The design of sic gate driver circuit is designed to work at 0 V as gate turn off voltage. Due to some undamped ringing at turn off, I want to change the gate turn off voltage at -2 or -4V which is generally recommended by all manufacturers to avoid false turn on due to low gate source threshold voltage. I wanted to ask how I can achieve that? Turning in the Rgon and Rgoff does not have any effect on this so option left is tuning in capacitor value but in this gate drive circuit, the only external capacitor added is between gate and source so should I tune that to achieve negative turn off voltage? Thanks
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1\$\begingroup\$ The naive way would be to provide an auxiliary negative supply. But I guess one could also employ a second negative-going bootstrap capacitor perhaps, because the Switch node often does swing a little bit negative. Just a thought 😊 \$\endgroup\$– tobaltCommented Dec 4, 2023 at 8:26
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\$\begingroup\$ If youbare using a pulse transformer then you can get some backswing for free \$\endgroup\$– AutisticCommented Dec 4, 2023 at 8:34
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1\$\begingroup\$ Can you include your existing gate drive circuit? \$\endgroup\$– colintdCommented Dec 4, 2023 at 8:57
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1\$\begingroup\$ @tobalt for medium/higher power isolated SiC MOSFET/IGBT drivers adding a negative supply is usually the way to do it. \$\endgroup\$– Andy akaCommented Dec 4, 2023 at 8:59
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1\$\begingroup\$ @tobalt you don't need the switch node to swing negative, just to have a negative slope at some point. Just like how you don't need the switch node to swing to higher than Vcc for a conventional bootstrap charge pump. \$\endgroup\$– HearthCommented Dec 4, 2023 at 14:20
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Simple approach is to drive gate through series capacitor C1 with capacitance so with R1 the Vx voltage at the end of Ton falls to about half.
During Toff you see the Vx voltage goes negative and would reach the 0V for very long time (much longer then Toff). So Vx stays in safety negative zone.
Note: The voltage levels in diagram are for standard mosfets so for SiC use datasheet values.
answered Dec 4, 2023 at 9:07
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\$\begingroup\$ This is highly duty cycle dependent because Vx must have a mean value of zero. So I doubt this can work in the simple way it is drawn here. \$\endgroup\$– tobaltCommented Dec 4, 2023 at 11:02