- #36
cabraham
- 1,181
- 90
I know this is an old thread, but I just found it. Interesting, it is. Anyway, I just thought I'd chime in.
An inductor, in the pure sense, cannot be current-driven, but it must be voltage-driven. Otherwise, an infinite voltage would be required at time t = 0. When a constant current source, dc, is connected to an open switch, & the other side of the switch connected to the 1-turn inductor loop, what happens when the switch is closed at t=0?
If the ccs (constant current source) current value is Ics, & the inductor initial current is 0, then we have a problem. We would get an infinite voltage at t=0 when the switch closes, since changing the inductor current from 0 to Ics in zero time span results in infinite voltage per v(t) = L*di(t)/dt.
But as we know, an ideal pure inductor cannot exist. There is always a little capacitance presented at the input. The surfaces of the input terminals present a very small capacitance which shunts the input. The ccs charges up this capacitance at t=0. The voltage across the input terminals increases very rapidly, & the current energizes the inductor. As the inductor energizes, the voltage falls. Eventually, the voltage at the input terminals is zero.
The difficulty w/ this problem lies in the fact that the inductor current cannot instantly change from 0 to Ics w/o infinite voltage & infinite power. But, the tiny capacitance at the input terminals alleviates the infinite voltage/power conditions.
The steady state voltage is zero since the loop has no resistance. The ccs is outputting a current of Ics, a voltage of 0, & a power of 0, in the steady state. Did I help at all?
Claude
An inductor, in the pure sense, cannot be current-driven, but it must be voltage-driven. Otherwise, an infinite voltage would be required at time t = 0. When a constant current source, dc, is connected to an open switch, & the other side of the switch connected to the 1-turn inductor loop, what happens when the switch is closed at t=0?
If the ccs (constant current source) current value is Ics, & the inductor initial current is 0, then we have a problem. We would get an infinite voltage at t=0 when the switch closes, since changing the inductor current from 0 to Ics in zero time span results in infinite voltage per v(t) = L*di(t)/dt.
But as we know, an ideal pure inductor cannot exist. There is always a little capacitance presented at the input. The surfaces of the input terminals present a very small capacitance which shunts the input. The ccs charges up this capacitance at t=0. The voltage across the input terminals increases very rapidly, & the current energizes the inductor. As the inductor energizes, the voltage falls. Eventually, the voltage at the input terminals is zero.
The difficulty w/ this problem lies in the fact that the inductor current cannot instantly change from 0 to Ics w/o infinite voltage & infinite power. But, the tiny capacitance at the input terminals alleviates the infinite voltage/power conditions.
The steady state voltage is zero since the loop has no resistance. The ccs is outputting a current of Ics, a voltage of 0, & a power of 0, in the steady state. Did I help at all?
Claude
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