I am trying to make an indicator that shows whether or not there is a load in the circuit, and if the mosfet is working properly.

I have come up with the following circuit, but I am not sure current will continue to flow through the LED if the MOSFET opens.

R9 may be an inductive load (with diode of course) with resistance about 100 Ohms and R10 should be lower.

I think when the MOSFET fully opens it will short-circuit the LED.

Will current still flow when the MOSFET is fully turned on?

enter image description here

  • \$\begingroup\$ Run us through your best guess of what will happen with the MOSFET gate held high and the MOSFET gate held low. \$\endgroup\$
    – TimWescott
    Jun 4, 2019 at 16:01

1 Answer 1


If the mosfet gate voltage is low, the leakage current will go into the uA-nA range (or maybe even lower!). This means when the mosfet is off, it's like a resistor with a value well above 1MΩ. Then check the current through the LTV847 to make sure there is enough current to turn on the LED. You can check this in the datasheet here. Right now the LTV847 has rougly 10k of resistance in series with it. That means if the LED has a 2V drop only 2.2mA would be going through the LED.

  • \$\begingroup\$ But if I change the restance values to lower and the MOSFET is fully turned on, will the LED light fully with the right resistor of course? \$\endgroup\$ Jun 4, 2019 at 16:23
  • \$\begingroup\$ Yes, I would ditch R10, it isn't needed. Then size R9 to meet the current needs of the LTV847. Also, make sure you check the power on R9, if you have a 1/4 or 1/8 watt resistor, make sure the power does not get higher than that. The circuit above with the mosfet on draws P=I^2*R or 28mW \$\endgroup\$
    – Voltage Spike
    Jun 4, 2019 at 16:25
  • \$\begingroup\$ Thank you very much for the help! So if I change R9 with an inductive load in the range of 18-100 Ohms, I would still need some kind of current limiting resistor in place of R10. I mean when the MOSFET is fully opened will it short circuit the LED, stopping it from working? \$\endgroup\$ Jun 4, 2019 at 16:30
  • \$\begingroup\$ Yep, the mosfet has really low resistance when on, and most of the current will flow through the mosfet, you can think of it like a voltage controlled resistor meta.stackexchange.com/questions/126180/… \$\endgroup\$
    – Voltage Spike
    Jun 4, 2019 at 16:31

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