replacing a pot, would that change impedance?

[Dimitree]

hello everyone,
this is a bit taken from the electro harmonix deluxe memory man schematic:



My challenge here is to replace the 1M log pot (level pot) with a 100K log pot (please don't ask why  ).
Of course I would need to lower R3 and R2 in order to get the same gain from the inverting opamp,
but changing R2 would change impedance too, right? so what should I modify in order to get the same results as the original schematic?

[pappasmurfsharem]

My challenge here is to replace the 1M log pot (level pot) with a 100K log pot (please don't ask why  ).

I have to ask why now..... :-)

[Dimitree]

because I can't find a 1M log pot with the mechanical specs that I need

[digi2t]

because I can't find a 1M log pot with the mechanical specs that I need

Is it because it`s a 24mm pot with long angled pcb mount pins?

The reason why I ask is because I recently refurbished an older DMM, which used said pots. All the pots needed to be replaced, but I couldn`t find the same pots (not at earthly prices). What I did was cut the old pots out with a pair of flush cutters, right at the bend, leaving the legs soldered to the board. I then used normal 24mm solder lug pots, bending the lugs to 90 degrees, overlapping to the old legs still on the board, and soldering them together. In the end, it gave me the same height off the board as well. I finished off by touching up the solder joints of the legs to the board, just in case there was any mechanical seperation.

Worked like a charm, and retained the same size pots as well.

[anchovie]

Tack a transistor emitter-follower set up for 100k impedance on to the input.

[Mark Hammer]

As near as I can tell, the pot itself is intended to allow the unit to be fed line level or various instrument-level signals.  With a 100k pot and the 22k resistor in series with it, it will have a max gain of 1.22x.  That, by itself, is not bad, but you better hope the input signal is hot enough.

[slacker]

You can use a 100k pot and keep the same input impedance by rewiring the pot so it forms a voltage divider. Connect the top of R3 to the wiper of the pot, connect the anti clockwise lug of the pot to the output of the opamp and connect the clockwise lug of the pot to whatever voltage "C" is through a 2k2 resistor.

When the pot is turned anti clockwise you get minimum gain with R3 in the feedback loop of the opamp like now. As you turn the pot up it reduces the signal/current that is fed back to the inverting input, so the gain goes up to compensate. The maximum gain should be the same as it is now. I don't know what this arrangement is called, I picked it off R.G. but I've occasionally seen it elsewhere. Not sure how the maths works either I just simulated it to get 2k2.

[PRR]

> I would need to lower R3 and R2 in order to get the same gain

R2 goes to 10K. And C1 has to go more like 1uFd.

A 10K input impedance on a guitar *really sucks*. (100K sucks too, but less.)

If you must use a low-value pot at VR1, the least brain-pain plan may be to let this stage be low-Z, but put a buffer in front. Any JFET or a TL072 can drive the new 10K impedance, while presenting =>1Meg to the guitar or other source.

James' transistor buffer can also work.

If you are working off a stock PCB, you can remove C1 and run wires from its pads to your buffer and back, caps in and out.

Ian's re-thinking of the gain network is also possible but perhaps non-obvious. It really wants to be Reverse-Audio taper. (And no: in this case you can't get there with "loading".)