Timeline for VOM1271 as isolated power supply for gate driver
Current License: CC BY-SA 4.0
8 events
when toggle format | what | by | license | comment | |
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Aug 8, 2023 at 16:43 | vote | accept | Jenny | ||
Jul 24, 2023 at 20:40 | comment | added | Jenny | The simulteanous conducting seems to be caused by the treshold voltages: -1V for the P-FET and 1V for the N-FET, which leaves about 7V in between where both are conducting. | |
Jul 24, 2023 at 20:28 | comment | added | Jenny | I tried it using the inverted method as I will be constantly switching and if I stop switching I will have time to charge it up again before starting operations again. However, on the first low-pulse both FETs are conducting at the same time which discharges the capacitor immediately. I am not yet sure why that is though. I have modeled my voltage source as an actual 9V voltage source with a series resistance of 200000 Ohms, that might not be accurate for the low-pulse switching speed. | |
Jul 24, 2023 at 19:47 | history | edited | Bravale | CC BY-SA 4.0 |
added 263 characters in body
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Jul 24, 2023 at 15:24 | comment | added | Bravale | If you use P and N channel , I think you will get a inverted signal. That is an issue if you want a 100% output (no power available), but will work if you are always switching. Regarding second question, you are right, first send some pulses and charge capacitor. | |
Jul 24, 2023 at 11:42 | comment | added | Jenny | Thanks, the first option looks really good! I would assume that instead of using bipolar transistors I could also use a P and N-Channel FET there? What exactly do you mean by "you should send a pulse train"? I was thinking about just constantly turning it on to allow the capacitor to charge, then turn it off again and start normal operations then. | |
Jul 21, 2023 at 3:39 | history | edited | Hearth | CC BY-SA 4.0 |
mS is millisiemens, you meant milliseconds which is ms. Capitalization matters!
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Jul 20, 2023 at 18:19 | history | answered | Bravale | CC BY-SA 4.0 |