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I have built several 200W AC dimmers using Microchip 8-bit CPU and the illustrated back-to-back MOSFET circuit.

I have seen on Internet schematics with the D1 diode in parallel with the R1 resistor, or without.

Assuming it should be a regular diode (tried 1N4148), I connected a diode and compared it on oscilloscope, but did not see any difference.

Can anyone tell me, why to use a diode at that location?

Update: with 200W load of 4 pieces of 50W bulbs, without the diode we often destroy the MOSFET, pretty much immediately after first power-on (transistor becomes short). With the diode, it seems to work even after many hundred power-ons.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ It depends on where your earth signal node (aka 0 volts aka ground) connects to and what your CPU's reference node is. \$\endgroup\$ – Andy aka Feb 21 at 12:55
  • \$\begingroup\$ @Andy I added the line to indicate, that the MOSFET ground and the CPU ground is the same. CPU is powered by Zener diode / NPN transistor 5.5V power supply directly from the same 240VAC (no transformer) \$\endgroup\$ – EmbeddedGuy Feb 21 at 13:36
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Given R3, R4 & C1 that diode is doing mostly diddly squat.

Generally you add a diode in parallel with the driving resistor at a gate of a Mosfet to accelerate turn-on or turn-off by providing more drive to the Miller-multiplied Drain-Gate capacitance.

In this particular case the diode seems to be there to bypass the resistive divider and provide more voltage drive to the gates. Which is then wasted in the other resistors and capacitors.

What they might be doing is to provide for a relatively slow turn-off while not compromising too much on turn-on time. Perhaps it’s part of a zero-crossing turn-off circuit?

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  • \$\begingroup\$ Turn off is generally timed by a timer in the CPU, which generates square PWM. With slowing the turn-off, what could one gain? I would say if there is a desire to speed up the turn-on, wouldn't it be better to simply lower the R1 resistor, even drastically to 1 kohm? \$\endgroup\$ – EmbeddedGuy Feb 21 at 16:28
  • \$\begingroup\$ @EmbeddedGuy slowing turn-off (or turn-on) could reduce RF emissions which are related to fast switching edges. If turn-off happens near a zero-crossing its speed is mostly irrelevant. Reducing R1 could put a lot more load on the micro controller. I am not agreeing with the design decisions (particularly adding a diode AND a capacitor!!), I’m just explaining what they might have intended to do. \$\endgroup\$ – Edgar Brown Feb 21 at 16:46
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    \$\begingroup\$ Given 22Kohm and 22,000 pF (what with Miller Multiplication) is 0.4 milliSeconds time constant, this looks like a fine method for zero-crossing detection/switching. \$\endgroup\$ – analogsystemsrf Feb 21 at 19:00
  • \$\begingroup\$ Update: with 200W load of 4 pieces of 50W bulbs, without the diode we often destroy the MOSFET, pretty much immediately after first power-on (transistor becomes short). With the diode, it seems to work even after many hundred power-ons. \$\endgroup\$ – EmbeddedGuy Feb 22 at 19:52
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    \$\begingroup\$ @EmbeddedGuy that suggests insufficient gate drive on turn-on causing the FETs to turn on too slowly (BTW: in this condition it can be much more than 200W depending on the type of bulb). Which can be improved simply by removing R1 and/or reducing the value of R4 & R3. \$\endgroup\$ – Edgar Brown Feb 22 at 20:06
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That diode will make the FETs turn on faster and turn off more slowly. It's probably done to achieve some advantageous compromise between switching losses and reduced RF interference.

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