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I tried to make a simple latch with a PNP transistor and a MOSFET. The idea is that initially it is off, when you press the button it turns on and stays on.

But it doesn't stay on.

Here is the circuit: enter image description here

I've tried it with both 12v and 5v, it doesn't work.

I also tried increasing the pullup and pulldown resistors R1 and R4 to 4.7K, in which case the LED is initially on. (Not sure why!)

I attached a picture of the actual implementation, I don't know how much you can make out from it but here it is: enter image description here

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  • \$\begingroup\$ The schematic looks OK, so maybe you have a wiring error. Can you show us your actual implementation? \$\endgroup\$ – Dave Tweed Aug 28 '17 at 1:59
  • \$\begingroup\$ the switch resistance can be the problem. It should be tied to base not drain using same as R2 \$\endgroup\$ – Sunnyskyguy EE75 Aug 28 '17 at 2:02
  • \$\begingroup\$ @TonyStewart.EEsince'75 I connected it to the base, still the same problem. I attached a picture of the breadboard.. \$\endgroup\$ – CaptainCodeman Aug 28 '17 at 2:11
  • \$\begingroup\$ What's the PN of your MOSFET? \$\endgroup\$ – The Photon Aug 28 '17 at 2:16
  • \$\begingroup\$ @ThePhoton It's an IRFZ34N, data sheet is here: infineon.com/dgdl/… \$\endgroup\$ – CaptainCodeman Aug 28 '17 at 2:18
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You could put a cap to ground at/near your mosfet gate. Perhaps across \$R_4\$ in your circuit.

Try this as an option:

schematic

simulate this circuit – Schematic created using CircuitLab

\$R_3\$ limits sudden currents to \$C_1\$ and the gate of \$M_1\$ and oscillations, but still also allows \$C_1\$ to discharge pretty fast through the LED load when the power is removed. \$R_4\$ is needed to keep trace currents from charging the mosfet gate and to also further discharge \$C_1\$ when the LED isn't conducting much anymore.

The above circuit is set up for an LED current of about \$20\:\textrm{mA}\$ as an example. If you plan some monster LED later on, the values and part selections would need to be changed.


EDIT:

I'd much have preferred the use of another BJT rather than a mosfet. In that case, the following circuit is an easy derivative, robust, and doesn't need \$C_1\$:

schematic

simulate this circuit

You could also get rid of one (or perhaps even both) of the two \$33\:\textrm{k}\Omega\$ resistors and further simplify it. The LED load is plenty, already, so I don't see much real need for \$R_4\$. And given that, the whole thing should power up in the right state also without the need for \$R_2\$. But this applies so long as there is a load. Otherwise, I'd keep at least one of them.

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  • \$\begingroup\$ Thanks for the detailed response. What is the purpose of adding a capacitor? \$\endgroup\$ – CaptainCodeman Aug 28 '17 at 12:21
  • \$\begingroup\$ @CaptainCodeman Well, I'd have much preferred using a BJT rather than a mosfet. But with the mosfet in your arrangement I'm mostly bothered by \$C_{GD}\$. \$\endgroup\$ – jonk Aug 28 '17 at 14:48
  • \$\begingroup\$ @CaptainCodeman Let me know if it works for you. I'd be interested to know. \$\endgroup\$ – jonk Aug 28 '17 at 15:08
  • \$\begingroup\$ Hi! yes, it worked, thank you! I believe the problem with my original design was that the pullup and pulldown resistors were too small! \$\endgroup\$ – CaptainCodeman Aug 30 '17 at 8:16
  • \$\begingroup\$ (By the way the reason I used a MOSFET is because I need it to drive a motor!) \$\endgroup\$ – CaptainCodeman Aug 30 '17 at 8:17

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