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?
Vcc=5 V, Vs=12 V, LD1=ordinary LED.
Datasheets: MOSFET, MOSFET driver.
Update
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…
Best Answer
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.
edit
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.