0
\$\begingroup\$

Let's say I have three phase voltage source inverter i.e. following circuit

enter image description here

I have a suspicion that one or more IGBTs are damaged. To confirm this hypothesis I have decided to use multimeter in ohmmeter setting and I have measured the resistance between the collector and emitor terminals of the individual IGBTs. I have measured resistances between:

  • terminal 30,31,32 and terminal 2 for IGBT in upper row on the left
  • terminal 30,31,32 and terminal 6 for IGBT in upper row in the middle
  • terminal 30,31,32 and terminal 10 for IGBT in upper row on the right
  • terminal 27,28,29 and terminal 4 for IGBT in lower row on the left
  • terminal 24,25,26 and terminal 8 for IGBT in lower row in the middle
  • terminal 21,22,23 and terminal 12 for IGBT in lower row on the right

My multimeter displayed follwoing values of the resistance

  • OL
  • \$22.95\,\mathrm{k}\Omega\$
  • \$23\,\mathrm{k}\Omega\$
  • \$7\,\mathrm{M}\Omega\$
  • \$5.7\,\mathrm{M}\Omega\$
  • \$4.8\,\mathrm{M}\Omega\$

Does it mean that all of the transistors are ok (because none of the measurements displayed \$0\,\Omega\$ or some value close to that) or is there any mistake in my measurement (I expected that all the measurements should be OL in case all of the transistors are ok)?

EDIT:

I have done another measurement with multimeter configured in diode test mode and I have measured voltage drop between

  • terminal 2 (positive probe) and terminal 30,31,32 (negative probe) for IGBT in upper row on the left
  • terminal 6 (positive probe) and terminal 30,31,32 (negative probe) for IGBT in upper row in the middle
  • terminal 10 (positive probe) and terminal 30,31,32 (negative probe) for IGBT in upper row on the right
  • terminal 4 (positive probe) and terminal 27,28,29 (negative probe) for IGBT in lower row on the left
  • terminal 8 (positive probe) and terminal 24,25,26 (negative probe) for IGBT in lower row in the middle
  • terminal 12 (positive probe) and terminal 21,22,23 (negative probe) for IGBT in lower row on the right

My multimeter displayed follwoing values of the voltage drops

  • \$0.03\,\mathrm{V}\$
  • \$0.374\,\mathrm{V}\$
  • \$0.374\,\mathrm{V}\$
  • \$0.374\,\mathrm{V}\$
  • \$0.374\,\mathrm{V}\$
  • \$0.000\,\mathrm{V}\$
\$\endgroup\$
3
  • \$\begingroup\$ I find resistance measurement dubious when it comes to semiconductors. Set your multimeter to diode test mode and fill in the values for that too in the table above. Please note that it’s polarity sensitive. \$\endgroup\$
    – winny
    Jun 30, 2021 at 12:08
  • \$\begingroup\$ @winny thank you for your reaction. I have just attempted to do the measurement you have suggested. Please see the edit in my question. \$\endgroup\$
    – Steve
    Jun 30, 2021 at 12:38
  • \$\begingroup\$ Thanks. I can’t correlate which one is which but 0 V = shorted and 0.37 V = body diode a d probably fine. \$\endgroup\$
    – winny
    Jun 30, 2021 at 12:59

1 Answer 1

2
\$\begingroup\$

Something is wrong with your upper left and lower right IGBTs; both of them appear to show no antiparallel diode. I'm not sure why they don't show low impedance in the opposite direction, though. When you say you measured the resistances between two nodes, do you mean you put the positive probe on the first and negative on the second? It does matter with semiconductor devices.

Another test that's good to make is the gate to emitter and gate to collector impedance--if it's anything less than hundreds of MΩ, something is probably wrong. If it's less than a kΩ, something is definitely wrong.

\$\endgroup\$
3
  • \$\begingroup\$ I would say probably something is wrong. Gate drive and any CE snubbers may put you below 100 Mohm with nothing being wrong. The schematic above is just a simplification. \$\endgroup\$
    – winny
    Jun 30, 2021 at 15:00
  • \$\begingroup\$ Okay, fair--I should clarify that a little. \$\endgroup\$
    – Hearth
    Jun 30, 2021 at 15:23
  • \$\begingroup\$ Much better! :-) \$\endgroup\$
    – winny
    Jun 30, 2021 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.