Quote:
Originally Posted by MPaulHolmes
On their calculator, what do they mean by the inductor on and off voltage? If input voltage was 200, and output was 600, would I use those 2 values?
edit: I guess inductor ON voltage would be the pack voltage. And off voltage would be the voltage across the inductor while the switch is off? How awkward.
I was looking at some SiC half bridges on Mouser. They had a 300amp 1200v 5mOhm RdsON mosfet half bridge, but it was expensive, and the switching losses weren't small. They had 200KHz listed as a sample switching frequency, but then looking at the losses when it was an inductive load, each switch would have about 1500w of waste heat. And the peak power dissipation was 1600w.
The CH-200 I bought awhile back is under $300 in quantity 10. Maybe that is good enough. 200amp continuous
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I am using the PFC calculations. Which one are you using?
Just make sure the maximun input voltage is in AC, so for DC divide by 1.41 (144V ~ 100VAC).
Those values for the mosfet seem correct. I'm assuming they tested at 600V with a 100A load or something. Whats the RDS(on)? IGBT is probably best in terms on conduction losses at high currents, but switching losses are a bit higher as it is slower.
So, we have conduction losses and switching losses:
Assume 200Amps for the inductor
PD(conduction) = VCE/VGE(sat)*Io*Duty(on)
So for a 200A and a voltage drop of 1.7(assumed) and a duty = 50% that would be 170W
Now switching, depends on the frequency and the RDS(on)
PD(sw) = (Vd*Io*F(sw)*(t_rise+Tfall))/2
PD(sw) = (144*200*20KHz*200ns)/2
PD(sw) = 28.8W
here I'm assuming a rise and fall time of 200ns
At 200Khz, that jumps to 280watt! But the resistive and core losses on the inductor probably will go down twice as much.