LED strip: DC, 4W

Mosfet: IRF740 (datasheet)

To Do: Use wall socket (AC) to run DC led strip and also turn it on/off with MOSFET and arduino.

AC to DC is done with rectifier and transformer. So this part is excluded from the diagram.

Circuit Diagram


simulate this circuit – Schematic created using CircuitLab

When I send a LOW signal to Gate, the led strip is still faintly on. I have read about why does this happen(stackexchange link)

What I was wondering is how can I stop this? Will a diode help? Is a MOSFET driver what it needs? Or should I increase the load (a pretty bad hack)?

I know the mosfet in use is not a TTL but the mosfet does turn on at low gate voltages (1.5~2V). I am able to fade the led strip but could not get it to completely turn off using the MOSFET.

Also, not sure if this is a concern or not, but the supply voltage (24v) decreases to 18v when the MOSFET is turned 'off'.

Any help is appreciated. Thank you.

  • \$\begingroup\$ If you disconnect the Arduino, do the LEDs turn off? \$\endgroup\$
    – HandyHowie
    Mar 12, 2019 at 7:53
  • \$\begingroup\$ No, they remain faintly on. \$\endgroup\$ Mar 12, 2019 at 7:56
  • 1
    \$\begingroup\$ IRF740 has a max leakage current of 0.1 uA. This may be enough for high intensity LEDs to be very faintly on and visible indoors or in the dark (particularly red LEDs). 24V dropping to 18V is a concern, something is getting shorted in your circuit, but it probably does not concern these LEDs. \$\endgroup\$
    – Indraneel
    Mar 12, 2019 at 7:57
  • \$\begingroup\$ Also make sure you have a gate resistor to uC, to save the pin from high current transients at switching, and also to protect the pin against mosfet failure. Depending on the connection quality of the arduino ground, it is also possible that gate does not reach ground when switched off. \$\endgroup\$
    – Indraneel
    Mar 12, 2019 at 8:04
  • \$\begingroup\$ Will look into what possibly could be the reason for the short circuit. Also will a resistor help with the leakage current? (Also, as I am currently away from the workbench, it will take me quite a while to update you guys. But Update I will. Thank you all.) \$\endgroup\$ Mar 12, 2019 at 8:04

2 Answers 2


To solve the issue that the LEDs are still glowing when they should be off I would try this:


simulate this circuit – Schematic created using CircuitLab

I tidied up your schematic a bit by moving the supply to be more "what we're used to" and adding a ground symbol. But the main fix is R2. Indeed the leakage current of the MOSFET will probably be the issue. You could use a "better" (less leaky) MOSFET but there might be a simpler solution. If only we can provide a path for that leakage current that is such that the voltage will be too low for the LEDs to conduct the current. That path is provided by R2.

You measured 18 V across the LED strips when they're (supposed to be) off. That means some current is still flowing because if no current was flowing you would measure 0 Volt. At 18 V across the LEDs the current through R2 would need to be around 0.75 mA. The leakage current of the MOSFET should be much lower than that 0.75 mA. So much less current will flow. That means there will be less current through R2 and also less voltage across the LEDs. That means they should turn off.

The disadvantage of adding R2 is that when the MOSFET conducts, there will be 24 V across R2 so some current will flow through it (1 mA) which is "lost" as it does not flow through the LEDs. But 1 mA isn't much so compared to the 167 mA of the LEDs.

Try R2 = 24 kohm first, if the LEDs still aren't off, try 10 kohm. If that still in not low enough then maybe your MOSFET is damaged.

If at R2 = 24 kohm the LEDs are off, you can try increasing the value of R2 to 47 kohm or 100 kohm and see if that is still OK. That would lower the 1 mA that is "lost".

  • \$\begingroup\$ OP needs to clarify where he measured the 18V \$\endgroup\$
    – Indraneel
    Mar 12, 2019 at 9:24
  • \$\begingroup\$ @Indraneel it is clear to me: when the MOSFET is off, there's still 18 V across the LEDs. Ideally there would be 0 V across the LEDs and 24 V across the MOSFET but leakage and multimeter impedance will result in different (non ideal) numbers. \$\endgroup\$ Mar 12, 2019 at 9:43
  • \$\begingroup\$ The way I remember it, it was 18V across the supply terminals not the load (LEDs). But I doubt myself now. I will check that and update with the results. Hopefully your circuit fixes that. Thank you for answering though! \$\endgroup\$ Mar 12, 2019 at 10:38
  • \$\begingroup\$ Ah yes! I had noted it down: Signal = HIGH Vin = 24v, Vo = 22v. Signal = LOW Vin = 13v, Vo = 8v. Vin is the supply voltage. Vo is the voltage across load. \$\endgroup\$ Mar 12, 2019 at 10:43

I have the same issue with my motion detector.I replaced the main IC but no luck. I replaced the FET that drives the array of LED with transistor and it works fine. FET is only good for relay or filament bulb, but not on a low current LED's


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