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Below is the datasheet of the IGBT https://www.semikron.com/dl/service-support/downloads/download/semikron-datasheet-skm75gb063d-22890050

It has 4 Control Pins , Number 4,5,6,7 . Generally a MOSFET has a single gate which is used to trigger the saturation mode (Short Circuit Collector and Emitter).

enter image description here

I'am confused here , can someone help me out on how to perform ON , OFF switching with these 4 pins available .

Thanks

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    \$\begingroup\$ "ON , OFF switching" with a phase leg dual IGBT does not make much sense. Draw a circuit diagram. Which IGBT do you want to turn on? You turn it on by applying +15 V on 4 relative to 5 and the same for pin 6 and 7. \$\endgroup\$
    – winny
    Nov 21, 2016 at 8:51
  • \$\begingroup\$ @winny will this work ? drive.google.com/file/d/0By00NCw9AfibdlRxUTN3SzFETEE/… I want to use both IGBTs . Basically , I want to control the AC load using PWM and I'am restricted in using this option due the rated current of this component . It will be used to control 10KW dummy / ballast load in Electronic load controller for MHPP \$\endgroup\$
    – T.ALi
    Nov 22, 2016 at 12:10
  • \$\begingroup\$ Certaily not! Your body diodes will conduct 50 % of the time so you can only control 50-100 % duty cycle. If you want to build a normal IGBT dimmer you need a diode bridge or anti-series connected IGBTs. You have a phase-leg/series connected IGBTs. See here: electro-tech-online.com/attachments/… People in general here are a bit too anti DIY when it comes to mains power but if you ask these kind of questions, you are on deep water when it comes to burning your expensive IGBTs or killing yourself. \$\endgroup\$
    – winny
    Nov 22, 2016 at 13:17

3 Answers 3

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From a drive point of view, an IGBT is very similar to a MOSFET

enter image description here

An IGBT,like a MOSFET is a 3 terminal device. It is controlled via the Gate-Emitter (gate-source for a MOSFET) voltage

To turn on a MOSFET you must raise the GATE voltage with respect to the SOURCE above the threshold voltage. Likewise to turn on an IGBT you must raise the GATE voltage with respect to the EMITTER above the threshold voltage

You are probably used to seeing the SOURCE of a MOSFET ground/earth/0V referenced and thus this key driving requirement appears as just a "gate voltage."

This particular IGBT module you are linking is an inverter leg made from 600V IGBT's. You will most certainly an isolated driver circuit that is capable of at least 600V+ between the primary & secondary which can also tolerate at 1500V/us dv/dt

When the voltage between pins4 & 5 (gate and emitter) exceeds 5.5V the left and IGBT will facilitate current flow from terminals 3 to 1

When the voltage between pins6 & 7 (gate and emitter) exceeds 5.5V the left and IGBT will facilitate current flow from terminals 1 to 2

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  • \$\begingroup\$ In your last statement , you mean pin 6 and 7 and not 6 & 2 ? .Coming to my question , I want to switch an AC load using this and I want to use both IGTBs because I want to make sure that this device perform at its rated current , 60 A . If I use individual IGBT 1 -> 3 and 1->2 current flow , the ratings will be divided . Is it possible to short terminals 2 and 3 , so that external circuit is connected . See this drive.google.com/file/d/0By00NCw9AfibdlRxUTN3SzFETEE/… . Is this correct . \$\endgroup\$
    – T.ALi
    Nov 22, 2016 at 12:22
  • \$\begingroup\$ That is correct, typo 6-7 (corrected). It would work 6-2 as 2 is the power emitter BUT you will run into the issues that I mentioned in another comment (the stray inductance in the power emitter causing unexpected switching events) \$\endgroup\$
    – user16222
    Nov 22, 2016 at 12:26
  • \$\begingroup\$ W.r.t. your image in your comment, no it isn't. There is potentially 600V difference between the two emitters. You have the emitters using the same reference point. Let me draw something \$\endgroup\$
    – user16222
    Nov 22, 2016 at 12:28
  • \$\begingroup\$ Also , other guy suggests that Gate emitter Voltage should be +15 ? . \$\endgroup\$
    – T.ALi
    Nov 22, 2016 at 12:29
  • \$\begingroup\$ I want to switch this device via MCU , the microcontroller pins will be optocoupled with this device using 4n25 \$\endgroup\$
    – T.ALi
    Nov 22, 2016 at 12:29
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The easy answer: An IGBT needs a gate-emitter voltage in order to turn on (that is, a voltage between the gate and the emitter) and having it set up this way makes it easy to access both of the important connections with a simple positive and negative wire. The SEMIKRON module you posted contains two IGBTs in a half-bridge configuration, which means you can switch the IGBTs alternately to get different output connections (connect the load to ground or connect it to a voltage source). You could also use two of them together to get an H-bridge, which can alternate the current through the Pin 1 connection. Either way, it makes it much easier to switch the IGBTs individually when you can access their own gates and emitters directly.

