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@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
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