Here's the issue we're considering: What impact does a current-limiting
resistor have on power supply regulation, when the resistor is placed
between the rectifiers and the filter capacitor?
Gary K4FMX, Bill W6WRT and Ron KA4INM all asserted that my earlier comments
were wrong when I concluded that the effect of a resistor (I used 25 ohms as
a typical value, with a amplifier load resistance of 3000 ohms) was
negligible. The error in my reasoning, these gentlemen concluded, was that
I had neglected the fact that the peak current through the resistor needed
to replenish the charge on the filter capacitor was much higher than I
believed. As Gary put it, "When the conduction time is shortened the
current must increase in order to supply the full amount of energy that has
been removed from the
capacitor. That will make the peak current much higher than the DC current
out of the power supply. That high peak current thru the 25 ohm resistor
will make for a large voltage drop across it."
In thinking about this point, I now believe Gary, Bill and Ron are correct:
I had not allowed for the high peak current through the series resistor.
However, I believe my basic conclusion is nevertheless true. Despite the
high peak current that occurs during a small part of the conduction cycle,
the series resistor has a negligible impact on power supply regulation. The
reason is that the power supply regulation is only affected by the DC
voltage drop across the series resistor and not the peak voltage. The high
peak voltage occurs over such a short fraction of the conduction cycle that
it has a minor impact on the average DC voltage drop across the resistor.
To test my reasoning, I bread-boarded up a power supply this afternoon and
took a series of measurements. I used a series resistor (Rs) of 22 ohms, and
a load resistor (Ro) of 2200 ohms, both of which are close to typical HV
power supply values. I measured DC voltages and AC peak voltages across both
resistors, as I changed the filter capacitance from C=0 (no capacitor), to
C= 10uF, and C=100uF. I also looked at the waveforms on a Tek 2465B
oscilloscope. The results are quite interesting and here they are:
(Actually, I've multiplied all the results below by 100, since I used a 30V
supply to simulate an HV supply with a no-load output of 3000V.) In the
table below, Vs(peak) is the instantaneous peak voltage drop across the 22
ohm series resistor. Vs(DC) is the DC voltage drop across the 22 ohm
resistor. Vo(DC) is the DC output voltage of the power supply with the 2200
ohm load, Vo(p-p) is the peak-to-peak ripple voltage across the 2200 ohm
load, and Io(DC) is the DC current through the load.
C Vs(peak) Vs(DC) Vo(DC) Vo(p-p) Io(DC)
0 uF 31V p-p 18.2V 1820V 2920V
0.87A
10 uF 165V p-p 25.8V 2580V 660V
1.17A
100 uf 156v P-P 27.5v 2750v 78v
1.25a
Here's what these measurement mean. First, note how the power supply
regulation improves when we add filter capacitance, and also note how the
ripple voltage decreases. (Keep in mind that with no load the power supply
provides 3000V.) Of course, this is no surprise. Also note how current drawn
by the 2200 ohm load increases as C increases. This increase is because
Vo(DC) is increasing and we're assuming the load doesn't change.
Now look at the DC and AC voltage drop across the series resistor. The DC
voltage drop only varies from 18.2V to 27.5V. This isn't surprising, because
Rs is 1% of Ro, so the DC voltage drop across it is 1% as great as that
across Ro.
But now notice how the instantaneous peak voltage across Rs, Vs(peak) jumps
from 31V to 165V when we vary C from 0 to 10uF, an increase of 530%. On the
scope, this peak occurs over a short part of the conduction cycle. This is
exactly what Gary, Bill, and Ron were saying: the peak voltage (and hence
the peak current) is much higher than the DC voltage in order to replenish
the capacitor charge.
The surprise, however, is that the peak voltage drops when we increase C to
100uF. At still larger capacitance, Vs(peak) declines even more (I didn't
show it here, but I increased C to 10,000uF in my tests), and it is clear it
would vanish completely in the limit of C= infinity. In other words, the
instantaneous peak voltage across Rs reaches a maximum as C increases, and
then begins to decline. The reason for this maximum, I believe, is that
while the conduction angle grows shorter as C increases, which tends to
increase the peak current, the larger capacitance is able to supply more
charge to the load. This is starting to get a bit technical, but the
charging time constant is RsC, while the discharge time constant is RoC. As
C increases, it takes longer for the load to deplete the charge on the
capacitor.
The bottom line, for those who waded through all this, is that a
current-limiting resistor in front of the filter capacitor has a negligible
impact (only tens of volts) on power supply regulation, even though the
instantaneous voltage across it may be substantial. The benefit of such a
resistor is partly that it helps limit inrush current to the capacitor bank
when the power supply is turned on, and also that it limits the peak current
from the power supply if the capacitors should short circuit to ground. The
resistor does NOT protect adequately against an arc in the HV supply
elsewhere in the amplifier, since it will not limit the current surge caused
by the stored charge in the capacitor bank. For that protection a fuse
resistor or additional current-limiting resistor is needed.
73,
Jim W8ZR.
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