Bypass current of parallel conected inductor/diode

  • Thread starter hsinhui
  • Start date
  • #1
3
0
Hello, good day!

I am stuck on the bypass diode current calculation on superconducting coil
The connection is as following.

->------Inductor(L)---Resistor(R)------>-
| |
-->------Bypass Diode---->----

The resistor R is the increasing quenched zone resistance.
The inductor L is the superconducting coil self-inductance.

If there is no bypass diode and coil is in persistent mode.
L*(di/dt)+i*R=0, i:coil current, t: time
i=i0*exp(-Rt/L), i0: coil initial current @t=0

But if the bypass diode is connected.
L*(di/dt)+i*R=VD, VD: diode voltage
i=i0*exp(-Rt/L)*(exp(Rt/L)-1)*(VD/R)

I simulated the coil current with the equation above.
The coil current is bypassed to diode instantly.
But this scenario is not true for the energy conservation.
The inductor stored energy (1.25MJ) can not be consumed so fast.

I miss something in my model.
If anybody can help me to figure out what happens?

Thanks a lot!
 

Answers and Replies

  • #2
bobob
Gold Member
149
82
I simulated the coil current with the equation above.
The coil current is bypassed to diode instantly.
But this scenario is not true for the energy conservation.
The inductor stored energy (1.25MJ) can not be consumed so fast.
Right. Your resistor is the dump resistor if the magnet quenches and "dump resistor" might be something like drums of water connected by a steel bar to dissipate the energy. It is true for energy conservation because the dump resistor is where the energy is dissipated.
 

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