I recognized the answer because I have used a similar circuit for a cheap/simple constant-current capacitor discharge circuit. Great riddle, Thanks for sharing!
An other solution would be to exchange the N-MOSFET for a type with a larger Vgs-th. But this requires a higher gate drive voltage, complicating the drive circuit. Yet an other solution would be to connect an NPN transistor with it's emitter to the source, it's base to GND with a resistor and a diode and the collector to the gate, shorting Vgs (driver must have some resistance) while the source is below GND. This solution requires no external negative voltage gate drive.
Yet, if you are keen on reducing the BoM, ditch the diode, replace the MOSFET with a NPN, tie R7 to the positive pin of V4 and to the base of the NPN. Thus, while induction energy exists, the NPN is in saturation/amplification.
Nice riddle, thank you! If I made R7 larger it would work the same way. But if it were too large or infinite, would it still work? Without considering parasitic capacitances it should, as L2 would draw a current from M2's source . But having all those capacitances there, I'm worried about oscillations. Are my worries substantiable?
Thank you. If the gate of mosfet will be driven with isolated source (for intance isolating transformer), in that case the transistor will stop conductig almost imadately and will behave similar to sink configuration of circuit, correct?
This is an excercise to test electronics attitude. The circuit may have advantages in certain applications as part of a DC driver to an inductor trferred to ground. I can only presume that you have not seen all schematics on earth. I didn't.
@@sambenyaakov leave irony aside, what i said in here is true, and your explanation, even if vague, is good, i appreciate. anyway, as an academic, you could try to formulate the second model of what is parallel configuration, for example. you forgot to do that. i need that paper, in order to rate it
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I recognized the answer because I have used a similar circuit for a cheap/simple constant-current capacitor discharge circuit. Great riddle, Thanks for sharing!
Thanks for comment.
An other solution would be to exchange the N-MOSFET for a type with a larger Vgs-th. But this requires a higher gate drive voltage, complicating the drive circuit. Yet an other solution would be to connect an NPN transistor with it's emitter to the source, it's base to GND with a resistor and a diode and the collector to the gate, shorting Vgs (driver must have some resistance) while the source is below GND. This solution requires no external negative voltage gate drive.
Yes. Thanks for participating.
Yet, if you are keen on reducing the BoM, ditch the diode, replace the MOSFET with a NPN, tie R7 to the positive pin of V4 and to the base of the NPN. Thus, while induction energy exists, the NPN is in saturation/amplification.
@@codures That might work as well with a resistor in series with V4.
@@foobarables that was what I was mentioning, sorry for not being clear. It's kind of hard to describe circuit changes in YT comments.
@@codures This is an excercise, a question on THIS circuit.
Thanks for presentation. Bit after bit man is building some knowledge.
👍😊🙏
Nice riddle, thank you! If I made R7 larger it would work the same way. But if it were too large or infinite, would it still work? Without considering parasitic capacitances it should, as L2 would draw a current from M2's source . But having all those capacitances there, I'm worried about oscillations. Are my worries substantiable?
Yes it will
Thank you. If the gate of mosfet will be driven with isolated source (for intance isolating transformer), in that case the transistor will stop conductig almost imadately and will behave similar to sink configuration of circuit, correct?
Yes, but the questions were on THIS circuit.
Thanks professor.
👍🙏
I have some doubts shall we connect??
You can write to sam.benyaakov@gmail.com
The topology with the inductor in the source, what is it for? I did not see schematics using it.
This is an excercise to test electronics attitude. The circuit may have advantages in certain applications as part of a DC driver to an inductor trferred to ground. I can only presume that you have not seen all schematics on earth. I didn't.
@@sambenyaakov leave irony aside, what i said in here is true, and your explanation, even if vague, is good, i appreciate.
anyway, as an academic, you could try to formulate the second model of what is parallel configuration, for example. you forgot to do that. i need that paper, in order to rate it
👍🙏❤️
🙏👍🙂
Gate drive to S1 is incorrect. It should be Gate to Source not Gate to Return !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
You missed the point!!!!!!🙂The questions are to explain the behavior of THIS circuit.