Continue to Site

Welcome to

Welcome to our site! is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] PWM MOSFET excessive power dissipation losses?

Not open for further replies.


Junior Member level 1
Apr 14, 2006
Reaction score
Trophy points
Activity points

I have problems with MOSFET overheating, while switching resistive load (resistor and LEDs). For driving a MOSFET (I'd tried with two different ones - FDD8444 and NTD3055L104) I'm also using a driver MCP1416. I've made some images with oscilloscope. Can someone explain me what is going one? Why is there a curve on the top of the square signal? Why doesn't MOSFET stays fully opened?



A breadboard circuit?

Show us the schematic. With values like voltages and frequency..

A scope picture without schematic tells nothing.



It's not a breadboard circuit. Everything is in SMD technique on PCB

PWM schematics.png

Looks like you have the load in the source pin of the MOSFET.

Consider where the voltage to 'open' the MOSFET has to be applied - between the gate and source pins. With the load in the source pin, when you 'open' the MOSFET the source voltage rises (as you intended it to so the LED lights up) but the gate voltage stays the same so the MOSFET tries to turn off again. It will try to reach an equiibrium where it's own source voltage limits it's VGS which will put it somewhere in it's partially conducting region, that's why it runs hot.

Move the load to the drain side of the MOSFET and it should work fine.

  • Like
Reactions: msmiha


    Points: 2
    Helpful Answer Positive Rating

Brian is absolutely right.

For you to remember for the future.
Dissipated power P = U x I.
A FET used as a switch is designed to generate low heat,means low power dissipation.
When it is OFF, the voltage across the switch is high (about supply voltage), but current is zero, therefore power is zero also.
When it is ON, then current is high, but voltage is near zero, some mV. Therefore power dissipation is low also.

In your design, when switched ON voltage across the FET will be some volts. It is not near zero. I expect it to be 3..4V(without reading datasheet). Multiplied with the current you get the power.

A small SMD part can withstand maybe 0.1W, a big one maybe 0.5W.
1W on a SMD device without heatsink is about impossible. You will see extreme hot spots.

  • Like
Reactions: msmiha


    Points: 2
    Helpful Answer Positive Rating

Thank you Brian an Klaus for explaining what is going on. I will put my load on the drain side of a MOSFET and report the results.


I have done as told but the results are not quite what I was hoping. The MOSFET is still hot but much, much less then before. Now there is a ringing at switching off? Why is that? Also the switch on characteristic is having some issues. I've tried to add 9.1 ohm resistor between the gate and the driver. But it looks like it does not affect the ringing. Look at the attached picture. Any more help would be appreciated.


First define your LED specs. If this is a 12V power LED ok. if not ..hmmm

It makes no difference whether you choose Anode or Cathode side drivers.
What matters is the RdsOn or Rce equivalent resistance compared to ESR of LED's ( incremental V vs I)

The Ripple could be your PSU Voltage in response to a step load....

speaking of which what are your design specs?
Current max, avg. Vf vs I etc.

LED ESR is typically Watt*ESR~1 WΩ

Ensure driver is ~1% of ESR for minimal loss and heat rise or add a series power R to limit LED current.

What are your LED specs please?

The turn off spikes tend to confirm it's either the PSU or the wiring to it that causes the problem. Try this: connect a 100nF capacitor across the incoming supply, one side to ground and the other to the 12V line. If it changes the ringing and/or overshoot you need to invest in proper supply decoupling.


I forgot to manged it. The load of the last oscilloscope photo are not the LED's but are resistors 3x6.8ohm in parallel. For now we can forget the LED's. I'm doing testing with resistors. However there will be approximately 70 LED connected to this dimmer. All connected in parallel (3.3V, 20mA). I believe it will be best to add one resistor in series with all the LED that are connected in parallel, since I can't add single resistors to each LEDs. I will check power supply and report. If this is the problem.

This is a bad idea.

