Still learning on my own so I am really not an expert.

I built a small circuit and can’t get the expected behavior.

Here is the schematic:

enter image description here

This is pretty simple. The PNP on top is connected to 5V on the emitter and the collector is connected to a current limiting resistor and an LED (so I can see when there is current.)

The base of the PNP is connected to the collector of a NPN and the emitter goes to ground.

The collector of the PNP provides currrent to the base of the NPN so when the PNP is on, it keeps the NPN on and the LED stays on.

I also connected the base of the NPN to a switch button that is connected to ground.

When I press it, the NPN base is grounded so it turns off and the PNP base gets 5V (from a resistor connected to 5V and the PNP base) and the LED turns off as expected.

Where I don’t get it is that as soon as I release the button, the LED goes back on.

Like if the NPN base was getting 0.7V from the PNP collector (that is supposed to be off.)

Here's a picture of my breadboard circuit:

enter image description here

I was looking at building something like this and wanted to test if it was working prior to adding the second npn (and using a pnp instead of a mosfet)

enter image description here

  • \$\begingroup\$ where's the problem? it looks like an inverter to me, at first glance (relying on leakage or capacitance to start up during turn on) \$\endgroup\$
    – Pete W
    Feb 6, 2021 at 20:41
  • \$\begingroup\$ Were you planning on 2 momentary switches with an SCR like latch ? Change 100k to 47k or less then add a switch with 100k from gnd to PNP Base for ON. That's either a high current rated NPN transistor or really leaky >1uA. BUT you are driving more current into PNP base than the LED, so add 100K in series to PNP base to NPN collector or scrap this design \$\endgroup\$ Feb 6, 2021 at 23:47
  • \$\begingroup\$ I added the « where I was aiming at » 😉 \$\endgroup\$
    – Dominique
    Feb 7, 2021 at 0:23
  • \$\begingroup\$ Are you trying to make a 1 shot?, that wont cut it \$\endgroup\$ Feb 7, 2021 at 1:42
  • \$\begingroup\$ @Dominique Are you looking for a press-on, press-off circuit? Or what, exactly? What's the design goal? \$\endgroup\$
    – jonk
    Feb 7, 2021 at 3:48

1 Answer 1


All transistors leak - maybe much less than a microampere, but more than zero.

Your transistors are not different. You may short the CB leak with the switch, but when you release the switch the rightmost transistor gets some base current as CB leakage ==> the leftmost transistor surely gets some base current through rightmost transistor. That makes the leftmost transistor conducting and outputting base current to the rightmost transistor. It's like a flipflop. You have the reset switch. There's no set switch, but CB leak autotriggers it.

You probably find a resistor which in parallel of the switch eats enough the CB leak and the flipflop doesn't get autotriggered as soon as you release the switch. I would try at first 3300 Ohm.

The set -switch would connect via a resistor either the base of the leftmost transistor to GND or the base of the rightmost transistor to +5V.

IMPORTANT: There should be a resistor somewhere in the path from +5V through the transistors to GND - preferably between the base of the PNP tr. and the collector of the NPN transistor. I would insert at least 2 kOhms. Otherwise the transistors can burn due no current limit.

BTW. No leak probably would be even needed. The sparse wiring collects easily enough noise and hum to trigger the circuit. Small enough resistor in parallel with the switch eats also noises that the wiring collects.

ADD after the edit of the question: The mosfet version behaves oppositely when compared to the original circuit. The collector current of T2 reduces the drive of the N-Mosfet T1. The results with the original circuit are useless with the mosfet version.

  • \$\begingroup\$ Doesn’t the npn base need 0.7V in order to turn on? That seems like a lot of leakage isn’t it? \$\endgroup\$
    – Dominique
    Feb 6, 2021 at 22:12
  • \$\begingroup\$ There's available the whole 5V through leakage resistances. BTW no strict 0,7V treshold exists. We only have used to have something between 0,6 and 0,8V in practical circuits because usual PN junctions can have a voltage in that range in a very a wide current range. \$\endgroup\$
    – user136077
    Feb 6, 2021 at 22:32

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