# Vary equivalent resistor linearly between two values

I'd like to use a potentiometer (with a value like 1kOhm) in a resistor network to achieve a close to linear variation between 35 and 250 Ohm. I've tried several configurations, but I'm having a hard time finding a possible configuration.

I've tried the following circuit:

R2 needs to be around 220 Ohm to allow this variation. Not only there aren't potentiometers with this values, as the response curve is very non-linear:

The response doesn't need to be super-linear, but needs to allow the user to vary it continuously.

• You are kind of stuck with standard pot values. (1) If you explain your application (in the question - not in the comments) you might get an answer that solves your problem in a different way. (2) Note that you can embed an editable schematic from CircuitLab when you use the toolbar button. You don't have to take screengrabs. (3) What values are available in the parts you have selected. Commented Oct 27, 2017 at 16:00
• (1) I may try to do that later on. (2) Thanks for the tip! Newbie here. (3) All values in the market (like .1% resistances). We can buy them as we need. Commented Oct 28, 2017 at 16:26
• (3) What values of potentiometer are available to solve this problem? Commented Oct 28, 2017 at 17:58

So you need a 215R POT. If you can buy a 250R pot it may be close enough with a 1.54K in parallel.

simulate this circuit – Schematic created using CircuitLab

NOTE: You can get pots closer to 215 than 250, to use with a larger resistor and get closer linearity, but they are uncommon and cost more.

simulate this circuit – Schematic created using CircuitLab

Figure 1. Using a stereo potentiometer to double the standard resistance gets you fairly close.

Figure 2. 2-gang or stereo potentiometer.

• Thank you. I thought of using stereo pots, but didn't know what configuration to use it in. One question? Is this pot. logarithmic (probably, if it's for audio)? I'll check on this possibility. Commented Oct 28, 2017 at 16:31
• You can get 2-gang pots in linear and log. Again, if you explain what you're doing we can stop guessing solutions for you. Commented Oct 28, 2017 at 17:59