-1
\$\begingroup\$

Lately I've been tinkering with an SSTC half-bridge driver. I've ran a couple of tests using relatively low input voltages ~50V and the IGBTS seemed to do fine, there was no noticeable heating at all. When I used rectified mains to power up the circuit it works fine for a few minutes and after that the transistors violently explode.

The gates should be pretty well protected: 15V Zenners, 10k pull down resistors, 10 Ohm resistor to prevent ringing and diode for discharge. The driver IC is an IR 2184.

I am wondering if the culprit for the failure at high voltage might be the lack of a DC blocking cap across the primary thus in the event that both IGBTs are on it shorts the power supply creating huge current. Heat is not a problem, the thermocouple showed them at 50 degrees C right before the explosion.

Any help is greatly appreciated, I don't know what else might be wrong with the circuit and I want to ask for a second opinion before plugging in my last 2 IGBTs. Thank you very much for your time.

This is the circuit I use, it's a combination between Great Scott's driver and Loneoceans SSTC2

Circuit used

\$\endgroup\$
  • \$\begingroup\$ Just to point out that the 1N4007 diodes across the 10 ohm gate resistors should be the ultra-fast 1N4148 4ns diodes. The 4007 series is very slow so maybe both MOSFET's are briefly ON at the same time, allowing shoot-through currents. \$\endgroup\$ – Sparky256 Dec 12 '18 at 22:47
  • \$\begingroup\$ I missed that, thank you very much for pointing it out! Gonna swap those asap. \$\endgroup\$ – Rpreda Dec 12 '18 at 23:06
  • \$\begingroup\$ You refer to IGBTs but IRFP260 are MOSFETs. \$\endgroup\$ – Steve Hubbard Dec 12 '18 at 23:48
  • 1
    \$\begingroup\$ @Sparky256 UF4007 is the ultrafast "version" of the 1N4007. 50ns or so trr vs. microseconds, so they're not really comparable. The part name similarity is a clever marketing gimmick. \$\endgroup\$ – Spehro Pefhany Dec 13 '18 at 17:01
  • 1
    \$\begingroup\$ @SpehroPefhany Thanks for this, I need to update my answer then \$\endgroup\$ – DerStrom8 Dec 13 '18 at 17:02
0
\$\begingroup\$

A few comments:

  • DVCC1 and DVCC2 are not in the correct location. They need to be connected directly across the IGBT collectors and emitters. Where they are located now makes them entirely useless - they are on the wrong side of the "output".
  • Addition of a ~4.7uF MKP high-voltage DC-blocking capacitor would definitely be a good idea as well, to prevent any damage in the case of latch-up, meaning the high voltage DC supply is shorted directly through the primary coil to ground.
  • You should be using a gate drive transformer (GDT) between the driver IC and the IGBTs. For a half-bridge, this is usually a single transformer with one primary and two secondaries. Look at loneocean's design. You can also use Mads Barnkob's design (from Kaizer Power Electronics) for reference: http://kaizerpowerelectronics.dk/wp-content/gallery/2011_08_14_-_kaizer_drsstc_ii/kaizer_drsstc2_bridge.jpg
  • What are you using for a mains rectifier and filter capacitors? Both will need to be quite beefy and capable of handling high current spikes. Adding them to the schematic would really be helpful, rather than just calling the high voltage DC supply "VCC".
  • You do not say what you are using for heat sinks on the IGBTs. You will need very beefy heat sinks, and probably forced air cooling too.
  • You should consider adding capacitance to the inputs of the 5V and 12V regulators. Any inductance in the connections between their inputs and the controller's DC power supply will cause instability
  • Your IGBTs are relatively slow, especially when driven from only 12V. You really should be driving them at around 15-18V. The IR2184 should be able to handle up to 20V, which is the absolute maximum VGE of the IGBTs, though I don't recommend driving them at maximum rating. If you do increase your drive voltage, make sure you also adjust the values of DZ1 and DZ2 appropriately. Since the maximum VGE stated in the datasheet is 20V, you should shoot for 20V bidirectional zeners (you are showing unidirectional, which are a no-no).
  • When driving an h-bridge or a half-bridge, it is very important that the first transistor shuts off before the next one turns on. If the first one doesn't turn off completely, you present a dead short of the full supply voltage directly across them, which causes them to explode. Knowing the timing of your switching is crucial.
  • Get an oscilloscope and measure across the gate-emitter connections and look at the waveform. This can be done with a low-voltage input (~30V). If the gate signal is underdamped (lots of oscillations), then your gate resistance (RGH and RGL) may need to go up, but that will slow down the switching speed. If the signal is overdamped (a slow, smooth waveform) then you need to reduce the resistance. The goal is to get as close to a square wave on the G-E connections as possible. See Richie Burnett's GDT page for more information: http://www.richieburnett.co.uk/temp/gdt/gdt2.html . Make sure your oscilloscope is isolated from mains through an isolation transformer, otherwise probing the high-side transistor could damage it.
  • You should be following loneocean's design exactly, not GreatScott's. Scott's design is for a full H-bridge, which behaves differently from the half-bridge. You should not be using it for reference. The only exception is the addition of DZ1 and DZ2, which was a good idea.
  • As always, be EXTREMELY CAREFUL when probing high voltages, and observe all safety precautions. Make sure you know EXACTLY what you are doing. We will not be held responsible if you go and get yourself injured or killed - Please be smart.
\$\endgroup\$
  • \$\begingroup\$ Thank you so much for the answer. For rectifying mains I am using an old big full-bridge rectifier and 1.5mF of 450V electrolytic caps. Heatsinks used are old computer cpu heatsinks that were drilled. Thanks again for all the points brought up, I'll make the necessary adjustments and study some more. \$\endgroup\$ – Rpreda Dec 13 '18 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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