EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: strawberry on May 17, 2018, 12:20:44 pm
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FGH40T65SPD Td off = 38nS 1370pF
IKW40N65ES5 Td off = 184nS 2500pF
Do Infineon make slower transistors or is it the way they test them?
How to estimate "dead time" ?
How much gate drive current is sufficient for 15nF IGBT gate capacitance?
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The testing conditions are different but it looks to me like the Infineon part is slower. There is a trade off between storage time due to minority carriers in the bipolar element and collector-emitter saturation voltage so the Motorola part is faster with a higher saturation voltage and the Infineon part is slower with a lower saturation voltage.
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FGH40T65SPD Td off = 38nS 1370pF
IKW40N65ES5 Td off = 184nS 2500pF
Do Infineon make slower transistors or is it the way they test them?
How to estimate "dead time" ?
How much gate drive current is sufficient for 15nF IGBT gate capacitance?
How much gate current you need depends on the switching frequency
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Isn't that Infineon count antiparallel diode Trr into Td off?
For 100kHz (rise time 100nS), Vge = 12V, Cge = 6nF estimated driver current 0.72A seems low
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By the way, note why Td(off) is slower in the first place: when Vge is rising, it's going from 0 to 15V (or whatever they tested at -- I didn't look, it's usually this), and the Miller step falls at, say, 6V. So the gate charges through the total gate resistance (including driver, resistor and internal R_G) from a difference of 15-6 = 9V. When Vge is falling, the Miller step falls at maybe 3 or 4V, so the gate discharges through only 3-4V, or taking about three times longer.
You shouldn't drive IGBTs with Vge(off) = 0V, anyway -- besides slow turn-off, they're prone to shoot-through during positive dVce/dt. Better to use -5 to -15V Vge(off), which ensures sharp turn-off as well. :)
Tim
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Isn't that Infineon count antiparallel diode Trr into Td off?
For 100kHz (rise time 100nS), Vge = 12V, Cge = 6nF estimated driver current 0.72A seems low
Cge? Don't use that, that's a small signal parameter -- useless. Use Qg(tot). I(drive) = Qg * Fsw, simple as that. :)
(Note that Qg(tot) goes up if you drive more voltage swing, as I recommended above.)
Tim
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Qg(tot) is for loss calculation
turn off current must be 3*0.72A=2.16A
Is it possible to design dead time circuit without delayed signal?