# core size for low frequency transformer

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?

• Tesla is flux density, not flux. But I agree with Richard Crowley. 2 Hz @ 12 W is pretty radical to consider. You are talking about 2 Webers per half-cycle! Seriously? Core volume with laminated steel would be measured in liters! Like "one or two" liters! – jonk Nov 13 '16 at 21:53
• If you want 12 W at 2 Hz, you would need a minimum core size for 360 W at 60 Hz. If the windings would not fit, a larger core might be necessary. – Uwe Nov 14 '16 at 14:09

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.

• Thanks very much for your answer Andy, it has taken me a while to digest - most likely owing to the fact that I am mathematically “challenged”. – EddieP Nov 28 '16 at 15:38
• I wonder if I could just ask you about the point you made regarding the turns ratio? I have been wondering how to arrive at the correct value for the induced voltage of the secondary coil. I have researched and found two formula given to calculate the voltage induced in the secondary coil and was just wondering if you could comment on whether I have used the right formulas, and whether my thinking has any glaring, basic mistaken assumptions. – EddieP Nov 28 '16 at 15:39
• The first formula was (Voltage Out/Voltage In)=(turns on secondary coil/turns on primary coil) The other was (Voltage in Primary/turns on secondary coil)=(Induced Voltage in Secondary/turns on primary coil) .. My primary coil has 400 turns & 8.05429 Voltage and my secondary coil has 50 turns. Each formula listed above gave the induced Voltage in the secondary as 1.00678625 Volts Have I done the right thing here in calculating the voltage induced in the secondary? This is all very new to me. Thanks again and sorry if I have posted too many comment/questions! – EddieP Nov 28 '16 at 15:40
• If you have a turns ratio of 400:50 and the voltage on the primary is 8.05429 volts, then the voltage on the secondary is 1.0068 volts so, it sounds like you have done this correctly. – Andy aka Nov 28 '16 at 15:44
• Thanks Andy - much appreciated. As someone with almost no science education, sites like this, and people like yourself are a blessing. – EddieP Nov 29 '16 at 1:41

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.

• Thanks for your comment Richard. As I am sure you can see, I am very much a novice when it comes to this area (and many others!) When you say "transformer technology rapidly approaches impractical as you decrease the frequency" - what kind of frequency is considered the minimum to be practical? This is a Power application, at this point speculative but I would like to apply it to a practical design. – EddieP Nov 15 '16 at 14:40
• You can probably go as low as 10Hz, but you will need enormous and HEAVY iron for power at very low frequencies. It is simply impractical in the Real World and you will probably find NO examples of people doing such things. Back 100 years ago they used 25Hz for electric trains. But this is the 21st century and we don't have to do that anymore. Higher frequencies are MUCH MUCH MUCH more efficient. Even back 50 years ago they used 400 Hz in aircraft, and today switch-mode power-supplies (SMPS) use frequencies up in 10s or 100s of KHz. Using very low frequencies just seems very bizarre. – Richard Crowley Nov 15 '16 at 15:00

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.