capacitors for a voltage divider circuit

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Robert Gotts
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capacitors for a voltage divider circuit

Post by Robert Gotts » Tue Feb 27, 2007 12:59 pm

I would like to use 180uf, 400 volt electrolytic capacitors connected to 120 V AC line to provide 24V AC, through a voltage divider circuit to a 2.4 ohm (total resistance) 30 foot heating cable.
Are there reasons why this is not a viable alternative to a high current 24V transformer?
Robert Gotts

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Bob Scott
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Post by Bob Scott » Tue Feb 27, 2007 1:23 pm

Hi Robert,

Is the capacitor rated for AC or just DC operation? Can the capacitor withstand the power generated by the arithmetic product I^2* ESR?

For what purpose are you using this heating cord?

Without the transformer, there is no electrical isolation. If the circuit is installed in a building it will be unsafe and not conform to any electrical code.


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Post by haklesup » Tue Feb 27, 2007 1:29 pm

I don't see too many capacitive dividers but it is allowed and should work.
I'm not sure about the long term reliability though.

You will need more than just the one value of 180uF to get a 5:1 reduction. Use the formula on the link above to determine the right values. A rating of 400V for the dielectric is good for 120V use. The only spec you lack is ESR. Since those capacitors will be passing a significant amount of current the ESR will be important so that the caps don't get hot and fail prematurely and so you don't waste too much energy in the divider. As with any voltage divider, it is inefficient because you will be throwing away a lot of power across the caps (P=I^2*Z). It will cost far less than a big transformer though.

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Chris Smith
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Post by Chris Smith » Tue Feb 27, 2007 1:36 pm

A single cap is a reducer.

The cap value drops the voltage/ current value.

Dual resistors are a divider.

I don’t know why they don’t use caps over large amounts of copper wire, but at least they have gone to switching PS's.

180 sounds wrong, the formulae is on the web somewhere.

Fluro fixtures that drop the heater value to around 5 volts, use a 7 Uf cap.

Even my water pump start circuit [1-1/2 hp] has less than 80Uf.

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Post by MrAl » Tue Feb 27, 2007 1:53 pm

Hi there,

To add to what hacklesup said, and i agree that ESR is an issue here...

First off, you need two capacitors of the same value connected
back to back (+ to + or - to -) in order to work with AC current.

Second, 180uf total capacitance isnt enough to push 10A though
your load of 2.4 ohms, assuming you want 24vrms across this resistance.
220uf total would do it, but since you need two in series (for AC)
and two capacitors of the same value in series results in a total
capacitance of one-half of the value of one of the caps, you actually
will need two 440uf caps in series, each one rated for 400v.
You also dont want to wire the two caps as a voltage divider,
instead wire them both in series (back to back) and that in series
with the load. This provides for the highest efficiency and lowest
value capacitance required. The reactance of the cap reduces the voltage
across the load which keeps the current at 10 amps.

Third, you need two high current diodes connected across each
one of these caps to prevent too high of a reverse bias, which would
destroy the cap quickly. The diodes are connected so that they
conduct if a reverse voltage tries to appear across the
other words, cathode to (+) terminal of each cap.

Fourth, the ESR (as hacklesup was good enough to mention) has to
be low enough to prevent high power dissipation in the cap which
will overheat the device and greatly shorten it's life.

Fifth, and very important, is the ripple rating of the caps. This rating
must be at *least* as great as the normal current which will be flowing
thought each cap (10A in this case), but even with that rating you
might only expect a two year life expectancy, so i would say at least
double this rating (to 20A or better) so that you get some decent life
out of this circuit, unless of course you either dont intend to use it
for that long or it will be used on an intermittent basis. The total
accumulated hours of life will be approximately 2 years.

Sixth, probably without the need to mention, a fuse to protect against
overcurrent in the event one cap should short out (a very typical failure
mode with electrolytics of any size). This might be a little tricky
to pick out, as the input surge during turn on could be as high as
71 amps, so perhaps a slow blow 15 amp fuse.

During the first run up, monitor the caps temperature rise. If either
one starts to heat up too much there is a chance the efficiency is too
low and this means the cap will fail sooner than expected.

BTW big caps like that can explode with super force when they
fail, so a blast shield over the top of the cap is probably a good
idea. In any case, stand back the first time you run it up and
watch eyes/wear protection.

As a last note, you might also look into the surplus market for
two 12vac 10 amp transformers and wire them in series aiding.
This would be a long term and safe solution. I would bet you could
find two for 20 dollars (or so) each.
Of course a 24v 10 amp transformer would do it too, or better yet
a 24v 15 amp transfomer.
If it were me doing this project i would never use electrolytics
in this way, i would pay the price and go with the transformer
even if it costs twice as much. I dont want to tell you what
to do though :-) Keep in mind that if the caps fail, you'll still
be looking for a transformer only by then you would have
already put out money for two very large caps.

