0
\$\begingroup\$

I need to drive a high side N-channel IGBT continuously. The circuit is essentially a passthrough, and the only thing I want to do is turn this pass through line on or off (usually if not always on) using a MOSFET. The line is 500V and running around 20A.

Using an external power source of 15 volts to drive a FAN73711MX driver, will I still be looking at the Cboot discharging and causing disturbances in my power flow on the line, or will this work at a zero switching frequency. If not what can be done?

Continued: So I Have added a charge pump circuit to the design, as well i have switched from MOSFET to IGBT because it makes more sense.

At this high of a voltage I have not been able to find any other solutions to this problem. Will this circuit work or is there another option that I should look into.

The power source is coming from an off the wall 15v supply, and the 500V lines are being stepped down from a PV array.

enter image description here

\$\endgroup\$
7
  • 1
    \$\begingroup\$ You cannot design a high power switch without full specs on source, line & load impedance , surge current, and immunity requirements with test acceptance criteria for all ingress and egress disturbances otherwise possible failure \$\endgroup\$
    – Tony Stewart EE75
    Feb 16, 2019 at 18:54
  • \$\begingroup\$ What is the power source? \$\endgroup\$
    – Bruce Abbott
    Feb 16, 2019 at 19:34
  • \$\begingroup\$ Post a schematic showing the FAN73711MX and the FET. I'm guessing from the wording of your question that you're powering the FAN73711MX from a ground-referenced 12V supply, which means that the gate voltage will only come from the Cboot. In that case, yes, Cboot will discharge, and the thing won't work. You need an isolated gate power source that swings with the FET source voltage. \$\endgroup\$
    – TimWescott
    Feb 16, 2019 at 20:11
  • \$\begingroup\$ @TimWescott sorry for the delay, I have been working on a solution and posted it here. Hoping that this circuit will work maybe? \$\endgroup\$
    – Les
    Feb 18, 2019 at 7:31
  • \$\begingroup\$ @SunnyskyguyEE75 You are absolutely right about that, and there are several factors that are going into the design to meet the required specifications of the power source. My question was about the boot capacitor discharging over time as i need to run the switch continuously. \$\endgroup\$
    – Les
    Feb 18, 2019 at 7:37

1 Answer 1

Reset to default
0
\$\begingroup\$

VB is a power source for the gate of the IGBT, and must always be higher than the emitter by enough to turn the gate on. The type of bootstrap you have drawn in depends on VS going to zero periodically so that your +15V can charge CBoot. Otherwise Dboot is always back biased so you will turn off the IGBT after CBoot discharges. You can make a simple power supply for VB with a transformer. Set the turns ratio so that you get at least 10-12 volts or so on CBoot. This can be a free-running oscillator because VB can always be present; the gate is always controlled by the input.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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