Choke coil sizing

[Peet19]

Hi everyone!

I would like to ask for some help with the sizing of the choke coil.
The picture shows that in the simulator, a 300mH coil reduces the initial high current of the capacitor to 717mA. This is perfectly fine for me, I'm just asking for help on how to calculate this. This was calculated by the simulator, but I would like to calculate it myself and learn how to size the choke coils.
Thank you in advance for your help!

 

[Lightium]

It's a pulse, so (2 * pi * F * L) = Lx.

 

[Peet19]

The formula is fine, but I'm not looking for the coil's reactance, but its inductance.
Eg: how do I calculate that the value of the coil is 300mH?

 

[Lightium]

You can configure a coil with this: https://coil32.net/download.html

 

[rjenkinsgb]

The formula is fine, but I'm not looking for the coil's reactance, but its inductance.

The formula gives the inductance in, in Henries?

 

[Peet19]

No.

 

[Peet19]

Lightium: Thank's.

 

[rjenkinsgb]

No.

????

 

[Peet19]

2*pi * F * L = XL
The last element of the formula (L) is the inductance.
The result is inductive reactance.
I am looking for the value of (L), i.e. the inductance and not the reactance.

 

[Nigel Goodwin]

2*pi * F * L = XL
The last element of the formula (L) is the inductance.
The result is inductive reactance.
I am looking for the value of (L), i.e. the inductance and not the reactance.

You simply rearrange the formula to give the inductance as the answer.

However, exactly what are you trying to do?.

 

[Peet19]

Yes, XL / (2*pi * F) gives the value of L, but the problem is that I don't know the reactance either .
I want to do something very simple like a soft starter, and I thought I would put a choke coil on the input that would reduce the high starting current for a short time.

 

[Nigel Goodwin]

Yes, XL / (2*pi * F) gives the value of L, but the problem is that I don't know the reactance either .
I want to do something very simple like a soft starter, and I thought I would put a choke coil on the input that would reduce the high starting current for a short time.

It also greatly increases cost, weight, and losses - but your answer is still very vague, do you have an EXACT use you're thinking of, and why would you think a choke might help you?.

 

[Peet19]

There is a 12V input that I regulate down to 5V with an L7805. At its output, there are 4 1000uF capacitors in parallel, which cause a large current surge when switched on. I want to reduce this to less than 1A while the capacitors charge.

 

[Lightium]

Other wise what are the caps for?

 

[rjenkinsgb]

I want to reduce this to less than 1A while the capacitors charge.

What's the maximum operating current?

You could add eg. a 12 Ohm resistor at the input, which would limit the switch-on current to under 1A peak, but also the working current to under 500mA. A higher value resistor could be used if the load current is lower.
That would also take some dissipation away from the 7805.

Or a simple single-transistor constant current circuit, which would limit both the startup and working currents at the same value.

There are various other possibilities, with varying complexities.

 

[Peet19]

This is what I'm planning.

 

[Nigel Goodwin]

There is a 12V input that I regulate down to 5V with an L7805. At its output, there are 4 1000uF capacitors in parallel, which cause a large current surge when switched on. I want to reduce this to less than 1A while the capacitors charge.

Well don't - there's so much wrong with that - you're struggling because your entire premise is wrong.

There seems no reason to add 4000uF on the output of a 7805, but if you did the IC would limit the current anyway - adding a pointless choke would be very expensive, achieve nothing, and lose you considerable voltage from the supply.

What current are you expecting to draw from the 5V supply?, and why did you think 4000uF was a good idea? - if for some bizarre reason you need 4000uF?, then place it at the input (C1), and allow the 7805 to do it's job and maintain 5V at it's output. You should bear in mind that you're likely to need a VERY substantial heatsink on the 7805, as at 1A it will be dissipating 7W.

 

[Peet19]

I thought of a 1000uF capacitor for the output of the fans to help the maximum current of 1A.
Should I make a DC-DC converter to produce the 5V instead?

 

[Nigel Goodwin]

I thought of a 1000uF capacitor for the output of the fans to help the maximum current of 1A.

No, the power still comes from the 7805 - capacitance on the input is useful if it's fed from unsmoothed DC, as a reservoir capacitor - but I'm presuming it's a 12V battery?.

Should I make a DC-DC converter to produce the 5V instead?

Why not just use 12V fans? - but certainly a DC-DC converter will be much more efficient than a 7805 - with a 7805 you're wasting more power (58%) than you're using.

 

[Peet19]

I understand,
Thank you.
+12VDC comes from a switching power supply.
I control a high-current circuit with the IRFZ48, but the 5V is needed on another board where the microcontroller will be.
The problem is that I don't trust the available DC-DC converters.
I have already bought and used a few pieces, but they all broke. They are unreliable. I'd rather make one myself.