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Old 2nd September 2010, 09:49 AM   #1
kall is offline kall  Sweden
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Default Compare IGBT-losses and MOSFET-losses

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
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Old 3rd September 2010, 03:06 AM   #2
star882 is offline star882  United States
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In general, MOSFETs are better at low voltages and high frequencies while IGBTs are better at high voltages and low frequencies.
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"Fully on MOSFET = closed switch, Fully off MOSFET = open switch, Half on MOSFET = poor imitation of Tiffany Yep." - also applies to IGBTs!
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Old 5th September 2010, 06:13 PM   #3
kall is offline kall  Sweden
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yes, thats true...but i need to make some calculations on this aswell...

so... how do u calculate accurate mosfet-losses (cond- switching-losses) with the data that is usually given in the datasheets for mosfets...
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Old 5th September 2010, 10:17 PM   #4
star882 is offline star882  United States
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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.
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Old 6th September 2010, 05:33 AM   #5
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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.

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Old 6th September 2010, 02:19 PM   #6
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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.

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Old 6th September 2010, 05:31 PM   #7
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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.
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Old 6th September 2010, 06:21 PM   #8
kall is offline kall  Sweden
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Quote:
Originally Posted by darkfenriz View Post
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??
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Old 7th September 2010, 03:26 PM   #9
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Quote:
Originally Posted by sawreyrw View Post
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
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Old 7th September 2010, 09:53 PM   #10
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Tim,

The BJT in the IGBT does not saturate, because the collector base junction is never forward biased. Thus the traditional BJT storage time is near zero.

Rick
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