I'm using FGH60N60 IGBT as a switch. The circuit is basically a capacitor-discharging circuit. I have a 10 mF capacitor charged to 310 volts and connected to 2 Ω of load via this IGBT (I connect 4 FGH60N60 in parallel. All emitters connected to each other and collectors connected to each other. The gate voltage is applied to all of them with the same length wire triggered from a microcontroller and an optocoupler). I apply 12 V to the gate of the IGBTs for three seconds and the capacitor should discharge in 30 ms. As mentioned in the Datasheet, this IGBT can handle 400 V 60 A (even the tests in datasheets were reported in this condition). So four of them can handle 240 A, which is 54% higher than the 150 A that will flow in a couple of milliseconds from them.

However, after switching, the resistance between the collector and emitter decreases to 5-8 Ω and basically shorts the path. So i assume that they are burned.

Can I connect the IGBTs in parallel? If I can do so, why do they keep burning despite applying in range current and voltage?

  • 3
    \$\begingroup\$ They could be oscillating, or there could be a layout problem. Can you show schematic and layout? How much inductance in the load? \$\endgroup\$
    – bobflux
    Commented May 28 at 9:07
  • 2
    \$\begingroup\$ For "sharing" current, use a "source" resistor ... for each IGBT. \$\endgroup\$
    – Antonio51
    Commented May 28 at 10:49
  • 2
    \$\begingroup\$ Please post Vge, Vce and Id oscillograms. \$\endgroup\$
    – winny
    Commented May 28 at 11:24

3 Answers 3


can I connect the IGBTs in parallel?

Yes, but you have to take certain precautions.

You have two problems:

  1. Current sharing among all four (4) IGBTs is not perfect, particularly at the instant of turn-on. If one IGBT turns on before the others then it carries the full 150A. The failure of one IGBT may trigger the others to fail.
  2. The IGBTs are stressed beyond their Safe-Operating-Area (SOA).

Solution for Problem 1:
Give each IGBT its own collector resistor.
An 8Ω resistor per IGBT will give similar discharge time to what you have now with the 2Ω resistor. This will ensure better current sharing, which is independent of variations in Vgs(th) and gate-drive characteristics for all points in time. This limits the maximum drain current to just 41A per IGBT at the instant of turn-on, regardless of what the other IGBTs are doing.

In addition, ensure each IGBT has its own resistor in series with its gate. This also ensures better current sharing at the instant of turn-on, which is the worst-case in your application.

Here is a suggested schematic, note that the schematic editor did not have the IGBT symbol so I used the MOSFET symbol instead.


simulate this circuit – Schematic created using CircuitLab

Solution for Problem 2:
Make sure the gate drive is sufficiently fast that the SOA limit is not breached. Refer to Fig 11 of the datasheet, copied here below; this shows that provided the fall-time of the collector-emitter voltage is below 10us then each IGBT will easily handle the 330V and 41A that will occur at the moment of turn-on.

enter image description here

Link to datasheet:

A Vce fall-time of 10us is quite easy to achieve provided the gate driver is capable of supplying the high current to charge the IGBT input capacitance. The datasheet states input capacitance is about 3nF; to change 3nF by 10V in 10us requires a current of:
i = C dv/dt = 3nF x 10V / 10us = 3mA.

However, that is a very tiny gate drive current compared to what is usually found for an IGBT of this size, and does not consider the Miller effect; I would suggest selecting a gate driver that is capable of supplying at least 100 times this per IGBT, so 0.3A per IGBT. Note also that the datasheet presents all the switching data with a gate resistance of 3Ω, which means an instantaneous gate current of over 3A.

Don't Forget About the Humble Resistors
The resistors in this application will be subjected to enormous power for a brief instant of time. 330V on 8Ω ==> 13.6kW!

Of course, this is not continuous, but you must select a good-quality resistor with high pulse power ratings; otherwise it will explode and possibly catch fire.


Yes, you can use paralleled IGBT's.

Do a "current sharing" ... with a source resistor, like this.
Be sure that the "resistors" are quasi-equal, "power share" wiring.
NB: you can use also "transformer-sharing" at the drain's (2 by 2).

enter image description here

Update: with the OP IGBTs ... Note the "log" scales. Only One pulse.
Must verify that IGBTs are "well" chosen.
Curve GREEN is the instantaneous power in the load,
curve BLACK is the instantaneous power in one IGBT,
curve BLUE is the instantaneous power in one resistor emitter.

enter image description here

And "dispersion" of currents with the resistor emitter value (40m, 45m, 50m, 55m Ohm).
And voltage across one IGBT.

enter image description here

  • \$\begingroup\$ Given that the goal of the circuit is to dissipate energy in one or more resistors, splitting the load-side resistor into four as shown in Fabio's answer would be a better way to ensure load sharing. Using source-side resistance will result in a larger portion of the energy being dissipated within the IGBTs rather than resistors. \$\endgroup\$
    – supercat
    Commented May 28 at 22:07
  • \$\begingroup\$ The OP question was" : "Can I connect the IGBTs in parallel?". It is what I did ... Dissipating energy in "one load" or in "one load and 4 resistor sources" does not "matter" ... \$\endgroup\$
    – Antonio51
    Commented May 29 at 8:43
  • \$\begingroup\$ Draw a graph of power dissipation in the IGBTs with your approach. I would be concerned that adding resistors to the source rather than the drain might improve current balance, but increase the worst-case power dissipation, which would defeat the whole purpose of trying to balance things. Adding drain resistors wouldn't do as much to improve balance, but shift heat dissipation away from the IGBTs. \$\endgroup\$
    – supercat
    Commented May 29 at 15:02
  • \$\begingroup\$ I added in the answer. Note that I have used the OP IGBTs. Must verify that they are ok. \$\endgroup\$
    – Antonio51
    Commented May 29 at 17:27

You haven't shown the layout, so here's a hypothesis:

enter image description here

If it is wired exactly as drawn, in the top schematic, the left IGBT has much more trace length resistance in series than the one on the right, so current will not be shared evenly.

On the other hand, in the bottom schematic, trace length is equal for all IGBTs, so we can expect much better current sharing.

Given the amount of current you're using, every milliohm counts.


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