I'm working on a circuit that should close a reed relay (normally open) when TI/COUT is low and open said relay when TI/COUT is high (6V).

I'm using a P-FET low side transistor to ground the relay & status LED, but it's working the opposite way of what I expected: LED is lit and relay closed when TI/COUT is high.

What did I do wrong, and what would be the necessary component or circuit changes needed?

Original design: Circuit Diagram

Mosfet Datasheet

Revised design: Updated/fixed design


3 Answers 3


If you want the relay to be 'off' when T1 is high, you have to wire it like this:


simulate this circuit – Schematic created using CircuitLab

This will only work in if the control voltage for 'off' is similar to (or a bit higher than) the relay supply voltage.

  • \$\begingroup\$ T1 is 6V and VCC_5V. Is 1V too much of a difference? I can change the VCC to be 6V so they're both 6V? \$\endgroup\$ Sep 12, 2023 at 21:27
  • \$\begingroup\$ Great! Hopefully I've translated your CL schematic correctly: !Revised Design \$\endgroup\$ Sep 12, 2023 at 21:50
  • \$\begingroup\$ Looks good, though you don't need the 1N4148 if your relay includes a diode, which the xx10 relay appears to include. \$\endgroup\$ Sep 12, 2023 at 23:01
  • \$\begingroup\$ Got it! Thank you 🙏 \$\endgroup\$ Sep 13, 2023 at 18:12
  • 1
    \$\begingroup\$ (minor correction) It’s fine with 6V, some positive Vgs is okay", with a P-channel MOSFET that tends to turn the MOSFET even further 'off', negative turns it 'on'. \$\endgroup\$ Sep 13, 2023 at 18:14

Your PFET will conduct all the time because it has a body diode that is forward biased in that orientation.

enter image description here

If you use an NFET in place of the PFET, as suggested in another answer, the circuit will "work", but the logic will be reversed. That is, the relay coil will be active when the gate is high, and the relay coil will be de-energized when they gate is low. However, you stated:

I'm working on a circuit that should close a reed relay (normally open) when TI/COUT is low and open said relay when TI/COUT is high (6V).

Spehro Pefhany beat me to the punch of getting a working CircuitLab simulation. But he has the right idea. The source pin of the PFET needs to be connected to the positive supply, making the switch a high-side switch.

One problem you may face with the particular PFET you are using is that its threshold voltage is only 0.4-1.0 V (per the CJ2301 datasheet). Therefore, to turn the relay off reliably, you need to bring the gate voltage to no less than the supply voltage minus 0.4 V. You might have better luck with a FET with a slightly larger threshold voltage.

  • \$\begingroup\$ Would e.g datasheet.lcsc.com/lcsc/… give me a significantly better threshold? -0.7 - -1.3 \$\endgroup\$ Sep 12, 2023 at 21:37
  • \$\begingroup\$ Yes, 0.7 volts is good. You could go more. It really depends on how close to the drain voltage your gate driver actually goes. Of course, you can overdo it, and choose a threshold voltage that is so large, a low gate voltage will not turn the MOSFET on completely. That should be avoided. \$\endgroup\$ Sep 12, 2023 at 21:44

To switch a load on the low side, you need to use an n-FET instead of a p-type. Then a normal logic high on the gate will turn the FET on.

Fortunately, for SOT-23 FET the p- and n- type pinouts are the same. So just change the FET to a suitable n-type. Then it should work.

What went wrong with the p-FET? Two things:

  • Body diode is forward biased (it will be reverse bias with n-FET)
  • Gate-source bias will never turn on the FET (will be ok with n-FET.)

Result? The p-FET conducts all the time, with a forward bias of one diode drop, and will not shut off.

One more thing. Add a flyback diode across the relay coil to catch the spike when the FET turns off. Otherwise your FET can be damaged.


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