Interesting comments, Jim. That's pretty cool software you're using. Thanks.
Pse see my detailed responses below. Basically, I get the same results from
my 30V test that you do, re peak currents and voltages. In my 30V test, I
used a transformer that was stiff enough that I could ignore the primary and
secondary resistance, as well as all the effects of the AC line and the pole
pig, etc. I understand that you believe that for my 30V test I should
dropped the load resistance from 2200 ohms to 22 ohms. I respectfully
disagree, and would be most interested in hearing your reasons. My reason
for keeping the 2200 ohms and merely scaling up the voltages and currents is
that all the circuit components (resistors and capacitors) in the power
supply are linear (setting aside issues of core saturation in transformers).
If you double the current through a resistor, you get double the voltage. If
you double the AC current through a capacitor, you get double the AC
voltage, but the same phase shift and frequency response. Everything scales
exactly with voltage. If the power supply has ten percent voltage regulation
at 3000V, it will also have ten percent regulation at 30V. So long as all
the circuit components are linear, all the voltages, currents, and waveform
shapes scale exactly. Of course, once non-linear elements are introduced
(transformer core saturation being the biggie) then the scaling breaks down.
Had I reduced my load Ro to 22 ohms as you suggest, the characteristics of
the power supply would have changed enormously. The discharge time for the
load (RoC), which is the rate at which the load drains stored charge from
the capacitor would drop by a factor of a hundred. This would change
everything and render the results of the experiments invalid. Now a few
detailed comments:
Date: Tue, 9 Feb 2010 04:37:46 -0800
From: "Jim Thomson" <Jim.thom@telus.net>
Subject: [Amps] resistor between FWB/FWB and HV caps.
To: <amps@contesting.com>
Message-ID: <3B934965A9EA46C19428C2B9F99AA0B2@JimboPC>
Content-Type: text/plain; charset="iso-8859-1"
## IE: peak current is aprx 5 x what the dc load is.
I get the same results in my 30V test. With a 100uF cap, the DC load current
was 12.5 mA and the peak current was 70.9 mA, an increase of 5.67. Scaling
up by a factor of one hundred makes the DC load current 1.25Amps and the
peak current 7.09 Amps. That's about same as you see.
## Now with peak currents being 4-5 times average current.... you end
up with one helluva big PEAK V drop on the 240 line.
This is a valid point, but I think it overestimates the problem. In my 30V
test, the xfmr could easily handle the peak current. In a real-life HV power
supply, the inductance in the transformer secondary will help maintain the
voltage over the msec or so that the peak currents are being supplied.
## Try sucking 4-5 times the dc load current through that 25 ohm
resistor, and the huge voltage drop across it is nothing to sneeze at.
That's a 125V drop with a typ ham amp drawing 1A dc plate current What
it really does... is impede current to the caps! The caps only get
'topped up' 120 times per second as is. The conduction angle is really low,
like 1.4 msecs at the bottom of the base line... and narrowing down to
just .4 msec at the top of the graph.
You're right that the instantaneous peak voltage through the 25 ohm resistor
is 125V, but the DC voltage across the resistor (which is basically the
voltage averaged over the full conduction cycle) only changes by 25 Volts.
In terms of power supply regulation, only the DC voltage drop matters.
You are also right that with the 25 ohm resistor the transformer
and diodes can't "top up" the charge in the capacitor as quickly, because
the resistor limits the peak current. However, you've not allowed for the
fact that the conduction angle increases in order to compensate for the
reduced peak current. In essence the diodes conduct a longer time during
each cycle. The reason the conduction angle increases is because the voltage
on the capacitor is proportional to the charge stored on it. (The definition
of capacitance is C=Q/V, where Q is the stored charge. As the Q drops, the
voltage drops proportionately). Thus if charge isn't being replenished
adequately, then V will drop, thus turning on the diodes for a longer time.
When one works out the details, the net effect of the reduced peak current
and the larger conduction angle is exactly what one would expect from
considering only the DC voltage drop across the 25 ohms, which is only 25
volts.
## BTW. to make that 30 vdc test valid... you would have to drop the load R
in proportion, other wise you are not simulating anything. Install a 22
ohm resistor for the load on the output.. then re-run your test.. you will
get an eye opener!!
Here, as I said earlier, we disagree. I'm keeping an open mind
on this point, but will need to hear your reasons why I should install a 22
ohm resistor, and also why my justification for keeping it at 2200 ohms is
wrong.
I'm sure this whole discussion must seem pretty boring to many list members.
However, it's important that we come to agreement, since the outcome
potentially affects every power supply that an amp builder puts together.
73,
Jim W8ZR
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