Please help me with math..

dynamo

So I’m not good at this (among plenty of other things lol).. if someone could help that would be cool.

If I have a source plugged in (single ended rca) to a “passive preamp” (so a 10k pot), plugged into a 40k input impedance amp, roughly what input impedance is the source seeing? I assume this may change with the 10k pot position, so maybe what range.

Maybe this is an easy answer, either way I always appreciate the help.

dcibel

8k, impedance doesn’t change with pot position.

dynamo

Thanks! Am I correct in saying the impedance stays constant since it is like an L pad?

Also, would you explain how you arrived at 8k so I know how to figure it on my own in the future?

dcibel

Sorry, bad answer. Impedance does change with pot position.

At 0% volume end of the sweep, you have the input of the amp essentially shorted to ground, so everything that the source sees is just the 10k pot.

At 100% volume the input of the amp receives full voltage from the source. The source sees 10k in parallel with 40k. Calculation can be searched easily as "resistors in parallel", so 1/((1/10000) + (1/40000)) = 8000 ohms.

At 50% volume (assuming linear sweep) source sees 5 k ohm in series with 5 k ohm in parallel with 40 k ohm. Calculation is 5000 + 1/((1/5000) + (1/40000)) = 9444 ohms.

So impedance the source sees varies from 10k to 8k over the sweep range.

dynamo

Awesome thanks and that makes total sense to me thanks to your explanation!