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?

  • \$\begingroup\$ Transformers require AC, which results in a changing flux. That in turn induces a current in the secondary. \$\endgroup\$ – Roger Rowland Apr 14 '15 at 19:44
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    \$\begingroup\$ the peak value of the flux is constant regardless of load. But the actual value is continually changing and reversing with the AC waveform. \$\endgroup\$ – Brian Drummond Apr 14 '15 at 20:25

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".

  • \$\begingroup\$ Ah HA! So my making and eating flux intuition wasn't completely mistaken then, is that right? \$\endgroup\$ – scanny Apr 14 '15 at 22:11
  • \$\begingroup\$ No, I think that intuition is quite good. The formula you quote not containing current is just about the flux in the core, which is the flux made but not eaten. \$\endgroup\$ – Michael Karcher Apr 14 '15 at 22:30
  • \$\begingroup\$ I expanded my answer to contain a paragraph about the part about the "unused" flux that remains in the core. \$\endgroup\$ – Michael Karcher Apr 14 '15 at 22:36
  • \$\begingroup\$ I re-read the second paragraph, and I will stick to it, possibly its unclear what I meant. Yes. If the current in the primary increases, the flux generated by the primary increases. But that only happens when the secondary current increases, too, which increases the counter-flux. The sum of flux and counter-flux stays the same, though, and that is what I call the "flux in the core". Also, you are right that the turn determines the voltage ratio. This is because for each turn, U=-d/dt(phi) is valid, with phi being the "flux in the core". \$\endgroup\$ – Michael Karcher Apr 15 '15 at 5:54
  • \$\begingroup\$ But if the flux does not take place in the energy transfer, how does energy get transferred? The flux amplitude has to do with current not voltage. How can "flux amplitude is proportional to the primary voltage" apply if the voltage does not change and flux does? The flux in the core will be in phase with the current since it produces it. If the load is lagging it will lag, etc. Give me a formula. As an engineer, I can never accept the magic happens. \$\endgroup\$ – StainlessSteelRat Apr 15 '15 at 11:44

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 :)


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

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    \$\begingroup\$ Welcome to EE.SE, Tom. Your answer could do with some paragraph breaks (2 x enter) and fixing some random capitalisation. It's a rather difficult read as is. Signatures are discouraged as they automatically appear beneath your post. \$\endgroup\$ – Transistor Jul 11 '17 at 12:35

Again, after posting a reference to the use of the Poynting Theorem in my previous statements, it can be equally shown the the iron core transformer can transfer energy from primary winding to secondary winding with ZERO "LEAKAGE" Flux. To prove this to myself, I once wound a power transformer using a "Coaxial"winding set up. To do so I wound many turns of a regular RF type coaxial cable around the center leg of my test transformer and simply used the center lead as the primary winding and the shield as the secondary. I detected a very small leakage flux directly next to the winding. When placed on load the small flux I had previously measured almost completely disappeared! That is a fully loaded iron core transformer with essentially NO leakage or stray flux! I realize that such a "co-axially would" transformer would be somewhat impractical except in special applications, but it raised some interesting questions in my mind regarding the transfer of energy from primary to secondary...


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