How do I utilize my capacitors capacity 100%

[richngreen]

Ive got 3 capacitors in series, and 4 in parallel, 3x4 grid of capacitors, and my power supply is a little 30volt 3amp 90 watt supply.

Im positive im not utilizing the capacity of the capacitors, whats my best option to reach full capacity in all of them?

 

[JimB]

What is the rating of your capacitors, how many micro-Farads, what is the working voltage?

What do you use the capacitors and power supply for?

JimB

 

[richngreen]

the capacitors are 35v 100uf. smallishmediumsizish. Because theres 3 in series, and 4 in parallel, i think im only hitting a 12th of what i would be getting if i were just charging one. why is that?
And 2 in parallel seems to make them not charge, as well as series, Im having to double the power going into them, and im not sure whats going on.

Also, would reducing the size of the capacitors reduce this "load" thats happening on the power source?

 

[Visitor]

The capacitance of capacitors in series:

1 / Ctotal = 1 / C1 + 1 / C2 + 1 / C3

If the capacitors are of equal value, the total capacitance of the string is 1/3 the value of each capacitor.

The value of the parallel strings of capacitors is the sum of the parallel strings.

So the array of 4 parallel strings of 3 capacitors in series is

Ctotal = 4 × (C/3) = 4/3 C, assuming all the capacitors have a capacitance of C.

Capacitors are normally used in series to achieve a higher voltage rating, so the voltage rating of the array is 3× the voltage rating of each cap (again, assuming all the caps are the same).

 

[Visitor]

The total capacitance of your array = 4/3 × 100 = 133uF.

The voltage rating = 3 × 35v = 105v.

 

[richngreen]

Thanks for that, even tho I should show my circuit cause its got a few resistors, but thankyou I have worked out:
----> if i use a cap with a lower voltage rating, this 3x voltage would be lower. yes, i have to use smaller capacitors.

... er... but what about little ceramic caps with high voltage ratings??? isnt that just the break down voltage? Because they would be effortless to fill right... this "electric field charging equalization voltage" is different. ah damn im confused again.

 

[JimB]

the capacitors are 35v 100uf. smallishmediumsizish. Because theres 3 in series, and 4 in parallel, i think im only hitting a 12th of what i would be getting if i were just charging one. why is that?
And 2 in parallel seems to make them not charge, as well as series, Im having to double the power going into them, and im not sure whats going on.

Also, would reducing the size of the capacitors reduce this "load" thats happening on the power source?

Sorry, but I did not understand any of that.

JimB

 

[richngreen]

What if having 3 capacitors in series is a 1/3 of the capacity, UNLESS you put 3x the volts in then it fills them 100%??

 

[Nigel Goodwin]

Most of what you're mumbling on about makes no sense? - perhaps you might like to tell us exactly what you're trying to do (because nothing you've mentioned helps us).

 

[alec_t]

yes, i have to use smaller capacitors.

No, you don't have to. It's fine to use a 35V rated cap with a 10V supply. In fact, it's not good practice to charge up a cap very close to its rated voltage. Always allow a decent working margin if you want your cap to have a long and happy life.
Btw, electrolytic caps in series can't be relied on to share voltage equally, because they have such a wide capacitance tolerance., typically 30% or worse. Your nominally 100uF cap may actually be 80uF or 150uF, for example.

 

[sagor1]

Capacitors in series should have equalizing resistors across each one, to help distribute the voltages equally on each capacitor. The resistor value depends somewhat on the voltages being applied, and how many capacitors are in series.
For example, I have 8 caps in series for a 2700V power supply, and each cap has a 100k ohm resistor across it. Power rating of the resistors also depends on total voltage and total current draw through the resistors. In my case, I draw about 3.4mA through the string. Power dissipated is just over 1.1W for each resistor, so I use 3W resistors.

Even if just using 10V, you should have some way to equalize the voltages across your 3 capacitors in series. Now, when you parallel more banks, you have to make sure all caps are identical. Those 3 resistors "should" equalize the banks that are in parallel.

