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Re: electronic PWM variac



Original poster: "Bob (R.A.) Jones" <a1accounting@xxxxxxxxxxxxx>

Hi,

If you think about it. The series element must chop up the current
symmetrical so it has no DC component. Then it can be applied to an
inductor.
One way is as follows:
The load to be controlled is put in series with a bridge rectifier. The DC
output of the rectifier is fed to a switching H bridge in parallel with the
appropriated (small at 60Hz and large at the switching frequency) smoothing
C.
The H bridge has an inductive load,.  By varying the pulse width or
frequency of  the H bridge the current thru the inductor and hence the
series impedance to the load is varied.

If we assume the H bridge is operating at 60kHz its inductor load only
requires 1/1000 (first guess) of the inductance that would be required
to be put in series with the load to control it at 60Hz.

Robert (R. A.) Jones
A1 Accounting, Inc., Fl
407 649 6400

----- Original Message -----
From: "Tesla list" <tesla@xxxxxxxxxx>
To: <tesla@xxxxxxxxxx>
Sent: Saturday, August 20, 2005 4:23 PM
Subject: Re: electronic PWM variac


> Original poster: Brady Hauth <bhauth@xxxxxxxxx> > > On further thought, you can put a high current low inductance inductor > in series with the pole pig, and then put a small PP snubber cap in > series with a high current low inductance inductor across that > assembly, with the IGBT connecting between the two high current low > inductance inductors. Properly tuned, the IGBT will charge the > capacitor through one inductor and run the pole pig through the other. > The capacitor will then push the charge back through its charging > inductor, according to the standard rules of LC resonance. The current > through the inductor in series with the pole pig then comes from the > capacitor for a little bit, at which point you can do a soft switch > with the IGBT. The two inductors then start charging the cap in an LC > resonance with the inductors and the pole pig in series, charging the > cap and dropping the current in the inductors. The IGBT then turns on > again, and since everything in front of it is through an inductor > there's no significant energy loss. > > One thing I'm not sure about is how much of a problem core losses in > the inductors would be. > > > > Putting in a small capacitor across the pole pig will prevent heat > > > buildup in the IGBT. > > > > That would be true, but then you get heat in the cap instead, which is > > probably worse. Stupid me. If you were using enough power to have a > > problem with hot IGBTs, you'd need to put a inductive element in with > > a resistor of something that would give you a voltage differential > > less than the IGBT's rating with your current draw when pulsing on > > running through it across it that can handle dissipating that much > > energy. > >