0
\$\begingroup\$

I am attempting to design a powerful, yet efficient soft switching fullbridge converter. Achieving that requires high side switching with an N Channel MOSFET.

Now while I have succesfully achieved a drain to source voltage drop of but 0.2mV when a stable dc voltage is applied. However, when a switching signal of 100kHz is applied to the gate of the MOSFET, via a Darlington transistor (see diagram below), strange things start to happen. With strange things I mean:

This is my first attempt at high side switching with an N Channel MOSFET and I am completely lost on what I should do to fix this. Why is it not a pretty square wave like the signal? My guess is that the gate is charging too slow.

As additional information, this is the circuitry of the test setup: circuit

I hope someone can help me with this. Thanks in advance!

\$\endgroup\$

1 Answer 1

Reset to default
3
\$\begingroup\$

The only means you have of charging the gate source capacitor (possibly somewhere in the realm of 1 nF), is a 4 kohm resistor. Go work how long this takes to charge. You need a push pull driver for high speed switching.

\$\endgroup\$
18
  • \$\begingroup\$ A push-pull driver, ofcourse! I'm going to try it out now! Thank you! I'll come again if it works! \$\endgroup\$
    – Azagro
    Feb 9, 2017 at 21:24
  • \$\begingroup\$ Hold on, the 30 volt supply for the gate is too high. Got for 15 volts or you'll pop the mosfet \$\endgroup\$
    – Andy aka
    Feb 9, 2017 at 21:27
  • \$\begingroup\$ I may have rounded the voltages to their nearby tens, but i'll make sure not to cross the +/- 20V difference! Thank you. \$\endgroup\$
    – Azagro
    Feb 9, 2017 at 21:36
  • 1
    \$\begingroup\$ what is being suggested is what the datasheet permits and there is a very high chance it is 20V max and thus 15V is recomended \$\endgroup\$
    – user16222
    Feb 9, 2017 at 21:42
  • 1
    \$\begingroup\$ ok I follow you. You need a bootstrapped push pull driver. Fairchild and international rectifier make them but there will be others. They steal a bit of output power and create a higher supply rail automatically. \$\endgroup\$
    – Andy aka
    Feb 10, 2017 at 10:14

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.