Hello, I would like to analyze the circuit below, we know that BEMF produced by the inductor is V = -L * di / dt. When we have a flyback diode, di / dt will be less than without flyback. In the simulation, the VL to the flyback diode was about 6V I believe it is due to the fact that (VDC + Vd) -> (5 + 0.7)
For the circuit without Flyback Diode, we have values as 1kV...

Idk if i made a correct simulation, My question is about these values, they are consistent?

 

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Hello, I would like to analyze the circuit below, we know that BEMF produced by the inductor is V = -L * di / dt. When we have a flyback diode, di / dt will be less than without flyback. In the simulation, the VL to the flyback diode was about 6V I believe it is due to the fact that (VDC + Vd) -> (5 + 0.7)
For the circuit without Flyback Diode, we have values as 1kV...

Idk if i made a correct simulation, My question is about these values, they are consistent? View attachment 113492

Hi,

Yes they are, if you consider that the transistor does not blow out with the 1kv case. In real life the transistor would blow out and probably short collector to emitter as well as possibly blow out the driver circuit.