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How does the num. of turns & voltage on the primary translate to an ideal voltage on the secondary? Answered

In electronics, one of the things that has been the most confusing to me is the physics behind transformers. I have learned of that stupid formula a while back, where the ratio of turns is proportional to the voltage ratio and inversely proportional to the current, however, I know from reality that it is no where near that straight forward. As far as I can tell, it is a bunch of BS.

Anyway, I want to know what the REAL story is, how the AC current going through the transformer is related to the magnetic flux through the core, and how that relates to the output of the secondary. What seems counterintuitive to me is that more windings on the primary should yield to a stronger magnetic field in the core for a given current, which means for magnetic flux (???) which seems like it should yield to more induced current in the secondary. Is the answer related to inductive reactance (more turns = more inductance = more inductive reactance = higher impedance at a given sinusoidal freq. = less current per unit of voltage?)

Note: If you have a good physics and calc background and understand it to heart, pls do explain. However, I am currently only nearing the end of calc 2 w/ absolutely no formal education in physics, so I may have some difficulty with wank terms and vocab. (a.k.a. do not try to show off how many terms you know attempting to explain things in the most complicated and technical way possible.) Also, any good tutorials online that go into depth into the topic?

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The reason I want to know these things is that I wasn't to build a good 6A 0-+/-15V dual isolated channel linear power supply with many transformer tabs so I can keep the voltage drop on the pass elements as low as possible at variable output's so huge $$$ heatsinks are not necessary due to lower low power dissipation. (I have mostly figured out the LPS circuitry, but have not found a suitable Xformer online.) I can get the laminated magnetic cores needed relatively cheaply, however if I have to do some custom windings, I want to make sure I do not choose the wrong number of windings for the core. (obviously one turn on the secondary and 5 turns on the primary will pop a breaker @ 60Hz... )

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steveastrouk
steveastrouk

6 years ago

You can only get current to flow in a load, unless you consider the load you can't get the right concepts.

The Ideal transformer has several properties.

All flux in the primary winding links all turns in the secondary.

The resistance of the windings is zero

The iron does not experience saturation, or hysteresis losses.

The interwinding capacitance is also zero.

....the more you pay, the nearer ideal the transformer is, and the closer the approximations are to reality. Bear in mind that a perfect transformer is lossless, so the energy in = energy out.

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-max-
-max-

Answer 6 years ago

OK, I guess I figured out what is going on mathematically (in the ideal case), and intuitively too!

Now the question is how to determine losses in flux for real materials...

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-max-
-max-

Answer 6 years ago

I understand what you mean by loading. What would occur under minimal load (open ended), and while loaded down with a full wave bridge rectifier with capacitor on the end (which will result in highly non-linear loading w/ respect to AC voltage from the Xformer, assuming 6A load)

Anyway, I was looking for something a little better than that formula mentioned. Like I said, if I was to make one from, IDK, this core, http://www.ebay.com/itm/like/181261347797?lpid=82&...

Obviously if I do not wind enough turns for the main's primary (which I wonder how hard or time consuming it would be anyway...) so that it does not draw too much reactive power (or whatever it is called when power is bounced back and forth) but what happens if I wibd too much? ESR becomes too high?