X model the tone controls are bypassed when centered and at this position the output impedance is not affecting the tone circuits, but I want to consider the case when the tone controls are moved off center as well.
So the issue I am pondering is this, if I wanted to actually *use* the tone controls on a PAS 3X, what would I need to do as far as circuit modifications?
Probably it will never be possible to drive a 10 K sub properly, but even driving the VTA70 may require modifications because the stepped attenuator drops the input impedance all the way down to 75 K and this is too low for an unmodified PAS. It is a simple matter to fix this though. We can either remover the stepped attenuator so it is not messing up the load (it is not necessary with the PAS anyway since presumably the PAS is being used to control volume), or we can change the PAS output resistors (62K and 510K) to values that maintain the 50K internal load the PAS wants to see.
It seems that ideally, to use a PAS with a VTA driver board, one should do both. First remove the 100K stepped attenuator (if present), and then remove the 510K fixed internal resistor from the PAS. This would make remaining internal load almost exactly 50K and keep the PAS happy. Does this sound right?