I want to re-use the reverse current that a coil will produce when current will turn off (in the coil), in order to charge directly capacitors.
You can look at the circuit in this video , it will be likely the same ( timecode: 0:51 ) :
But as I know, reverse current generated from a coil being switched off can be at very high voltage.
My question is, in that circuit, the reverse current coming from the coil, is it going to be too high voltage for recharging the capacitor ? it will not recharge all the energy the capacitor already gave but a only part of it, but is the temporary high voltage dangerous for the capacitor itself ?
The guy in the video shows that this reverse current regenerates the capacitor and the circuit has better efficiency, so for him it doesn't seem to be a problem to recharge capacitor with this reverse current coming from the coil.
When I was younger, I power supplied a MOT transformer (1:10 turns ratio) with 12V DC , and suddently switch off the power supply, and it generated +1000Volts spike around the secondary. with 12V AC it would just give 120V AC, but this phenomenon seem to exist when the magnetic field of a coil collapse and generate a reverse current from the inductor (maybe i'm not using the good words i speak only french normally)
this is why people use flyback diodes, but my question is, if i use diodes and capacitors, is the voltage spike dangerous.
I've seen few minutes ago someone saying that the voltage doesn't go very high if there's immediately a way for the electrical current to go through (example: with diodes). It only goes up to 1000V etc if necessary in order for the current to pass somewhere. it seems logic, I don't know if it's totally true but maybe it is.
Yes i didn't use the correct terms, sorry. (i normally speak only french )
i've seen someone on a forum saying that the reverse voltage (coming fromthe coil absorbing its collapsing magnetic field) is at very high voltage only if there's no other path for it to pass through. Otherwise, if there's a clear path (in this example diodes + capacitor) the voltage will not need to go very high.
So, if it's true, in my case it will probably be at a voltage acceptable for my capacitors.
Thank you for your answer !
Yes, that's exactly the point. The current through the coil cannot be switched off immediately - it will continue to flow in any case. But if there is no path for the current, the voltage increases until the current finds a path - and if through a spark.
Thank you for the answer.
So, if I understand correctly I can remove the "high side FET" and just replace it with wire in that circuit below (timecode 0:51 ) ?
The "low side FET" will work properly to discharge the capacitor into the coil (and the diodes will allow the reverse coil voltage to go back into the capacitor, as the transistor will be activated a very short period of time) ?
Thank you again
If I don't make a mistake in my understanding, in that circuit, if I remove one of the transistors (example the "high side FET") and replace it with wire, when the coil will give voltage in reverse, this wire will go to the diode D1 output and do a shortcircuit of the coil reversed/return voltage instead of going into the capacitor ? what do you think ?
Let me explain a bit more. An inductor opposes a change in current flowing through it. When we force current though an inductor, a magnetic field develops around it. If we switch off the current abruptly, the magnetic field collapses. This causes a voltage to be induced in the inductor with reversed polarity to try to keep the same current flowing through it. Now the inductor becomes a current source and its terminal voltage is determined by the resistance of the path that induced current flows through. If we use a diode to clamp that voltage, the terminal voltage will be low. If we provide no path, the voltage will be very high.
You cropped off important parts of that circuit, but its looks as though a transformer is being driven, not an inductor, so the rules are very different. (note the 'T' visible over the winding, for transformer).
The circuit on the secondary side of the transformer needs to be posted in order to evaluate this issue. Normally transformers are not inductive, they simply reflect the load impedance back to the primary with a scaling factor. If the load is resistive there's no inductive kickback...
The circuit shown does already have diodes, suggesting the load isn't simply resistive.
And there is a discrepancy between the diagram in the video's image frame and that presented inside the video, so there's a big question there... I suspect there is a transformer involved as you need high voltages to get fast high power drive from an electromagnet.
I think you are missing that the circuit is regenerative and dumps the energy stored in the inductor into the capacitor. As far as I can see this can't work with one MOSFET and one diode. If I'm wrong can you show how it would work.
Yes, Perry, I see your point. Using one MOSFET and one diode will waste the energy stored. As I see it, in order for the given circuit to regenerate, both MOSFETS must be switched off at the same time. The circuit is difficult to analyze because the inductor is driven through the rds of both MOSFETS but the capacitor is charged through two diodes. What about the body diodes?
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