Hi Jim,
I didn't quite follow the whole thing as I don't think that you are using
the correct values for all components when multiplying by 100 on some. As
Bill said, if you are going to divide down from 3000 volts to 30 volts then
the load needs to be divided by the same amount.
Also you need to know what the transformer resistance is, both secondary and
primary as they are effectively in series with each other and in series with
your series resistor.
Also the "high peak current thru the resistor for part of the cycle" is not
just part of the current it is the TOTAL current that is a narrow pulse. All
of the capacitor charging current is in the form of a narrow pulse (fraction
of the cycle) as the diodes start conducting only near the peak of the
supply voltage wave and turn of at the very peak.
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 7:05 PM
> To: amps@contesting.com
> Subject: Re: [Amps] Resistor before the capacitor REDUX
>
> 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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