Power Supply circuit and inrush current

[matte87]

Hi everybody,
I’m focusing on a power supply circuit consisting of a LC filter and an opto Mos as switch controlled by a logic signal. Someone proposed me this circuit to power a capacitive load and assuring me it was a good solution to control the inrush current at startup (Since at the startup the load seen by the power supply is a “short circuit”, all capa discharged )

What is not clear to me is the role of the LC filter. I mean, I know that the inductance acts as a electric inertial mass making the current grow slower, while capacitors act as filter and energy storage. But I was thinking this was valid only if the inductance is discharged at the beginning of time. But in this case at the instant t=0, when the logic signal is high (the photomos conduces) both capa C1 and inductance L1 are already charged and I would expect that the inrush current would be greater since the contribution of these charged elements.

So I’m trying to test the real circuit measuring the load current with a current active probe in two cases, with and without inductance.

What I get is that the presence of the inductance really reduces the inrush current since in the first case the current is two times lower than the second case.

So, what is wrong with my reasoning?
I post a schematic view of the circuit.

Thanks

 

[goldsmith]

Dear matte
Hi
Where is the input and out put of your circuit ?
and why this way ? if you want limit the inrush current there are some simplest way . but befor that , tell me what do you want to do accurately , perhaps i can help you at better way .
Best Wishes
Goldsmith

 

[matte87]

Dear matte
Hi
Where is the input and out put of your circuit ?
and why this way ? if you want limit the inrush current there are some simplest way . but befor that , tell me what do you want to do accurately , perhaps i can help you at better way .
Best Wishes
Goldsmith

Thank you for your interest, goldsmith. In the picture (sorry for the low quality), the input is the +12V voltage source and the 100uF capacitor is the capacitor "simulating" the load. The photomos is represented by a simple switch.

I need to power a very capacitive load which means very high inrush current. Considering the worth case, I put a capacitor as load and I want to look into the transient startup current.

Thank you,
matte

 

[goldsmith]

Well , at first let me ask why not a simple inductor ? why t you added another capacitor ? i think a simple inductor can have enough opposition instead of inrush current .
And another thing , i thing you need a freewheeling diode ( or fly back diode ) to damp bad impulses of inductor ( due to the cutting the key ) .
Best Wishes
Goldsmith

 

[matte87]

Actually, this circuit doesn't come from my mind. Starting from zero, I would have used a simple inductor after the switch and not before the switch. In this way the inductor is discharged at t=0 acting as current limiter.

 

[goldsmith]

As we know according to the lenz law and according to the linear density ( inertia ) it will has opposition by fast and instantaneous changing at current . and as we know at transient time the inductor ( at t=o ) is open circuit and after 5 time constant it will going to be short circuit ( instead of DC signal ) .
So a simple inductor can help you simply and don't forget to apply that freewheeling diode .
Best Wishes
Goldsmith

 

[matte87]

So, there is no difference using the inductor behind or forward the switch?

 

[goldsmith]

I don't thing there is any difference . however if you use that inductor after switch you can use the freewheeling diode to recover the stored energy ( not loss it )
Good luck
Goldsmith

 

[matte87]

I don't agree with the fact that at t=0 the inductor is an open circuit. It is fully charged, so di/dt=0 and it is a short..

 

[goldsmith]

VL=L*di/dt dt is zero at the time that t=0 and then el can considered about vi ( it means that it is open circuit , and you can simulate it simply .
Respect
Goldsmith

---------- Post added at 13:49 ---------- Previous post was at 13:46 ----------

And according to the inertia , the inductor wants to keep that stable voltage ( before giving current ) so it has to create a voltage that it can be high . and in the polarity of supply to avoid the flowing current .

 

[mtwieg]

The inductor can help to reduce inrush current from the source, but there will still be a large surge of current in those two caps when the switch is closed. Putting the inductor in series with the switch will reduce that inrush. But keep in mind using a simple inductor as an inrush limiting device will cause a large oscillation and overshoot on the load, which can be damaging. You should apply some kind of clamp circuit (like a diode in antiparallel with the inductor) to limit the overshoot.

 

[goldsmith]

Hi again
Or here are another ways to do it , such as using a resistor and then a triac or thyristor or mosfet to short circuit that , slowly , it will do the action such as soft starter , simply , but and inductor and diode are more better .
Best Wishes
Goldsmith