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kalee20 · 3rd Aug 2018, 10:08 am

As drawn, the waveform at the RHS will be exactly a square wave, no ringing (assuming perfect coupling between the two halves of the winding, and a 1:1 duty cycle drive.

At the moment of the MOSFET turning 'on' the waveform will be just about to collapse anyway. 12V across the winding for a defined period will charge the core, followed by 12V in the opposite direction (assuming the Zener diode us exactly twice the supply voltage) will exactly discharge it again.

If however on the RHS you add a bit of downstream, uncoupled inductance, and stray capacitance, you'll get approximately the waveform you have drawn. In this case, when the MOSFET is turned 'on' for the first time, the extra LC circuit will be shock excited. One side will be jerked up from 12V to 24V; the other side will ring with 12V peak amplitude, which is likely with 'realistic' losses to die out by the time the MOSFET is asked to change state.

When the MOSFET is turned 'off' the whole thing will ring. If the coupled winding has very small inductance compared with the leakage inductance (an unlikely condition in practice), it will largely drive the situation and the leakage will be jerked down from 24V to zero. So you'll get 24V peak amplitude of ringing. And then 24V in the first direction... ad infinitum.

If however, the coupled winding has inductance which is not small - and in practice it is likely to be larger than the added leakage - then the whole thing will ring at a considerably lower frequency on MOSFET turn-off. I'm on a local train right now so can't sketch anything out, worse luck (picture being worth 1,000 words and all that) but hopefully you can follow the above?

3rd Aug 2018, 10:08 am	#10
kalee20 Dekatron Join Date: Feb 2007 Location: Lynton, N. Devon, UK. Posts: 7,088	Re: Vibrators Solid State Replacements As drawn, the waveform at the RHS will be exactly a square wave, no ringing (assuming perfect coupling between the two halves of the winding, and a 1:1 duty cycle drive. At the moment of the MOSFET turning 'on' the waveform will be just about to collapse anyway. 12V across the winding for a defined period will charge the core, followed by 12V in the opposite direction (assuming the Zener diode us exactly twice the supply voltage) will exactly discharge it again. If however on the RHS you add a bit of downstream, uncoupled inductance, and stray capacitance, you'll get approximately the waveform you have drawn. In this case, when the MOSFET is turned 'on' for the first time, the extra LC circuit will be shock excited. One side will be jerked up from 12V to 24V; the other side will ring with 12V peak amplitude, which is likely with 'realistic' losses to die out by the time the MOSFET is asked to change state. When the MOSFET is turned 'off' the whole thing will ring. If the coupled winding has very small inductance compared with the leakage inductance (an unlikely condition in practice), it will largely drive the situation and the leakage will be jerked down from 24V to zero. So you'll get 24V peak amplitude of ringing. And then 24V in the first direction... ad infinitum. If however, the coupled winding has inductance which is not small - and in practice it is likely to be larger than the added leakage - then the whole thing will ring at a considerably lower frequency on MOSFET turn-off. I'm on a local train right now so can't sketch anything out, worse luck (picture being worth 1,000 words and all that) but hopefully you can follow the above?