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I was reading about how AC to DC converters work with a step-down transformer and then a diode bridge to convert the lower, stepped down AC voltage into DC. What I don't understand is since the input AC appears to be connected to the primary coil of the transformer, how does the DC load affect the power used from the AC supply?

Does the DC load somehow feedback and lower the resistance of the primary coil so that more power can be drawn?

When there is no load on the DC side, does power still flow through the AC primary coil, and if so, why doesn't it just melt?

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    $\begingroup$ May I add, that technically the DC isn't DC, it's just AC sine wave that has been prevented from going below 0V. Also, technically, the diode bridge would still be drawing some power, since there's voltage drop across it . $\endgroup$ – Sergiy Kolodyazhnyy Feb 9 '15 at 19:54
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    $\begingroup$ "... the diode bridge would still be drawing some power, since there's voltage drop across it." This is incorrect. $ P = VI $ and if I is zero then P is zero. $\endgroup$ – Transistor Mar 17 '18 at 12:55
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Does the DC load somehow feedback and lower the resistance of the primary coil so that more power can be drawn?

Yes. It would be simpler to analyze an AC load though. The diodes are not central to your question:

AC voltage source connected to transformer connected to load resistance

The impedance of RL is also transformed, so if you have a 10:1 transformer and RL is 2 Ω, the AC source will see the transformer as a 200 Ω resistor ($10^2⋅2$)

As the current in a coil changes, it creates a changing magnetic field. In the case of a transformer with a load, however, the change in magnetic field creates a current in the secondary, which immediately creates its own changing magnetic field in the opposite direction, cancelling out the primary's field. People tend to forget that an ideal transformer has no magnetic field while operating. Any change in either coil's field is immediately cancelled by a change in the other.

The "feedback" is caused by the same effect. The primary causes the secondary to change, and the secondary causes the primary to change in return.

When there is no load on the DC side, does power still flow through the AC primary coil, and if so, why doesn't it just melt?

With nothing connected to the secondary side, the secondary coil is open circuited and does nothing. It's just some metal that happens to be nearby. The circuit is now just an AC source driving the primary coil, which behaves as a lone inductor:

AC voltage source connected to an inductor

Ideal inductors do not consume any power; they just store energy temporarily in one half of the cycle and return it to the supply on the other half. Real coils are not made of perfect conductors, though, and have some resistance, so the power consumed by the primary coil will be determined by the resistance of the wire.

Also, it's not quite right to say "power still flow through the AC primary coil". "Current" is flowing through the primary, and the resistance of the primary to that current causes it to "dissipate energy" (or power) into the room. "Power" is actually the rate at which energy flows, and energy actually flows through the empty space between the wires, not in the wires themselves. Once you understand this, a lot of things make much more sense.

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A transformer offers resistance to AC current flow due to the magnetic field produced by the current flow. This "AC resistance" is termed "impedance" and is a function of number of turns, core material, air gaap in core , core dimensions and more.

When there is no load the applied AC voltage will cause "magnetising current" to flow. This will cause some losses due to eddy current losses in the core and copper losses due to resistance in the winding ("I squared R losses" as power = Current^2 x Resistance).

These losses are relatively small compared to full load power but not trivial at rest. A few percent of full load power would usually be good.

When a DC load is applied it loads the AC secondary circuit which is tightly couple by the core's magnetic fields to the primary winding. So the DC load resistance appears as if it is an AC impedance load on the primary side and input power increases to meet the load.

If you apply DC (rather than AC) to a transformer winding there is no ongoing magnetic field change, there is no impedance due to the varying magnetic field and current is limited by the resistance which is low compared to the impedance that should be being generated. If the DC supply has enough muscle power the transformer "just melts".

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Energy delivered to the primary goes to:

  1. The secondary load, off coarse, zero if no load,

  2. Copper losses: both primary and secondary IR losses of winding resistance. If the secondary has no load, that part of the loss is zero.

  3. Iron loss: A. To spin magnetic flux one way and the other the iron needs a magnetizing current. This current generates part of the IR loss in primary loss,

3B. Magnetic properties of iron are "sticky" in that residual magnetism remains when magnetized, and energy has to be spent to remove it before it reverses its orientation. The cycle is hysteresis loss, becoming heat.

3C. Magnetic flux induces 'eddy currents' circulating along the circumference of iron core ending as IR loss, the R being resistance of iron along the cross-section. Lamination of the core increases the effective resistance, since now the induce voltage on 'thin' laminate is smaller, the path of flow is longer.

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If a transformer has no load on the secondary, there is no current consumption. Maybe some leakage but this is minuscule. If you see the transformer as an inductor, this will imply that the transformer winding blocks AC and passes DC. Versus capacitance that blocks DC and passes AC. So an inductor is simply an AC resistor. If you do the ohms law math, your voltage is constant so the resistance of the coil is what changes when you add a load to the secondary winding. It is like completing the circuit allowing more current to flow.

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