# Calculating the core area for a low frequency transformer part 2

I posted a question a while ago about calculating the core size of a transformer when the frequency is relatively low (<50Hz). Thank you to everyone for the comments and feedback. I have done some more research and would like to post a follow up question outlining my thinking. I would really appreciate any comments.

I am interested in calculating the core size for a 20 Hz transformer. Single phase, step down. This is a theoretical question - I understand that low frequency transformers require core sizes that make them impractical, but I am just trying to understand how the core size would be calculated. For the purpose of understanding this, I have treated the transformer as a “perfect” transformer with no core losses - the thinking being that core losses could be factored in after I understand how the core size is worked out. (if this is a mistake, please let me know)

My primary coil has 400 turns & 8.05429 Voltage. My secondary coil has 50 turns
The formula (Voltage in Primary/turns on secondary coil)=(Induced Voltage in Secondary/turns on primary coil) gives the induced Voltage in the secondary as 1.00678625, with a turns ratio of 400:50=8:1

I wanted to then use the following formula-

induced voltage = 4.44*fNA*B

Where f= frequency in Hz

N= number of turns in the coil

A= cross section of the core area in Meters

B= flux density in the core in Tesla

(I have done some FEMM simulations that give me a value for flux density in the core)

My questions:

1. I know the saturation point of the core material, so can I switch the formula

induced voltage = 4.44*fNA*B

to

B=induced voltage/(4.44fNB)

and then play around with plugging in values that deliver a value for B that is safely below the saturation point of the core material(e.g. 1.8T)?

2. How does the turns ratio of 8:1 influence the values for the Induced Voltage and the Number of Turns that I need to put in to the above formula? At first I thought I should just use the primary coil values, then I thought I should add the primary and secondary together. I have chased my own logic around like a dog chasing it’s tail and now lie in a dizzy heap! If anyone has any advice here, I surely would appreciate it. Thanks

I know the saturation point of the core material

Core saturation has nothing to do with secondary current or even the extra current drawn by the primary when the secondary is supplying current to a load. This is because the H fields generated in the core from load currents in primary and secondary totally cancel each other. It doesn't sound intuitive but it's the truth that matters!

Core saturation is to do with the small current that flows into the primary when the secondary is, in effect, removed.

So you don't need to switch the formula because whatever you do, and under whatever load conditions, the flux in the core remains the same for no-load and fully loaded conditions.

In fact, due to leakage inductance and copper losses, the core flux reduces on higher secondary load currents.

How does the turns ratio of 8:1 influence the values for the Induced Voltage and the Number of Turns that I need to put in to the above formula?

It doesn't affect core saturation

At first I thought I should just use the primary coil values, then I thought I should add the primary and secondary together.

It's just the primary (driven winding) that produces flux and hopefully, that should be clear by now.

Nah, I don't like formulae like 4.44fNAB. It may be correct, but in what units, and where does it come from, and what about square wave, or other assumptions? I prefer going from first principles.

Your core flux swings from +ve max to -ve max. You say saturation for your core material is 1.8T, so that max must be < 1.8T. How much lower? It depends on what losses you want to accept, let's say 1.5T to be quite generous.

Your 400 turn primary takes 8v, let's assume that's rms, so we have 20mV per turn rms, or about 28mV per turn peak.

Let's assume we have a sinusoidal voltage, and corresponding sinusoidal flux. The maximum rate of change of flux is 2.pi.f * the peak flux (simple calculus, or draw a graph of a sine wave and look). For 20Hz and 1.5T peak, that's a rate of change of 2xpix40x1.5 = 377 T/s.

The peak rate of change of flux 377T/s generates the peak voltage of 28mV/turn. The scaling factor between those is the core area, which is 28m/377 = 74u m2, or about 8mm x 9mm.

Are you sure you have that volts per turn right for the primary, 8v and 400 turns?

You will notice I have completely ignored the secondary, or turns ratio. They have no effect on the calculation, at least to first order.

• Simple! Only for sinusodial wavesforms and everything in SI. Otherwise it's good old Vt=NAB! Nov 30, 2016 at 20:15
• How you got 40 there , 2xpix40x1.5 = 377 T/s. Frequency is 20Hz. wB=2*pifB. Nov 4, 2021 at 15:28