Electronic – What’s the real inrush current in the AC/DC rectifier

capacitorcurrentfusesinrush-currentrectifier

I'm working with rectifiers and power supplies at the moment and I'm trying to reconcile simulations with real world observations about my circuit.

It's a full-bridge diode rectifier (4x 1N5821's) with 4 parallel aluminium 470uF, 63V capacitors. Input AC peaks at 33.9v and the circuit is loaded with a 330R 5W resistor.

The input AC comes from a 240v-24v transformer and has a fast-acting 1A fuse on the 24V side.

In LTSpice modelling the diodes with the built in 1N5819's and capacitors with an ESR of 0.1, there is a large current drawn from the source when first started, peaking at 18.6A.

So why doesn't the fuse blow every time I turn the thing on?

Edit: Update – I measured the slope of output voltage when I first turn on the rectifier, from which I calculate the current by i = 1/c * dv/dt. I found that I is approximately 4A during that first charge cycle.

The answer about fuses is on the right track I feel but I'm going to leave this open for a bit.

Best Answer

The inrush current is a very short (in time) event. A fuse is a (comparatively) slow device, the wire in the fuse needs to heat up. It needs to collect a certain amount of energy over time to heat up and melt. If the inrush current pulse integrated over time does not exceed a certain value, the fuse will not blow. Did you ever notice that fast and slow fuses exist ? But even the fast fuse might be too slow in your case. Which is good, you only want it to blow of course when something is seriously wrong.

Additionally the 18.6A might be optimistic. I mean that in practice it will probably be lower. Did you take into account the DC resistance of the secondary side of the transformer ?