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Short answer

It's a Half-Bridge IGBT with a Gnd routed to the low V side with tandem CE contacts for internal Kelvin current shunt conductors to pins 5&7 for external I_sense amplifiers .

enter image description here


Long answer

**Why two active Drivers ** 2 IGBT's are better than 1

  • Less power dissipated, faster switching times more modes of switching for different load types

enter image description here

Why so many pins? Ease of Kelvin current sensing from Low Voltage side with Rs of Emitter Current sense bond wires, which must be calibrated, as no specs are given.

enter image description here

How is used in industrial applications?

enter image description here

How flexible? What switching modes?

enter image description here

  • Neutral Switch (NS), Zero Voltage Resonant Switch (ZVRS), Zero Current Resonant Switch (ZCRS), Zero Voltage Switch (ZVS), Zero current switch (ZCS), Hard switch (HS) (e.g. PWM)

Since this was a newb question, I thought some may benefit from the reasons for this topology. ( and some may resent it)


  • Have you ever looked at TTL output stage? It has similarities.

enter image description here

  • Although the name is fading out of favour, it was called a Totem Pole or Push-Pull driver. This is just for historical similarity in topology except 1,667x more current capacity.
  • (joke) It looks like the pole came down in their schematic. enter image description here
  • the separate 100A Free Wheel Diodes have equal current capacity to the output stage and offer additional protection or flexibility in design.
  • In MOSFET topology , it's also called a half-bridge that allows you keep impedance low for each half cycle rather an open drain topology with a T=L/Rs discharge time for value of Diode Rs similar to 20 mOhms of IGBT specs.
  • For historical reference CMOS has reduced Vcc, 18Vmax down to 3.6Vmax with scaled outputs for Vgs, RdsOn now also down to 50 or 25 Ohms in different families.
  • except for SMPS the impedance of charging low ESR caps must be driven by Rce values of equal or lower values which determines tradeoff between match impedance for low ripple or lower source impedance for high efficiency but higher peak ripple currents.

  • so although this power half-bridge has very little resemblance with a Schottky gate or a CMOS gate they are all Push-pull drivers, yet the Transmission Line and Characteristic Impedance Rules still apply to both for well-controlled pulses with minimum ringing where ESR and ESL (inductance ) of cabling to load is also critical as well as Common Mode (CM) noise.

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  • \$\begingroup\$ Pls comment your personal bias \$\endgroup\$ Nov 21, 2016 at 23:32
  • \$\begingroup\$ not that I voted you down... I would dispute your edit where you reasoned that the many pins is flexibility. The Kelvin Emitter connection is provided when the power emitter inductance is substantial enough during load switching to cause false switching events. Likewise to permit the creation of small inductive loop gateleads. I can do all those switching schemes you list (hard,soft, reso...) with a Semikron with a kelvin connection or an Eupec module that doesn't \$\endgroup\$
    – user16222
    Nov 21, 2016 at 23:57
  • \$\begingroup\$ good point I'll add that \$\endgroup\$ Nov 22, 2016 at 0:30
  • \$\begingroup\$ @JonRB I appreciate your experience and comments far greater than the trolls that silently vote down , contributing zero intelligence \$\endgroup\$ Nov 22, 2016 at 5:01

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