Jim,
Regarding the placement of your 25-ohm surge resistor between the full-wave
rectifier and the filter cap: Do you have the resistor placed between those
components in the HV(-) or the HV(+) line in your amp? Can you think of any
protection benefit of dividing the R value say in half and insert 12-ohm
resistors in both the (-) and (+) leads in front of the filter C?
Paul, W9AC
----- Original Message -----
From: "Jim Garland" <4cx250b@muohio.edu>
To: <amps@contesting.com>
Sent: Monday, February 08, 2010 12:44 PM
Subject: Re: [Amps] Resistor Before the Capacitor? (long)
>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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