How does a transformer transmit power from the primary to the secondary?

Question

I'm trying to get my head wrapped around transformer operation and in the process regretting the times I snoozed in my Electromagnetics class as a EE student back when I was a lad :)

I'm looking for an intuitive understanding, but not just an analogous one. I'd like it to be grounded in the actual physics of what's happening. I've found several excellent sources on the web, but they all seem to skirt this question.

I've come across a few interesting hints and am now tantalizingly close, I think, but still yearning :)

Fact 1: Although varying sinusoidally, the "peak-to-peak" flux, so to speak, in a transformer's core is essentially constant (for a given voltage applied to the primary), regardless of the load.

My intuitive hypothesis was that variation in the "strength" of the flux was what transferred the energy from the primary to the secondary, but this fact would seem to contradict that theory. I had thought that the primary makes a bunch of flux based on the current flowing through it and the secondary sucks it up to make current of its own. No dice, it seems.

Then of course there's the fact that the formula for flux involves only voltage, time (frequency), and turns :)

Fact 2: The current in the primary is (approximately) 90 degrees out of phase with the voltage at no load, and approximately aligned in phase at full load.

This fact seems very promising and also curiously satisfying. It would imply that the Volt-Amps (VA) of the primary is constant and only the power factor changes as the current load on the secondary increases.

But I still don't get how the energy is actually being transferred. It seems vaguely like the flux is just there as an energy conductor or something and some other phenomenon is actually doing the energy transfer bit.

Can someone see what I'm missing and explain what's actually happening in there?

Answer 1

The answer has been totally rewritten to fulfill the request of a mathematical, format treatment, as asked for by StainlessSteelRat. The bold-face part in the center of this answer is the actual answer to your question. The remaining part is to show how I get to that formula and to connect your two facts (which I consider true) to the answer.

Take a look at the following formulas are important to understand the transformer:

• Faraday's law of induction (for a coil with \$n\$ turns): \$U=-n\frac{\mathrm d}{\mathrm dt}\Phi\$.
• Hopkin's law (aka magnetic Ohms law): \$\Phi = \frac{nI}{R_m}\$

Faradays law describes how the rate of change of the magnetic flux \$\Phi\$ through a single turn of a coil is related to the voltage \$U\$ induced in that coil. Hopkin's law describes the connection between the current \$I\$ applied to a long coil with \$n\$ turns and the flux \$\Phi\$ caused by that current. The value \$R_m\$ (magnetic resistance) depends on the geometry of the core, so it is a constant value for a given transformer.

If a sinusodial voltage \$U(t)=U_0\sin(\omega t)\$ is applied to a coil, induction of faradays law yields

\$\Phi(t) = -\frac{U_0}{\omega n}\cos(\omega t)\$ (The integration constant has been chosen in a way that the average flux vanishes)

In a transformer, there are two coils on a shared core. The resulting flux in the core \$\Phi_c\$ is the sum of the flux generated by both coils on the core:

\$\Phi_c = \Phi_p + \Phi_s\$

The law of induction relates the derivative of total flux through a coil (whether it is caused by the current of that coil or is caused by external fields) to the induced voltage. So you can write faradays law for both the primary and the secondary winding as

\$U_p = -n_p\frac{\mathrm d}{\mathrm dt}\Phi_c\$; \$U_s = -n_s\frac{\mathrm d}{\mathrm dt}\Phi_c\$.

Note that while the voltages on the primary and secondary side (as well as the turn counts) are different, the shared total flux \$\Phi_c\$ appears in both equations. It can be eliminated yielding the well-known transformer equation

\$U_s = -\frac{n_s}{n_p}U_p\$

More step-by-step: By integration the \$U_p\$ equation (as already shown for sinusodial voltage above), the flux \$\Phi_c\$ caused by a voltage source on the primary side can be calculated. By differentiating \$\Phi_c\$ again in the \$U_s\$ equation, the secondary voltage caused by the flux can be calculated.

Note that up to now, the current did not really take place in the discussion of the transformer. The equation (and thus the integrated \$\Phi_c\$) is valid for all load conditions of an ideal transformer driven by an ideal voltage source. (This is your "Fact 1")

To add the effects of the load, consider the two parts that make up the resulting flux:

\$\Phi_c = \frac{n_p I_p}{R_m} + \frac{n_s I_s}{R_m}\$

As we assume the primary side is an ideal voltage source, there is nothing known about the the current \$I_p\$, as the source would deliver every current needed so that the primary voltage is as required. If we assume a simple ohmic load, the load current is known, though, it is \$I_l = U_s/R_s\$, with \$R_s\$ being the resistance of the load. At this point, signs get very important. If I call one terminal of the secondary "ground" and the other one "live", the voltage between the "live" and the "ground" terminal has a clearly defined sign. On the other hand, the current has to flow from "ground" to "live" inside the transformer as it flows from "live" to "ground" in the load and the other way around. So if secondary voltage and load voltage are defined as the voltage from the "live" to the "ground" terminal, they are obviously equal-signed, whereas the currents, measured as "current flowing into the respective live terminal" are oppositely signed. This means \$I_s = -I_l\$, so the formula for the flux can be written as

