could you explain what it is that causes harmonics in transformers?
I've read that these are the reasons, but it still doesn't give me a proper explanation.
- Saturation of the core
- Magnetizing current
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2\$\begingroup\$ Take a step back, assume we have a black box with an input and an output. We apply a pure sinewave at the input. 1) if the transfer function of the black box was ideally linear like: y = A * x (where x = input, y = output and A is any real number you like), would there be any harmonics at the output? 2) Now the black box behaves slightly non-linear for example: y = A * x + B * x^2 (where B is quite small but not zero). Will there be harmonics at the output? \$\endgroup\$– BimpelrekkieFeb 23 '21 at 21:20
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2\$\begingroup\$ Once you thought about what I wrote above, suppose that the black box is a transformer, what is the transfer function of a transformer? Is it ideally linear ( y = A * x ) or not? Is the relation input voltage => magnetic field ideally linear? Hint: en.wikipedia.org/wiki/Saturation_(magnetic) do magnetic materials saturate or not? \$\endgroup\$– BimpelrekkieFeb 23 '21 at 21:22
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\$\begingroup\$ @Bimpelrekkie So if I understood you correctly, in 1) we would have no harmonics at the output because the black box is ideally linear. However, in 2) we have distortion caused by the saturation of our non-ideally iron core, which will output harmonics. \$\endgroup\$– gripenFeb 23 '21 at 21:27
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\$\begingroup\$ Have a look at tex.stackexchange.com/questions/127375/…. You'll see a highly distorted waveform and then the Fourier transform showing the level of the harmonics. \$\endgroup\$– TransistorFeb 23 '21 at 21:31
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\$\begingroup\$ @gripen So if I understood you correctly... Yes, you understood correctly. The non-linear transfer of the transformer's magnetic core introduces harmonics. \$\endgroup\$– BimpelrekkieFeb 23 '21 at 21:48
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could you explain what it is that causes harmonics in transformers?
A transformer is only approximately linear, and that only within limits.
The net current in a transformer (the magnetizing current) creates an "H" field proportional to the instantaneous current. The "H" field induces a "B" field, but the relationship between H and B is non-linear. Changes in the "B" field, create induced voltage in the windings, and this voltage is proportional to the rate of change of "B".
The relationship between "H" and "B" becomes especially non-linear when the core is "saturated". That is, when "H" is very high, a change in "H" causes very little change in "B" (compared to the change in "B" when "H" is small).
All this entails that if the voltage is a perfect sine wave, the current will not be a perfect sine wave, but a distorted sine wave. Every distorted sine wave is equal to the sum of a sine wave of the fundamental frequency and sine waves of harmonic frequencies.
answered Feb 23 '21 at 21:29
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A bit more detail: because of core saturation, the output signal will vary linearly with the input signals only for small signals. When saturation first appears the predominant nonlinearity will be cubic, so for sinusoidal input sin(x) (where x = 2 pi f) the output will have one term proportional to sin(x) and another one proportional to sin(x)^3. The cubed term contains sin(3x):
$$\sin^3(x) = \frac{3\sin(x) - \sin(3x)}4$$
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\$\begingroup\$ Marcus thanks for the type setting. I could not get it to work in-line, but I see that it does work for displayed equations. \$\endgroup\$– 10ppbFeb 28 '21 at 15:59