Multiplying a potentiometer - Page 1
Forum section: electronics / projects

author: smbaker
author_rank: Regular Contributor
timestamp: July 26, 2018, 06:32:34 pm
content: I have a 10k potentiometer that I'd like to behave like a 500k potentiometer.The application is that I'd like to build a custom game controller for a vintage arcade console. The console I think measures the analog joystick position by charging a capacitor and timing the delay through the pot. It's setup to use a 500k pot. The pot I need to use is 10k. I'd like to do this without modifying the console.Console looks kinda like this:Iopin --> 0.047 cap to ground --> 1.8k resistor --> wire to controller --> pot to 5vYou can sort of get away with using a 100k pot by adding some additional capacitance at the controller. It's not perfect as there is a low value resistor inside the console, and resistance in the cable. Trying to do this with a 10k pot was not successful.There's a digital solution using an adc together with a digital pot, and I currently have a prototype pcb on order that will use that idea. However, I expect it to introduce digital artifacts and lag. What I'd really like is an analog solution.I'm pretty sure I could use an opamp and a transistor to convert a voltage (from my 10k pot) into a current. I think what I want though is to convert a voltage into a resistance. Any pointer?

author: necessaryevil
author_rank: Regular Contributor
timestamp: July 26, 2018, 07:48:16 pm
content: Is it not an option to use an even larger capacitor?

author: smbaker
author_rank: Regular Contributor
timestamp: July 26, 2018, 10:09:07 pm
content: Quote from: necessaryevil on July 26, 2018, 07:48:16 pmOn the controller? Tried that. Couldn't make it work. If I pick a capacitor so that the maximum position on the pot reads correctly, then it's only reading 2/3 of what it should at the midrange. If i pick a capacitor so that the midrange reads correctly, then the reading at maximum pot is too high.The capacitor technique worked fine with a 100k pot by adding a 0.22uf capacitor, but at a 10k pot with a 3.3uf or larger capacitor starts to get wonky.

author: gbaddeley
author_rank: Regular Contributor
timestamp: July 27, 2018, 10:06:48 am
content: The 1.8K resistor is probably having an effect. Patch in a much lower value directly across it, eg 10R.Are you putting the bigger cap directly across the 0.047uF ? It may need to be about 10uF to get the correct scaling of time constant, if you are changing a 500K pot to a 10K pot.

author: Brumby
author_rank: Unknown
timestamp: July 27, 2018, 11:19:15 am
content: Quote from: gbaddeley on July 27, 2018, 10:06:48 amNo. He's going the other way.Bottom line - you are not going to be able to get 500K of resistance variation through a 10K pot. I would be looking at other ways, but it would be helpful if we had a schematic of the circuit in question.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 27, 2018, 12:06:02 pm
content: It depends on the circuit used. If the pot is towards ground, used as a simple variable resistor, they might be a chance to use a simple OP circuit:Use on OP to buffer the voltage and than divide is by the required factor (e.g. 50). Than use a second OP and a MOSFET with the 10 K pot at the source to get the smaller voltage over the pot. The drain current would than simulate the resistor current. So no ADC / DAC needed.The response could still be tricky if the readout is fast. It could require quite accurate and fast OPs and maybe extra limiting of saturation.

author: Ian.M
author_rank: Super Contributor
timestamp: July 27, 2018, 01:10:22 pm
content: Kleinstein is on the right track, but no MOSFET is needed.Its actually trivial to do in the analog domain provided you've got 0V available at the controller. All you need is a dual low input bias current true rail to rail precision OPAMP (that works at 5V) configured as two unit gain buffers, and two precision resistors, one 49x the other. Use one OPAMP to tap off a buffered copy of the voltage at the bottom end of the 'pot' The resistors form a divider between its output and +5V, so as to get 49/50 of the voltage between it and the +5V rail. Buffer that with the other half of the OPAMP and feed it to the top end of the 'pot'.Result: the voltage across the 'pot' is reduced by a factor of 50, so from the point of view of the console, the 'pot' appears to be 50 times its actual resistance.It doesn't need very fast OPAMPs, because it only has to handle the slew rate of the worst case 1K8 x 47nF RC charging curve + settle in whatever the minimum time the I/O pin is taken low to discharge the capacitor, which is not likely to be very brief.LTspice sim attached.