I think some commenters may have overestimated the negative impact on power
supply regulation, if a "glitch" resistor is placed between the diode
rectifiers and the filter capacitor of an HV supply. Here is an easy way to
estimate the impact of such a resistor on power supply regulation.
Consider an HV power supply that is designed to deliver, with no load, 3000V
DC. Further, let us suppose that the total load on the power supply by the
amplifier is 3000 ohms, which is a reasonable value for an amateur legal
limit amplifier.
The question we want to answer is this: What effect does a series resistor
have on power supply regulation, if it is inserted between the diode
rectifier bank and the filter capacitor? Let's assume we're talking about a
25 ohm resistor, a typical value.
To answer, we'll consider two limiting cases. First, we'll assume that there
is no filter capacitor at all, i.e., C=0. Next, we'll consider the opposite
extreme, an infinitely large filter capacitor, i.e., C=infinity. The
answers we get then will bracket the answer we would find when we have a
real filter capacitor of, say, 50 uF. In essence, the C=0 case is the worst
case scenario, and the C=infinity is the best case scenario. To keep things
simple, we'll also suppose that our plate transformer is so large that it's
"equivalent series resistance" can be ignored. This assumption means that
any degradation of the power supply regulation can be attributed purely to
the 25 ohm resistor.
Now consider the case of C=0. In this limit, our power supply would be
delivering unfiltered rectified AC. The total load on the power supply is
3025 ohms, because the 25 ohm series resistor is in series with the 3000 ohm
load of the amplifier. (Obviously, we're neglecting also any bleeder
resistors, and assuming full wave rectifiers.)
The voltage on the load would thus range between 0 and 3000V, and would be a
rectified sinewave with a repetition frequency of 120Hz. The average value
of the rectified sine wave would be 3000V divided by the square root of 2,
or 3000/1.414 = 2121 Volts. In other words, with no filter capacitor at
all, the average DC voltage provided by the supply would be 2121VDC, with
3000V peak-to-peak ripple superimposed on it.
The average (DC) current drawn by the 3025 ohm total load is then
2121Volts/3025 ohms = 0.7011 Amps. The peak current drawn by the load,
which occurs when the rectified voltage is at its peak, is then
3000Volts/3025ohms = 0.992 Amp. The voltage drop across the 25 ohm resistor
thus has a peak value of (25 ohms)x(.992Amps)=24.8V. To repeat, this is the
worst case scenario. No matter what value of filter capacitor we use, the
peak voltage drop across the 25 ohm resistor cannot exceed 24.8V and the
peak current cannot exceed .992 Amps.
Now let's consider the best case scenario, when we have an infinitely large
filter capacitor, which we'll assume is charged to 3000VDC. (We won't worry
about how to charge an infinite capacitor, but just assume we've somehow
done it!) In this limit, the DC voltage on the capacitor is 3000V, with
zero ripple, and the current drawn by the amplifier is 3000V/3000ohms = 1.00
Amps. Since the peak voltage provided by the diode rectifiers is 3000V, and
since the DC voltage on the infinite capacitor is also 3000V, there is no
voltage drop across the 25 ohm resistor, so that no current ever flows
through it. Intuitively, this means that an infinite capacitor can supply
whatever current the amplifier needs and never needs to be replenished by
the transformer.
To summarize, the worst case scenario is that the 25 ohm resistor will have
only a 24.8V drop across it, and the best case scenario is no voltage drop
at all. Thus we can conclude that the effect on voltage regulation of a
series 25 ohm resistor is negligible, in that it cannot cause more than a
24.8 V degradation of the power supply output.
Now, let's take up the practical question of the most desirable value filter
capacitor one should use? In a theoretical sense, the answer is the largest
capacitance possible, since the more the capacitance the stiffer the
regulation and the closer the DC voltage is to the peak voltage. But a whole
host of practical considerations argue for a lower value of filter
capacitance. Lower capacitance values mean smaller and cheaper and more
reliable capacitors, even if the voltage regulation isn't optimal. When all
these practical tradeoffs are factored into the decision, most designers
believe that a 10 percent drop in regulation (from no load to full load) is
acceptable. Thus a 3000V no-load supply would drop to 2700V under full load.
Typically, this means a filter capacitance of about 20 to 40 uF. More
capacitance is better, of course, in terms of voltage regulation, but once
one moves to higher capacitance values these other problems begin to crop
up. One problem with very high capacitance is the current inrush, since an
uncharged capacitor presents a dead short to the transformer and rectifier
bank at startup. The bigger the capacitor, the more current inrush has to be
allowed for.
73,
Jim Garland W8ZR
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