I'm trying to find this part online, to replace it, as it broke. However, using the text on the part, I wasn't able to find anything.

Could someone help me? The board itself would be 200$ to replace, so it'd be way better if I could just use the old board and repair it.

  • \$\begingroup\$ What is the device it comes from? What is the resistance? What are the dimensions? \$\endgroup\$ – Rocketmagnet Apr 30 '12 at 22:27
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    \$\begingroup\$ It's from my car A/C & Heating control unit, it looks like a 10K, not sure, dimensions are 60.00mm x 8.00mm x 7.00mm. \$\endgroup\$ – jValdron May 1 '12 at 11:10

It's a 10k\$\Omega\$ slider potentiometer. The "B" suffix indicated it's a logarithmic one. Linears have the suffix "A".

  • \$\begingroup\$ So the "10KB" at the end means it's a 10kΩ? \$\endgroup\$ – jValdron May 1 '12 at 11:08
  • \$\begingroup\$ @jValdron - Yes, "10K" means 10k\$\Omega\$, "B" means logarithmic. \$\endgroup\$ – stevenvh May 1 '12 at 12:46
  • \$\begingroup\$ I've read different things about this. One version says my answer is the European convention, in Japan it's the other way around. Another says it's the old convention, and that they changed the codes (when?). So it may depend on the age and/or the origin of your device. \$\endgroup\$ – stevenvh May 22 '12 at 4:43

It looks like a slide pot without the slider. You said it broke. Could it be that the slider broke off? There seems to be a slot where the slider would fit in.

Finding the exact right slide pot replacement could be difficult. Electrically any pot of the same resistance would work. You can replace it with a much easier to find rotary pot if you're willing to run wires to it and turn a knob instead of pusing a slider.

Use a ohmmeter to measure accross the ends of the pot. You will need to know this value to find a electrical replacement, whether it fits mechanically or not.

  • \$\begingroup\$ +1. also make sure it's linear or logarithmic. If logarithmic, it has polarity, so check the "direction" (e.g. which end you slide the slider towards in order to get about half the resistance -- it should be close to one end for a log pot.) \$\endgroup\$ – Jason S Apr 30 '12 at 23:04
  • \$\begingroup\$ Yep, sure is a slide potentiometer, the slider part of it had to be removed to remove the pot from the case. It looks like a standard 10K pot which looks like it could be replaced with the PP1045SB-ND from Digikey. \$\endgroup\$ – jValdron May 1 '12 at 11:17

Suppliers like Digikey have a category for slider potentiometers. There are only 155 items in that one. If you know the resistance of your broken part (you can measure it, also check if it's linear or logarithmic), it will narrow the search a lot. When you arrive at the short list, measure dimensions of the pin pattern of the old pot. Look through the datasheets for a compatible pattern.

If you don't get lucky, you can do surgery on the board. Your PCB by itself is very primitive: single sided, throughole components. In principle, you could drill new holes for a brand new slider pot and fly wires to it.

By the way, what kind of instrument are you repairing?


It appears to be a fader resistor. \$ 10k\Omega\$ appears to be a common value.

If you cannot get replacement parts from the board manufacturer, you may have to go the route of a 'compatible' fader and try your luck:

Bourns pro audio

Panasonic EWA series

  • \$\begingroup\$ Also, I believe the B next to the 10K means that the side pot has a linear as opposed to a logarithmic or audio taper. \$\endgroup\$ – Hristos Apr 30 '12 at 20:05
  • \$\begingroup\$ @Hristos - it's the other way around. \$\endgroup\$ – stevenvh May 1 '12 at 7:14

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