Gary Schafer wrote:
On 1 Feb 2005 08:06:26 -0800, "Mike Silva"
wrote:
Yes, both C1 and C2 are charged with 1/2 wave power (one pulse every
full cycle). However, it appears to me that the difference is that
they only discharge during one half-cycle of every full cycle. On
the
charging half-cycle they are not also discharging, but they only
discharge on the opposite half-cycle. This is different from the
2x2
doubler, where both caps are charged on alternate half-cycles, but
both
caps discharge during the entire full cycle. If I've analysed this
correctly, it seems that there should be some improvement in voltage
regulation over the 2x2. But I wanted others to look at the circuit
and offer their ideas on it as well.
73,
Mike, KK6GM
It is true C1 and C2 are not discharging while charging. But when
they
do discharge on the other half cycle they are being discharged more
than they would be in a regular doubler circuit.
They still each have to supply 1/2 of the output load at that time to
recharge the output capacitor. The output capacitor is getting
recharged through C1 and C2 only and not directly from the diodes and
transformer. So in effect you have 3 capacitors in series for the
load
rather than 2 in a regular doubler.
Although at first glance it would seem that the output capacitor is
getting part of its charge through some of the diodes, it only does
so
on the first cycle at startup . After that the output capacitor is
charged to a higher voltage than the transformer supplies directly.
That keeps the diodes directly from the transformer to the output
capacitor reverse biased so no current flows directly from the
transformer to the output capacitor.
So the output capacitor ends up only being charged by C1 and C2 on
alternate cycles. So it would seem that performance would be worse
than a standard doubler circuit. Unless much larger capacitors were
used.
73
Gary K4FMX
I ran simulations for both circuits. With the basic assumption that
the output voltage being developed is large compared to the diode
drops, I can't find any advantage to the more complex circuit.
In both circuits the transformer supplies energy on each half cycle.
The ripple at the top of the single capacitor in the complex circuit is
the same as the ripple at the top of the 2 capacitor string in the
simple circuit. The ripple at the output in both cases is 120 Hz for a
60 Hz supply with identical magnitude. The ripple at the junction of
the two capacitors in the simple circuit happens to be 60 Hz. I varied
the load and the regulation appears identical for both circuits and is
mostly a function of the source resistance of the transformer (not
terribly surprising). The series resistance added by the diodes would
generally be an order of magnitude smaller for high voltage supplies
and not a big factor in either case assuming these are high current
diodes.
The simple circuit has the advantage of being able to easily supply
both a double voltage 120 Hz ripple output and a standard voltage 60 Hz
ripple output.
In a power supply I built for a Heathkit SB-101 a long time ago I took
advantage of the two outputs from a simple doubler. The 850 V output
went to the finals. Then from the 425 V output I added a divider and
additional filter capacitor to provide the 300 V supply. By providing
a load resistance across the upper cap similar to the load resistance
of the divider on the lower cap the equally shared voltge was
maintained at the two filter capacitors. The divider on the lower cap
allowed me to tap off at a 300 V point to which I added a little
additional filtering (remember that point has 60 Hz ripple). By doing
this I didn't need a separate winding or transformer for the 300 V
supply. I used an old 300 VRMS TV transformer for both my 850 V and
300 V supplies. That supply is still in use with original components
35 years later.
For the simulation of the complex circuit I used a 50 uF output filter
and for the simple circuit I used two series 100 uF capacitors to
provide the equivalent filtering (i.e., 50 uF).
So, all I can tell is that the complex circuit takes one extra 50 uF
capacitor, and 4 extra diodes to do the same job as the simple circuit
with no difference in efficiency that I can find.
Curtis Eickerman
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