I've made a circuit with a touch switch module turning on/off a LED strip via a transistor, and the whole thing works, but in the "off" state the LED strip still shines a bit. touch switch led circuit Touch switch module I'm using: link to aliexpress (I've configured it to work in latch mode)

The touch switch module is supposed to have 0V in off state, so the transistor should be in a "closed" state. Why is it allowing some current to flow through emitter-collector?

What can I do to make it turn off completely?

In the "on" state everything is fine, and touch-switching works well.


1 Answer 1


Your sensor is trying to drive the transistor's base to 5V in the off state (because with a 5V supply the sensor can't output more than 5V), and its emitter is at 12V. (Given that the transistor is not on fire, it's safe to assume the emitter-base voltage is not actually 7V, and the output resistance of the sensor is limiting the current. This should not be relied upon; you should put a resistor in series with the base.) This forward-biases the emitter-base junction, driving the transistor into saturation or forward-active (depending on circuit conditions) and allowing current to flow.

If you want to use a 5V signal to switch 12V power, you'll want to use an NPN transistor on the low side; this transistor as it is won't turn off until you apply 12V to it.

  • \$\begingroup\$ Do you mean I should use NPN instead of PNP? I didn't understand what is "the low side". So if I keep using this transistor - can I fix the circuit by adding some resistors? So the base-emitter voltage should be less than 1.4V for transistor to shut down? Didn't realize PNP is so different from NPN, thx! :) will read more about it :) \$\endgroup\$ Commented May 13, 2017 at 0:33
  • \$\begingroup\$ The emitter-base voltage should be less than 0.6~0.7V for it to turn off. Likewise, in an NPN transistor, the base-emitter voltage should be less than 0.6~0.7V to turn off. It's easier to treat them as current-controlled devices though. No, you can't fix it with resistors, you would need a level shifter of some sort which is more complicated than you need. "the low side" means put the transistor on the low-voltage side of the LEDs; NPN transistors are typically used with their emitters at or near ground potential. \$\endgroup\$
    – Hearth
    Commented May 13, 2017 at 0:36
  • \$\begingroup\$ One way to think about it is like this: ignore the collector of the transistor and just treat the emitter-base (or base-emitter) junction like a diode, directed in the same direction as the arrow in the symbol. If that diode is forward-biased, the transistor is either in saturation or forward-active mode, and it is conducting. \$\endgroup\$
    – Hearth
    Commented May 13, 2017 at 0:39
  • \$\begingroup\$ Can you recommend some popular NPN transistor with 1A or more max current? :) To replace this TIP32. \$\endgroup\$ Commented May 13, 2017 at 0:49
  • \$\begingroup\$ Sure, I'd suggest a TIP31 or if you want to really over-spec it, a D44H8. But you can also always just use Digi-key's parametric search, instead of having to ask around! \$\endgroup\$
    – Hearth
    Commented May 13, 2017 at 0:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.