It's much worse than you think. If you had a transformer with Lpri = 10H, rewinding with HALF the turns would reduce Lpri to 2.5H, and the core would be pushed so far into saturation that the unloaded (but energized) exciting current would be so large that the transformer would likely catch on fire (if the circuit breaker didn't pop!). The non-linearity induced by saturation is strong, and the increase in exciting current due to moving well into saturation is more than one might think. To see this, use a variac to increase the applied primary voltage and watch how fast the current rises. If you were to connect a 50 VA, 115 VAC rated transformer to a 230 VAC source, this would be equivalent to halving the primary turns. The current would be so large that it would likely pop the house circuit breaker for that circuit.Thankyou for that...great points, thats very interesting and worth looking in to....
If i have a 50Hz iron transformer with enough turns to give Lpri = 10H.....then say i reduce the turns (on the same core) such that Lpri now becomes 200mH. Then are we saying that putting it across the mains 240VAC again would now result in B(sat) being exceeded, and also "delta B" being pushed up to a point that the core loss grew much higher?
(Note, when the turns were reduced, lets pretend that the wire diameter was increased such that winding loss remained the same, as when Lpri was 10H)
Do we say that Ferrite is "linear" in that the B by H curve is a straight line up until almost saturation?
However, 50Hz iron core material is "non linear", and its B vs H curve is not ever straight?(unless of course looking over a vanishingly small section)
Thanks...i know i am missing it...why the magnetising current is not sinusoidal?...with a V (sine) across an inductor, you would think the current in the L would be the integral of V w.r.t. time.....ie, another (co)sinusoid......so why is it linear?, if you look at the magnetising current in the typical Tx it is zero at the mains peaks and then rises linearly but peaks at the mains V zero x-ings
If an inductor consisting of a coil of wire were wound on a non-ferromagnetic material (the core), such as wood, plastic, glass, air, vacuum, the B-H loop wouldn't be a loop, but would be a straight line through the origin. Then the relation between B and H would be perfectly linear. Given a magnetic field in the "core" of an inductor, the flux density (B) is proportional to the integral of the applied voltage. So the variation of B with time will be a sinusoidal wave (if the applied voltage is sinusoidal), and the corresponding H will also be sinusoidal if the "B-H curve" is just a straight line. Since H has units of ampere-turns, the current will be proportional to H.Thanks...i know i am missing it...why the magnetising current is not sinusoidal?...with a V (sine) across an inductor, you would think the current in the L would be the integral of V w.r.t. time.....ie, another (co)sinusoid......so why is it linear?
I know what the green and yellow curves are, but what is the third (purple?) curve?I remember an old thread discussing transformer magneizing current and inductance behavior https://www.edaboard.com/threads/transformer-inductance.302652
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I have an old no-load measurement related to the discussion. It's made at a cheap 12 VA print transformer, you see that the transformer is designed with relative large core losses, about 10 % of rated power and respective high magnetizing current. A 50 VA transformer as discussed in the initial post cab be expected to have better efficiency.
View attachment 172664
- it appears to be the instantaneous VAthird (purple?) curve?
Sure, product I*V, the same as in post #24. It's average value is transformer loss.it appears to be the instantaneous VA
Very small transformers often are designed to be "short circuit proof".notice how low the losses are; only 278 mW compared to 1.57 W for the previous one:
Neither is short circuit proof. The first one is very ordinary, laminations likely not even as good as M6 material. The second one has very low loss core material, probably M3 or equivalent.Hi,
Very small transformers often are designed to be "short circuit proof".
This has something to do with coupling, but maybe with loss, too.
Is one of the above transformers a "short circuit proof" one?
Maybe it's just saving cost by reduced wire cross section and reduced core size in the 5W transformer.
Klaus
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