Your total capacitance is about 133uF and rated for up to 105V (105V only with equalizing resistors). You can use that for 5V, 10V, 24V, or any other voltage under 105V. That said, as mentioned already, never run caps near their upper voltage limit. Even with equalizing resistors, some voltages may vary on individual caps. So, you may be ok in the 75V range, but I would be careful going much above that without measuring each cap and the voltage on it with no load.

 

[schmitt trigger]

A complete schematic, please!

Complete with component values and ratings, input voltages, whether the source is a rectifier or a battery, the output load current, etc, etc, etc.

 

[AnalogKid]

What if having 3 capacitors in series is a 1/3 of the capacity, UNLESS you put 3x the volts in then it fills them 100%??

No.

First, changing the voltage across a capacitor does not change its capacitance value. OK, for the purists, it changes it a small amount, but not enough to matter in the context of your questions. It does change the amount of energy stored in the capacitor, but that is not the same thing. The total energy stored in a capacitor is

1/2 x C x V^2

The total energy (in watt-seconds) equals 1/2 times the capacitance value (in farads) times the square of the voltage.

This means that for a given capacitor value, if you double the voltage across the cap the stored energy increases by 4x.

Second, what do you mean by "full capacity"? If you mean storing the maximum amount of electrical energy (in joules or watt-seconds) then the voltage source that is charging the capacitors must be equal to the rated voltage of the array. In your case of 3 x 4 capacitors, the voltage source would have to be 105 Vdc - AND - you have to have a way of guaranteeing that that voltage is distributed evenly in each serial string, 35 V per capacitor. Because electrolytic capacitors have relatively poor tolerances (-20% to +80% is a common value), this cannot be done without some external components.

ak

 

[Reloadron]

As mentioned and explained when capacitors are in series frequently "voltage Balancing Resistors" are used. I suggest you read "Voltage Balancing Resistors" and understand what is going on.

Ron

 

[richngreen]

Thanks for the help. I still dont get how a capacitor can magically change its capacitance due to its wiring in the circuit, wouldnt it be the same in a way no matter the configuration?

 

[Reloadron]

Well because we let resistors do it then it's only fair to let capacitors do it.

OK think of it this way. Capacitors consist of plates separated by a dielectric. The symbol even reflects that. Think of the plates as having an area. If I take two pieces of aluminum foil and each is 2" x 2" then each plate is 4 square inches. What depending on the space between the plates and the material between the plates (the dielectric) I have capacitance. Anyway just focus on the area of my plates. What happens to that area if I place another pair of identical plates in parallel with it? Heck, forget about capacitors. If I take two pieces of paper and each is 2" X 2" what is my total area?

Ron

 

[AnalogKid]

Thanks for the help. I still dont get how a capacitor can magically change its capacitance due to its wiring in the circuit, wouldnt it be the same in a way no matter the configuration?

There is no magic.
Really.
The capacitance of each individual capacitor does not change.

The effective value of a group of interconnected capacitors is based on a mathematical combination of the individual capacitor values. The mathematical combination varies depending on the physical interconnections. In the most simple case, look up the equations for two capacitors in parallel and two capacitors in series. See that for the same individual capacitor values, the total capacitance ***and the voltage rating for the combination*** change depending on the connections.

ak

 

[ChrisP58]

The equation for the internal mechanics of a capacitor is \(C=ε_0 {\frac{A}{d}}\)
\( ε_0 \) is the dielectric constant of the separator.
A is the area of the plates.
And d is the thickness of the dielectric material.

The construction of a typical capacitor is {plate1-dielectric-plate2}. Two capacitors in series is {plate1-dielectric-plate2-plate3-dielectric-plate4} If we ignore the two inner plates, you now have {plate1-dielectric-dielectric-plate4}

So, two equal capacitors in series is the same as one capacitor with twice the dielectric thickness that the single capacitor had. So the capacitance is half due to the double thickness of the dielectric.

**broken link removed**

 

[Pommie]

So, two equal capacitors in series is the same as one capacitor with twice the dielectric thickness as the single capacitor had. So the capacitance is half due to the double thickness of the dielectric.

And because the dielectric is double thickness it can withstand twice the voltage.

Mike.