Always match your LED string voltage to the available supply and higher the better.

With better LEDs the Vf is 3.0V @20mA which is all I use. 16,000 mcd at 30Deg Neutral White 4500'K

Then you can use a string of 4 LEDs in series. Each 65mW LED is ~16 Ohms (=1/65mW) ESR thus each string , ESR is 64 Ohms approx. @ 20mA
then 72 LEDs is 18 parallel strings. If they come from the same batch, they will be matched otherwise you need to add 5 Ohms or so to normalize the ESR, which also makes it dimmer for a string of 4, so then choose 3 LEDs in series or a slightly higher V+ or choose 24V or 48V or a Laptop charger of 19.2V or a universal Laptop charger of similar V.

Thus compute the ESR by the S/P or series / parallel ratio like 4/18 * 16 Ohms = 3.56 Ohms

Your idea of 1S/70P would use 8.7V drop over 20mA ~ 435 Ohm and thus only 28% efficient.

If you wanted to use 3.3V it is like 1/70*16=0.228 Ohms which would work with a 3.3V supply only with 70*20mA or 1.4A @3.2V so 0.1V drop using a series R of 0.1V/1.4A = 70 mOhm or a length of AWG 30 wire, but if mismatched (likely in your case) by more than 0.1V then you need 70 R's each dropping 0.1V or 0.1V/0.02A= 10 Ohms from 3.3 V.

TO make the high side or low side switch work , I would choose lowside Logic level N-FET with 1% of net ESR.

I know it's a bad idea but. I don't have any other options, since it's impossible to rewire LEDs.

Regarding ripple in rising characteristic the problem is really in step response of power supply. I add capacitor 4700uF on the output of power supply and the ripple is completely gone. However spike at turn off of MOSFET is still present. Adding 100nF capacitor across power supply does not help.

Did you add the 100nF capacitor directly across the circuit or at the power supply end of the power wires? The only possible cause of reverse polarity spikes is inductance in series with the load or supply. I'm not sure what probes you are using to monitor the waveform but it would also be a good idea, if they incorporate a divider network, to check they are properly compensated.

In general, connecting LEDs directly in parallel is not a good idea but in short strings, each with a series resistor is fine, uses fewer components and is more efficient.



If I add the capacitor 100nF across the load, ringing at turn off starts to disappear... However falling time also increases...

Not across the load - put it across the wires from the power source but located as close as possible to the circuit rather than at the power supply end of the wires.

In order to get spikes (aka. ringing) like that you need an inductive component in your circuit yet you have none. That only leaves the inductance of the wiring itself which although small, may still be enough to create the spikes. A capacitor will help to absorb and release energy where it is needed and should reduce the inductive effects.



I have tried to put as close as possible to the supply source of the circuit but it does not have effect, unless I put it across the load. What would happen if I put resistor between the driver and the gate? I guess it would affect rising and falling time do to change in RC constant. But could this also change behaving of a ringing?

Regards, Miha

You need a snubber across your fet if you want to reduce ringing in a simple circuit like yours, also you are turning the mosfet on and off quite quickly, this will make and ringing and switching noise worse, sounds like you need a fet with lower RDSon, to keep it cooler, as above if you parallel leds you are asking for trouble, you need a small resistance in series with each one to avoid failures...

Another way to reduce ringing is to treat the load like a buck inductor and have a good diode from the fet drain to +Vcc, this way the current will circulate in the load for a short time as it decreases to zero, and the Vmax on the fet will be Vcc, need a good quality high speed diode.

Thank you Easy peasy. Your were absolutely right, adding diode reverse to the MOSFET helped. The ringing is almost completely gone. I don't know why I was convinced there is a diode already integrated inside MOSFET. Probably because of the symbol in datasheet. Why is symbol in datasheet drawn like there is a parallel diode already integrated with MOSFET?

Thank you all for good advice.

Not open for further replies.

Part and Inventory Search

Welcome to