Converter In-Rush Thermistors etc, UPDATE -4
UPDATE on 17 Nov, Update 2 on 22 Nov, Update 3 on 16 Jan 14, Update 4 on 19 Oct 14. --------------- In a recent thread there was mention of how you can wreck your gizmo (in my case a converter) ...
Forum Discussion
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.
Salken 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.
