I've been studying this transistor "latch" circuit over the past couple of days and I think I've grasped most of it, but I'm not sure about the purpose of capacitor C1.



If I understand correctly, the operation of the circuit is as follows: When the input is high Q1 turns on which in turn causes current to flow through the base of Q2 to the collector of Q1, turning Q2 on. Feedback to Q1's base from Q2's collector keeps Q1 turned on, which in turn maintains Q2 on.

As for the capacitor, I notice that if I don't include it sometimes the circuit won't latch, but I'm not 100% on the reason why. I thought perhaps the purpose of the capacitor was to allow Q2's base current to flow briefly in the event that Q1 is momentarily turned off?


1 Answer 1


You are very close to being correct. C1 creates a time delay so the feedback loop has time to work. It also has a 'debounce' effect on a change in inputs, thus a predictable change in states.

Transistors are so fast that without a time delay the contact noise of your switch could make it toggle many times, resulting in an unknown final state.

In addition as Dave Tweed mentioned in comments C1 makes sure the circuit powers up in an 'unlatched' state by briefly keeping Q2 OFF.

C1 will also limit the speed at which the latch can change states, so a noisy input will not make the relay chatter.

  • 4
    \$\begingroup\$ It also helps make sure the circuit powers up correctly (unlatched). \$\endgroup\$
    – Dave Tweed
    Commented Apr 9, 2018 at 22:02

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