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60Hz transformers are smaller than 50Hz transformers for the same power rating. Transformers designed for operation in the kHz range are even smaller. Why does transformer size decrease with frequency?

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    \$\begingroup\$ "If we increase voltage in kHz". What does this mean? \$\endgroup\$ – Leon Heller Dec 17 '14 at 20:51
  • \$\begingroup\$ 1,$s/voltage in/frequency into/ \$\endgroup\$ – Wouter van Ooijen Dec 17 '14 at 22:28
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Each AC cycle electric energy is converted to magnetic, and back again. The amount of magnetic energy that a transformer can 'store' is more or less linear in its mass. At a higher frequency, more of these cycles occur, hence the same transformer would transform more power, or the same power can be transferred by a smaller transformer.

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The other answers so far have given an intuitive explanation. I'd like to show you how the equations work if we model a transformer.

If we simplify the transformer by assuming the no-load resistance drop is very small, then we can say that the induced EMF in the transformer is equal to the applied voltage. If we assume that there is no load on the transformer and we assume that the applied voltage is sinuoidal, the induced EMF is sinusoidal and the flux is sinusoidal, we can say that the induced EMF in the primary is \$e_1=N_1 \frac{d\phi}{dt}\$, where \$e_1\$ is the induced EMF, \$N_1\$ is the number of turns in the primary, and \$\phi\$ is the flux in the core.

As I assumed above, \$\phi\$ is a sinusoid so we can write \$\phi = \phi_{max} sin(\omega t)\$. Then we can say that \$e_1=N_1 \frac{d\phi}{dt} = \omega N_1 \phi_{max} cos(\omega t)\$. If we rearrange that and also remember our assumption that the induced EMF is equal to the applied voltage, we get \$\phi_{max}= \frac{V}{\sqrt2 \pi f N_1}\$.

Basically what this equation says is that our peak flux is proportional to the applied voltage and inversely proportional to the frequency of our applied voltage and the number of turns in the primary of the transformer. The higher your flux is the more steel you need in your transformer in order to keep the flux density at a reasonable level, so that means higher frequency transformers can be smaller.

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The length of time between cycles where the transformer is charging the iron core decreases with the increased frequency.

Imagine trying to move a baseball 1Hz between your hands, then try it 1000x faster... it might be possible with a smaller ball, but its still difficult.

I has to do with the amount of magnetic flux that is being stored into the metal in the transformer core. The faster the switching, the less time it has to discharge/charge and therefore the correct device will account for this.

airplanes use 440hz transformers and 440hz AC for most systems, since they are smaller/lighter and weight is an issue in airplanes.

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  • \$\begingroup\$ Nitpicking : Large aircrafts traditionally use 115VAC, 400Hz, 3 phases, when they have a constant frequency generators (which is a complex electrohydromechanical device). Some other use variable frequency generators (which are simpler, more reliable) and mains frequency depends on engine speed, varying around 300Hz-700Hz. As these frequencies are straight into audio frequencies, radio/audiocommunication wires must be well isolated and placed far from power circuits and high power transformers can be quite noisy. \$\endgroup\$ – TEMLIB Dec 17 '14 at 22:08
  • \$\begingroup\$ sorry I don't understand this: High frequency cannot be transformed correctly with small transformers, did you forget a "but" at the beginning of this sentence? \$\endgroup\$ – JinSnow Apr 29 at 13:24
  • \$\begingroup\$ reading it now, 5 years after i wrote it, i don't know. I'll remove it for clarity. \$\endgroup\$ – Jeff Wurz Apr 30 at 15:50
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The transformers which handle high power have their size more or less proportional to the frequency as the power lost in iron increases with the frequency and therefore the Xmer gets heated up more rapidly. Therefore for its efficient cooling the surface area has to be increased which demands for larger Xmers. Whereas for low power Xmer temperature rise is not that big of an issue and its size is governed by the flux it has to handle (lower the flux, smaller is the Xmer). And the amount of flux depends on the length of the cycle.

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from emf equation of X'mer E=4.44fNAB (http://en.wikipedia.org/wiki/Transformer)

where E=voltage f=frequency A=Area N=number of turns B=magnetic flux density in general we can say A=E/(4.44fNB) for constant value of E,N,B if we increase F,then Area of core will decreases means size of transformer will reduce.

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    \$\begingroup\$ Can you make your answer more readable? Use double space for new line..for ex.: \$\endgroup\$ – Umar Feb 3 '17 at 15:49
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    \$\begingroup\$ Learn how to use the shift key \$\endgroup\$ – laptop2d Feb 3 '17 at 17:12
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Shorter answer, no math. AC energy transfers across transformers via induction. Induction takes place as magnetic field lines of force cut across conductors. Magnetic fields in AC expand and collapse at the frequency rate. Higher frequency means more lines of force cutting conductors, more energy transferred.

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The transformer permeable material (iron, ferrite, etc) that helps couple the primary and secondary can only handle so many volt-seconds before that material saturates. When the transformer material saturates, the presence of the iron goes away, the windings shows very low inductance and ends up shorting out the primary source. A lower frequency at a given voltage across the windings and therefore across the core material, applies more volt-seconds because it is positive or negative for a longer period of time.

So, increase the frequency and you can decrease the transformer size and volt-seconds by decreasing the number of turns of wire that makes up the primary and secondary.

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  • \$\begingroup\$ Please put personal info in your profile page, not in answers. Thanks \$\endgroup\$ – laptop2d Feb 15 '17 at 23:25
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That's because you need less inductance (hence transformer size) at higher frequencies.

I (inductor) = V/2pifL P = IV = V^2/2pifL

Thus, to provide the same power, the following couple needs to remain the same: (fL)_1 = (fL)_2 i.e. if you divide the frequency by 2, you need to multiply the inductance by 2. Decreasing the inductance means decreasing the size.

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