Parallel resistor with logarithmic pot?

[MonarchMD]

Okay, so it's been a long time since I've posted but I came across something I can't seem to find concerning a resistor in parallel with a potentiometer.

This all comes from a repair that landed on my bench. One of those silver boxed Vox V830 Distortion Boosters. Opened it up and 2 of the pots are damaged. Most notably, the drive pot. It's a 9mm dual 200kA pot. Nobody is stocking them. One gang of the pot controls the gain of the first op-amp, the classic TS style drive section. The other controls a second non-inverting op-amp gain stage with no diodes in the feedback loop.

First off, the pot is hard as heck to find, but I can get a 250kA with a solid shaft that should be just fine.

I know you can put a resistor in parallel with lugs 1 and 2 of a linear potentiometer to make it behave more logarithmically (and lower the effective range depending on the size of the resistor) but what affect does the parallel resistor on an already logarithmic pot have?

[crane]

You can model the behaviour of a resistor in paralel to log pot in excel or any other similar software.
The trick here is to know the taper of the pot (I know you are saying it's log - but you know that usually pots are not true log but consist of two "straight lines")

[crane]

ok I did play a little in excel.
it's hard to understand where is your parallel resistor connected but if I understand you corrctly - depending on it's value it straightens the log out.

[Elektrojänis]

Depending on the position in the circuit, difference between 200kohm pot and 250kohm pot would most likely be something between maybe noticeable to practically indistinguishable. I'd try with just the 250k and see how it works.

It's also good to remember that pots typically have a tolerance of 20%. That means your 200k  might be anything between 160k and 240k and still be in spec. The nubers for the 250k would be 200k to 300k.

[ElectricDruid]

+1 what Petri says. I'd certainly *try* just sticking the 250K pot in and see how it goes. The chances are it won't be a huge difference. Maybe you get a little bit "more X" at the far end of the range? Since people choose their setting from the range available, all you've done is make that range slightly wider.

There are situations where this isn't true, but it's most likely.

[Elektrojänis]

I found a schematic on anothe website... If that and my Q&D calculations are correct, the max gain of the first stage is about 0.86dB higher with the 250kohm pot. The second stage will be 0.85dB hotter when maxing the gain. Combined that makes about 1.7dB.

I wouldn't worry about that too much. I wouldn't even say this one goes to 11.

[MonarchMD]

Thanks everyone. Dang I forgot how much I like this forum. I did not expect the responses to come in so quickly and early. Not sure where y'all are but I posted my question just before midnight on central daylight time.

You can model the behaviour of a resistor in paralel to log pot in excel or any other similar software.
The trick here is to know the taper of the pot (I know you are saying it's log - but you know that usually pots are not true log but consist of two "straight lines")

Don't tell me how to do this...yet. I'm decent with excel and noticed in some searching for answers before I posted that this was possible but I didn't dig too much into it since most of what I was finding was folks adding the parallel resistor to a linear pot.
I'll chime back in if I have trouble wrapping my head around the data.

It's interesting that I'm able to find linear taper pots easier for this than log. Wondering what the process was in using log if just to make it more linear like with a parallel resistor. The parallel resistor is 1Meg so it's not really lowering the overall variable resistance if I'm doing my equations correctly.

Off to order the 250k. It's solid shaft, which I hate, but it's what I'm gonna do. I always have back luck with knobs on solid shafts never getting tight enough or staging tight.  The original pots are D-shaft.

Thanks again everyone!

[MonarchMD]

Okay. I ran the google sheets version just for fun. Assuming the stock pot was A15 based in the following statement from potentiometers.com's FAQ, "Most off-shore manufacturers use a 15% audio taper for their audio taper designs," the 1Meg resistor in parallel with this pot gives it a more A20 taper, which translates to about 20% of the resistance range at the halfway point. What I find interesting is that I can't believe that would make much discernible difference in the taper for most folks.

I found a schematic on anothe website... If that and my Q&D calculations are correct, the max gain of the first stage is about 0.86dB higher with the 250kohm pot. The second stage will be 0.85dB hotter when maxing the gain. Combined that makes about 1.7dB.

I wouldn't worry about that too much. I wouldn't even say this one goes to 11.

I wonder if throwing in a 500k would be fun. It's a decent 5.7 dB increase on each stage if my math is right. You wouldn't get more gain really out of that first stage because of the diodes, it'd just clip sooner(?). And similar with the second stage? That stage is kinda "Rat-ish."

