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I am using the IXDP 20N60B IGBT as an AC switch. This IGBT has no internal diode. When the Gate Emitter voltage is ~10V the device is on and conducts in the forward biased region as expected, shown below.

enter image description here

However when the Gate Emitter voltage is set to 0V the device is off but there is some reasonable voltage in the reversed biased region as shown below.

enter image description here

Why is this the case? The voltage should be 0V when the device is off.

Thanks.

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  • \$\begingroup\$ What value load are you using? \$\endgroup\$ – Andy aka Jun 6 '18 at 15:05
  • \$\begingroup\$ @Andyaka - Your question prompted me to change loads. I was using a high load of 10K. I then I then reduced this to 180R. There is now no reverse conduction. Why does the load affect the operation of the device like this? \$\endgroup\$ – MXG123 Jun 7 '18 at 11:19
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Not all IGBTs are capable of blocking reverse voltage. If you need an IGBT for switching AC voltage for example you have to use one that is specifically rated for that application.

From this app note:

..the IGBT consists of a PNP driven by an N-Channel MOSFET in a pseudo-Darlington configuration. The base region of the PNP is not brought out and the emitter-base P-N junction, spanning the entire extension of the wafer cannot be terminated nor passivated. This influences the turn-off and reverse blocking behavior of the IGBT, as will be explained later. The breakdown voltage of this junction is about 10 to 50V and is shown in the IGBT symbol as an unconnected terminal (Figure 2). For this reason IGBTs have an undefined reverse conduction characteristic...

However, IGBTs with defined reverse blocking capability do exist. This paper explains how they work:

The primary difference in structure between the Fuji RB-IGBT & a conventional IGBT is that the former has deep junction isolation structure that limits carrier generation thereby providing the needed (higher) reverse blocking capability.

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  • \$\begingroup\$ There doesn't seem to be many with specific reverse blocking capability. Do you know of any that are in supply? \$\endgroup\$ – MXG123 Jun 7 '18 at 8:39
  • \$\begingroup\$ You might try IXYS, I think they are sampling a new device: ixys.com/ProductPortfolio/PowerDevices.aspx \$\endgroup\$ – John D Jun 7 '18 at 14:28
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enter image description here

Unless specifically shown in the data sheet you can assume that the reverse breakdown from collector to emitter happens at a few tens of volts. I see nothing in the data sheet that suggests otherwise.

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  • \$\begingroup\$ @MXG123 with a high impedance load like 10 kohm the leakage currents and impedances of the IGBT will be of a similar order hence they will from roughly a 50% potential divider and you will see some voltage. If that voltage began to exceed the reverse breakdown of the device then you will see a much larger voltage because the IGBT will have a much lower impedance. With 180 ohms (and not exceeding the reverse breakdown voltage) the potential divider is much much smaller hence you don't see much of a an effect across the load. \$\endgroup\$ – Andy aka Jun 7 '18 at 12:02
  • \$\begingroup\$ Thanks for this info. Another thing is when I increase to high voltage the devices fail. The datasheet says they are rated to 600V and 32A. I have left the gate tied to the emitter - device off. Connected to mains (230V) with a 13A fuse. The device is conducting even though it is off, gets real hot and then fails(low impedance between pins). The leakage current is small so why is this happening?? \$\endgroup\$ – MXG123 Jun 7 '18 at 14:50
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    \$\begingroup\$ Because it is not rated for negattive voltages and AC cycles positive and negative with peaks up to + and - 325 volts. \$\endgroup\$ – Andy aka Jun 7 '18 at 15:23
  • \$\begingroup\$ I was under the impression that non-punch through IGBT's can block reverse polarity voltages. It would be nice if that was specified in the datasheet. For my issue I suppose I could put a diode in series so the IGBT will never see the negative voltage. I wanted to avoid this as this will introduce more losses. \$\endgroup\$ – MXG123 Jun 8 '18 at 8:05

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