inductor's Ldi/dt property

Hi, can anyone help me the Ldi/dt effect of inductor, given the following 2 examples..

Originally, the switch is closed. How should the waveform at V1 look like when the switch is suddenly opened?

banh,
For an ideal inductor:
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In the 1st example, the voltage would rise to an infinite negative value, with a zero rise time and a zero width, ie, a negative impulse.
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In the 2nd example, the voltage would rise to an infinite positive value, with a zero rise time and a zero width, ie, a positive impulse.
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Of course, in the real world, ideal inductors do not exist. They have distributed capacitance, non-zero DC resistance and finite equivalent parallel resistance. The actual waveform for example 1 would be a damped sine wave with an initial negative slope. The actual waveform for example 2 would be a damped sine wave with an initial positive slope.
Regards,
Kral

Well first as kral said that happens, but his conclusion is wrong. Faraday law stablish that the indutor opposes to abrupt change in current. The problem is the switch that when opens pass from almost no resistance to a big one but gradually, not in an instant so when it is openning his resistance is low and electric arch occours, if the current stored is low a spark and a instant discharge occours. But if the current is high enough the switch will be destroyed.

banh,
As a practical matter, Miguel is entirely correct. I was assumming an ideal switch (infinite off resistance, infinite breakdown voltage) to match the ideal inductor. Of course, this ideal switch, as well as the ideal inductor, exist only in fantasy land. What happens is, the self-induced voltage in the inductor goes to whatever value is necessary to maintain the current that existed at the instant before the switch opens. Since, with the ideal components, the resistance through which the current must flow is infinite, the self-induced voltage would be infinite. This is all academic of course, since Radio Shack does not stock ideal inductors and switches
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My comment regarding the responses with a real inductor is correct, if you ignore the arcing.
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Thaks, Miguuel!
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Regards,
Kral

I'm agree with you Kral

In both the figures the current is flowing downwards from 12V to GND. But in Fig 1 when switch gets open the inductor tries to maintain the current. Hence a current sink through V1 occures. ie: a -ve signal. In Fig 2 too the same current flow is tried to maintain, so this will cause V1 to act as a current source. ie: a +ve signal there.

Hi everybody,

I would like to know about second example.

When the swith open, will inductor raise its induce voltage?
(I think when switch close, inductor have less than 12V. Current will rise to max amper.)

If voltage suddendly raise , How can I calculate.
( I think it need to use instantanious time , inductance of coil, amper etc..)

*** I want to produce about 400V at this time. What should I do? ****

Zaw

Hi
Yes the inductor will try to mintain a current flow outwards the V1 terminal. ie: there will be a +ve voltage.

If you want to raise a high voltage then you have to control the parameters like 'di' and 'dt'.
ie: v = -L dt/dt
Thus for high 'v' you have to provide high inductance ('L'), a huge change of 'i' ('di') in a small interval of time 't' ('dt').

zawminoo said:

Hi everybody,

I would like to know about second example.

When the swith open, will inductor raise its induce voltage?
(I think when switch close, inductor have less than 12V. Current will rise to max amper.)

If voltage suddendly raise , How can I calculate.
( I think it need to use instantanious time , inductance of coil, amper etc..)

*** I want to produce about 400V at this time. What should I do? ****

Zaw

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