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I have designed an MCU-based PWM dimmer for LED strips.

My circuit works fine but there is a problem: The MOSFET is unexpectedly hot (i.e. 65°C). So I scoped the drain voltage using an oscilloscope and I saw that the drain voltage was about 0.130V (as expected) when the MOSFET is on, but it is 9V --it should be 16V-- when the MOSFET is off. So I'm not sure but I think that this causes the MOSFET to heat up.

schematic

simulate this circuit – Schematic created using CircuitLab

Where is my fault? How can I solve this problem?

EDIT: Here are the waveform of VDS:

enter image description here

enter image description here

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    \$\begingroup\$ "So I scoped the drain voltage using an oscilloscope " no pictures it did not happen \$\endgroup\$ – Trevor_G Jan 24 '18 at 14:43
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    \$\begingroup\$ Looks like Q1 isn't turning fully on. What's the voltage across R6 As a start, try removing Q1 and R5, then wire-link the Q1 base pad to the R5 pad that goes to R6. Then look at the FET voltages again. \$\endgroup\$ – TonyM Jan 24 '18 at 14:44
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    \$\begingroup\$ You need to get together with this guy and combine your driver stages... electronics.stackexchange.com/questions/351842/… \$\endgroup\$ – Brian Drummond Jan 24 '18 at 14:58
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    \$\begingroup\$ "but it is 9V --it should be 16V" - no, it shouldn't. Your scope input has resistance which draws current though the LEDs to ground, and the FET has leakage current when turned off. Some 5mm white LEDs that I tested had a voltage drop of 2.2V each at 0.25uA. 20 parallel x 3 in series would drop 6.6V at 5uA. The AOD444 has maximum leakage current of 5uA at 55ºC. A 1 Meg scope input draws 9uA at 9V. \$\endgroup\$ – Bruce Abbott Jan 24 '18 at 15:43
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    \$\begingroup\$ With three LEDs and a 10-ohm resistor, you're putting something on the order of half an amp through each string, and 20 strings in parallel will be putting 10 amps through the MOSFET. That's 1.3W of dissipation -- do you have a heatsink on it? \$\endgroup\$ – Dave Tweed♦ Jan 24 '18 at 16:14
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The 9 volts is a red herring. 0.13 volts across a FET of on-resistance of 0.05 ohms is a power of 340 mW and this will generate heat. The junction to ambient thermal resistance is typically 50 K/W, so without a heat sink and at a local ambient temperature of 30 °C, the temperature will rise to 30 °C + 0.34 W * 50 K/W = 47 °C then factor in that the temperature rise might locally increase ambient another twenty degrees and you are at 67 °C. Not really a problem.

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  • \$\begingroup\$ It comes from the leakage current of the MOSFET. This leakage current causes about 2 volt drop across each LED. And yes, the numbers don't lie; 40°C of temperature rise is very logical according to calculations. \$\endgroup\$ – wooz Jan 25 '18 at 5:20
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Actually 65C, a 40C rise is not really out of the question with this circuit.

Assuming your numbers are right, a \$50m\Omega\$ on resistance with 0.13V that's 1/3W already. At 60C/W that's a 20C rise if it is turned on all the time.

Add in the switching losses and other errors it does not take much to get up to 2/3W. Improving the turn off time would help a little.

As Bruce Abbott mentioned in the comments, the 9V thing may be a "red herring" here.

Also, as I mentioned in a comment, I'd also be adding a fly-back diode to this circuit... That's likely a good whack of inductance with 20 of those LED strings in parallel.

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  • \$\begingroup\$ Your argument about high inductance due to parallel LED strings sounds wrong. Inductances in parallel are just like resistors in parallel. Total inductance is L/20, not 20*L. \$\endgroup\$ – DaveP Jan 24 '18 at 21:51
  • \$\begingroup\$ @DaveP yes indeed, I just meant it sounds like he has a lot of wiring. Diodes are cheap.. \$\endgroup\$ – Trevor_G Jan 24 '18 at 21:54

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