I'm building a multi channel power amp and I am having trouble with the power supply. I'm in Europe, so I've got 230V mains and I'm running into a transformer which brings it down to 60V.

The problem is that at no load, the fuses blows. As I've installed the circuit breaker, it gets tripped.

Am I screwing something up? Is it a problem of having no load?

Transformer datasheet (pdf)

My windings

  • \$\begingroup\$ The pdf is not readable as it is in european language. Try to operate the transformer at some load. You got that issue because of no load \$\endgroup\$ – Atom Sep 13 '13 at 11:54
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    \$\begingroup\$ What rating is the fuse? By "no load" do you mean that the secondary is open-circuit? Have you created a shorted-turn by bolting the transformer down incorrectly perhaps? \$\endgroup\$ – MikeJ-UK Sep 13 '13 at 12:00
  • \$\begingroup\$ Could be inrush current into the transformer. But without knowing the size of the transformer and the fuse, there's no way of telling. \$\endgroup\$ – Stephen Collings Sep 13 '13 at 12:35
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    \$\begingroup\$ If you don't have a variac (to bring the primary voltage up slowly to see what's going on), try the well known technician's trick of using an incandescent light bulb in place of the fuse. If there's an overload, the bulb will light brightly but will limit the current. \$\endgroup\$ – Alfred Centauri Sep 13 '13 at 12:58
  • \$\begingroup\$ I had first tried it with one of the power amps and tripped the breaker. Then with no load, and there isn't any sort of short. Then no load and the two primaries switched. Each time the breaker is tripped or fuses blown. The transformer produces 30v 1000va from 230v mains \$\endgroup\$ – Michael Tuttle Sep 15 '13 at 11:44

Inrush current is greatest when operating no-load on secondary

Is it a problem of having no load?

Yes, it might be significant.

Most likely the problem is "inrush current" and this is due to saturation of the transformer core. At this point some folk may be shaking their heads and not believing this and they may point out that "surely the current will be greater when the transformer secondary is fully loaded and surely this will produce more saturation in the core?"

The magnetization of the transformer core is due to the inductance \$X_M\$shown below (with a blue box around it: -

enter image description here

This is the equivalent circuit of a transformer and shows other stuff like iron loss and copper losses. When the transformer secondary is open circuit, the copper losses are extremely small and can be ignored; the upshot is that \$X_M\$ receives the full supply voltage because there is minimal volts dropped across the copper loss components and leakage inductance.

This means that the magnetization of the transformer is at its highest because when secondary current flows (due to load), the volts on \$X_M\$ reduce slightly due to \$R_P\$ and \$X_P\$.

But why should the core saturate? Here's what the magnetic field looks like when power is applied when \$V_P\$ is at the top of its voltage cycle: -

enter image description here

The voltage waveform is in black and the magnetic flux (or current) is in red. Forgive my drawing ability. Two scenarios and one produces normal peaks of current (as if the supply had been connected forever) and the other scenario causes the current to hit a big high-spot then settles down.

There's nothing mystical about this - this is what would happen in an inductor if you simulated it (look at the current).

If the magnetizing current peak is bigger, then so is the driving force of the magnetic field. If you looked at what the flux density is like when the ampere-turns are bigger you are likely to see the flux reach saturation and, when this happens you also draw a lot more current which in turn makes the problem bigger. This settles down after a few cycles of the power waveform of course.

What you will probably find is that sometimes the current limit trips and sometimes it doesn't - it's all down to where on the voltage cycle you applied power to the transformer.

  • \$\begingroup\$ +1, excellent answer with an equivalent transformer schematic that is seen too rarely. In case you have a source for the schematic, could you include a link for further reading? \$\endgroup\$ – zebonaut Sep 14 '13 at 7:50
  • \$\begingroup\$ @zebonaut it's one of my drawings I think. Originally it came from the web but I modified it. Right click and save-as is only option. \$\endgroup\$ – Andy aka Sep 14 '13 at 9:33
  • \$\begingroup\$ Great answer! So, if it's an inrush current problem, how do I calculate the specs for the necessary components for an inrush current limiter? \$\endgroup\$ – Michael Tuttle Sep 15 '13 at 11:56
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    \$\begingroup\$ @MichaelTuttle I think that inquiry is worthy of another question on this site. I'll try and answer it here though... you need an inrush current limiter and these are usually devices called thermistors with a negative temperature coefficient. When "cold" (at power up) their resistance is several ohms and enough to limit the peak currents but they rapidly get warm and their resistance lowers so that in normal operation (seconds after power up) they are a fraction of an ohm. That's the short story! \$\endgroup\$ – Andy aka Sep 15 '13 at 12:06

This is a guess, but you probably have the two primaries connected out of phase instead of in phase. Flip the connections of one of the primaries and see if that fixes the problem.

  • \$\begingroup\$ I tried that, didn't make a difference. I wired it up (minus the fuses) like the picture above. After a little more research I believe it may be inrush current which is the problem. \$\endgroup\$ – Michael Tuttle Sep 15 '13 at 11:38

I'm wondering what type of transformer you are using. If you are using a toroidal type, on an earthed chassis & have also earthed the securing screw with a separate wire, you will get a short, which will blow the input fuse, with or without a load on the secondary.


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