[SOLVED] Computing bootstrap capacitance

Hello,

I'm trying to calculate the bootstrap capacitance for my MOSFET driver, but my result seems way too low. I'm using application note AN978 from IR, which gives this equation:

C = 2(2Qg + Iqbs/f + Qls + ICbs/f)/(Vcc - Vf - VLS - Vmin)

The values I'm using are:
Qg - MOSFET gate charge - 65 nC
Iqbs - Max VBS quiescent current - 230 uA
f - Operating frequency - 10 kHz
Qls - Level shift required per cycle - 5 nC
ICbs - Capacitor leakage current - 0 A (tantalum capacitor)
Vcc - Supply voltage - 12 V
Vf - Voltage drop across bootstrap diode - 0.7 V
VLS - Voltage drop across low-side MOSFET - 0.06 V
Vmin - Min voltage between Vb and Vs - 0.3 V

Putting it all together, I get a capacitance of about 0.03 uF. This is much less than any bootstrap capacitor I've seen in a MOSFET driver. Do any of my values seem off? Are there better guides to use for determining the capacitor's value?

Thanks.

You misunderstood the meaning of Vmin. Vmin must be at least higher than the maximum Uvlo threshhold, e.g. 9.4 V for IR2110, otherwise the driver can switch-off unintentionally.

Thank you. That increases the computed capacitance from 0.03 uF to 0.17 uF.

Still, the capacitance values I've encountered are usually between 5 uF and 30 uF. Is there a justification for these large values?

Thanks again.

Larger values would be used if the PWM frequency is lower, or if the low side may be inactive during a longer period, e.g. with "flat top" modulation schemes or if overmodulation can be expected.