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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
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!
Jul 20, 2023 at 18:19 history answered Bravale CC BY-SA 4.0