Current Transformer Deisgn

[MorbidFractal]

Yes. I got my sum wrong. The equation is right. I probably went number blind. It is kind of another example of where theory does not match practice or rather practicality. Generally cores have a Effective Core Area and Winding Area, Ae and Aw, which are roughly the same but not necessarily. ETD more or less. RM more core less winding. There are many sums that you might use to get a first guess only to find out that some other constraint spoils the result.

If Ae = Aw then you would have to fit 2 x 50 primary turns on the core, plus the others. If you assign half to the primaries, not an unreasonable guess, then 100 turns in 2mm^2 gives you 0.02mm^2 per turn, good for 80mA at 4A/mm^2 or a wire diameter of 0.08mm which would be impractical.

In the case of a push-pull power transformer supplied by a 12V battery your sums might suggest that you only need 1/One turn primaries but then you find out that the magnetising current equals the reflected load current which is not a very good idea so you start adjusting other parameters/constraints to arrive at a more meaningful solution.

There is no one sum fits all and you end up having to iterate.

 

[JLeon]

Yes. I got my sum wrong. The equation is right. I probably went number blind. It is kind of another example of where theory does not match practice or rather practicality. Generally cores have a Effective Core Area and Winding Area, Ae and Aw, which are roughly the same but not necessarily. ETD more or less. RM more core less winding. There are many sums that you might use to get a first guess only to find out that some other constraint spoils the result.

If Ae = Aw then you would have to fit 2 x 50 primary turns on the core, plus the others. If you assign half to the primaries, not an unreasonable guess, then 100 turns in 2mm^2 gives you 0.02mm^2 per turn, good for 80mA at 4A/mm^2 or a wire diameter of 0.08mm which would be impractical.

In the case of a push-pull power transformer supplied by a 12V battery your sums might suggest that you only need 1/One turn primaries but then you find out that the magnetising current equals the reflected load current which is not a very good idea so you start adjusting other parameters/constraints to arrive at a more meaningful solution.

There is no one sum fits all and you end up having to iterate.

Thank you, man. As a newbie, I also rely on self-study. I made a mistake, that is, I did not separate theory from actual engineering. If you didn't guide me so patiently, I might still be in this confused state. My English is very poor, I don't know how to express my gratitude.Anyway,thanks. HaHa

 

[MorbidFractal]

No worries. You would be right to assume that I am making educated guesses.

Unfortunately I've seen lots of sums that have not made much sense to me and have had to go back to basics to derive my own equally worthless ones. Then I haven't bothered extending things as much as I might or should. It's all bits and pieces.

It all starts with

L = UoUeAeN^2/Le

and

B = UoUeIN/Le

L inductance Henries
B flux density Tesla
Uo permeability of free space 4PiE-7 Dimensionless
Ue relative permeability Dimensionless
N Number of turns.
Ae Effective core area M^2
Le Effective core length M
I current in the winding

There is another common term called the specific inductance Al

Al = UoUeAe/Le

That's normally given as nano-henries per root turn but you have to be consistent with the overall dimensions. Ae might be M^2 but will be given as mm^2 so you have to adjust by 10E-6.

Substitute for Al in the equation for inductance.

L =AlN^2

Substitute for Al in the equation for flux density.

BAe = AlNI

I had to multiply by Ae on the left side for the Ae multiplying the right.

Multiply both sides by N

BAeN = AlN^2I

AlN^2 is L

BAeN = LI

Rearrange for N

N = LI/BAe

If you have a saturation flux density, Bsat, and peak inductor current, Ipk, then given a core area and target inductance you can calculate the required number of turns to avoid saturating the core.

Here's another one. Faraday's Law...

E = -LdI/dT

Shake it around a bit. Let's say you apply a voltage across an inductor.

dI/dT = V/L

The minus sign disappeared because you are driving the EMF rather than the inductor producing a back EMF. Assume the inductor is the primary of a transformer, Lmag the magnetising inductance, and V is the input supply, Vin, which you supply for a period of time, Ton. Also assume that the current starts at zero. At the end of Ton the current in the magnetising inductance is...

Ipk = VinTon/Lmag

We already have...

N = LI/BAe

Substitute and you get the minimum number of primary turns to avoid saturating the core...

Npmin = VinTon/BsatAe

It's all congenital, existing from birth or the base equations, or perhaps incestuous, interbreeding of the base equations.

This is kind of square wave SMPS specific and uses SI units. A term like VinTon is integral volt seconds. In mains power electronics using sine waves you will often see the term 4.44 crop up which has something to do with the integral of a sine wave voltage. Same sort of thing.