core size for low frequency transformer

Question

I have read that for transformers operating at low frequencies, a larger core is needed. I am trying to calculate the core size for a single phase transformer operating at 2Hz.

I do not have a physical example of a transformer, but I have done a simulation of a transformer using Femm that gives me a value for the flux in Telsa.

I would appreciate it if anyone could comment on, or correct my method for finding the ideal core size of the transformer. I thought this formula for finding the induced voltage in the secondary coil would apply:

induced voltage = 4.44fNAB

Where

• f= frequency in Hz
• N= number of turns in the secondary coil
• A= cross section of the core area in Meters
• B= flux in the core in Tesla

My primary coil is 0.5mm diameter wire with 400 turns

• P coil resistance: 5.3695271 Ohms
• P coil current: 1.5 Amps
• P coil voltage: 8.05429 Volts
• P coil power: 12.08144 Watts

I would like to assume for the moment that the transformer is a “perfect” transformer with no core losses, just make my understanding easier for the time being. My questions -

Is this the correct formula to use?

and

If the above formula gives me a value for the induced voltage in the secondary coil that equals a lower Watts value in the secondary coil than in the primary coil, does this mean that the transformer core is too small for the frequency?

Answer 1

I have read that for transformers operating at low frequencies, a larger core is needed.

Correct when comparing to a transformer operating at a higher frequency but with the same line voltage. This is because of core saturation problems. Generally, a 240 V AC 50 Hz transformer run at 5 Hz should not have more than 24 V AC applied to it (for example).

I thought this formula for finding the induced voltage in the secondary coil would apply

The voltage induced on the secondary is related to the ratio of the turns between primary and secondary with a couple of provisos: -

1. Primary leakage components (such as copper loss) is not significant enough to "drop" much voltage. Any vot drop here results in a lowering of the voltage actually applied to the "useful" turns in the primary.
2. The coupling between primary and secondary is near 100%. Loss of coupling means loss of secondary voltage and increased leakage components on secondary and this has more of an effect under load conditions.

Is this the correct formula to use?

No, you need to take account of the turns ratio and the other points mentioned above.

If the above formula gives me a value for the induced voltage in the secondary coil that equals a lower Watts value in the secondary coil than in the primary coil, does this mean that the transformer core is too small for the frequency?

It can also mean that copper losses are significant. It can also mean that eddy current losses are significant. It can mean that coupling is not great and yes, it can also mean the core might be saturating.

Answer 2

Conventional transformer technology rapidly approaches impractical as you decrease the frequency. It is not clear whether this is a practical question or simply theoretical/speculative? It is also not clear whether this a SIGNAL application, or a POWER application? If you need galvanic isolation from a 2Hz SIGNAL, there are more efficient and economic ways of accomplishing that than traditional iron-core transformers. And if you are talking about POWER at 2Hz, it may be more practical to rectify the power to DC as 2Hz is rather difficult to deal with.

Answer 3

Cores are sized by VA capacity and V/f is constant.

You might (wish) to rethink your solution and try something else like a class D switched transformer at 20kHz.