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I am using the 2N7000BU Enhancement MOSFET to control a 5 VDC signal to a relay with a 3.3 VDC gate voltage applied.

Vds max is 60 V and Vgs(th) is ~800 mV with a +/- 20 V max. and a max. drain current of 200 mA. I have it connected to a Solid State Relay with a 4-32 V input voltage and a maximum of 20 mA current draw.

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The diagram appears to show that left is the source and right is the drain. I tested the pins with nothing connected to the gate with a multimeter and it shows no connection between 1 and 3.

When I connect my +5 V to the drain and -5 V to the source with nothing at the gate there is no current flow initially, but after powering the gate with 5 V it does not turn off on its own: when the 5 V is removed it takes a few seconds to turn off. After that, even gently brushing the gate pin with my finger causes the relay light to turn on and remain stuck.

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I would really appreciate some help troubleshooting why this is not working as expected.

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2 Answers 2

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You are open-circuiting the gate of the very sensitive MOSFET. When you remove the gate voltage, the gate capacitance is retaining enough charge to keep the channel inverted and activate the relay. Adding a pull-down resistor (R2) will provide a path for this charge to return to ground and turn the MOSFET "off" when the button is released. R1 limits the current to the gate when the button is pressed. Moving the MOSFET to the low side of the relay gives you a more stable Vgs and will ensure that the relay turns on.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thank you I will try this \$\endgroup\$
    – Jman
    Nov 2, 2022 at 22:27
  • \$\begingroup\$ 10K was too high and kept the FET on but a 120 Ohm was enough to ground the pin and allow the signal to control it. Thanks! \$\endgroup\$
    – Jman
    Nov 2, 2022 at 23:01
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Sounds exactly like how a FET would behave when used like that. I think there is nothing to troubleshoot. It would help to know how you expect it to behave, the expectation may be wrong.

FETs are controlled by gate voltage. Not base current like BJTs.

FET gate is extremely high impedance and will store charge when disconnected from voltage supply and it will take very long for the gate charge to leak off.

Touching gate with finger would also likely charge the gate and turn FET on, and it will take long time to turn off. If it is not damaged by excess voltage from excess charge.

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  • \$\begingroup\$ I added a pull down resistor and that helped. Thanks! \$\endgroup\$
    – Jman
    Nov 2, 2022 at 23:02

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