I'm working on building a soft-latch power switch. Before I build the entire thing, I'd like to ensure that I can simply latch power on, without adding off functionality. I am wondering why the following circuit won't latch. If I press the switch, the LED lights up. But upon releasing the switch, the LED switches off.

I don't understand why the base of the bottom transistor doesn't 'see' the voltage at the emitter of the top transistor.

I am using ZTX457 NPN transistors. http://www.diodes.com/datasheets/ZTX457.pdf

Any thoughts?



2 Answers 2


The circuit shown does not have the transistors doing anything useful, or even turning on at all.

When the push-button is pressed, a circuit is completed via the connection between base and emitter, straight to the resistor. This turns on the LED. Relase the button, that circuit is broken, LED goes off.

The base-emitter junction does not see any voltage to become forward biased and turn on. You can take out the transistors and it'll work the same.

  • 1
    \$\begingroup\$ Wow, very stupid of me. That would do it. \$\endgroup\$
    – ryantuck
    Commented Feb 4, 2013 at 18:24
  • \$\begingroup\$ Hehe, you beat me to it this time - +1 \$\endgroup\$
    – Oli Glaser
    Commented Feb 4, 2013 at 18:27
  • \$\begingroup\$ @OliGlaser Next time, your turn! :-) \$\endgroup\$ Commented Feb 4, 2013 at 18:34

Currently the LED is lighting because your switch provides a direct path, it's not the transistors working. There is no means of supplying current to the bottom transistors base with the switch open (the emitter output will always be lower than the base, so it cannot provide current)

You need something like this (M1 can be a PNP transistor if desired):

Latching Circuit

When the switch is pressed, current flows through Q2s base, so Q2 turns on, pulling M1s gate low and also turning it on. Once M1 is on, R4 now provides the base current to keep Q2 switched on when the switch is released.



  • 3
    \$\begingroup\$ you could also use a single SCR for this if you were so inclined \$\endgroup\$
    – akohlsmith
    Commented Feb 5, 2013 at 2:48

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.