I have designed a 50 W, 12V-5V buck converter operating at switching frequency 200kHz.This is just for understanding practical converter circuits, so I made a totem-pole bjt driver circuit for IRF640N.MOS

Can someone help me out with the isolation required for the MOSFET? At first I had placed the switch in the return path, but this design is for closed loop control, so I think I might have trouble with feedback noise if I don't place it in the go-path. I know there is pulse-transformer isolation & digital isolation, but I'm not sure how to incorporate that.

EDIT: Parallel to trying to do the isolation part, I was looking for a gate driver IC, but I'm stuck at how to choose one.Any inputs on that front is helpful too.

  • \$\begingroup\$ You only have Vgson = 3V on the IRF640 (15-12). You need 6V or more - see datasheet || Isolation is not neceeesary - as long as you protect your uC against driver failure (or don't care). The "easy" way by far is to use a P Channel high side FET. You then need no highside supply and the driver is a pull down - which we can help with if or interest. \$\endgroup\$ – Russell McMahon Jul 23 '20 at 12:12
  • \$\begingroup\$ The second circuit in my Jan 2010 SE answer here was provided by Olin Lathrop. It both level shifts and controls the Vgs swing and level of the high side gate. Not complex but be sure you understand its 'clever trick'. \$\endgroup\$ – Russell McMahon Jul 23 '20 at 12:17

One option that I have successfully used is a bootstrap MOSFET driver or a half bridge driver with the HIN input controlling the high side MOSFET. e.g. LM5109 from TI. This will allow you to use a high-side N-MOSFET. An example circuit is shown below using an half-bridge driver. To modify this for a buck converter, forget the LIN input and can omit the power MOSFET Q2 during normal operation. Use a power diode in place of Q2.

When the diode is conducting, i.e. the high side switch is off, then Cboot gets charged up to 15V (you could also use a lower voltage than 15V, just ensure that it is higher the the gate threshold voltage of the MOSFET).

When the high side switch is on, the supply for the MOSFET gate drive comes from the boot capacitor, which has a potential of 15V higher than the source.

The only caveat with this circuit is the maximum duty cycle is limited to approx 95-98%, since the boot capacitor must have time to charge up. You also need a small npn instead of Q2 that you briefly turn on at power up to precharge the boot capacitor.

enter image description here

  • \$\begingroup\$ I understood the idea, but I don't quite get what I need to replace Q2 power MOSFET with.....I think a diode would do, why an npn though? \$\endgroup\$ – SM32 Jul 23 '20 at 15:02
  • \$\begingroup\$ Also, I think this should work as a gate driver UCC5304(ti.com/lit/ds/symlink/…) Could you check it out and tell me if I should change it? \$\endgroup\$ – SM32 Jul 23 '20 at 16:26
  • \$\begingroup\$ A diode is simpler rather than a MOSFET for Q2. You just need to generate one signal. Otherwise you need to generate 2 signals and ensure interlocking, i.e. turn off Q1 before turning on Q2. The npn (or small MOSFET) is just in parallel with the diode and allows the system to be kickstarted \$\endgroup\$ – mr_js Jul 23 '20 at 20:29
  • \$\begingroup\$ @SM32 - If you use the UCC5304 you need to provide an isolated supply for Vdd on the isolated side of the driver. \$\endgroup\$ – mr_js Jul 23 '20 at 20:32

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.