2
\$\begingroup\$

enter image description here

I want to drive a MOSFET with a 50Hz, 50% duty cycle clock pulse. The voltage to be switched is as high as 300V DC. The section inside the dashed lines will stay electrically isolated from the rest of the circuit. I am planning to use an optocoupler for isolation (I'm open for any other suggestion).

enter image description here

Best solution I can find is using a voltage divider network (as seen in the second schematics) and control it with the optocoupler. But I don't think that this is an optimal solution, because there are two major problems.

  1. I have to use a bulky power resistor for R1.
  2. Switching losses. Gate capacitance of a typical MOSFET is about 5nF. Time constant for a 5nF capacitor charging over a 20k\$\Omega\$ resistor is 100\$\mu\$s which is 0.5% of the switching period. The maximum current through RL will be 10A, and the Rds,on resistance of the MOSFET will be 50m\$\Omega\$ at most. So the switching loss will approximately be
    \$(10A)^2 \times 50m\Omega \times 0.5\% = 25mW\$.

I can increase the values of R1 and R2, but this time the switching loss will increase. There is a trade off. Either way there will be a heat source in the circuit, and some energy will be lost.

What is the efficient way of driving a MOSFET like this?


EDIT: What about this circuit?

enter image description here

With enough dead-time between on states of the two optocouplers, the required rated power of the R1 resistor may stay below 1W. Do you see any problem in this new circuit model?


EDIT2: I decided to put external darlington transistors for supplying more current to the MOSFET gate.

The gate voltage will drop down to 1.4V at low state, at which most MOSFETs are completely turned off. Is this new circuit feasible?

enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ Switching loss calculation looks suspect : Rds(on) will be 50mohm WHEN FULLY ON, but MUCH higher during the 100us switch-on period. \$\endgroup\$ – Brian Drummond Apr 6 '14 at 11:54
  • \$\begingroup\$ Here's another problem with your circuit: you have drawn a phototriac, which will never switch off with the DC load you have drawn. \$\endgroup\$ – Phil Frost Apr 6 '14 at 12:13
  • \$\begingroup\$ Also, you can search "isolated gate driver" for hundreds of canned solutions. \$\endgroup\$ – Phil Frost Apr 6 '14 at 12:18
  • \$\begingroup\$ Re your new circuit - driving the FET with the output from an opto might work but optos are limited on how much current they can sink/source and you still have the basic problem to overcome of charging the gate-source capacitance of 5nF. \$\endgroup\$ – Andy aka Apr 6 '14 at 15:04
  • \$\begingroup\$ Your charge time calculation has at least one error. The Thevenin equivalent source resistance of the voltage divider is 952 ohms, not 20K ohms. Big difference. Secondly, you should use gate charge in nC for rather than capacitance. This could also be a big difference (the other way). \$\endgroup\$ – Spehro Pefhany Apr 6 '14 at 17:36
4
\$\begingroup\$

Further down I'm suggesting something that should work in a lot of applications but maybe not yours (due the the high gate capacitance of the MOSFET). Because of that I'd consider using a DC-DC isolation module to provide isolated power to gate circuits at the MOSFET. Here's a circuit that is 90% indicative: -

enter image description here

B2 would be supplied by the DC-DC converter and U1 can be isolated as well although in this particular diagram they show both sides of the coupler grounded (directly below U1). DC converter like this would be OK: -

enter image description here

It can continuously supply 250mA and this can be used in conjunction with a reasonably sized electrolytic capacitor to provide the high current surge needed to charge the gate.

I'd also consider the simplicity of using a photovoltaic opto isolator such as the Vishay VOM1271.

enter image description here

It can switch on in 53 us into a 200pF load and produce a drive voltage of about 8V making it suitable for a lot of MOSFETs. Of course if the MOSFET gate capacitance is 2nF then it will take about 0.5 milli seconds to turn on.

\$\endgroup\$
0
\$\begingroup\$
  • In your diagram you show a optocoupler-triac, which is not appropriate here (usefull with AC only).

  • Note that whatever you have in your optocoupler, the secondary must be able to handle the full 300V!

  • For your charging calculation, you must assume that the 5nF is charged from a (1/21)*300 = 14V source with a 1//21 ~= 1k resistance. You seem to assume a 20k resistance. Note that you have to take the discharge into account too!

  • A 'discrete' solution could be to create a 14V supply at in the HV area, with a big capacitor, and use a conventional optocoupler and gate driver.

You don't mention the frequency you want to achieve, which might be an important factor in the design.

\$\endgroup\$
0
\$\begingroup\$

Have you considered using a transformer? At high freqeuncies, you could use a small pulse transformer, but at 50 Hz, you'd want a good 1:1 transformer designed for audio frequencies.

Connect the scondary between the source and gate of the MOSFET. Drive the primary with the signal you want to apply to the gate. Also connect a diode from source to gate to limit the negative swing on the gate.

\$\endgroup\$

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.