transformer frequency fluctuations.

FOIWATER

How does a transformer react to a changing frequency

Obviously the inductive reactance of the primary changes, which causes a decrease in the current for an increase in reactance, then so too shouldn't the output voltage be less?

This seems to contradict Vs/Vp = Ns/Np

What I am saying is, intuitively, I'm thinking a frequency increase should increase the output voltage.

But arithmetically, I wouldn't get that increase

I was thinking that the output voltage DOES change, but at a multiple of the input voltage for a change in frequency so the ratio remains constant.

But this doesn't make sense since neither Vp, Np, or Ns is changing, therefore Vs cannot change either for a change in frequency.

I know there are more losses, I have a document which explains why they use different frequencies in different parts of the world although I can't really find anything pertaining to how frequency effects output!...

This is probably going to be one of those duh moments..

V = 4.44*phi*f*t so here it appears that it is frequency dependent

I now keep reading that the frequency increase will increase the output proportionally, because of this equation..

ahh, does an increase in freqequency change Vp magnitude by increasing it? thereby keeping the ratio constant because it will also increase Vs?

FOIWATER

surely someone can help me here!

psparky

I'll take an educated guess to keep the ball rolling.....take the higher frequency case.

I think the voltage on the secondary would be the same....because the voltage across the primary doesn't care how much resistance is there. Therefore the secondary is just going to react to the turns ratio.

However, the current would be much smaller through the primary due to the extra w in your JwL.....therefore lowering the current on the secondary side.

Therfore I would guess the effective KVA of your xfrmer would be lower with higher frequencies.....and your effective KVA would be higher with lower frequencies than spec'd.

Just a ball park guess.

jim hardy

Until other effects come in to play, all that happens for a frequency increase is this:

Flux is integral of applied voltage
assumiing sinewave, remember when you integrate sin(wt) your result includes a 1/w cos(wt) term.

Since w got bigger with frequency increase, flux got smaller. So its rate of change will stay same, which is necessary to produce same voltage.. n X dphi/dt....

Applied voltage stays same, flux gets a little smaller, rate of change of flux stays the same, secondary voltage is the same, the transformer works just fine.
Since flux got smaller , the magnetizing current necessary to make that flux got smaller, which is in accordance with inductive reactance getting larger.

If we get the principles straight in our head the formulas fall out naturally.

So long as you dont take frequency up to the point the core losses become significant that mental model should work.

Think it through and c'mon back.

old jim

Bob S

This is correct. The output voltage remains constant (independent of frequency) as long as the transformer iron does not saturate. More specifically the limit is on the transformer volt-second rating. Faraday's law is
[tex] V=-\frac{d}{dt}\int N \space B\cdot n \space dA [/tex]So
[tex] \int V dt=-\frac{d}{dt} N \space B\cdot A=\omega N A \space B_{max} \text { volt-seconds}[/tex]
Exceeding the volt-second rating is always a consideration when using a 60-Hz transformer on 50 Hz, for example.

psparky

What exactly do you mean when you say the "iron core saturates"?

What is the iron core doing before it saturates?

Bob S

See both the first two plots in http://www.electronics-tutorials.ws/...ysteresis.html
In transformer iron, if the field anywhere in the iron exceeds about 1.4 Tesla (see first plot), the hysteresis loop (second plot) widens, which represents loss of inductance and increase of primary excitation current (both real (losses) and imaginary (inductive) components).

FOIWATER

OK, the fact that flux decreases makes sense to me, not only through the formula but practically, since a frequency increase will cause more coil impedance, limiting current and thereby flux.

So you are saying that as the frequency increases, and the flux decreases, the dphi/dt seen by the secondary is the same as if the frequency had not changed? Let me explain my thinking by way of example... let's assume a 60hz and 30hz system. Is the frequency is halfed, and assume the flux doubled, the change in flux with respect to time of the secondary remains the same, is this what you are saying? Because although it crosses less frequently, it is more concentrated?

So this is the relationship that stays the same, interesting.

Now, as the iron core saturates and does not provide an increasingly permeable flux path, less flux now combined with less frequency yields less output voltage.

I think I have it thanks guys!.

FOIWATER

i think everyone should read this thread it's important and im sure not too widely understood.

so bob, the equation for voltage, t is what?

that equation still seems to lend itself to output voltaage changing, unless I don't understand its use as an equation

FOIWATER

this would explain higher losses at a lower frequency as well, since more current is drawn at the lower frequency.

psparky

Does anything make sense in all this "speel".....or is this all crappola?

FOIWATER

maybe someone else is more geared to let you know, I think the only part that isn't clear is the fact that you said the voltages don't care about the resistances...

Jim explains the reason the output voltage doesn't change for a change in input frequency, is because the frequency effects the primary inductive reactance right? It will increase it for a increase in frequency. So now, you have less current drawn into the primary. And you have less flux produced because of it.

So - that last paragraph summarized, you have MORE frequency, and LESS flux, as the frequency increases, the flux decreases proportionally, the net result is the dphi/dt term in E = N*dphi/dt remains the same, so the voltage (E) remains the same as well.

I think.

Bob S

In both the following equations, t is time. For [itex] \frac{dB}{dt} [/itex], this is the time derivative; e.g., d[sin(ωt)]/dt = ω cos(ωt). For [itex] \int V dt [/itex], this is the integral of voltage over a half cycle.

Bob S

Generally, the KVA rating of a transformer is lower at lower frequencies below its design lower frequency f_lower (volt-second limits), and lower above another higher frequency f_higher due to lamination losses (eddy currents). So there is an optimum design frequency range, f_lower to f_higher.

jim hardy

Nice work, guys. Great math, and great symbiotic thinking.

old jim