What is the voltage of capacitor in series?

Thread Starter

dante_clericuzzio

Joined Mar 28, 2016
246
Let's say my power source is 12v and I have a super capacitor of 10F but the voltage is 2.3v which is too low against the power source voltage. Now if I join 10F @ 2.3v capacitor with 4700uF @ 60 v capacitor in series will the total voltage threshold increase to 62.3v with less capacitance? I need simple and definitive answer without going around the bush. Thanks in advance.
 

MrSalts

Joined Apr 2, 2020
2,767
Let's say my power source is 12v and I have a super capacitor of 10F but the voltage is 2.3v which is too low against the power source voltage. Now if I join 10F @ 2.3v capacitor with 4700uF @ 60 v capacitor in series will the total voltage threshold increase to 62.3v with less capacitance? I need simple and definitive answer without going around the bush. Thanks in advance.
Yes, that will be the total allowable voltage. But I don't understand why you would want to put a relatively small capacitor in series with that big one, your total capacitance will be less than the smaller capacitor. 4697uF.
 

Thread Starter

dante_clericuzzio

Joined Mar 28, 2016
246
Yes, that will be the total allowable voltage. But I don't understand why you would want to put a relatively small capacitor in series with that big one, your total capacitance will be less than the smaller capacitor. 4697uF.
There is a reason for that...I have 10 pieces of supercapacitor 10F @ 2.3v but if i joint them in series i get lower capacitance so the trick is I will parallel all to get 100F then joint with with high voltage cap with low capacitance...this will get more capacitance compare to putting all the super cap in series. Thx for the answer..Salute to you!!
 

Thread Starter

dante_clericuzzio

Joined Mar 28, 2016
246
The total value of two capacitors in series will always be lower than the lower capacitors value. So it doesn't matter how high you make the supercapacitor (10F, 100F, 1000F), the resulting capacity will allways be lower than 4700uF
Ok i might be wrong i calculated again using digikey it is indeed very low. Perhaps i should parallel in two like 20, 20, 20 , 20, 20 then join with another super cap with 1F @5V x 5 pieces in series. That will make up total capacitance around 2.2F
 

StefanZe

Joined Nov 6, 2019
191
There is also the potential problem that the supercaps are not 100% identical, so the voltages will not be the same at all capacitors. Modules use voltage balancing to solve this problem
 

BobTPH

Joined Jun 5, 2013
8,922
If the caps are at 2.3V, you need six in series to get to 12V. Five will only get you 11.5V.

I would recommend you use a module with balancing and protection. You are likely to destroy your supercaps if you don’t.
 

WBahn

Joined Mar 31, 2012
30,012
Let's say my power source is 12v and I have a super capacitor of 10F but the voltage is 2.3v which is too low against the power source voltage. Now if I join 10F @ 2.3v capacitor with 4700uF @ 60 v capacitor in series will the total voltage threshold increase to 62.3v with less capacitance? I need simple and definitive answer without going around the bush. Thanks in advance.
No. You can't just add the voltage ratings of the two capacitors. Not even on paper.

In series, and assuming no leakage resistance, both capacitors will have the same charge, Q. The voltage on each capacitor will then be:

V1 = Q/C1
V2 = Q/C2

V1 = V2 * (C2/C1)

So the smaller capacitor (let's call that C1) will have a voltage that is a multiple of the voltage on the larger capacitor. In this case, that multiple is 10 F / 4700 µF = 2127.

That means that if you apply enough voltage to charge the 10 F cap to 2.3 V, the 4700 µF cap would be charged to nearly 4.9 kV. Going the other way, if you apply enough voltage to get the 4700 uF cap to 60 V, the 10 F cap will only be charged to 28 mV.

So your effective voltage rating is only 60.28 V.

This is on paper, where everything behaves ideally. In reality, capacitors have leakage currents and this means that they will not have the same charge on them. If the leakage currents are not the same, the one with the lower leakage current will take on more voltage. The result is that you run a real risk, even if operating at much lower voltages (say 12 V) of having your 10 F capacitor exceed it's rated voltage.

When you put capacitors in series, it is best to use capacitors that are individually rated for the full voltage. This is particularly important when putting large-valued caps in series (which is usually not a good idea in the first place because the parasitics generally make them behave in a way that defeats the intended purpose of doing so).
 

MisterBill2

Joined Jan 23, 2018
18,414
There is a reason for that...I have 10 pieces of supercapacitor 10F @ 2.3v but if i joint them in series i get lower capacitance so the trick is I will parallel all to get 100F then joint with with high voltage cap with low capacitance...this will get more capacitance compare to putting all the super cap in series. Thx for the answer..Salute to you!!
The assumption has been incorrect.
If you put the 10 ten farad/ 2.5 volt capacitors in series then you will have one farad capacitance at 10x 2.5 volts, which id VASTLY more than the 6700 MICROfarad capacitance.
 

WBahn

Joined Mar 31, 2012
30,012
There is a reason for that...I have 10 pieces of supercapacitor 10F @ 2.3v but if i joint them in series i get lower capacitance so the trick is I will parallel all to get 100F then joint with with high voltage cap with low capacitance...this will get more capacitance compare to putting all the super cap in series. Thx for the answer..Salute to you!!
Not the way you think it will.

If you put a single 10 F capacitor in series with a 4700 uF capacitor, the combined capacitance will be 4697.8 µF.

If you put ten 10 F capacitors in parallel and then put those all in series with a 4700 µF capacitor, the combined capacitance will be 4699.8 µF.

As you can see, the combined capacitances in either case is just the 4700 uF. The supercaps have NO effect at all. This is really driven home by the fact that the tolerance on that 4700 µF, 60 V capacitor is probably something like -10%, +50%,
 

MisterBill2

Joined Jan 23, 2018
18,414
Like I stated: If you put the 10 ten farad/ 2.5 volt capacitors in series then you will have one farad capacitance at 10x 2.5 volts, That greatly exceeds 4700x 1/1000,000 farads. Understand that a micro-farad is 0.000001farad Unless I have got it wrong, which is possible.
 

MrChips

Joined Oct 2, 2009
30,780
Like I stated: If you put the 10 ten farad/ 2.5 volt capacitors in series then you will have one farad capacitance at 10x 2.5 volts, That greatly exceeds 4700x 1/1000 farads. Understand that a micro-farad is 0.001farad Unless I have got it wrong, which is possible.
1 microfarad = 0.000001 farad

i.e. 1 micro is 1/1000000 of the parent unit (one million times smaller).
 
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