Electrolytic capacitors in series...

[Externet]

Hi.
Have a few 50 Farad at 2.5 Volt capacitors and want to put them in series inside a led flashlight, charging them with 14V.

What are the wiring considerations to equalize voltages ? Cannot remember how was it.

Miguel

[Joseph]

I have heard of using voltage divider resistors across each capacitor. I have not used resistors much based on the idea that as one cap begins to get too much voltage, it will tend to leak and re balance the whole thing. For an LED flashlight it might work well enough that way.

[MrAl]

Hi there,


Is there any advantage to putting them in series?

You could also use a boost converter to convert 2.5v up to
3.5v to run an LED.

[Robert Reed]

Many high voltage power supplies(i KV to 3 KV) use two 'lytics in series as part of the wiring scheme. The voltage is balanced out across them with high ohmage resistors ( aproaching and beyond megohms) of equal value. However these are line operated and extreme energy conservation is of no consequence here. BTW. whats with the 14 volt charge?

[Chris Smith]

Put them in series not parallel [= 5 v total]

Set up a voltage devider to deliver the actual voltage from the supply to the cap that you need for the LED, not the max v.

[X. point what ever volts the led requires]

Charge the caps to the right voltage value, and use the led at the appropiate starting voltage. [With out the need for a dropping resistor]

Caps will be full to the right Led voltage at the start up, and drain accordingly.

[russlk]

So, what is the purpose of the capacitors?

[dyarker]

6 50uF 2.5V caps in series will give a 15V rating, with a capacitance of 8.3uF.

Cheers,

[Forrest]

A flashlight with 50 Farad capacitors? For who - King Kong? A 50 Farad capacitor is going to set you back $200 to $400 each and it looks like they weigh 10-20 lbs a piece!

[Chris Smith]

Have a few 50 Farad at 2.5 Volt

6 50uF 2.5V caps in series will give a 15V rating, with a capacitance of 8.3uF.


Some one is ASSUMING instead of just answering?

Just answer the question, let them make the mistake.

[Externet]

What a mess i have triggered !...

Joseph: resistors as voltage dividers can work, but would discharge the capacitors - unless I did not get it.

Mr. Al: If they were in parallel, a boost chip to pump up to ~3.8V for a white led would stop working at about 1.2V, with still plenty of Joules that cannot be used inside the capacitors, being that about half of the charge wasted.

Robert: 14V is a car electric system.

Cris: You say "Put them in series not parallel [= 5 v total]" That is what I said but you are assuming only 2 capacitors to get to 5V.

russlk: instead of using batteries.

dyarker: yes, six is good, but not 8.3µF --> 8.3F !

Forrest: What? They are about an ounce each, diameter of a nickel, less than 2 inches long; I have all the flavors in this picture:

http://www.cooperet.com/products/produc ... capacitors

And am trying to do this just to play with, to put those parts to good use.

Miguel

[ecerfoglio]

You may use a zenner diode (instead of a resistor) in paralell with each capacitor to act as a safety limit.

Choose zenner with a voltage that is slightly less that the capacitor's maximum rating.

Add a current limiting resistor between the caps/zenners series and the power supply (the car's electrical system) in case the supply's voltage exceeds the sum of the zenner´s voltages.


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This reminds me of a topic some time ago about charging (Li ion?) batteries in series, but the capacitor's voltage varies (as it charges or discharges) much more than the cell's one, so the zenner's leakege current will appear only while the capacitor is "fully charged".
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[Chris Smith]

You put them in series like flash light batteries, then charge it up to the exact Led voltage using the voltage resistor circuit.

at 2.5 Volt capacitors times two

Un plug it and the LED will start at the exact voltage value with out the need for more dropping [wasting] resistors to drive the circuit.

However, 50 uf , .5 F , 5 F and 50 F are all different things.

Two 50 uF caps wont power the led for much time, where as .5F or 5 f will.

After two caps are in parallel, series is needed to keep the “pairsâ€

[MrAl]

Hi again,

Im not saying that there is no other way, but if you do put them in
parallel they make boost converters that work down to 0.8v, which is
1/3 of the top voltage. If you put them in series (say 6 of them)
you will have your 14v, but when they drain down to 0.8v you will
still only have 4.8v, so depending on what series resistor you
choose to drive the LEDs you will have much less output anyway.
If you put any LEDs in series, 4.8v still wont light them up, so
either way you waste *some* energy stored.
I just think that if you use a boost chip you wont have to worry about
one of the caps charging up too far and blowing out, then shorting,
then causing the next cap to charge up too far, blowing out, then
the next one, etc., until all the caps are blown out.
It's up to you of course which way you want to do it.

The series connection has the advantage that you dont need a boost
chip (or circuit) to drive the LEDs, but then you have to pay close
attention to how the voltage distributes amongst the six caps...
good luck doing that.
If you know the tolerance of the caps perhaps later today i can
do a calculation to see what the required parallel resistor would
have to be for each cap to keep the voltage correct when charging,
but then, that would be assuming each cap started out from zero
volts, which also *is not* going to be the case. Each cap will start
at whatever voltage it had previously, if any, and depending on
the leakage of each cap it could be very very hard to predict the
starting voltage, if not impossible.
If you try to use zeners, you'll have to see what leakage they add
when the caps are not charged up to 2.3v each.

BTW, did you calculate the run time you will get when you have
the LEDs turned on?

Also, how did you plan to CHARGE the caps...through a resistor?

[Chris Smith]

Caps in series through the voltage divider will equalize them self naturally.

They cant go higher than the given voltage divider, and if they discharge faster than the surrounding one, the other one will drag the parallel one with them to that exact lower value voltage because they are “chained togetherâ€

[Externet]

Hi all !

-ecerfoglio : I may use your suggestion, with 3 x 1N4001 silicon diodes in series paralleled to each capacitor; would safely limit charge to ~2.1 V and minimal leakage.

-Chris : the plan is to regulate the 14 V of the series of 6 capacitors down to ~3.6 V to feed the led; avoiding use of the booster chip, and no limiting resistor.

For your example of 2 x 50F in series, that is 25F at 5V
25 Farads x 5 Volts = 125 Coulombs
125 Coulombs ÷ 0.02 Amperes = 6250 seconds = 104 minutes !

-Mr Al : Right; you got the picture of avoiding remnants of unused charge. If the boost converter can work down to a 0.8V supply, that improves the idea of using one, and six could be wired in parallel, for a beefy 300 Farads at 2.5V

300 Farads x (2.5V - 0.8V) = 510 Coulombs
510 Coulombs ÷ 0.02 Amperes = 25500 seconds = 7 hours of led shine! BUT to charge 510 Coulombs in just 10 seconds would take
510 Coulombs ÷ 10 seconds = 51 Amperes ! ... a little too much to play with a flashlight
Yes, charging in about 10 seconds via a few ohms resistor to 14V.

Miguel