Quickie Gain

[Doc B]

I can't think of any reason not to do it to any pot. Sealed pots should, in theory, not need it (or at least need exercise less often). Why not do it anyway?

I have just installed a couple of CTS conductive plastic sealed pots (283 series) in my new guitar. Small, quite nice feel and very quiet. Interesting to note the rated lifetime for the similar 282 series is 2 million cycles. I'm guessing 283 is log and 282 is linear, but not sure about that.

 

[adamct]

If my DAC puts out a 2V signal, and I turn the Quickie up all the way, what will the signal to my amp be?

Best regards,
Adam

 

[Paul Birkeland]

The answer to this question depends on a few other parameters, and they are (I'll probably miss one or two here!)

1.  The load seen by the Quickie (input impedance of your power amp(s).
2.  PJCCS, Choke, or Resistor plate load
3.  Freshness of batteries
4.  How much THD is acceptable?

Under ideal circumstances (PJCCS/no load), I think you'd get maybe 10dB of gain out of the Quickie, which would get you 6V out.  This seems tight in terms of compliance within the PJCCS, but I would want to test it myself to see what happens.  With a stock Quickie, that would drop to ~4V.  With decreasing load impedance, the output impedance of the Quickie will become a factor, and the voltage present at the input of your amplifier will decrease more. 

The good news is, you can play a 60Hz tone through your DAC and measure the AC voltage out with your meter.  This won't tell you much about distortion, but it's a start!

Having made too many assumptions, I just performed the measurements. I can get 11dB of gain with 0.7V in and 1.5% THD with the PJCCS and no load.  This drops to an even 1% at 8dB of gain.  Unfortunately, my testing setup won't generate 2V of signal, and our low-noise/low-THD testing amps aren't finished yet, so that's all I can report. 

 

[adamct]

Thanks, Paul. Much appreciated. I can't find input impedance specs. for my headphone amplifier. I will be using the Quickie with the PJCCS.

But you're already answered by question, so I'm all set. Thanks!

Best,
Adam