Need to check calculations of my inductance

[argens]

Hello folks,


am new here and already seeking for help.
Found this forum thanks to my schoolmates. Hope to get some answers and in the future also help to find some solutions.

My problem is as follows:

I have such a schema (please note the trafo on the schema is in reallity a Coupled Inductor, like https://uk.farnell.com/bi-technolog...lftr/inductor-33uh--20/dp/2192082?ost=2192082):



And need to calculate the needed inductance (L).


I know the following:
U = 30 V
I = 3 A
f = 70 kHz

And was counting as:

Z = U / I = 10 Ohm

|Z| = sqrt((R^2)*(Xl^2) |R=0

|Z| = Xl = 10 Ohm

L = Xl/w

L ~= 22,7 uH

Am doing it right?
Thanks for any help!

 

[KlausST]

Hi,

Your circuit seems to be wrong.
Because you feed your inductance with rectified signal, there will be a DC (component) in the inductor voltage.
This generates high DC currents, and thus saturation of your inductance core.

Your calculations are for pure sinusoidal AC voltage without DC across the inductance.
This is not the case here...

Klaus

 

[brushhead]

Hi,
If your coupled inductor is meant to be a transformer, it needs to go into the AC side before the rectifier.

Rob.

 

[argens]

Hi,

Your circuit seems to be wrong.

Klaus

Hi,
If your coupled inductor is meant to be a transformer, it needs to go into the AC side before the rectifier.

Rob.

First of all thank you for your replies.

Of course, that you are right. I made a mistake in the schematic. Sorry for this.

Here is the right schematic:



now, have I calculated it in a good way?

 

[brushhead]

Hi,
Is this meant to be a common mode filter?

Rob.

 

[argens]

Yes, just a common filter to reduce the ripples.

 

[KlausST]

Hi,

this common mode filter will not (significantely) reduce voltage ripple in your output capacitor.
It is more a filter to be high impedance for HF to the right side circuit referenced to the left side circuit.


--> To reduce ripple in the right side capactior you need a (one or two) non coupled inductors.

 

[brushhead]

Hi,
Normally you would remove C1 and have a non common cored inductor, then the smoothing Cap on the right side across your output terminals.

Rob.

 

[andre_luis]

Common mode inductors are not designed to meet to a certain inductance, but to operate saturated, becoming lossy above a specified frequency range.

.