[Amps] Resistor before the capacitor REDUX

[Jim Garland]

> -----Original Message-----
> From: Gary Schafer [mailto:garyschafer at comcast.net]
> Sent: Monday, February 08, 2010 7:02 PM
> To: 4CX250B; amps at contesting.com
> Subject: RE: [Amps] Resistor before the capacitor REDUX
(From Msg 1)
>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.

(From Msg 2)
> 
> In looking at your post again I think that I see one problem with this.
> There is no "DC voltage drop across the series resistor" per say. The
actual
> voltage drop across that resistor is the AC voltage drop that you see or
> nearly so as the conduction angle is so small. You have short DC pulses
that
> are measured as AC. There is no steady state DC value across the series
> resistor.
> If you are measuring it with a meter the meter will not give a true
> indication.
> 
> The DC charging current , voltage drop across the series resistor, is
going
> to be pretty much the same as the AC.
> 
> 73
> Gary  K4FMX

Interesting points, Gary. You're absolutely correct about the transformer
resistance adding to the series resistance. I allowed for that in my test by
using a bit filament xfmr that had only about an ohm secondary resistance.
Since my series resistance was 22 ohms, I felt justified in ignoring the
transformer resistance.

Your second point, about the peak current being the total current, is
correct in the sense that the diode rectifiers are supplying no current at
all during most of the conduction cycle. This means, as you noted, that the
average current over a cycle is entirely produced by the short current
pulses.  However, that fact doesn't mean that there is no DC current through
the resistor. One can always decompose an AC current whose average value is
not zero into a DC component and an AC component whose average IS zero.
That's why a complex waveform viewed on a scope will seem to shift up or
down, with no change in shape, when one changes the scope's input coupling
from DC to AC. The scope basically offsets the trace by the DC component of
the complex waveform.

In my tests, I measured the DC component of the current pulses both with my
scope and also with a Fluke 87-V multimeter.  Because the  Fluke DMM
averages over several periods of the waveform, it give the correct DC
reading (as I verified by comparing it with the scope measurement). Thus,
there really IS a steady state DC value of voltage and current
across/through the resistor, even though during most of the conduction cycle
no total current (AC +DC) is flowing at all. Strange but true!

Returning to the point about my not scaling the load resistance by dividing
it by 100, as I did the voltages.  Had I done so, I would have changed the
waveforms by shortening the RC time constants of the circuit, and that would
have drastically changed the relative waveforms from those seen in a
full-voltage supply. By lowering the voltage applied to the filter from 3000
to 30, and leaving everything else the same, all I did was lower the current
by a factor of 100. All the measurements of peak current, average current,
peak voltage and average, scaled down precisely by the same factor of 100.
To get the corresponding values from a "real" 3000V supply, one has only to
multipy the voltage and current measurements by 100.

73,
Jim W8ZR