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I am driving an inductive load with an inductance \$L = 500\mu H\$ with a steady-state current of \$I = 300A\$. I would like to switch this current off quickly, and I plan on using an IGBT to do so. Assuming that the time constant of the load \$ \tau = L/R \$ is much longer than the switch off time of the IGBT, we can assume that upon switching off the IGBT, the stored inductive energy \$ E = \frac{1}{2}LI^2 = 22.5J\$ will be almost entirely dissipated into the IGBT.

My questions are:

1) Is my logic above correct, i.e. in assuming that in this case there will be \$E = 22.5J\$ dissipated into the IGBT.

2) What spec in a specsheet determines what the maximum value of \$E\$ is before damaging the transistor?

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    \$\begingroup\$ If the IGBT is completely off, then the voltage across the inductor will rise until something gives, either the igbt or the insulation on your wiring. Don’t do that. \$\endgroup\$ – C_Elegans Apr 1 '18 at 2:00
  • \$\begingroup\$ Every load has finite inductance, so in principle the circuit I am describing above (a transistor switching a load) is extremely generic. I would think every transistor has specs as to what inductance it can switch quickly before "don't do it" becomes good advice. I would like to know how to find these specs. \$\endgroup\$ – user2640461 Apr 1 '18 at 2:14
  • \$\begingroup\$ You are creating a 'boost' regulator of sorts, but with no series diode or load to clamp the rise in voltage. I do not know what your goal is other than finding out the peak voltage limits of a IGBT. \$\endgroup\$ – Sparky256 Apr 1 '18 at 2:31
  • \$\begingroup\$ It depends on the type of IGBT-s used , diodes and if 2 level or 3 level and what snubbers if any are used. More details please. Obviously dumping 22.5J into waste heat into IGBT is not desireable, so choose a better design. \$\endgroup\$ – Sunnyskyguy EE75 Apr 1 '18 at 2:32
  • \$\begingroup\$ IGBTs may not be suitable as their turn off speed is much slower than a MOSFET. Read this paper as a starting point: ixyspower.com/images/technical_support/… \$\endgroup\$ – Jack Creasey Apr 1 '18 at 2:42
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A general bit of advice for reading data sheets. If you're looking for (let's say) the power consumption, which will be in amps, or the gain bandwidth product, which will be in Hz, it's far easier to run your eye quickly down the righthand 'units' column, looking for the relevant unit, than reading down the left hand measurement description column.

This energy dissipation will be measured in Joules, and it's about the only specification in those units. It's called either the 'Unclamped Inductive Switching Energy' (UIS) or the Collector-Emitter Avalanche Energy.

As the inductor current is interrupted, the voltage will rise. The IGBT structure is fairly robust, and avalanches stably at the breakdown voltage, as long as the energy dumped (assumed adiabatically (that is instantaneously, without any heat loss to the surroundings)) into the junction is limited to limit the temperature rise.

This example IGBT is a fairly small one, 20A 400V, designed for car ignition. It has a single pulse CE avalanche energy of around 400mJ. You might expect the spec to more or less scale with IGBT current, so it looks like you're looking for a >>1000A part before you start seeing 10s of Joules specified. This might be reasonable if you have a 300A load.

If you want to switch the current off quickly, then you need to allow the voltage to rise, and the higher the better. There are alternatives to dumping the energy in the IGBT junction.

The simplest is to put a resistor plus diode in parallel with the inductor. Arrange the resistor value R2 to develop a bit less than the IGBT's breakdown voltage at the inductor current, so that it limits the voltage across the IGBT. The current will not stop as quickly as for using a constant voltage breakdown mechanism, but heat in a big resistor is a lot easier to handle than heat in a small semiconductor junction.

A faster way (more time-averaged voltage provided) is to charge a capacitor C3. If it's big enough, the voltage will be fairly constant during the pulse, and it can be steadily discharged by a low power route, R3.

schematic

simulate this circuit – Schematic created using CircuitLab

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Yes your math is correct and only the Mitsubishi hybrid IGBT’s can handle this non- repetitive 300A current rating. So if the pulse is repetitive then it must not exceed T-junction max.

In this latter example external snubbing is used to redirect and share the Discharge energy

e.g. 3ph hybrid http://www.mitsubishielectric.com/semiconductors/content/product/powermodule/mosfet/a_series/fm600tu-2a_e.pdf

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Your question sounds like you would use the IGBT as a breaker. In this case, there is the problem of overvoltage u=L di/dt which is will probably destoy the IGBT. In solid-state breakers one adds a parallel varistor to limit the overvoltage to a defined value, and the varistor takes a larger part of the energy of 22.5J.

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