I built the top circuit in the schematic below. First I breadboarded it and it worked great. I swapped the 10k resonance pot for a 5k as I didn't have any 10k trimmers handy and used 1000pF cap instead of a 500pF for ground.

The breadboarded circuit worked great. It's a voltage controlled low pass filter, like in a synthesizer. Plugged a guitar in and it sounded great, good sweep from the 100k frequency pot, resonance range with the 5k pot was good. The 1M input trim was plenty to take the higher output of a synthesizer down to a level which wouldn't overload the filter.

Then, I created the circuit on veroboard. The only thing I changed here was switching the little ground capacitor from my 1000pF to a 680pF. As the schem recommended 500pF and I'd used 1000pF with decent results, I figured a value between that wouldn't be too bad.

I built the circuit. Connected a battery, and heard sound. It sounded good. Then when I moved the 100k frequency pot (at this point a small trimpot), it started smouldering and smoke came out. I immediately unhooked the 9v battery. I thought maybe I had a bad pot so I switched it for another - this time a proper 16mm - Same problem. I saw red light inside the pot and lots more smoke this time.

This wasn't long before I had to go to bed, so I unplugged everything and left it. Can anybody tell me what I might have done wrong in order to make this happen? It worked fine on the breadboard. I've built a lot of circuits and I'm used to debugging circuits that don't work at all or sound crackly, but I've never had a fire before!

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1 Answer 1


It sounds like you have wired the pot with the V+ or the ground going to the wiper (the middle connector) instead of one of the outside connectors. When you turn the pot to the minimum position, you will be shorting out the supply.

  • \$\begingroup\$ Thank you - this might be what I've done. I've done some reading from people who have had similar problems. Looks like I put the circuit input on pin 3, V+ on pin 2 and ground on pin 1. Could someone explain to me what the actual, mechanical difference is between putting V+ in 2 and in 1? In my head, it seems like there's current flowing through these two points regardless of 'polarity'. Perhaps there's an example or diagram? \$\endgroup\$
    – TCassa
    Commented Jan 12, 2017 at 14:58
  • 1
    \$\begingroup\$ With the supply positive on pin 1 and the supply negative on pin 3, the pot acts as a potential divider, with a varying voltage on pin 2. There will always be 100k across the power rails. With the positive on pin 2 as you turn the pot, so that the wiper (pin 2 with positive on it) moves closer to pin 3 (the supply negative) the resistance will get lower and lower until a large current starts to flow through the pot, which will burn it out. \$\endgroup\$
    – HandyHowie
    Commented Jan 12, 2017 at 15:12
  • \$\begingroup\$ Thank you so much! It seems to be one of those obvious fundamentals, which I hadn't quite understood - but now I do perfectly. I'll have another go tonight! \$\endgroup\$
    – TCassa
    Commented Jan 12, 2017 at 15:19
  • \$\begingroup\$ Let us know how you get on. \$\endgroup\$
    – HandyHowie
    Commented Jan 12, 2017 at 15:34
  • \$\begingroup\$ Re-boarded this circuit last night with the correct pot polarity and it works correctly. Thank you so much! \$\endgroup\$
    – TCassa
    Commented Jan 13, 2017 at 10:44

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