2
\$\begingroup\$

I understand why some transistors have a built-in (normal) diode from drain to source, so that when the transistor turns off, back-emf can flow back up, specially important in inductive charges, but why do some MOSFET like IRF540 have a zener? From what I understand, a zener diode conducts backwards with a specified drop voltage. I can only suspect that this zener has something to do with the absolute maximum voltage sustained between source and drain, specified in the datasheet, is that correct?

Also, I am trying to build this circuit: MOSFET gate drive using a transformer

But the transistor I am using as Q1 is currently IRF540 that has a zener diode instead of a normal one. As expected, it doesn't work very well. As proof of concept, with a normal diode (no Q1), Q2 I can successfully charge the gate and it lets current pass through.

I tried to put a normal diode in parallel with the intrinsic zener and it still doesn't work...

\$\endgroup\$
3
  • 1
    \$\begingroup\$ A diode in parallel won't be able to block current flow through the zener, unfortunately. You'd need to insert a diode in series, which is impossible as the zener is intrinsic. \$\endgroup\$
    – Jason_L_Bens
    Commented May 22, 2014 at 20:30
  • \$\begingroup\$ Ah, that's right, silly me. But why the intrindic zener in the MOSFET? \$\endgroup\$
    – rmarques
    Commented May 22, 2014 at 20:47
  • \$\begingroup\$ @Andyaka . The IR data sheet does show a zener of 100 volts. I suspect they used zener if the MOSFET and body diode don't have an avalanche rating. Thereby letting the zener absorb a transient and able to survive. \$\endgroup\$
    – Marla
    Commented May 22, 2014 at 20:55

2 Answers 2

Reset to default
6
\$\begingroup\$

I think you've got it mixed up. The body diode is a part of the structure of the MOSFET and does not help when switching an inductive load with a single MOSFET.

The zener in the symbol represents the avalanche rating of the MOSFET and is of use when switching an unclamped inductive load. You won't see it conduct unless you exceed the voltage rating \$V_{DS}\$ and get to \$V_{(BR)DSS}\$.

If you are actually getting +/-12V out of the pulse transformer, I don't see any obvious reason why that part would not work. Your negative swing out of the pulse transformer does have to be sufficiently large to turn the control MOSFET on, when loaded with both gate charges (-10V would be good).

\$\endgroup\$
6
  • 1
    \$\begingroup\$ +1 ... The zener symbol is often used for MOSFETs where the body diode is "Repetitive Avalanche Rated" \$\endgroup\$
    – Tut
    Commented May 22, 2014 at 20:53
  • \$\begingroup\$ I now know I am constantly frying Q1 from the schematic I show in my question. I am not exceeding maximum VDS, my power VCC is only 36V. Only one possible problem remains. In the primary of the gate driver transformer I'm putting 0V in one terminal and 5V in the other from an old computer power supply. I thought of not adding a capacitor as this was just to filter the signal, but maybe without the capacitor the voltage surge in the gate will exceed 20V for in a small instant, even if I supply the transformer with a 5V input... \$\endgroup\$
    – rmarques
    Commented May 22, 2014 at 23:43
  • \$\begingroup\$ You must not exceed the maximum gate voltage for even a small instant, though typically this kind of part won't actually die until far above the +/-20V maximum, so it's still a bit suspicious. If you're hitting it with 50V or 60V, that's your problem. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 22, 2014 at 23:52
  • \$\begingroup\$ @Spehro Will a capacitor in the primary make the signal safer, i.e. bounded to -5/+5V in the secondary side? \$\endgroup\$
    – rmarques
    Commented May 23, 2014 at 8:27
  • \$\begingroup\$ As I said above, you want about +/-12V out of the transformer for it to work properly. You could put a zener diode on the gate to keep it from getting to 20V. Something like 15V. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 23, 2014 at 10:18
2
\$\begingroup\$

The IRF540 has a body diode and not a zener AND the diode is a by-product of the manufacture and "internal wiring" of the MOSFET to its substrate. They weren't purposefully put in or fabricated alongside - it's a by-product an internal wiring thing that prevents another parasitic component (an NPN) causing irrational behaviour (from memory). See this.

\$\endgroup\$
2
  • \$\begingroup\$ Posted my comment in wrong place earlier. The IR data sheet does show a zener of 100 volts. I suspect they used zener if the MOSFET and body diode don't have an avalanche rating. Thereby letting the zener absorb a transient and able to survive. \$\endgroup\$
    – Marla
    Commented May 22, 2014 at 20:57
  • \$\begingroup\$ The IRF540 is repetitive avalanche rated, so the body diode does behave similar to a zener diode. The transistor will be fine as long as the avalanche energy doesn't exceed a few hundred millijoules (see figure 12c on the datasheet). \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented May 22, 2014 at 21:12

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.