I am going to design power drivers for a high power LED that is capable of high currents (up to several amps). I am using a power MOSFET for fast switching. I have designed a PCB for this circuit, but I have some problems with it.

I connected a dimmer circuit with the pcb. When I give 5V to the circuit the LED doesn't glow. I can check voltage switching between 5V at the gate of mosfet but there are also little voltages switching at drain and source and mosfet gets hot after some time.

Can anyone tell me the reason why there is no current at load, that is, why the LED doesn't light? Why is there some voltage at the drain and source of the mosfet, and why does it get hot very quickly?

enter image description here

Vcc=5 V, Vs=12 V, LD1=ordinary LED.

Datasheets: MOSFET, MOSFET driver.


I have removed the diode from the circuit but very low voltage is present there on the output that cannot drive my led. I am facing the same problem. the supply voltage is vcc=10V and Vs= 3V. and the resistor r1 is 80 ohm. can anyone tell me what could be the problem...

  • \$\begingroup\$ I'm guessing you're not driving the FET correctly, but of course we can't tell what you are doing and how things are hooked up without a schematic. \$\endgroup\$ May 3, 2012 at 12:56
  • \$\begingroup\$ You are obviously connecting things incorrectly in some way. You MUST provide a circuit diagram of what you are trying to do. Please also provide part numbers (LED, MOSFET, ...) and datasheet web links. || Trouble shooting: Apply power. Connect a 100 ohm resistor from drain to source on the MOSFET. If the MOSFET is the ONLY switch in the LED circuit then the LED must now light slightly if all is connected correctly. If not, check the circuit against what you have built until you find the mistake or accidental short circuit. \$\endgroup\$
    – Russell McMahon
    May 3, 2012 at 13:00
  • \$\begingroup\$ A normal LED does not stand 12 V. It may be fried. That's one thing. Then, T1 may be hot because either D1 or T1 are wrongly connected. Just remove D1, and tell us if T1 gets hot. \$\endgroup\$
    – Telaclavo
    May 3, 2012 at 13:54
  • \$\begingroup\$ i have connected these T1 and D1 correctly on the pcb. there are also little voltage switching at drain and source of mosfet \$\endgroup\$
    – dani
    May 3, 2012 at 13:56
  • \$\begingroup\$ @scorpdaddy should i give 10-15V at Vcc (means gate of the mosfet) instead of 5V. and what about Vs. is it okk. \$\endgroup\$
    – dani
    May 3, 2012 at 14:10

3 Answers 3


If you used an ordinary LED for your your first test it's probably gone to LED heaven: you only have 1\$\Omega\$ current limiting resistor, that's good for 12A (if your power supply can deliver that). Now if the LED broke shorting that's also the current through the FET. It has a very low \$R_{DS(ON)}\$, so it shouldn't get hot, even at 12A. That probably means the gate voltage isn't high enough. Increase the supply voltage of your driver.

Try again with a higher value for R1, like 470\$\Omega\$. That should get you 25mA. Good enough for an ordinary LED.

BTW, D1 is not necessary. Get rid of it.

On second thought it seems unlikely to me that an indicator LED, when broken, will conduct 12A. The thin bonding wire from the anode post to the die will break and leave an interruption. Then the next suspect is D1. If you mounted it wrong it would explain the high current, and why the LED doesn't light. One more reason to remove D1.

  • \$\begingroup\$ D1 is only in the ciruit for protecting the led (i will use power led later). and also i want to drive the current in several amps upto 40amp or more \$\endgroup\$
    – dani
    May 3, 2012 at 14:00
  • \$\begingroup\$ @dani You only need D1 if you have the LED far away from your circuit, so that its wires add up a non-negligible inductance. \$\endgroup\$
    – Telaclavo
    May 3, 2012 at 14:08
  • \$\begingroup\$ @Telaclavo if i remove D1 from the ciruit then what you think current will be switched to the led pin in pcb. Please tell me what this D1 causing problem in the circuit? \$\endgroup\$
    – dani
    May 3, 2012 at 14:17
  • \$\begingroup\$ @dani I don't understand your first sentence. Regarding the second one, no, D1 would be causing problems only if it was connected upside down. \$\endgroup\$
    – Telaclavo
    May 3, 2012 at 14:19
  • \$\begingroup\$ @Telaclavo do you think the whole problem is because of this D1? if i remove it from circuit then current will be switched to the load i.e led. and also there are some voltage at source and drain of mosfet. \$\endgroup\$
    – dani
    May 3, 2012 at 14:23

VGS must be >4.5 Volts at 40A. Remove R2 and test again.


If the FET is getting warm, there is current flow through the FET. If the LED is not turning on, there is insufficient current flow through the LED. The two most likely possibilities are:

1) Your LED needs more current to visibly light than the FET can deliver in this circuit. 2) Your circuit topology is wrong. Possibly you have confused the drain and the gate, and the FET is shorting across the power supply.

Impossible to diagnose without more information.

  • \$\begingroup\$ I have removed the diode from the circuit but very low voltage is present there on the output that cannot drive my led. I am facing the same problem. \$\endgroup\$
    – dani
    May 4, 2012 at 13:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.