0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Consider the above circuit, (the values are not correct). When I want to turn off the MOSFET, the output of the driver would be 0. If I have a large di/dt and a large parasitic source inductance, the voltage across L2 would be large and the driver won't be able to close the MOSFET, at least it will oscillate betwenn turn on and turn off until the di/dt be not enough significant for producing a voltage Vgs superior to Vdriver-vth ? Is this possible ?

It will help to turn off ...

Thank you very much.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ You could improve your question a lot by better annotating the parts and at least bring the inductances to a reasonable relation. Load inductance isn't probably the same size as wire inductance. And btw. which one is supposed to be the load? (I assume L1) \$\endgroup\$
    – Ariser
    Feb 15, 2020 at 12:05

1 Answer 1

Reset to default
1
\$\begingroup\$

The inductance in the source won't help to turn the MOSFET off; as soon as the gate voltage drops, the source voltage drops (due to L2 back emf and will fall below 0 volts) and, this ensures that the MOSFET stays on longer until the energy previously stored in L2 is depleted. It won't oscillate unless L2 has significant parasitic capacitance and Q factor.

\$\endgroup\$
1
  • \$\begingroup\$ Ok that was exactly what i thinked this morning and when I took my shower and I was telling that I was wrong and i came back to close the subject ... Thanks again ! \$\endgroup\$
    – Jess
    Feb 2, 2020 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.