# Amplify a potentiometer section

i've been looking for a replacement 5k potentiometer for a mobility scooter.

they have 312 deg mechanical rotation but only 45deg electrical spread, so the center 45 degrees are 4ohm-5kohm.

they are quite pricey.

would it be difficult to take a more standard (cheaper) pot and modify it, either with components to alter the resistance or maybe a 0 to 30k pot and mechanically adjust the mounting?

all input appreciated, thanks.

• If the pot is 5kohm how can the central 45 deg section also range up to 5 kohm? Sep 21 '15 at 14:02
• it goes 0 - 312 0-130ish is 4ohm (ffs my editing is awful) 130ish to 175ish is 4-5kohm 175ish to 312 is 4ohm sensing.honeywell.com/index.php?ci_id=151838 page 26 Sep 21 '15 at 14:07
• I'm betting that you mean, 0 to 130ish constant 4 ohms - 130ish to 175ish increases from 4 ohms to 5k - 175ish to 312 constant at 5k. It's an S-9 taper, per page 4. Sep 22 '15 at 2:00
• maybe im reading the sheet wrong but i thought it reads 4ohm at fully CW and CCW Sep 25 '15 at 10:16

I know it's a month old, but the existing answer seems to miss the question a bit as I read it, so here goes:

If it feeds an ADC or other device that expects a voltage input rather than resistance (the pot is simply an adjustable voltage divider), then you could do this:

simulate this circuit – Schematic created using CircuitLab

The Gain and Offset controls will have some interaction, but that's probably okay for a set-and-forget application, and it's hard to beat the parts count. Set them to provide the correct response given the available travel from the (now standard) Input pot.

Ideally, the opamp should be a rail-to-rail type. Others may have lower distortion, higher bandwidth, or some other attractive performance characteristic, all of which are easily unnoticed or subconsciously corrected for in a manual control, but they will clip before they reach the full supply. In this application, you need as much range as you can get.

I'm using 10k pots instead of the original 5k because the scooter's signal input doesn't see any of them. Their only effect from that perspective is their supply current. You could also use 100k or anything in between if you happen to have them.

From the linked datasheet, this pot is a 2W variety. And on page 26, the "tapers" listed are S-9 and S-10, which are fairly exotic. (A "taper" is how the resistance relates to rotation angle; linear is a linear relation between angle and resistance, log is a logarithmic relation, etc.) This is likely where the large price tag comes from, since most pots are 1/4W or less and have a common linear, log, or anti-log taper.

Now trying to decipher your measurements,

• 0° to 130° = 4Ω.
• 130° to 175° = ~5kΩ *
• 175° to 312° = 4Ω

*Now does this change gradually (and linearly), from 4Ω at 130°, up to 5kΩ at 152°, then back down to 4Ω at 175°?

This would make sense if terminals 1 and 3 were wired together AND the pot were a 10k variety. This is because terminal 2 is the "wiper", and when terminals 1 and 3 are connected, the maximum resistance would be seen at 152.5° - exactly middle of the rotation. Since there isn't one 10k resistance but two (one 10k from terminal 2-1, another 10k from 2-3), two 10k resistors in parallel = 5kΩ.

Now as for "amplifying" this or replacing it with some other device, well that depends on what it's connected to. If it's connected to a motor, then the motor will require more power than a standard 1/4W pot could provide so that would literally burn up. Such a pot can be used to control a motor but this is not done directly; a driver circuit must be used to take the weak currents coming out of the pot and "amplifying" them to drive the motor. The construction of such a device is rather involved, and the specs of the motor and power source must be known.

• i believe it just drops to 4 ohm at 175, like runs out of track of something... it's used to control the speed of a mobility scooter. i'd +1 your comment for being helpful, thank you. Sep 25 '15 at 10:18