[Amps] soft start, how to determine if needed?

[TexasRF at aol.com]

 Hi Alex, no, I didn't forget what inrush protection is; that is why I  
suggested the 25 ohm resistor and shorting relay.
 
You are the one suggesting  installing a one ohm resistor and leaving it in 
the circuit. Please go back and  read your email. I was simply responding 
to your suggestion and explaining why  it is a bad idea.
 
Also, on the idea of the  capacitors supplying current peaks during the 
peak of the AC cycle and  recharging them during the remainder of the cycle, 
please consider  this:
 
The capacitor filter and the  load resistance are in parallel. When the 
capacitor is charged as much as  possible, it reaches some given voltage. For 
example, let's say 3000vdc. When AC  voltage from the secondary drops below 
this value during most of the AC cycle,  the rectifier diodes prevents any 
secondary current from flowing into the  capacitor (and load).
 
While there is no current  flowing from the secondary, there is no 
capacitor charging. In fact, this is the  time that the capacitor provides current 
flow into the load. The capacitor  voltage decreases as it discharges into 
the load and this is what causes ripple  voltage. You can't see the ripple 
voltage on a dc voltmeter as it indicates the  average voltage from peak to the 
bottom of the ripple. Even if the meter could  respond to the 120 ripples 
per second, your eyes can't see  that.
 
When the AC cycle reaches the  point that the secondary voltage is more 
than the capacitor value plus the  rectifier voltage drop, the capacitor 
charging commences again. The load is also  supplied needed current during the 
charging period from the transformer. During  the short charging period the 
capacitor voltage builds up again to maximum value  determined by the 
transformer primary and secondary resistance and the cycle  continues as above.
 
If the load is removed, the  capacitor charges to full voltage and remains 
there as basically very little  discharge going on.
 
The amplitude of the ripple is  quite predictable. We know that one time 
constant will allow a 65% discharge a  capacitor. If we have a 50 microfarad 
capacitor and a 4000 ohm load, this  constitutes a 200 millisecond time 
constant. At 60 cycles per second, a half  cycle is 8.3 milliseconds. The voltage 
drop over this time is .65 X 8.3 / 200 =  .027. With our 3000vdc power 
supply, this is 81v peak. If the filter C was 25  microfarads, the peak ripple 
voltage would be twice as much and so on. This  is an approximation since the 
discharge period is actually somewhat less than  8.3 milliseconds, 
typically 20% or so less.
 
As can be seen, the transformer  has to supply all of the needed power for 
the load and charging during the short  charging period. If the charging 
period is only 20% of the cycle, then five  times as much current is required 
to make up the needed energy. Losses being  related to current squared, in 
this case, they are 25 times greater than a  purely resistive load. No wonder 
we see the drop in plate voltage from no load  to full load. The drop is 
also predictable but you have to know the transformer,  power cable and ac 
service loss resistance to calculate it.
 
None of this has anything to do  with turn on surges. That is a different 
but similar issue. Time constants are  involved in that as well and is also 
fairly predictable.
 
73,
Gerald K5GW

In a message dated 2/28/2010 1:43:39 A.M. Central Standard Time,  
alexeban at gmail.com writes:

you  forget what inrush current protection is!
when  switching on a power supply with empty capacitors, the zero impedance 
of the  capacitor bank is reflected across the primary so that at this time 
the  current can reach about 250 A worst case and gets worse the better the 
 transformer is (less internal resistance). at this time all you have as  
protection is that resistor and wiring resistance. what the mains sees is an  
R-C of 1 ohm and about 50uF network, with a time constant of about  
50-100usec. 
after  this time, even for a 20A average current you get a sag of 10% in 
the  voltage. actually it's less since the 20A is a peak current 220V x 20A 
equal  4400 Watts! during the time of the peaks, which are quite short 
duration ,  what comes into play is the dynamic regulation of the supply whereby 
the  capacitors supply the peak current and get recharged during the rest of 
the  cycle. that's why they are there!
Alex    4Z5KS