Another interesting mod could be a dual concentric pot if it'll fit. Have each gain stage independently controllable?

Now I'm just brainstorming.

[Rob Strand]

I ran the google sheets version just for fun. Assuming the stock pot was A15 based in the following statement from potentiometers.com's FAQ, "Most off-shore manufacturers use a 15% audio taper for their audio taper designs," the 1Meg resistor in parallel with this pot gives it a more A20 taper, which translates to about 20% of the resistance range at the halfway point. What I find interesting is that I can't believe that would make much discernible difference in the taper for most folks.

There shouldn't be much to calculate.

Target resistance at full: 200k // 1M = 167k
- To get 167k with a 220k pot => change 1M to 1/(1/167k-1/220k) = 690k  use 680k
or put a 2M2 in parallel with the existing 1M
- To get 167k with a 250k pot => change 1M to 1/(1/167k-1/250k) = 503k  use 470k or 610k
or put a 1M in parallel with the existing 1M.

What happens at 12 O'clock is out of your control.

If we take a 15% taper,
Pot                15% with parallel R
200k // 1M         29.1k
220k // 1M // 2M2  31.5k
250k // 1M // 1M0  34.9k

I've measured anything from 12% to 22% on A-tapers.  So if you had a 200k with 22% the total resistance would be 200k*0.22 // 1M = 42.1k.  So you can see the uncertainty in the taper is larger than the error at 12 O'Clock with a 250k 15% pot.

If you start tweaking the parallel resistance to match the 12 O'Clock resistance it will greatly affect gain at full boost - bad idea.

[PRR]

a resistor in parallel with a potentiometer.

This really depends how the "pot" is connected. Potentiometer, rheostat, or other?

And then: resistor end-to-end, or wiper-to-end(s)?

I took a wild poke in Bingle for an example schematic. This may not be the one you have.

If this is the plan, there is already a parallel resistor. Both are simple rheostat gain-set connections. You "can" scale the whole network to suit the available value.


However I don't think any refinement or adjustment is needed. Worst-case (bad layout), it will squeal on "10", don't do that, or adjust the parallel resistor smaller till it shuts up.

[MonarchMD]

Yes. I should've clarified that both gangs are rheostats. I'm just curious why they did a parallel resistor in the first place. Both of those gain stages are less max gain then their counterparts (the first TS style gain stage into the second RAT style stage) by quite a bit, and I wouldn't think the tiny little bit of dB boost without the 1M resistor and the slight change in taper would be discernible to 99% of users. With such low tolerance for these kinds of pots, it seems like two excess parts.

The taper change, according to the spreadsheet I dialed up, seems to be from an A15 taper to something like A17.5 if they made such a pot. Lol.

[Elektrojänis]

I wonder if throwing in a 500k would be fun. It's a decent 5.7 dB increase on each stage if my math is right. You wouldn't get more gain really out of that first stage because of the diodes, it'd just clip sooner(?). And similar with the second stage? That stage is kinda "Rat-ish."

The gain will not really translate to more volume on either of those stages, as the diodes clip it off at some point anyway. Gain control on these is more of a distortion control anyway. Guitarists tend to use the word gain to mean distortion (or overdrive, which is not really different).

I quess 500kohm might make an audible difference in max gain, but you would have to try it to see. 5.7dB per stage might seem like quite a bit, but the max gain of each stage is already over 30 db with the 200kohm pot and it will be distorting quite a lot with a typical guitar signal. At that point the differences tend to start sounding less than you might expect. Or that's my experience anyway, but I've never played with that Vox Distortion Booster.

Edit: You could just temporarily wire in two 500kohm (or some close standard value like 470kohm or 510kohm) regular resistors in place of the gain pot to test what the max gain would be like. It would probably be easy to try that when you are changing the pot anyway.

[MonarchMD]

I thought I found it. The one I was looking for has a 300 degree fully CCW position when I actually need a 30 degree fully CCW position if I was gonna end up using a D Shaft in order to keep the original knobs. Those part numbers can get quite complex for some of these.

 Found a distributor with the correct shaft position in stock! Went to put a few in the cart, but damn it, minimum order was there entire stock of 252 pots!

Proceeding with the 250k. Thanks again for all the input!