Another idea is to look for an ac cap with value 220uf.

Another idea is if you happen to need heating in another area,
another solution would be to wire a standard 120vac 1000 watt
heater in series with the heating cable. This might get you enough
current though the cable, but of course would use much more
LEDs vs Bulbs, LEDs are winning.

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Post by haklesup » Tue Feb 27, 2007 4:30 pm

Chris Smith Wrote:
A single cap is a reducer.
I get that, it's still a divider but the other impedance is the heater. With one part being capacitive and the other being all resistive, the divider formula would get a bit more complicated.

If you do it this way (loop), you need to drop 96V across this cap and 24V across a 2.5ohm load. You didn't say the wattage of the heater so I'll guess its 100W and now you need 10A through that cap!

I suggest you download a free copy of electronics workbench (or whatever) and simulate this tiny circuit.

Come to think of it, a 120V light bulb in series would work and contribute its own heat. Probably too big and fragile though.

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Chris Smith
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Post by Chris Smith » Tue Feb 27, 2007 4:42 pm

Technically its not called a voltage divider.

Technically, the whole circuit could be referred that way, but that’s out there?

The easiest way to not consider all the ramifications of the cap is simply to purchase the “Dâ€

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Post by dyarker » Tue Feb 27, 2007 7:22 pm

Reducing the voltage isn't very "green". Why draw 10A from the AC mains when you can draw only 2A with a transformer. Step down transformers don't only step down the voltage, they also step up the current!

Dale Y

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Chris Smith
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Post by Chris Smith » Tue Feb 27, 2007 7:52 pm

Why not draw what you need, instead of the wasted transformer, energy and copper?

Save on copper, heat, draw, etc.

I realize that most don’t have any idea about the energy, but caps is the lost art, the art of actual money.

[rollover in your grave republicans, it works and no milking the public!]

All is not about profit for the dain brammaged!

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Post by rshayes » Wed Feb 28, 2007 4:22 am

The actual capacitance needed would be 226 uF instead of 180 uF. Two of these capacitors could be connected in series (positive to positive or negative to negative) as MrAl pointed out. Diodes across the capacitors would allow each capacitor to be active only on opposite half cycles, so the value would not have to be doubled. The diodes would have to be rated for at least 5 amps and preferably 10 amps.

The current would be 10 amps RMS. Since current flows through each capacitor only half the time, the capacitors would have to be rated for a minimum of 7 amps RMS of ripple current. The lifetime under these conditions would probably only be a few thousand hours at best.

An electrolytic capacitor with this high a current rating may not be available. The 220 uF, 400 V capacitors that Digikey lists are only rated at 1.5 amps RMS, a factor of five too low.

It might be possible to parallel a large number of some other type of capacitor (possibly polypropylene) to meet the current rating, but this would probably be more expensive than a suitably rated transformer.

Some electrolytic capacitors are sold as motor starting capacitors. These are not designed for continuous duty and would probably fail in a few hours in this application.

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Post by MrAl » Wed Feb 28, 2007 8:13 am

Hi again,

Yes i thought you were right in that two 220uf cap could be used,
thanks for pointing that out...but, when i did a transient analysis
the only way to get 10 amps rms (14 amps peak) through the
2.4 ohm resistance of the cable was to use two 440uf caps,
even with the two diodes installed. The reason for this is that
after the circuit runs for a few cycles the caps have charged
up to some nominal dc value, and the two diodes no longer
conduct. This means effectively the two caps are still in series
which means they have to have twice the value of a single
cap. The diodes only conduct for about three cycles, and
only the first cycle puts high current through the diodes...
the remaining two cycles have low current though them.

I dont agree though that the ripple current rating should be
7 amps, or even 10 amps, because as mentioned i believe that
would not provide a long enough service life.
220uf caps with high ripple current ratings are out there,
it's just a matter of finding them and then <insert chuckle>
paying for them.
I too was worried about the motor starting caps, regardless of
what construction they are.

In the final analysis, my opinion stands in that i am always
going to be in favor of using a transformer, except in the
case where there is some additional heating to be done and
a 1000 watt (or so) heater can be wired in series with the
cable, thus supplying the cable with the needed current and
also heating the secondary area. This may require a heater
rated 1500 watts though, but something in between that
(say 1200 or 1300 watts) may work almost as good).
A smaller heater of 200watts or even 500 watts wont work
very well however.

