Consider a transformer primary driven by sinusoidal mains.

In my circuit I have an NTC thermistor in series with the primary which I short out after a few cycles to limit the inrush current at start-up.

This transformer is a very large torriod in my case 1800VA rated.

Now modelling this transformer as ideal (no losses) with a magnetising inductance \$L_m\$ in parallel with the primary we can work out the flux swing in the core.

$$v = N \cdot Ae \dfrac{\text{d}B}{\text{d}t} \Rightarrow \Delta B = \dfrac{1}{N \cdot Ae} \int_{- \frac{T}{2}} ^{\frac{T}{2}} V_{rms} \cdot \sqrt{2} \cdot \sin(2 \pi \cdot f \cdot t) \text{ d}t $$

$$\Delta B = \dfrac{V_{rms} \cdot \sqrt{2}}{N \cdot Ae \cdot \pi \cdot f}$$

Now because the transformer was allowed to saturate during start-up with current limited by the thermistor \$B_{max} \approx \dfrac{\Delta B}{2}\$ and \$B_{min} \approx - \dfrac{\Delta B}{2}\$.

Now we consider the transformer to have been running for several hours and we have a single negative half cycle drop-out (to 0V). Just prior to the drop-out flux density \$B\$ was at it maximum value ,and because the voltage has remained constant zero over the missing half cycle, has not changed. The next positive half cycle causes \$B\$ to increase and the transformer may saturate again. In my experience most commercial transformers do saturate.

Now let's change the scenario: Again we start the transformer but instead of having a half cycle drop-out we just turn it of precisely at the end of a positive half cycle. The flux density is at a maximum value. Does this stay the case forever as this simplified version of the maths suggests or is there a mechanism to reduce the flux density over time?

If so what is it and how would you estimate the flux density at some later time \$t\$?

  • \$\begingroup\$ If you switch off at maximum magnetising current you must remember that switching off is opening the primary circuit not dropping the supply voltage to zero, this causes a large voltage spike (V=L*di/dt and all that), and dissipates the energy stored in the field in arcing the contacts (or in better designs in a snubber resistance). Flux is (ignoring iron losses, not a safe assumption!) proportional to current, so no current, no flux. \$\endgroup\$
    – Dan Mills
    Dec 17, 2017 at 12:52
  • \$\begingroup\$ @DanMills Here I am explicitly setting the mains to zero volts not opening a switch this is to understand dips and surges. I am aware if I break the connection there will be a voltage spike as the magnetising current has to go somewhere. And will cause the switch to arc if we don't provide another path. \$\endgroup\$ Dec 18, 2017 at 20:10

2 Answers 2


There is a mechanism for resetting the flux in the core eventually, and that's the mechanism you explicitly chose to ignore in your question, the copper winding resistance.

With a truly lossless inductor, at a certain flux (which means at a certain current), with zero volts on the input, the flux and current will remain at those values indefinitely. Consider what happens in a superconducting magnet, when shorted by its internal superconducting switch.

I'll wager that your 1800VA toroidal is wound with copper rather than superconductor. What happens is the current through the winding resistance generates a voltage, which reduces the current, with a time constant of L/R. Plug that into the normal exponential relation to find flux and current as a function of t. Of course L/R is infinity for the superconducting case.

A dropped half-cycle is just equivalent to more volt.seconds than usual, so would be expected to saturate the inductor.

It's interesting how the recovery from a switch-on saturation occurs. Although the average DC voltage applied from the supply is zero, and so you would not expect the flux to slew cumulatively in either direction, in saturation the transformer draws a heavy current, creating a larger voltage drop in the supply and winding resistance. This provides the net average voltage which gradually resets the flux to mean zero. With a lossless transformer, and stiff supply, the transformer would continue to saturate for a period every cycle. The same argument can be used to explain why a unidirectional (diode connected) load on the secondary of a transformer can drive it into saturation.

  • \$\begingroup\$ For clarity are you asserting that although the secondary current has fallen to zero because there is no voltage to drive it the magnetising current though small is at a maximum. If this is the case then when I open circuit the supply I should momentarily see some voltage on the primary. This makes logical sense and I will test tomorrow when back in the lab. I will up-vote and accept this answer if (as expected) confirmed. \$\endgroup\$ Dec 13, 2017 at 19:29
  • \$\begingroup\$ I'm saying nothing about the secondary current, apart from my throwaway at the end. I thought we were talking about the primary inductance and the magnetising current? The secondary current and its scaled analog in the primary can be linearly superposed on the primary voltage/magnetising current relationship (at least in the lossless winding case) and so ignored. Core flux is entirely due to primary voltage (in the lossless case) so we still ignore load currents for flux calculations. Load/magnetising currents do interact through voltages generated in the winding resistance. \$\endgroup\$
    – Neil_UK
    Dec 13, 2017 at 19:39
  • \$\begingroup\$ I am testing this with a purely resistive load so no voltage on primary means no current on secondary. Sorry for any confusion -- I am almost certain this answer is correct but want to test before accepting. Thanks for your insight, I think you have covered the point I was missing. \$\endgroup\$ Dec 13, 2017 at 19:49

There is an exact relationship between remaining core flux density, time of switching, impedance, voltage,.... and transformer inrush current. This current is due to the phenomena you are describing, the flux is 90 deg out of phase with respect to the voltage. For example if the transformer is switched on at voltage zero, then the flux has to be at peak value (looking as sine wave) so the core gets magnetized by DC flux.

From now and onward the total flux in the core is a sum of superimposed DC and AC flux. The DC portion gets aperiodically vanished within few cycles.

Maybe the best representation of this phenomenon is the inrush current formula and the scope trace.

enter image description here

If you switch the transformer off, then this will have an impact on next switch on due to the remanence - remanent flux density.


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