I've always had this question and never heard an answer.  I hear
everyone say that turning the opposite Mosfet on is better because you then
don't have the ~.7V loss across the diode.  My question is, when running at
relatively high currents (>20A), don't you have a similar voltage loss
across the Mosfet?  20A * 0.03 ohms (Rds on) = 0.6V.  So, isn't it about
the same as just using the built in diode if you're running high amperage
(except a lot easier to control)?

Gary

At 01:17 PM 2/8/2002 -0800, you wrote:
>On Fri, 8 Feb 2002, Mitch Miller wrote:
> > Perhaps I missed part of this thread ... what does doing the following
> > acomplish?
>
>When you are doing PWM with an H-bridge, you turn on a positive transistor
>and a negative transistor, then turn them off, then repeat.
>
>Your motor has energy stored because of inductance, it releases that when
>the transistors turn off. That energy must be sent back to the battery(or
>else wasted).
>
>In a MOSFET, it has a backward diode and that conducts the energy back to
>the battery. That's great, except it's a typical .6 to 1 volt loss.
>
>At 50% duty cycle, there's a lot of loss because of those diodes.
>
>Spehro suggested turning the opposite pair of transistors on during the
>off period. Unfortunately in reality it leads to more loss than letting
>the diodes do their job. Unless you get really fancy and know just how
>long to turn them on for. Which varies depending on the motor load.
>
>Perhaps there would be a benefit of doing it even for a tiny speck of
>time, when you know the current will be greatest, and then switching them
>off and letting the diodes do the rest. Maybe I'll try that some time.
>
>You can also just toggle from one polarity on to the other polarity on,
>but from zero to low speed the losses are gigantic, huge, undesirable.
>
>Cheers,
>
>Bob
>
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"Imagination is more important than knowledge."
                                 A. Einstein

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