There is only one possibility left that i would favor that
we havent considered yet, and that is a switching regulator.
Since the heater cable can probably be run on DC as well as
AC, it is most likely possible to use a switcher that takes
120vac input and outputs 24vdc up to 10 or so amps.
This would have to be a full switcher however, not simply
a pulsing circuit, which means it would have the series
inductor on the output too, as well as some filter cap.
The filter cap on the switcher would of course be MUCH
smaller than that discussed for the non-switching circuit.
Perhaps 100uf at 35v or something like that. The inductor
would have to be suitable sized too, with a min sat current
rating of 15 amps. The pass transistor could probably be
a 10 amp MOSFET transistor but it's voltage rating would have
to be 200v or better. Luckily these are not too expensive
these days. A switching regulator controller chip (99 cents)
could be used to control the MOSFET, or it could be set
up to simply provide a set ratio from input to output similar
to a transformer ratio by driving it with a 555 and mosfet
driver chip or two driver transistors.
With the proper fusing, i would expect the switching regulator
to be a safer circuit than the series capacitor(s) would be.
It is even possible that the resulting circuit would be more
efficient than a transformer, with proper design.

The current may be 10 amps, but it's out of phase with the voltage.
This means the total power use is 240 watts, not 1200 watts.
This is because power in an AC circuit is E*I*cos(Theta) and not
simply E*I as in a DC circuit. Any resistance in series still drops
a voltage v=I*R however, so the transformer is still slightly
better in this respect.
LEDs vs Bulbs, LEDs are winning.

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Post by dyarker » Wed Feb 28, 2007 10:26 am

The power of the heater will be 240W. The total power will be less than 1200W, only by half. Yes "E*I*cos(Theta)", but the phase shift is more like 60°, not 90°. COS60° = 0.5.

Yes, for a DC supply a switcher beats a transformer/rectifier/cap/linear regulator hands down. For AC it is very hard to beat a transformer.

We still haven't heard from Robert Gotts on where he's using the heater. Till he says, I'm staying with suggesting a transformer for the isolation.

Dale Y

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Post by rshayes » Wed Feb 28, 2007 12:43 pm

The current drawn from the AC line has an in phase component of 2 amps and a quadrature current of 9.81 amps. The RMS current drawn from the line will be 10 amps, with a phase angle of about 78 degrees. The total load on the AC line would be 240 watts plus 1177 vars.

You would only be able to operate two of these heaters on a 20 amp circuit without blowing the fuse or tripping a circuit breaker. The power company also frowns on highly reactive loads like this.

I would also recommend a ripple current rating above 7 amps for the capacitors. This is the actual stress on the part, and the capacitor should be rated higher to give some safety margin.

A transformer or switching supply is probably a better option.

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Chris Smith
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Post by Chris Smith » Wed Feb 28, 2007 5:31 pm

Any and all draw will be converted into heat, the sole purpose of your gain.

Design your cap and resistor well to actually draw what you want.

Ill take a cheap cap over wasting expensive copper windings any day.

And there is less heat loss as well in a unit that doesnt give it back.

Transformers are supposed to be perfect but they are any where near this function.

Seen the latest price on copper?

Robert Gotts
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More information concerning capacitor voltage divider.

Post by Robert Gotts » Wed Feb 28, 2007 7:47 pm

Thanks to all who responded to my question.
The capacitors that I have are Nippon Chemicon CE (M), 105° C, 180 uF, 400V. I presume the voltage rating is for DC. They are 22mm diam. by 50 mm long with right angle radial leads. They have 13G208 as an identifier. A search for them in the Nippon Chemicon catalog was not successful so I don't know the rating of their ESR or their ripple rating. The price was right at $1.00 each.
The heater cable that I have is Actel waterproof heating cable, Oatley Electronics No. 099957. It has six 0.25 mm diam. wires and is insulated for use in wet areas and has an aluminum braid covering for grounding. I don't have a wattage rating. It's resistance is 0.2 ohms per meter which calculates to 2.4 ohms for the 39 foot length that I will use for a seed germination mat.
The divider circuit that I had in mind would have, for the series connected section, a set of four parallel connected pairs of series connected capacitors (+--+). Eight capacitors with a total of 360 uF for the section.
The parallel part of the divider circuit contains two parallel connected capacitors(+'s connected and -'s connected) for 360 uF total in one leg and the 2.4 ohm heater cable in the other leg. The calculated capacitive reactance for 360 uF, 60Hz, 120VAC is 7.35 ohms.
If I have done my calculations correctly, this circuit should provide approximately 24 VAC across the the pair of capacitors and the heater cable with a rms current of about 13A through the series capacitor group of the circuit, 3A through the parallel set of capacitors and 10A through the cable to provide about 240W for heating.
With ten capacitors sharing the current, my hope is that they won't overheat enough to be a problem.
I don't know how the diodes for reverse voltage protection suggested by MrAl would be connected with the circuit that I have in mind. Would you use a diode around each capacitor group?
I planned to have a fuse and thanks for the 15A slow blow fuse suggestion.
Also a GFCI would be used.

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