Compare IGBT-losses and MOSFET-losses

kall

Member
2010-02-01 1:35 pm
When it comes to calculate switching losses for IGBT it is fairly simple because manufacturers give you Eon and Eoff of the switch..

ex. Pon= Eon * U/Uref * I/Iref *fsw

This is not given for MOSFETs, how can u calculate the mosfet-losses in a way to have a good comparison against IGBTs
 

star882

Member
2007-03-19 8:41 pm
RDSon is the resistance of the MOSFET when it's on. Calculate that power loss exactly as for a resistor (P=I^2 * R).

Dynamic losses are harder to calculate and will depend on the circuit. You'll want to drive the gate with a low impedance to minimize the amount of time the MOSFET or IGBT spends in the half-on "Tiffany Yep" mode, but not low enough to cause destructive ringing. Driving the gate negative to turn it off can help accelerate switching by counteracting stray capacitance.
 
Oh, and if you'll be using MOSFETs anywhere near rated power, calculate junction temperature and find the resulting change in Rds(on). Transistors over 60V have a strong positive tempco (typically double at ~125C). This results in twice the power dissipation compared to the 25C figure, not something you want to forget.

Notably, transistors under 30V have a very low tempco (~1.2x at 125C), so if you're building a very low voltage power supply, it's worthwhile to get 20 or 25V transistors instead of 30 or 60V types.

Negative gate drive is useful for devices with stored charge, like IGBTs. Whereas MOSFETs turn off essentially as soon as the gate falls, IGBTs take tens of ns to turn off. After the delay, Vce can rise again, inducing miller charge into the gate, making it "chatter". Reverse bias keeps the gate further from Vge(th), preventing miller turn-on.

Tim
 
Tim,

You are confused about IGBT and MOSFET gate charge (input capacitance). Both devices have Ciss, Crss and, of course, exhibit the Miller effect. It takes current and time to charge and discharge the gates. Generally speaking IGBTs will have larger capacitances, because they are large devices.

Rick
 

kall

Member
2010-02-01 1:35 pm
A first approximation is:
1/2*ton*fsw*U*I, same for switching off.
Switching times are more dependant on gate drive, compared to IGBT which have some internal slowdown mechanisms like current tail.

is this really a good approx?, i found this very simple...

-------------

i have a case with 100kHz 20V 210A (i guess MOSFETs are the only option here?)

any rule of thumb what "size" to choose to have a good "safety margin", go up to 40V, 60V??
 
Tim,

You are confused about IGBT and MOSFET gate charge (input capacitance). Both devices have Ciss, Crss and, of course, exhibit the Miller effect. It takes current and time to charge and discharge the gates. Generally speaking IGBTs will have larger capacitances, because they are large devices.

Ah, you're confused about my meaning of charge-- gate charge is easy to deal with because you can suck it away with the gate driver. "Stored charge" is a junction effect, and the cause of "storage time" in BJTs. As a hybrid, IGBTs have both effects. Stored charge means you can't turn it off arbitrarily fast, even if you force the gate off in just a few nanoseconds. The MOSFET part turns off in that time, but the BJT part continues to conduct for a few hundred ns (depending).

Incidentially, IGBTs of the same ratings as MOSFETs have lower gate charge, because the junction is smaller (higher current density). Another advantage to IGBTs!

Tim
 
is this really a good approx?, i found this very simple...

-------------

i have a case with 100kHz 20V 210A (i guess MOSFETs are the only option here?)

any rule of thumb what "size" to choose to have a good "safety margin", go up to 40V, 60V??

Which such a low input voltage and high frequency MOSFETs are the practical way to go. Safety margin will depend on the topology. for instance for Push-Pull the switching transistors need to have at least twice the device voltage rating. It also depends on how much inductance leakage your transformer (or inductor) will create and how aggressive your snubbers are. It might make sense to use higher voltage rated MOSFETs during your initial testing until you are able to measure inductance leakage spikes in a working circuit.

I think your first big problem will be laying our your circuit that is capable of sustaining 210 Amps and not having layout problems picking up excessive switching noise. 210A is going to require a lot of copper. FWIW: I would recommend doubling or tripling the input voltage to reduce the input current demand. If you haven't noticed, Copper is pretty expensive these days. Its going to cost you a small fortune in copper to pass 210 amps!
 

kall

Member
2010-02-01 1:35 pm
Which such a low input voltage and high frequency MOSFETs are the practical way to go. Safety margin will depend on the topology. for instance for Push-Pull the switching transistors need to have at least twice the device voltage rating. It also depends on how much inductance leakage your transformer (or inductor) will create and how aggressive your snubbers are. It might make sense to use higher voltage rated MOSFETs during your initial testing until you are able to measure inductance leakage spikes in a working circuit.

I think your first big problem will be laying our your circuit that is capable of sustaining 210 Amps and not having layout problems picking up excessive switching noise. 210A is going to require a lot of copper. FWIW: I would recommend doubling or tripling the input voltage to reduce the input current demand. If you haven't noticed, Copper is pretty expensive these days. Its going to cost you a small fortune in copper to pass 210 amps!

thx for answering...

im using a fullbridge topology with input 400V, i have 20V on the low-voltage side... i was thinking of using sync. rect.(MOSFETs) instead of diodes for rectifying to reduce my losses..