Jim,
Here is the problem: while your calculations with no filter capacitor are
true, the peak current is only .99 amps but the ripple is 100%!
You can't use an infinite value capacitor as that implies it never needs to
be recharged. Not the kind of power supply we are talking about here.
With a practical capacitor of say 50 uF it is supplying energy to the load
between pulses from the rectifier to reduce the ripple which results in a
higher average voltage output.
The energy taken out between the ripple pulses has to be re-supplied to the
capacitor DURING the pulses from the rectifiers. Since the peak voltage from
the rectifiers can not increase beyond the 3000 volt level then the current
must increase in order to deliver the energy needed to recharge the
capacitor.
Also consider that when the capacitor is charged to near the full 3000
volts, or to say 2700 volts as in your example under load, the charging
current pulses are no longer sin wave pulses as the rectifiers will only be
conducting over a short length of time at near the peak voltage of the
transformer. That will only be a few degrees of conduction time between 2700
volts and 3000 volts of the rectifier pulse.
We can no longer divide by 1.414 to find average current. That value only
applies to sin waves.
Again, 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.
Maybe the other Jim VE7RF can give us some actual numbers on the peak
current from his calculator. I am too lazy to figure it out at the moment.
73
Gary K4FMX
> -----Original Message-----
> From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com]
> On Behalf Of Jim Garland
> Sent: Monday, February 08, 2010 12:44 PM
> To: amps@contesting.com
> 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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