I have built the following circuit as a static AC switch (no high frequency switching).

The IGBT is a dual common-emitter module (https://www.infineon.com/dgdl/Infineon-FF300R12KT3_E-DS-v02_00-en_de.pdf?fileId=db3a304313b8b5a60113b981ab7e0010) I am using a MGJ6 15V/10V DC-DC converter (https://www.mouser.com/datasheet/2/281/kdc_mgj6scdc-1019175.pdf) to power my gate driver, which is a 1EDC60I12AH (https://www.mouser.com/datasheet/2/196/Infineon-1EDCxxI12AH-DataSheet-v02_00-EN-1272149.pdf). 100nF decoupling caps on both modules and a 20μF bulk cap on each DC-DC output. 5 and 20R resistors on sink and source driver outputs respectively. 1μF-47R snubber network. I tried following the recommended setup using a bipolar supply for the gate driver the datasheet recommended.

Also, in case it isn't clear, the common emitter is tied to 0V coming out of the MGJ6.

The issue is, the IGBT is never in the off state. When I first built the circuit and measured it, I was getting a Drain-Drain voltage of about 80VAC. So it was sort of half-off? When I measured it again later, it didn't seem to want to turn off at all. I was getting about 1VAC across Drain-Drain, which would indicate to me that it was still completely in saturation. At the time I measure this value, the Gate-Emitter voltage (VGE) is -8V! When I switch them "on", VGE becomes +16V. Same voltage drop across Drain-Drain. I cannot figure out why I cannot get them to turn off.

A theory--am I destroying the IGBT module somehow? Am I not providing the gate with enough current? Do I need different cap values on the 15V and -8V rails? Any help would be appreciated. I'm rather stumped.

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2 Answers 2


You seemingly have thought to switch AC in a clever way: For +cycles of the mains AC one IGBT is active switch and another works passively through its protection diode. For negative cycles the roles of the module halves are changed.

Both IGBTs seem to get the same drive signal (Vge = +15V for conduction and -8V for OFF-state). It's a puzzle why this doesn't work. Neither of the IGBTs is attempted to use with negative Ic and the capacitive feed through the IGBT from the main AC circuit to the gate should be neglible with low frequency sinusoidal AC.

Only some guesses can be given without having your parts in my hands. One possibility is that your power supply leaks, there's a substantial, maybe capacitive leak between the mains AC input and the DC output GND. That can be in the filter which try to protect other mains AC users from getting your high frequency switching noise.

But that guess looks nonsense if your driver power supply really is the one which is designed for low (=few pF) coupling between input and output as your datasheet claims.

So, start debugging. Try at first to get it operating with DC power and DC load to see if there's something broken, for ex. you've fused your IGBTs permanently ON with a high capacitance load (=too high peak current )or with inductive load (= too high voltage). Then you can also check the drive. Multimeter can be used if you have no oscilloscope because at DC everything is static.

You sould have in tests RL which makes specified leaks through IGBTs neglible. Without a load your multimeter can show whatever between 0...mains AC voltage.

Sorry for no better ideas. Except one: If your parts are destroyed consider to get a solid state relay. It has all in one block.

