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I was reading a paper regarding the IGBT driver circuit based on M57962AL IC. I haven't found better information regarding the use of this IC in the datasheet therefore I had to read this. I am confused about the following paragraph in the paper:

M57962AL is equipped with a protection function, the short-circuit fault can be judged by detecting the voltage drop between C and E. When the fault occurs, the collector current rises rapidly to desaturate and the voltage drop between C and E rises. The short circuit protection is triggered to reduce the gate voltage and send a protection signal.

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  1. I am not sure which CE are they talking about. Is it about the CE of the IGBT?
  2. Secondly which collector current rises to desaturate and which C and E are they referring to in this case as well? Kindly explain to me whats happening in the IC basically.
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  • \$\begingroup\$ "CE" is the voltage between the collector and emitter of the IGBT. \$\endgroup\$ Commented Aug 16, 2023 at 22:23

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To answer your questions briefly:

  1. Yes, the "C" and "E" referred to in the paragraph correspond to the "collector" and "emitter", respectively, of the IGBT connected to the M57962AL.
  2. Yes, the "collector current" referred to is that of the IGBT connected to the M57962AL.

You can find datasheets for this chip here, however, please note that these are not an exact match for the part number you quoted:
https://www.pwrx.com//pwrx/docs/m57962cl-01.pdf

and here:
https://www.pwrx.com//pwrx/docs/M57962L-71R-02.pdf

The datasheets show how to connect the driver IC to the IGBT, refer to the "test circuit" diagram copied from the datasheet below.

Note that there is a diode between IC pin 1 and IGBT collector, this is very important to protect the IC from the very high voltages that occur on the IGBT collector; without this diode the IC would be destroyed.

Also note that the IGBT emitter is not connected directly to IC pin 6; this may be confusing at first, but this is done to provide a negative voltage on the Gate pin of the IGBT (with respect to the Emitter pin). This negative voltage on Vge ensures the IGBT has good turn-off speed (good == fast), and reduces the chance of an unwanted turn-on.

I hope this helps.

Test circuit, copied from datasheet

EDIT: Please note:

  1. The OEM is now listing this product as being obsolete, although it seems variants of it can still be purchased, refer:
    https://www.pwrx.com/Product/M57962L

  2. I could not find the datasheet specifically for the "AL" suffix.

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  • \$\begingroup\$ The short in this case refers to the short circuit between the IGBT collector and emitter. Or any other terminal? \$\endgroup\$ Commented Aug 17, 2023 at 4:33
  • \$\begingroup\$ Secondly, if I attach the detect pin at the emitter instead of the collector. Will it make any difference?? since it just has to check for the high current? and current will be approximately the same in the collector and emitter. \$\endgroup\$ Commented Aug 17, 2023 at 4:49
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    \$\begingroup\$ @kam1212 No, the IC must sense collector potential, and the short is presumed to be the load itself, not a short between C and E. While the IGBT is on and saturated, its \$V_{CE}\$ should be close to 0V. If then the load (at the collector) becomes a short to the positive supply, both collector current and potential will rise while the emitter remains grounded. A shorted load will therefore raise \$V_{CE}\$, which the IC detects by monitoring collector potential. \$\endgroup\$ Commented Aug 17, 2023 at 5:18
  • \$\begingroup\$ @SimonFitch Means when the collector voltage is equal to the VCC of main or the collector end joins the VCC then the short signal is generated? Secondly suppose my voltage to the IGBT is 500V. If the signals is shorted the pin detect will recieve 800V. Woun't it damage the IC for getting such a high voltage. Secondly how much voltage at the collector is assumed to be a short for it. 100V,200V etc because all of them are greater then the 15V or VCC of the IC \$\endgroup\$ Commented Aug 17, 2023 at 8:23
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    \$\begingroup\$ @kam1212 "Woun't it damage the IC for getting such a high voltage." As mentioned in post above: "Note that there is a diode between IC pin 1 and IGBT collector, this is very important to protect the IC from the very high voltages that occur on the IGBT collector; without this diode the IC would be destroyed." \$\endgroup\$ Commented Aug 18, 2023 at 3:48

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