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IGBT power consumption

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Actually, most of the power will be due to how the ON resistance of the FET effects the driven internal bipolar transistor at any given frequency and power. People seem to think of IGBT's as a simple device, when they are in fact compound devices consistance of the driving FET internal biasing and the BJT itself. The formula to calculate IGBT power losses is probably isanely complex, and would require detailed knowledge of both the materials and structure of the IC itself, IGBT's are IC's not "trasistors" in the classic sense. Much the same as a Darlington pair, only the initial input stage is a FET instead.
 
The conduction losses are defined as follows (see ploss_igbt.png) :
 

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That equation doesn't take the FET itself into considering and it's switching losses and how it's linear range effects the BJT. So it's only valid if the IGBT is used as a basic on/off switch at very low frequency. As frequency increases past low audio your real losses will increase dramatically from that equation.
 
Of course, I referred to a simple basic application. (The conduction losses...)

It is well known that power losses depends on frecquency, the switching losses are increasing proportionally to the frecquency, the actual limits to the switching frequency are set by the switching losses.

Total Power Losses = Static Losses + Switching Losses + Driving Losses
Static Losses = On-state Losses + Blocking Losses
Switching Losses = Turn-on Losses + Turn-off Losses

Forward blocking losses and driver losses may usually be neglected.

IGBT total power losses = On-state power dissipation + turn-on power dissipation + turn-off power dissipation.
 
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