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Looking at different datasheets of SMPS transformer, I found that the manufacturer always specifies voltage in a certain frequency range. Why is this parameter important?

For example this transformer:

https://www.weonline.de/katalog/datasheet/750311564.pdf

voltage is only specified for 40 V @100kHz.

Is this voltage-frequency parameter somehow related to volt \$\times\$ second or saturation?

EDIT: Removed the insulation part. I went to fast through the datasheet. So, the part of the insulation voltage is quite clear, I am interested in the relation between voltage-frequency and the magnetic properties.

Does it mean that if If I would apply 80v at a higher frequency (>200 khz)that the core would not saturate since the increased voltage is compensated with lower “seconds”

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The manufacturer always specifies voltage in a certain frequency range. Why is this parameter important?

Because these are the parameters the transformer has been specifically designed for.

As depicted, this transformer's primary input of 10-40v @100kHz will produce X to 5v/3A at the output. (Since the input ranges from 10-40vAC, then 10 is 1/4 of 40, so the output at (input=10V) would be about a quarter of 5V/3A, or 1.25V/0.75A.)

The 500v rating is an insulation-test between the windings, to ensure that it will not break down during service (and accidentally inject 40v pulses into your 5v device.)

What would happen if I apply 200V at a certain frequency ?

If you applied 200V at any frequency between windings (such as one lead to the primary winding and one lead to the secondary winding), then nothing will happen as this transformer has already passed it's 500V isolation test.

If you applied a slightly higher voltage than 40v (@100kHz) to the primary winding, the output would rise linearly from 5V to match. However, the transformer would run hotter than designed and may fail prematurely. Attempting the primary at 200vAC@100kHz (5x the voltage it is rated for) would far exceed the saturation limits (explained below) and likely burst into flames.

Is this voltage-frequency parameter somehow related to volt.second or saturation?

This transformer uses a ferrite core material which enables the magnetic flux to change orientation at a very rapid rate (in this model, 100,000 times per second or 100kHz.) This fast rate is desirable because, for a given power rating, the faster the rate-of-change, the smaller the core needs to be. This means a smaller and lighter transformer.

Saturation is the inability of the core material to "hold" any more magnetism. Ferrite is very good at "collecting" or "holding" this magnetism, but has limits. Attempting to put more power into (or take out of) the transformer than it is designed for will result in exceeding the saturation limit. When this happens, the core cannot perform it's function any longer, reducing efficiency and generating heat. Slightly exceeding the limit infrequently, for a short time, may be ok; just realize that exceeding it is bad and should be avoided.

As for frequency, this is limited by the composition of the core material. If the transformer is rated for 100kHz, then faster alternations than this will incur losses, as the core material is only "so fast." Drastically exceeding this (such as 200kHz) will be very lossy, resulting in reduced output and excess heat generation because the core material is too slow. It is possible to make the core material glow red hot by doing this, just like an induction heater.

Going lower in frequency is interesting. Faraday's law of induction states that the voltage across a coil is equal to negative the time rate-of-change of flux, times the number of turns of wire. For a sinusoid wave, the time-rate-of-change is 2*pi*frequency. So for a fixed flux capacity (the core material) and fixed number of turns (the winding), reducing the frequency means the voltage must reduce by the same factor. i.e. for a 40v primary at 100kHz, but using 10kHz, you can apply at most 10kHz/100kHz*40v = 4V. This means that using 4V@10kHz will produce the same saturation as if it were powered from 40v@100kHz! Not very efficient...

In short, high-frequency transformers are carefully designed for a set power, input range, and frequency. There is a little wiggle-room in these specs but for the most efficiency and longest life, stick to the datasheet specifications and be careful of exceeding them.

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  • \$\begingroup\$ Great answer! Exactly what I was looking for. So, I guess the maximum of 100 KHz is related to the core losses for this particular material?is it correct to say that if the core losses would be negligible then It would be possible to apply a higher voltage at a higher frequency? \$\endgroup\$ – Navaro Nov 19 '19 at 17:19
  • \$\begingroup\$ That is correct. However ferrite, as good as it is, has it's limitations. There are different grades of ferrite also, so it may be possible to find a similar-performing transformer in a 250kHz or even 1MHz version, while they become progressively smaller (and more challenging to design for.) \$\endgroup\$ – rdtsc Nov 19 '19 at 18:47
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The 500V is a voltage used for insulation test between different coils, not voltage over a single coil.

Otherwise the transformer works best in the frequency and voltage range it is designed for. If the volt × second is higher then the core starts to saturate.

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  • \$\begingroup\$ You are basically not answering the main question \$\endgroup\$ – Navaro Nov 19 '19 at 15:28
  • \$\begingroup\$ @Navaro, he is. You've misunderstood the spec. The 500 VAC spec is not for a voltage applied across one coil. It's for voltage applied common mode to one coil relative to another coil. The 40 VAC spec is applied across a single coil. Therefore you can't compare them to each other and there's no contradiction between them. \$\endgroup\$ – The Photon Nov 19 '19 at 16:03
  • \$\begingroup\$ Ok I will change my question \$\endgroup\$ – Navaro Nov 19 '19 at 16:44

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