NJFET and breakdown avalanche

[bartinla]

Hey every one,

here is my first post, and it seems very dumb I know, but I got that question, and could not answer it completly, so the teacher asked me to think about it, and to come back to tell him what I found.

Anyway I attached a file with the circuit. he asked me to draw him the voltage Vd and the current iL... so for me at the beginning when the NJFET is off the current is null thus Vd = 12V and iL=0 A.

when we turn the transistor ON, the current starts to grow through DS... for me iL = iDS and since VL = L diL/dt, it increases proportionnaly until it reaches the Isat or a constant value of Ids, then VL = 0, and thus iL is constant.
concerning Vd, it decreases since it equals 12v-VL, after what it goes back to 12V since VL --> 0V.... well it seemed that my problem was there cause for him Vd is 12V when the NJFET is OFF, and directly after we turn it ON Vd = 0V ???

Anyway after that I think I got the avalanche breakdown right, and the a way to avoid it was to put reduce the iD by using a Resistor.

thanks guys for your help

Jay

 

[Roff]

Things to think about:
1. Why do you think VD ramps down when the transistor turns on?
2. To avoid breakdown, think about providing an alternate path for the inductor current when the transistor turns off. The diode is a big clue.

 

[bartinla]

Hey Ron, thanks for your reply

well when the transistor is turned on, the current through the transistor is increasing, thus IL as well => VL increases and since VD = 12V - VL ... VD ramps down until ID becomes constant, in that case VL = L*dIL/dt = 0 since the current is a constante

In addition is VD was equal to zero, ID will be equal to zero too, isn't it?

Concerning the breakdown, if I can reduce the current through the Drain by adding a resistor at the drain, won't it be ok?

thanks for you help

Jay

 

[Roff]

Hey Ron, thanks for your reply

well when the transistor is turned on, the current through the transistor is increasing, thus IL as well => VL increases and since VD = 12V - VL ... VD ramps down until ID becomes constant, in that case VL = L*dIL/dt = 0 since the current is a constante

In addition is VD was equal to zero, ID will be equal to zero too, isn't it?

Concerning the breakdown, if I can reduce the current through the Drain by adding a resistor at the drain, won't it be ok?

thanks for you help

Jay

1. If you accept the fact that the current is a ramp when you turn on the transistor, and V=L*di/dt, ... what is the derivative of a ramp? Keep in mind that as the current increases, the drop across the transistor will go back up, eventually to +12V, which you recognized. You just have the voltage waveform time-reversed right after turn-on.

2. Adding a series resistor may help reduce the flyback voltage, but you are still faced with di/dt being a huge number, because you are still interrupting the current by turning the transistor off. I might as well tell you that the standard way of preventing breakdown is to connect the diode in parallel with the inductor, anode to drain. The inductor current will take a considerable amount of time to decay through the diode, because the time constant (L/R) is long, which in turn is because the diode resistance is low. You can balance inductor current decay time against flyback voltage by adding resistance in series with the diode.

 

[bartinla]

Thanks again Ron,

I can believe I missed that... I guess that it's time for me to try to spend more time in laboratory than in the classroom where the only thing we have learned so far is theory...well I guess that it's also my fault for not studying it deeper :-( anyway thanks again.

Jay

 

[Roff]

Thanks again Ron,

I can believe I missed that... I guess that it's time for me to try to spend more time in laboratory than in the classroom where the only thing we have learned so far is theory...well I guess that it's also my fault for not studying it deeper :-( anyway thanks again.

Jay

De nada.