Bizarre Power-MOSFET speedup notion...

Discussion in 'General Electronics Chat' started by Nik, May 20, 2006.

  1. Nik

    Thread Starter Well-Known Member

    May 20, 2006
    Okay, this notion is so daft that I'd like confirmation...

    One problem driving power-MOSFETs in switched mode is they have a significant gate capacitance, and don't begin to switch on until ~ 4 Volts G_S.

    Getting a steep edge on the capacitance-loaded drive pulse has all sorts of problems with ground-currents and EMI etc...

    What if you put a comparable varactor ( reversed-biased specialist diode with voltage dependent capacitance ) between G_S ?? At low GS, the varactor has a modest capacitance, swallows the slow rise of the driver. As V GS rises, the incremental capacitance falls sharply, current is switched to MOSFET gate improving speed.

    At switch-off, the falling GS voltage causes the varactor's capacitance to rise and suck charge from MOSFET gate, improving speed...

    Yeah, right...

    Please, help me shoot it down !!
  2. hgmjr

    Retired Moderator

    Jan 28, 2005
    Unless I have missed something in your description, it appears that you are connecting one end of your varactor to the gate and the other end of your varactor you are connecting to the source. The result is that the capacitance of the varactor is being place in parallel with the capacitance present at the gate of the FET. Since capacitance in parallel is additive (ex. 1uFd in parallel with 10 uFd gives you 11 uFd) then the hookup you suggest is actually increasing the capacitance into the gate of the FET.

    Hopefully, I have not misinterpreted you suggestion.

    The most effective way to drive the gate of an FET is to do so with a driver which has as low an output impedance at possible. That way the effect on the switching signal of the capacitance at the input to the gate will be minimal. The low impedance driver will provide an additional benefit in that it will minimize the effect of any rapid change in voltage at the drain on the gate drive signal caused by coupling back into the drive circuit through the FET's gate-to-drain capacitance.

  3. Nik

    Thread Starter Well-Known Member

    May 20, 2006
    Sure, at least to a first approximation !!

    Yes, I'm happy with KISS 'fast drive' strategy.

    But that finite rise time hitting varactor's voltage-dependent capacitance combined with the VGS threshold ties my brain in knots...

    Much baffled googling found arcane transformer-driven switching circuits that 'tune' their MOSFET gates for 'critical' damping: D'uh, not on my breadboard !!!
  4. richbrune

    Senior Member

    Oct 28, 2005
    Are you referring to an N-channel enhancement mode MOSFET? Have you considered an IGBT with soft recovery diode?
  5. n9352527

    AAC Fanatic!

    Oct 14, 2005
    In fact, this kind of idea (charge transfer) is already been used for quite some time. I've worked with BJT and IGBT where the respective base and gate capacitances were charged and discharged through charge transferred from external capacitors. These resulted in very fast turn-on and turn-off times.

    We also experimented with inductive based circuit to discharge the base or the gate capacitance quickly. It was better than the capacitance based circuit, but it was also more expensive.

    What you need to focus on is the turn-off time, tailing at turn-off is far worse than turn-on and most of the wasted power is associated with it. Especially for IGBT. Both BJT and MOSFET repond very favourably to turn-off speed-up circuit based on charge transfer or inductive back kick.

    Oh... your varactor idea, to change the capacitance of the varactor you have to change the voltage first. And to change the voltage you have to discharge the gate capacitance, plus the additional capacitance exposed by the varactor. So.. a varactor in itself across the GS would not improve the discharge/charge time, in fact it would make it worse. You need an external circuit to transfer the charge quickly to see any improvement.
  6. billbehen

    Active Member

    May 10, 2006
    Nothing wrong with a superlow impedance drive, if only you can get one that will do the current, which is anything but trivial. Also, there is an efficiency problem: All a voltage source driver can do is connect, via an 'ideal' switch the gate capacitance to an 'ideal' source. Unfortunately, such a scheme is limited to 50% efficiency, since charge must be conserved (analogous to the loss of energy in an inelastic collision in physics class!)

    For this reason, a current drive is better, since the power is transferred to the gate smoothly until it is adequately charged. The inductive charging (resonant) circuits mentioned use this idea! but are problematic with PWM converters, since they tend to resonate at a fixed frequency, which clashes with the need to turn ON and OFF at arbitrary intervals (lengths of time) as the duty cycle varies during regulation.
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    Billbehen, are you implying that current drive is more efficient than voltage drive?
  8. Papabravo


    Feb 24, 2006
    The voltage across a capacitor is controled by capacitence, current, and time. For a fixed capacitence the voltage is proportional to the integral of current times delta-t. If you want delta-t to be small then you have to make the current(i) large. You can do that any way you want, but that is the only approach with any chance of success.
  9. beenthere

    Retired Moderator

    Apr 20, 2004

    The thing that is fun about the gate on either IGBTs or power FETs is that they appear capacitive. Plus there's a +\- 20 volt limit G - S that will destroy the device if exceeded, so you can only have just so much voltage to drive charge into the gate.

    n9352527 is correct about an inductive setup being more efficient for gate drive. There is another benefit for turn-off. You can arrange a very low Trr FET between the main device's gate and the inductor. So, instead of just dropping the gate drive, you also turn on the smaller FET and crowbar the main FETs gate charge. The inductor lets you do this without also eating the drive source. Any cap goes flat pretty quick through .005 ohms.