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# Wiring Capacitors In series?

What are rules for wiring **multiple** capacitors in series? Especially when The voltages are different. If i wire two 80v's with a 200v what would my WV be?

5294Views10Replies

What are rules for wiring **multiple** capacitors in series? Especially when The voltages are different. If i wire two 80v's with a 200v what would my WV be?

## Discussions

BUT a lot depends on each capacitance value because a sudden application of the full voltage could over volt and destroy a capacitor

while charging current flows..

- Usually when series capacitors are attached the cap values we EEs use the same capacitance value.
- That guarantees an attempt to share voltage,
- Still you use balancing resistors
- And use the lowest voltage to double or triple the total WV of the caps together

AThanks all! I charged the 200v with two 80's to 360 and one of the 80's did start smoking heavily, which would be because i did not use balancing resistors :P. I think that instead of rigging all scavenged cap's together I should just order a couple bigger ones online. Anyone know of a good website that ships to Canada?

Where are you located? (I'm in B.C.)

BTW, don't forget afterwards, to select a "best answer". It marks your question as "answered" on the main page. ;-)

The voltage rating is the safe failure rating - Don't quote me on this (believe me, steve will correct me within about 8 minutes, on average ;-) )

Since putting 2 capacitors in series is like increasing the plate separation (2 layers of dielectric instead of 1) then: (from wikipedia:

Capacitors are combined in series to achieve a higher working voltage, for example for smoothing a high voltage power supply. The voltage ratings, which are based on plate separation, add up, if capacitance and leakage currents for each capacitor are identical. In such an application, on occasion series strings are connected in parallel, forming a matrix. The goal is to maximize the energy storage of the network without overloading any capacitor.)

Total capacitance in series diminishes -- its the inverse of the sum of the inverses of the capacitors.

1/C

_{total}= 1/C_{1}+ 1/C_{2}+ ...In parallel they all reach the same charge level, and the capacitance adds up

Ctotal = C1 + C2 + ...

2x80v + 200v = capable of 360v, but the capacitance will be less than any one of the single capacitors.

Voltage rating is maximum OPERATING voltage, nothing to do with failure....

There, fixed it for you.

lol... either you're correct, or Steve is late... ;-)

Merely asleep.....

The Balancing Resistors are ......... A

The voltage rating of a capacitor is an estimate of the maximum voltage you can put across the capacitor without breaking it. So basically if you have a capacitor rated at such-and-such a voltage, then you do not want to put so much charge into it that it exceeds that voltage.

So the question is: if you've got a number of capacitors in series, say C1,C2,and C3, with voltage Vtot across the whole string, how to you predict the voltage that will appear on each individual capacitor?

http://en.wikipedia.org/wiki/Series_and_parallel_circuits#Capacitors

To do this, you just use the definition of capacitance, Q=C*V, and also you assume that each capacitor starts out with zero charge (and hence zero voltage). Then the charge that flows into each capacitor is the same, because they are in series.

Q = C1*V1 = C2*V2 = C3*V3

So then the voltage across each is:

V1= Q/C1

V2= Q/C2

V3= Q/C3

Vtot is just V1+V2+V3:

V1+V2+V3 = Vtot = Q*( (1/C1)+(1/C2)+(1/C3))

The voltage on each capacitor, as a fraction of Vtot, is:

V1/Vtot = (1/C1)/((1/C1)+(1/C2)+(1/C3))

V2/Vtot = (1/C2)/( (1/C1)+(1/C2)+(1/C3))

V3/Vtot = (1/C3)/( (1/C1)+(1/C2)+(1/C3))

So you do the math, and then you check to make sure those voltages V1, V2,V3 are not exceeding the voltage ratings for C1,C2,C3 respectively.

As you can see from the equations, the largest voltage falls across the smallest capacitor, and the total capacitance the series string is always less than that of the smallest capactior. (BTW: This is the same math as for current through resistors in parallel. The largest current flows through the smallest resistor, and the total resistance is less than that of the smallest resistor.)

As others have pointed out,

the math is a lot easier if you can assume all your capacitors have the same capacitance. Then if you have N identical capacitors, then the voltage across each one works out to 1/N of Vtot. E.g. if you have 3 capacitors with the same capacitance, then you predict (1/3)*vtot across each capacitor in the series string.Note that in practice, capacitors cannot be expected to have

exactlythe same capacitance. For example, the tolerance on electrolytic capacitors is especially bad, something like the value on the can plus or minus 20%!As pointed out above, the capacitor with the smallest capacitance takes the largest voltage, and that capacitor is kind of like the weakest link in a chain. You sort of have to do your design based on that capacitor, expecting that smallest capacitor will be the one to fail first.

Also putting some large resistors across each capacitor can help enforce that assumption about each capacitor starting with zero initial charge on it.

It is for reasons like this, I usually avoid using series combinations of capacitors.

Capacitors in parallel is another story. That trick usually works out well, with few complications, assuming you just want more capacitance, without increasing the voltage rating.

If the capacity of each cap. is the same then the voltage rating it the total of them. If the caps are different capacities and used in an a/c circuit then the voltage rating will not be the total, it will have to be figured. Google "voltage divider rule".

The voltage drop across caps using DC will add up. Theoretically your could run 360 volts thru your 3 caps. BUT since they are different capacity and voltage the voltage won't distribute evenly thru them and you'll have a breakdown.

I don't combine caps often to get higher voltage so my theory may need a little tweaking. I'm sure that Steve will check and correct my work.