Converter In-Rush Thermistors etc, UPDATE -4

Let's look at it another way. During the first 1/4 cycle, the capacitor charged to 75V. At that point in time the ac voltage is at 170Vpk.

The current at peak voltage is:

I = Vin - Vcap / R
I = 170V - 75V / 3 ohm = 32A

I see that I made an error. Charge current does not remain constant. Still, I see no way the current can peak at 86A.

Sal

ken white wrote:

Salvo wrote:
That's wrong. Go ahead and put a scope on your converter. You'll find the capacitor charge waveform has a constant slope. In other words, during charge, dV/dt is constant. Since dV/dt is constant then the current charging the capacitor is constant.

Sal

ken white wrote:

Salvo wrote:
...I = C * dV/dt = 2400 uF * 75V/5ms
I = 36A...

Sal

That current you calculated is the average current through the diode bridge and windings.

The period for a 1/4 cycle is 4.167 ms which makes the average current approximately 43.2 amps and the peak current approximately 2 times that amount or 86.4 amps.

The capacitor waveform is actually a sinusoid for the 1/4 cycle charge period.

The peak current occurs when C is in a fully discharged state and drops to zero when C is in a charged state, or the peak of the 1/4 cycle.

This is why the current pulse looks like a right triangle and why the peak current is about twice the average value.