Why switching losses in IGBT become less when the voltage rate of the IGBT becomes less.

 

Why switching losses in IGBT become less when the voltage rate of the IGBT becomes less.

The longer the device is in the region between saturation and cutoff the greater the power dissipation. The faster the transition, the less time the power supply is "shorted"

 

For the manufacture of higher-voltage transistors, a semiconductor with a lower concentration of impurities and a larger thickness should be used. And such a region of the semiconductor will have a greater resistance. To restore small low resistance it is necessary to increase the size (area) of the transistor. This increases the parasitic capacitance. By the way. A smaller concentration increases the lifetime of carriers, which increases the loss on the "tail"

 

Thank you for your response.
Please note that the switching losses has relation with Ploss=fs*Vs*Is.
Take imagine in the first case study we have two parallel IGBTs in a circuit and in the second study we have two series IGBTs in the very circuit.
As you know the voltage of each IGBTs which are series with each other is half of the two parallels IGBTs voltage. However, the series IGBTs current is twice the value of each IGBTs current which are parallel each other.
in the series, IGBTs Ploss is fs* Vs/2 * is, and in the parallel, we have fs* Vs* i/2. Apparently the switching losses of each of the IGBTs, whether parallel or series, are equal. Notwithstanding, the switching losses of two parallel IGBTs are much more than two series IGBTs due to their extra voltage.
I am thankful if you say why the voltage rate is much more important than the current in switching losses?

 

You should realize that the derivative of the current can be very large in an open channel. That is a channel which is not "pinched-off". The derivative of voltage on the gate cannot change rapidly. It is constrained to small values due to the gate capacitance. Large currents can flow while the gate capacitance is slowly charged and discharged. You can also look at simulations to see that during charging and discharging the dV/dt can drop to very low values due to the Miller effect.

https://en.wikipedia.org/wiki/Miller_effect

 

Thank you so much. I found my answer. For calculating the IGBT switching losses we should notice on the IGBT Datasheet and we cannot use theoretical formulas. In fact, these formulas for calculating switching losses is correct for Mosfets.

Maybe it is due to the Miller effect, as our friend "Papabravo" said.


Regards