K7ITM wrote:
With regard to circuit simulation: have a look for LTSpice on the
Linear Technology website. Or if you don't mind text-based netlists
and classical Spice, see
http://www.rfglobalnet.com/IndustryS...hResults.aspx?
keyword=Spice&TabIndex=4&image1.x=0&image1.y=0 for some other sources.

With regards the circuit, it's fairly easy to do some mental arithmetic
to see if there is an advantage or not over the two-diode, two-caps,
floating-transformer "full wave voltage doubler" circuit.

In the old circuit, there are, if you will, four distinct time periods
in each cycle (though two of them are electrically equivalent, and the
other two are symmetrical). One is where the upper cap is charging,
and the load current may be considered to be through the two caps in
series -- or may be considered to be through the lower cap and the
transformer and upper diode. Following that is a period where the
transformer voltage is too low to charge the upper cap and too high
(not negative enough) to charge the lower cap. Then the load current
flows through the two caps in series. Following that, the transformer
charges the lower cap, and after that the transformer voltage is again
too low to forward bias either diode. So the net voltage across the
two caps increases twice each cycle, once when the upper cap is
charging and once when the lower cap is charging. The output (load)
ripple fundamental frequency is twice the line frequency. You can get
a reasonable idea of the ripple voltage if you assume the charging
takes place in a tiny fraction of the total cycle time; in fact, that
will be an upper bound on the ripple voltage. For simple analysis,
assume a 1Hz input so there are two charging pulses per second (one per
cap), and assume two 1-farad caps and a 1-amp load. Then the voltage
across each cap between the charging pulses sags at 1 volt per second,
and the output sags at 2 volts per second, but resets twice per second,
so the ripple is 1 volt peak to peak.

In the new circuit, you need to say something about the capacitor
values. Presumably the two "input" caps are the same value, but may
differ from the output cap. But what happens if you say that you are
going to use the same total energy-storage ability as you had in the
simpler circuit? Then the output cap might be, say, 0.25 farads, and
the two input caps might be 0.5 farads each, since the output cap must
be rated at twice the voltage of the caps in the original circuit.
Now...can you figure out an allocation of the caps that gives you even
the same performance as the original circuit, let alone a better one?
And is it worth using more caps and more diodes? Especially for low
voltage use, the original circuit has a distinct advantage of having
only one diode drop during charging.

You can also analyze transformer utilization and I think you'll find
that the "improved" circuit doesn't really offer any advantage, if you
set it up to give the same output ripple.

Any other viewpoints, backed by some analysis or simulation?

Cheers,
Tom

David J Windisch wrote:

Build s test ckt with a filament xfmr and *small* caps and a "heavy"

load,

and look at the oupt ripple with a scope. Or build it virtually.

OT: Which are the other good, (affordable!?) ckt sims?
Anyone have a transferable version of Electronics Workbench he'd like

to

sell?

73, Dave, N3HE
Cincinnati OH


Looks like about 100V ripple. Here it is for LTSpice...

Version 4
SHEET 1 880 680
WIRE -16 64 -96 64
WIRE -96 64 -96 208
WIRE -96 352 -16 352
WIRE -128 224 -128 208
WIRE -128 208 -96 208
WIRE -96 208 -96 352
WIRE 48 64 112 64
WIRE 176 64 208 64
WIRE 320 64 416 64
WIRE 416 64 416 352
WIRE 416 352 320 352
WIRE 256 352 240 352
WIRE 112 352 48 352
WIRE 112 160 48 160
WIRE 48 160 48 64
WIRE 112 256 48 256
WIRE 48 256 48 352
WIRE 176 256 208 256
WIRE 208 256 208 64
WIRE 208 64 256 64
WIRE 176 160 240 160
WIRE 240 160 240 352
WIRE 240 352 176 352
WIRE 416 400 416 352
WIRE 416 496 416 464
WIRE 416 352 528 352
WIRE 528 352 528 384
WIRE 528 464 528 496
WIRE -272 160 -272 0
WIRE 48 64 48 0
WIRE 48 0 -272 0
WIRE 48 352 48 416
WIRE 48 416 -272 416
WIRE -272 240 -272 288
WIRE -272 368 -272 416
FLAG -128 224 0
FLAG 416 496 0
FLAG 528 496 0
SYMBOL diode -16 80 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D1
SYMATTR Value MUR460
SYMBOL diode -16 368 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D2
SYMATTR Value MUR460
SYMBOL diode 256 80 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D3
SYMATTR Value MUR460
SYMBOL diode 256 368 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D4
SYMATTR Value MUR460
SYMBOL cap 176 48 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 200µ
SYMBOL cap 176 336 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C2
SYMATTR Value 200µ
SYMBOL diode 112 176 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D5
SYMATTR Value MUR460
SYMBOL diode 112 272 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D6
SYMATTR Value MUR460
SYMBOL cap 400 400 R0
SYMATTR InstName C3
SYMATTR Value 100µ
SYMBOL res 512 368 R0
SYMATTR InstName R1
SYMATTR Value 175
SYMBOL voltage -272 144 R0
WINDOW 3 -81 305 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 250 60)
SYMBOL res -288 272 R0
SYMATTR InstName R2
SYMATTR Value .5
TEXT -366 520 Left 0 !.tran 0 1 0 .1m


John
KD5YI

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

looks like a cross between a full wave bridge and a pair
of voltage doublers in parallel.

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

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

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

=============================

A simpler logical analysis -

The more complex circuit will have lower efficiency because it does exactly
the same job with 5 additional components to lose power in.

Back to the drawing board !

Many thanks for running that simulation. Sometimes more is better, and
other times more is just more!

73,
Mike, KK6GM

Mike Silva wrote:

Many thanks for running that simulation. Sometimes more is better, and
other times more is just more!

73,
Mike, KK6GM

I think a way of looking at the difference is that one is two half
wave doublers in series and one is two half wave doublers in
parallel. You would have a hard time using the parallel version if
you wanted an 800 volt output and intended to use electrolytic
capacitors, because each of the 3 would have 800 volts across them but
it may have some utility for low voltage outputs. There may be a
better way to connect the diodes, but I haven't got that far, yet.
--
John Popelish