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I am trying to drive a MOSFET using a TL494 IC. The controller IC is separated from the power circuit using a transformer. LTspice is unusually slow while simulating this circuit.

This is my schematic:

enter image description here

Following the secondary of the transformer, there is a DC restoration circuit and a Zener diode to limit the voltage pulses to nearly 15 V. Before connecting the output of this Zener diode to the MOSFET gate-source, I am getting good pulses as shown below (at the output of the Zener diode before interfacing the converter):

enter image description here

After the converter is interfaced, there is initially an output but it ramps down to zero after some time. The output pulses at the Zener diode (AB) also behave abnormally. I don't know what the reason behind this might be. This is the output after interfacing:

enter image description here

In the above figure, the green coloured waveform is the output of the Zener diode and the blue coloured one is the output of the buck-boost converter. All my files are available here (see 'Final Circuit').

Jonathan_the_seagull
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    MOSFET gate-sources are capacitors which means they do not empty on their own. – DKNguyen Dec 29 '22 at 16:03
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    It'll work much better if you didn't try to create an asymmetric 0V/15V waveform. Make it symmetric (-15V/+15V) instead and connect the output of your gate drive transformer directly to the MOSFET source and gate. Then update your post since you'll likely encounter further problems (ringing). – Jonathan S. Dec 29 '22 at 16:05
  • @JonathanS. Thank you. I will try that. – Jonathan_the_seagull Dec 29 '22 at 16:10
  • There's no ground return path. The emitter pins only source. Consider using a "totem-pole" type like TL598, or use a pull-down resistor then a gate driver like TC4420. – Tim Williams Dec 29 '22 at 16:22
  • @TimWilliams Where is the return path needed? – Jonathan_the_seagull Dec 29 '22 at 16:32
  • @DKNguyen Isn't there a path available for the parasitic capacitor to discharge in the schematic? – Jonathan_the_seagull Dec 29 '22 at 16:38
  • `Where is the return path needed?` from C4 to L1. One alternative would seem to be to use the TL494 in pull-push mode. (almost misread ref for REF.) – greybeard Dec 29 '22 at 16:43
  • @greybeard Why is a return path from C4 to L1 needed? I thought the series capacitance is used to block the DC value of the generated PWM signal. – Jonathan_the_seagull Dec 29 '22 at 16:47
  • I see "the non-L1-end" of C4 pulled up by the paralleled transistors. When and how does it get pulled down? – greybeard Dec 29 '22 at 16:58
  • (Have a look at the voltage at the emitters during 1.6 to 1.8 ms.) – greybeard Dec 29 '22 at 17:18
  • @greybeard Thank you. Can I add a resistance in parallel with C4? – Jonathan_the_seagull Dec 29 '22 at 17:19
  • Counter-intuitive. How about a diode to GND? – greybeard Dec 29 '22 at 17:21
  • @greybeard I don't know if this is correct. I connected a diode in reverse bias between the emitter and ground. The pulses at the emitter appear good initially but after 3 ms, the pulses become distorted. – Jonathan_the_seagull Dec 29 '22 at 17:37
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    The problem with the TL494's output as you've connected it is that it can only source current, it can't sink any. That means C4 will just charge more and more until it reaches the supply voltage, and then it'll stop. You really have to operate the TL494 in push-pull mode for this to work. I missed that when I first looked at the schematic. C4 is also way too large, make it 3.3µF instead and add a small series resistor (3.3 Ohm, maybe). Example (not TL494): https://electronics.stackexchange.com/questions/640432/how-does-this-llc-converter-keep-50-duty-cycle-and-zero-dc-offset-with-ac-coupl – Jonathan S. Dec 29 '22 at 17:45
  • @JonathanS. Thank you. I will work on it. I have made a similar schematic but, using an optocoupler. It gives a good result compared to this. Can I use this single ended design instead of the push-pull configuration with an optocoupler instead? – Jonathan_the_seagull Dec 29 '22 at 17:53
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    Better use an isolated gate driver IC if you're going to ditch the gate drive transformer anyway. One part that'd likely work very well is the SI8751. – Jonathan S. Dec 29 '22 at 18:06
  • FYI, the "DC restore" network (coupling cap, clamp diode(s) on transformer secondary) is a fine way to do it; that's not the problem. The problem is reciprocity: the same must be true on the primary, but it's not, as shown. – Tim Williams Dec 29 '22 at 19:14
  • @Jonathan_the_seagull I didn't see one and I looked a few times. I could have missed it...where do you think the path is? The only place I can see for DC current to flow to discharge the gate-source capacitance is backwards through D2 or D3 which means they won't. – DKNguyen Dec 29 '22 at 19:58
  • @DKNguyen Understood. Thank you. Will adding a resistance in parallel or a diode to the gate-source suffice? – Jonathan_the_seagull Dec 29 '22 at 20:52
  • @Jonathan_the_seagull Probably not since if you want to turn it on fast, you probably also want to turn it off fast. The transformer also is a pulse drive so can't keep Vgs charged against a resistor. That's why a symmetrical transformer drive was mentioned previously. There, you don't try to discharge the gate. Instead, you take advantage of the fact you already have circuitry to prevent DC discharge so gate-source voltage can be maintained and instead charge the Vgs in the opposite direction to turn off. – DKNguyen Dec 29 '22 at 21:29

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