Inrush current sources in a conventional supply

[rtarbell]

With a conventional power transformer, bridge rectifier, and filter caps for + and - supply rails, is the main source of inrush current due to the energizing of the transformer windings, or is it due to the sudden current needed to charge the filter caps?

[zigzagflux]

With typical power transformers, inrush current varies between 5 and 12 times the nominal current rating of the primary winding. The actual value depends to a great degree on where on the AC sine wave the transformer is turned on. Being turned on at the zero voltage crossing is worst.

For most capacitor input filters, you could do the math with C, ESR, and xfmr impedance, but capacitance inrush could be greater. Keeping in mind, a dead short on a transformer secondary produces Inominal / % impedance. So for a 720V, 120VA xfmr at 10% impedance, you have Inominal = 120/720 = 0.167A. Therefore shorted secondary amps = 0.167/0.1 = 1.67A. The secondary winding cannot put out more than this. Reflect to the primary, and you have primary amps = 1.67*(720/120) = 10A. So the maximum primary inrush possible (neglecting magnetizing inrush) for any size capacitor would be 10A.

As I stated above, magnetizing inrush would be at most 12*(120/120) = 12A. So in this example magnetizing inrush is about equal to capacitor inrush.

These are completely approximations, of course, using superposition, which is not entirely valid. But hopefully you get the idea. And hopefully my math is correct.

[AndrewT]

Hi,
at initial startup the transformer acts as a simple resistor with just the primary resistance across the supply voltage.

If the primary has a typical 2r0 resistance and 240Vac just happens to be at peak of 340Vpk then expect a maximum current ~170Apk.
Inductance does not enter the process until the transformer starts to develop core flux.

Charging of the capacitors cannot start until after the transformer has started to work as a transformer, so the two initial currents, start up surge and charging, cannot happen at the same time.

[zigzagflux]

Originally posted by AndrewT
Hi,
at initial startup the transformer acts as a simple resistor with just the primary resistance across the supply voltage.

If the primary has a typical 2r0 resistance and 240Vac just happens to be at peak of 340Vpk then expect a maximum current ~170Apk.
Inductance does not enter the process until the transformer starts to develop core flux.

Have to disagree with you. Inductance is there immediately, regardless of core flux. See any textbook on transformer energizing. Time has nothing to do with it; the point on the AC wave where the switch is closing has everything to do with it. Simple AC wave with decaying exponential.


Besides, in North America, you would trip the branch breaker every time you energize this transformer at 170A peak. Knowing this does not happen, you are missing some parameters in the calculation.

[AndrewT]

Originally posted by zigzagflux
you would trip the branch breaker ............at 170A peak. Knowing this does not happen, you are missing some parameters in the calculation.

Yes, the time vs instantaneous current relationship of ALL breakers and fuses. Look up any manufacturers time/current curve to see the relationship.
Is it correct to call this curve the I^2S rule?
At 1uS the current passing ability of ALL fuses/breakers is enormous.

I measured a twin primary 800VA before my last post, just to remind myself of typical resistance values. R+R = 1r1. You could expect the 2r0 value I quoted to apply to typical 220/240Vac 300VA to 500VA toroids.
Each winding on a US supply will present 0r55 to the mains. each winding of half R and half peak voltage will potentially draw 200Apk and twin primaries in parallel draw 400Apk.
But at these enormous currents the impedance of the supply has become very significant. Adding ~0.3ohms and an unknown value of incoming cable inductance will reduce the instantaneous peak, possibly by half.