\$R_m\Phi_c = n_p I_p - n_s I_l = n_p I_p - n_s \frac{U_s}{R_s} = n_p I_p - \frac{n_s^2}{n_p}\frac{U_p}{R_s}\$

The first term of the right hand side shows the flux "created" by the primary winding, and the second term the flux "consumed" by the secondary side. The part of the flux consumed by the secondary side corresponds to the energy transmitted from the primary to the secondary. Solve that equation for \$I_p\$ to obtain:

\$I_p = \frac{R_m\Phi_c}{n_p} + \frac{n_s^2}{n_p^2}\frac{U_p}{R_s}\$

substituting a sinusodial voltage for \$U_p\$ and using the integrated term for \$\Phi_c\$, one obtains:

\$I_p(t) = -\frac{R_m}{n_p^2 \omega}U_0\cos(\omega t) + \frac{n_s^2}{n_p^2}\frac{U_0}{R_s}\sin(\omega t)\$

The first term on the right-hand side is out-of-phase with \$U(t)=U_0\sin(\omega t)\$, so it describes reactive power and is load-independent, while the second term is in-phase, so it describes active power and is load dependent. If the transformer is unloaded, the second term vanishes and the first reactive term remains, causing the 90° phase shift, while on a highly loaded transformer, the second term dominates so that the phase shift gets very close to zero. This is your "Fact 2".

Answer 2

From what I know, the delta in primary coil current (as per an AC system) is what builds up energy inside the metal core of the transformer, which in turn generates a magnetic field in the opposite direction of the current flow (so it is in effect resisting the current flow). But, at the same time, in the secondary coil this magnetic field causes induction, which generates a current flow in the secondary circuit, which is what gives the current there. The fact that two different principles are at play generates the phase difference.
But I'll let any expert poke holes in my explanation for my own learning and understanding :)

Answer 3

For a circuit, for example, there is a voltage source in a simple circuit like above, and the voltage source is followed by a resistance. For such a circuit, the energy transfer from the battery to the resistance is the energy flow corresponding to the Poynting vector in space.

But for two separate circuits, such as the primary coil of a transformer followed by the power supply, and the secondary coil followed by the load resistance, the corresponding energy transfer energy of Poynting vector is not all right. In this case, the energy flow corresponding to Poynting vector must be decomposed into three parts: (1) Poynting energy flow of primary coil electromagnetic field, (2) energy flow of secondary coil electromagnetic field, and (3) energy flow corresponding to mixed term. The energy flow of the primary coil and the energy flow of the secondary coil belong to the self energy flow, and the energy flow of the mixed term is the mutual energy flow. Because the self-energy flow is pure imaginary, it does not transfer energy on average. The mutual energy flow can be a real number, so it can be responsible for transferring energy. for details please see the reference:http://pubs.sciepub.com/ijp/11/1/4/index.html

In the following figure, the mutual energy flow have been used to calculate the energy flow from primary coil to the secondary coil. The self-energy flow (corresponding to the Poynting vector) failed to do so.

Answer 4

Power (energy) transfer in iron core transformer has never been given a thorough treatment outside of the math realm. An interesting fact is illustrated by te Poynting Theorem. In the case of the transformer on load, the core flux is almost constant, however the "LEAKAGE" flux is load dependent anc can reach staggering levels, that is flux intensity. That is flux outside of the core around the Primary winding while on load. that flux, is not dependant on core permeability, as even if the core had infinite permeability, the load flux (Called Leakage) would still exist. Many sources claim this so called "leakage" flux is essential for power flow thru the transformer. I agree, but I think it is fundamentally wrong to call it a "LEAKAGE" flux. The core GUIDES energy to the secondary. The core flux is 90 degrees lagging the primary voltage so the only energy associated with the core is purely inductive storage that gets cycled to and from the supply every quarter cycle and hysteresis and heating losses due to eddy currents in the core. It would seem that the core flux induces EMF in the secondary coil and the "LEAKAGE" flux surrounding the loaded primary and in phase with the voltage satisfies the poynting theorem for energy flow. As an illustrative example, consider the old constant current street lighting transformers of yesterday, where the secondary winding was free to move up and down away from the primary to keep the series current stable; when on low load the electromagnetic forces developed from those (LEAKAGE) fluxes were weak and the coils were physically close, when on heavy load, the secondary winding would be repelled far away from the primary which would reduce the energy flow thru the transformer. Textbooks explain by stating the leakage flux around each winding, now far apart choke current flow, if close together they would tend to cancel each other. I am saying, maybe the primary "LEAKAGE" flux cannot not be close to the secondary, thereby reducing the power flow. Just my thoughts, but of intense interest to me........ Tom Cosby