  • \$\begingroup\$ I think it could be more fundamental even. The datasheet says that VGE can be +/- 20V, and at the peak of a cycle VGE would be 120V right? Also the ground from the DC-DC to the emitter must be wrong. Fundamentally I believe this approach is flawed because at best I think all I could do would be to turn it off at a zero-crossing, and there are already better ways to do this I think (thyristors). A gate could turn off the IGBT but the diodes are always connected. \$\endgroup\$
    – usinjin
    Apr 7, 2020 at 13:47
  • \$\begingroup\$ IGBT isn't triggered, Vge affects as long as it's unchanged. If the voltage in the Ic circuit drops to zero but returns later no retriggering is needed to let Ic flow again - only let Vge stay positive and above the treshold. Think it as N-channel mosfet, but the current is possible only to one direction due the internal bipolar transistor. \$\endgroup\$
    – user136077
    Apr 7, 2020 at 14:02
  • \$\begingroup\$ @usinjin (continued) as long as your IGBT driver and its power supply do not leak there should be no Vge overvoltage problems. Check that your own connections do not make any connection over the isolation borders, not even capacitive except few pFs. \$\endgroup\$
    – user136077
    Apr 7, 2020 at 14:09
  • \$\begingroup\$ Power supply leakage does not make sense to me. BUT, I think what might be the key is how "far" I can turn the IGBT off, which is to about 80Vce. I read more about IGBT capacitance and it seems like there is a parasitic capacitance between C-G and between G-E. As Vce increases it forms a voltage divider and a voltage appears on Vge, keeping it on. To remained turned-off Vge needs to remain shorted at all times. Is there a possibility of turning the IGBTs off and on by either shorting Vge or disconnecting, and allowing the parasitic capacitance to keep them turned on otherwise? \$\endgroup\$
    – usinjin
    Apr 8, 2020 at 8:34
  • \$\begingroup\$ Reduce the resistor between the gate and OUT- to improve gate discharging for faster turn-off in case voltage at collector jumps. I am not sure how this could be the problem outside pulse switching applications, but try it. How your IGBTs work in low voltage DC switching? \$\endgroup\$
    – user136077
    Apr 8, 2020 at 8:55

EDIT: I have looked around for some time, checked more info on your setup and devices and I was wrong about IGBT (or a MOSFET) not being used to control AC.

The capacitances on your IGBT module are fairly large, and are definitely affecting the behavior: Gate charge alone is 2.5μC. Your driving requirements would be significantly reduced if you use a less powerful IGBT. Do you really need the 300-480A capability of this IGBT module? Sometimes too much good stuff can turn into bad.

Comparing your schematic to the datasheets and related materials, I have noticed a few things:
1.) Your gate driving resistors are switched around: the higher value resistor is supposed to be on the positive (OUT+) output. Also, the 1EDC datasheet suggests 10Ω and 3.3Ω, while you have used about double that. Couple their higher resistances (and their values switched around) with the very high capacitances of your IGBT module, and it's an invitation for problems.
2.) The application note DT 94-5A shows a resistor between the common gates emitters, but it doesn't mention its value. It is obviously there to reduce the parasitic voltages in such configuration. I would suggest trying from 10kΩ down to 100Ω.

  • Have you tested your IGBTs separately with some relatively high voltage (20-100VDC) and plain DC voltage control applied to gates to see if they are still good?
  • Have you checked your voltages/signals on an oscilloscope to see what is going on?
  • What are your loading and testing currents? Your IGBT has a 5mA cutoff or leakage current at 1200V Vce. Have you considered that?
  • \$\begingroup\$ Ah, of course. I understood how they are unidirectional devices in regards to collector-emitter voltage. The module is equipped with diodes across each IGBT to allow current to pass in the opposite direction. But I failed to account for when the switch was on--and suddenly gate-emitter voltage swings between large positive and negative values. My circuit was based on this idea: drive.google.com/file/d/1-2_WqkOMJRI_TBjC6rSgrUbeniF2iG5I/… They even mentioned this issue, but I don't understand how they are attempting to solve the issue with a photovoltaic isolator. \$\endgroup\$
    – usinjin
    Apr 7, 2020 at 10:55
  • \$\begingroup\$ I have worked with SSRs before and are aware they are a much easier (and probably better) at switching AC. But I was drawn to IGBT's speed and high power capabilities. \$\endgroup\$
    – usinjin
    Apr 7, 2020 at 11:02
  • \$\begingroup\$ @usinjin You should probably draw the antiparallel diodes in your schematic, then. \$\endgroup\$
    – Hearth
    Apr 7, 2020 at 14:19
  • \$\begingroup\$ @usinjin: I have edited my answer. Let me know if you have tried my suggestions and if it helped you. \$\endgroup\$ Apr 23, 2020 at 11:31

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