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I have a circuit with a high voltage source (rectifet 400VAC) (part of a capacitor discharge ignition.)

High voltage should only be available if normal power (low voltage DC power supply for the rest of the circuit) is connected.

That means if power is off there has to be a way to shunt/dissipate the power of the high voltage source. Turn circuit off --> kill high voltage.

I have developed a circuit that should do the trick

Maybe someone has feedback, a better idea or sees a failure point.

Circuit description:

The gate voltage of the MOSFET Q1 to the left should always be 1,8V below the source (high voltage) so it is conducting even if power is off connecting the high voltage to a resistor connected to ground for power dissipation.

Now if the second MOSFET Q2 is conducting it is shorting the three diodes to the right which would be after circuit powerup if the MCU switches Q3 over the optocoupler. The resistors in series to the diodes are picked very high so in normal “on” operation there is close to no high voltage supply power loss. This is because (in normal “on” operation) the high voltage is used to charge a capacitor and every bit of change is needed.

enter image description here

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  • \$\begingroup\$ Hard to say as there are no component types or values. Is the FET going to need more than 20V Vgs? And the optoisolator does not isolate. Did you simulate the circuit and did it work in simulation? \$\endgroup\$
    – Justme
    Jan 9, 2021 at 19:59
  • \$\begingroup\$ i added the name of the Q1 FET and deleted the opto cuppler Q2 is just a generic P-Channel type FET. but is more of a "is it a good idea" do do it like this type of question. why Vgs 20V? it is -3,5V max thats why i used the diodes to get the constant voltage drop. \$\endgroup\$
    – JonnyKing
    Jan 9, 2021 at 20:15
  • \$\begingroup\$ What happens if the 400VAC is still connected but the MCU crashes or otherwise loses power. Your discharge resistor will probably burn up unless powered continuously, at which point it is simpler and more reliable to power it continuously. \$\endgroup\$ Jan 10, 2021 at 4:33
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    \$\begingroup\$ In general, these kinds of systems usually just have a contactor on the AC side and a bleeder resistor to empty the capacitors in a power cut. Everything on the secondary side should be assumed to be live any time the power is on regardless of what the MCU does as MCUs are not safety-rated. \$\endgroup\$ Jan 10, 2021 at 4:34

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First, I notice that in your question, you refer to 400VAC. But your circuit only handles DC. So I am going to presume that you only need to short DC.

Second, the term 400VAC refers to an rms value, not peak. When it is rectified, it is considerably more than 400VDC. Peak value is about 400*1.4=560V. But in your schematic, you have 400VDC, so I am going to assume that 400VDC is actually what you have, not 560V. But that is something you need to check.

I offer the following basic circuit, which may need to be modified to suit your needs.

schematic

simulate this circuit – Schematic created using CircuitLab

R3 draws less than 1 \$\mu\$A from the high voltage rail. If \$V_{G1}\$ is above the threshold voltage of M1, then M1 will sink R3's current, and the \$V_{G3}\$ will be less than M3's threshold voltage, and M3 will be off. It may leak a little, and finding high voltage but low leak MOSFETs might be worthwhile.

If Vcheck goes below the threshold voltage of M1, that M1 will turn off, and this will cause the gate of M3 to rise to the point that M3 conducts.

M2 and R4 serve to limit the rise of M3's gate voltage to about 2 times \$V_{th}\$, and to limit the current through M3's channel to be approximately \$V_{th} / R4\$. If enough current passes through R4, the voltage across it will be enough to turn on M2. This will draw current away from the gate of M3. The circuit, as drawn, will draw about 4 mA of current from the high voltage rail when Vcheck is low.

I don't know how fast you need to discharge your capacitors. You may want to change the value of R4 to fit your requirements. Use smaller values if the capacitor discharge is too slow. Use larger values if the current is too high.

Note that I said that the circuit will provide a shunt if Vcheck goes under the threshold voltage of M1. I didn't say what voltage guarantees that the circuit does not shunt the high voltage. The IRF830 datasheet gives the threshold voltage as anywhere between 2V and 4V. If you need a more definite transition voltage, especially if you are using a 3.3V MCU or a 5V MCU with weak output pins, let me know.

The power that M3 must dissipate is 400 V \$\times\$ about 4 mA or about 1.6 W. A suitable heatsink for M3 should be used.

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  • \$\begingroup\$ First of thank u very much that was the circuit i was looking for especially because i had all the components at home.Sadly it is like you said the circuit basically works but the voltage is shunted slowly in even if M2 is not conducting. Just to clarify, the HV Source is an Exciter coil on a flywheel. I connected Vg2 directly to ground (just to check) but the leakage current through M2 seems to be enough to shunt the Voltage. \$\endgroup\$
    – JonnyKing
    Jan 10, 2021 at 13:43
  • \$\begingroup\$ moreover the capacitor (which is charged by the coil) is not charging to 400V anymore. \$\endgroup\$
    – JonnyKing
    Jan 10, 2021 at 13:50
  • \$\begingroup\$ If you connect Vg2 to ground, you are shunting through R3 (and whatever the leakage is for M2). In your circuit, you had a 5M\$\Omega\$ resistor to ground. I don't know how much current/charge you have, but increasing R3 to 5M would reduce your leakage. \$\endgroup\$ Jan 10, 2021 at 13:50
  • \$\begingroup\$ I have noticed that the datasheet for at least one manufacturer of IRF830 lists a max leakage current Idss of 1uA at 25C, and 10uA at 150C. But the datasheet for another manufacturer give a leakage of 0.2mA at 25C and 1mA at 125C. Since leakage is a significant concern in your application, choice of manufacturer will have a big impact. \$\endgroup\$ Jan 10, 2021 at 15:51
  • \$\begingroup\$ might get more peace of mind if you gate the mosfet through an opto -- if there is any doubt about the common ground \$\endgroup\$
    – Pete W
    Jan 10, 2021 at 19:16

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