Take a full bridge switching circuit connected to the primary of a transformer.


No load on secondary. Voltage into full bridge is DC. Duty cycle of switching is 50%.

At t = 0, full bridge connects voltage to primary, lets call it "positive".

Faraday's law says slope of flux will be opposite in magnitude, and flux will increase (negatively) linearly until half cycle is over and voltage flips. At that point, flux will increase (positively), and so on.

What I see here is that flux starts at 0, and reaches some negative peak, then heads back to zero. It never crosses into positive territory and never changes sign.

In typical full bridge SMPS design, how is flux made to be bipolar and actually change polarity?

I am assuming it is desirable and possibly required to make it do this in order to prevent flux walking and also because if current is to change polarity in secondary, flux must also change polarity.


1 Answer 1


What I see here is that flux starts at 0, and reaches some negative peak, then heads back to zero. It never crosses into positive territory and never changes sign.

This certainly happens and is a real problem with power transformers.

It happens because the current through the inductive element is the integral of the applied voltage and this is the main reason why transformer primaries can reach high core saturation levels when voltage is initially applied. Another name for it is "inrush" current.

If you simulated the circuit with initially a half width pulse at the beginning all would be fine and you would see flux equally distributed both negatively and positively.

For another answer last year I drew the following picture. It was related to a regular power transformer that was "switched-on" when the voltage sine wave was passing through zero volts: -

enter image description here

As you can see, the current (red) starts from zero amps and rises to a larger-than-normal value initially. This is the same effect you are describing and the maths totally underpins the situation. However, for a real transformer with losses, the current waveform can acquire an average zero value after a few cycles and that is what the picture is attempting to show. Eventually the current settles down to the lower image in the picture.

It's worth noting that instead of initially activating the transformer when the sinewave voltage (blue) was passing through 0 volts it was activated at the peak of the voltage, the current waveform would be naturally passing through 0 amps and you would not see this effect. And that is exactly the same as initially driving it with a half pulse as mentioned earlier.

In reality these problems die-down due to real losses in the core and copper. Try simulating with a small resistance in series with the primary or a load on the secondary and see what happens.

Related question

  • \$\begingroup\$ Thank you Andy. I've seen your other answers on here and they've been helpful. I suspected this issue was essentially transformer inrush current but my intent is more an SMPS application and I didn't want to muddy the waters. From what I can tell reading white papers on industrial transformers (like transformers at factories), this problem is common even today. However, I would think in an SMPS application, something like this would be easily addressed since the switching is controlled, but I can't seem to find anything on it. Like a full bridge SMPS feeding a primary. Any ideas? \$\endgroup\$
    – Not Really
    Commented Nov 25, 2018 at 19:14
  • \$\begingroup\$ Also I'd like to say, I suspected your answer was the way, but staring at the waveforms, I haven't been able to grok how it eventually stabilizes with real losses, as you describe, but thats my problem, I just have to keep thinking on it. I've seen some other people point out thats the solution as well. \$\endgroup\$
    – Not Really
    Commented Nov 25, 2018 at 19:16
  • \$\begingroup\$ I think several amps chips have current limiting especially those used in flyback controllers. \$\endgroup\$
    – Andy aka
    Commented Nov 25, 2018 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.