Why inductor's initial current have DC component ?

[rocky56hk1]

I am design a transformer gate driver. When I apply a AC source though a inductor, inductor current always have a DC component and last for a period until decay to 0. when I increase the resistance of inductor, the time increase. I apply a AC source to the inductor, its current should be AC, why there is DC component ?

I try to simulate using LTSPICE and NG spice. The results is the same.

Can anyone know why ?

 

[LvW]

I apply a AC source to the inductor, its current should be AC, why there is DC component ?

Should be? Why?
The current is ac only after all switch-on transients have disappeared (as you have noticed correctly). Thus, there is an ac current for steady state conditions only.

 

[FvM]

I guess you are applying a SIN source starting from zero voltage. If you look at the steady state solution, you'll notice that the current will be maximal (for a pure inductive load) or at least non zero for zero inductor voltage. You can assign an initial condition for the inductor current to make the simulation start without a "transient DC" current.

when I increase the resistance of inductor, the time increase

According to T = L/R, we can assume that the transient duration decreases with increasing inductor series resistance.

 

[rocky56hk1]

yes, i apply a ac source , 12v,100khz, 0v offset, to an pure inductor with inductance 500uH, transient start from 0-50ms, step 1us. start the simulation run and suprising find that the inductor current always shift above the 0 level. when i place resistor in series with it, the ac current will shift above the 0 and fall down slow to 0 level at the stead state

for the L/R explination, i still not understand. even it is, the ac current should swing between negative to positive across the 0 level and with larger envolpe at initial, smaller at steady state instead of always over the 0 level

 

[crutschow]

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for the L/R explination, i still not understand. even it is, the ac current should swing between negative to positive across the 0 level and with larger envolpe at initial, smaller at steady state instead of always over the 0 level

I don't quite understand your comment. If you asking why is there a DC component, that has been answered. It's due to the transient of suddenly applying the sine-wave. Fourier analysis and the Laplace transform will show the DC component of that transient (which varies depending upon where it the sine-wave the voltage is suddenly applied). The steady-state current only occurs after the sine-wave has been applied long enough for all transient effects to subside.

 

[FvM]

The steady-state current only occurs after the sine-wave has been applied long enough for all transient effects to subside.

Specifically with an ideal inductor you have R=0 and L/R=∞, so steady state will be never reached. It's a pure simulation problem, because inductors with R=0 don't exist (if we don't consider superconducting inductors).

 

[crutschow]

Specifically with an ideal inductor you have R=0 and L/R=∞, so steady state will be never reached. It's a pure simulation problem, because inductors with R=0 don't exist (if we don't consider superconducting inductors).

Good point. You need to add in some real world resistances so the transients will subside.

 

[rocky56hk1]

Wu ! Thank all your information. That mean apply a AC source to a superconducting inductor, the charging current will be a DC current. right ?
Now come back to my transformer gate driver, I am headache because when I apply a pulse width signal the the input, the whole waveform output always has low few component shift above zero level and last for few cycle before the steady state, making the mosfet on time longer that actual PWM's ON-OFF

 

[FvM]

I don't see how you conclude from inductor current to transformer output voltage. Average transformer DC output voltage will be always zero, causing problems at least with varying duty cycle of a gate transformer.

It should be added that gate transformers often need active pulse shaping circuits for acceptable switching behaviour. In this context, you can also make provisions for correct gate voltage level with varying duty cycle and during starttup.

 

[crutschow]

Wu ! Thank all your information. That mean apply a AC source to a superconducting inductor, the charging current will be a DC current. right ?
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No. A superconductor inductor will have a transient current and a steady-state AC current, the same as any inductor. But, since there is no resistance, the transient will never disappear.