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Wiring Capacitors in Series

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rcollege

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Joined 2010
#1
HI all...

I've gotten mixed answers elsewhere.

I know wiring caps in series decreases their capacitance but,
Does wiring Capacitors increase the voltage capacity?
Or is the capacitor Voltage rating remain the same..

I ask because I have some 430V and 405V transformers and some 330u 450V caps and plenty of 470u 385Vcaps.
I see plenty of circuits using 400V+ for their B+...but not many 500V+ caps.

Thanks!
 
#2
Well,

Caps in series need to have resistors across them to ensure the voltage across the caps is equal. Otherwise one could have more voltage than the other.

Next thing,

If one cap goes short then the other sees the full line voltage.

Leakage is another issue.

Regards
M. Gregg
 
#3
If you wire two 400 volt caps in series and the caps are identical in every way then in theory any applied voltage would equalise between the two. In practice that doesn't happen due to different leakage currents etc. If you wire two identical value resistors (say 470K of suitable wattage) across each cap then the voltage does equalise out.

(And the above doesn't imply its a technique I would recommend without knowing all the fine details).
 
#4
Thanks for the clarifications.

I may try to source some 600V caps.

Or just bring the voltage down before reaching the caps...and keep close eye on the plate and cathode resistor to be sure their values keep the supply voltage low enough.

I'm currently am using 2x 470u caps for my RC filtering with a 230V power supply...zero hum. Would a lower value or smaller caps have any benefit?
 
#5
This is quite common in guitar amps with higher HT, and seems to work OK...
just make sure that the balancing resistors are of proper wattage.
Downside is, of course, that you need more capacitors (4x) as opposed to a single capacitor.
 
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#9
AuroraB said:
...just make sure that the balancing resistors are of proper wattage.
Click to expand...

It's also worth checking the resistor values every few years. I had some balancing resistors change values over about 15 years, and they were 2W resistors dissipating maybe only 0.5W each. In each case one of a pair or series resistors went high by about 30%, that meant that the voltage balance was significantly out. I had specified capacitors capable of withstanding even greater imbalance so fortunately no harm was done before I discovered the problem.
 
#10
12E1 said:
It's also worth checking the resistor values every few years. I had some balancing resistors change values over about 15 years, and they were 2W resistors dissipating maybe only 0.5W each. In each case one of a pair or series resistors went high by about 30%, that meant that the voltage balance was significantly out. I had specified capacitors capable of withstanding even greater imbalance so fortunately no harm was done before I discovered the problem.
Click to expand...

That's an excellent point actually. Resistors have voltage ratings as well as maximum power dissipation values and so for any high voltage applications you should definitely obtain resistors suited to continuous high voltage operation. It doesn't matter that the dissipation might be low, the resistors fail because of the voltage.
 
#11
Capacitors in series adds (Or multiply if identical units) its Volatge rating.

Capacitor in parallel adds (Or multiply capacitance if identical units) its capacitance value.

Of course, the Q = C * V in each capacitor will be equal.

Capacitor in series need to be the same capacitance value otherwise voltage will be fractiones proportionally to their C value. The net capacitance will be the inverse of the summation of the inverse of each capacitor.

Capacitor in parallel will have equal voltage across it. Remember to add a bleeder resistor to discharge them as the apparatus is power down, so no dangerous voltage are present inside it, moreover during testing or modifying the unit.
 
#12
Thanks everyone for your responses.

I'm building a preamp. The voltage on other preamps I've seen use higher voltage Transformers and large anode resistor values.
I was using 200V+ plate voltage previously, but my Blu-Ray had a 2-3V rms. So no amplification was needed...it acted as a line driver.
The new Blu-Ray Player I have now has a 200mV rms. So 5x+ Amplification will be needed to reach 1V RMS.

This means I now will need higher B+ to get the same plate voltage.
Using Capacitors in series allow me to have the higher V on the capacitors leaving some current to get to the plate.
 
#13
For two caps of equal capacitance (C) in series, the total capacitance will be C/2. The total voltage rating will be 2*V, assuming that the charge is evenly shared.

To ensure equal sharing, place a resistor in parallel with each cap. All the resistors should be the same resistance.

Now... In high-voltage circuits, it's commonly advised (if not required) that a bleeder resistor is added in parallel with the supply capacitance. This is to discharge the cap when the power is turned off. What I do in my circuits is to make the charge sharing resistor the bleeder resistor. So I'll have, say, 100 kOhm 3 W across each capacitor and two 400 V 100 uF caps in series. This gives me a total of 50 uF with a working voltage rating of 800 V.

I tend to run about 2 mA through the bleeder resistor. This is enough current to power a high-efficiency LED. So I've taken to include a red LED in series with the bleeder resistor. This serves as an on-board power ON indicator. For debugging it's pretty handy to have a visual indicator of power on (do not stick fingers into amp!)

~Tom
 
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#14
When you pass charge current through 2 caps in series then they aquire a charge voltage across them according to their capacitance. So putting (for example) a 100uF and a 50UF in series is a bad idea, the 50uF will charge to 2 x the voltage that the 100uF will.

Electrolytic Capacitors generally have a tolerance of -30% to +70% so even using 2 identical capacitors you are VERY unlikely to get equal voltage sharing.

To address this you MUST use voltage share resistors across the capacitors.

If you are going to get seriously technical, the voltage share resistors should be sized to pass 3 to 5 times the capacitor leakage current, but Toms recommendation opf 2mA above will in general be good enough. His idea of adding the RED "Danger Live Voltage" LED into the voltage share resistor string is also RECOMMENDED - it certainly would have saved me quite a few nasty surprises in the 30 years I've been "messing" with tubes..

The other note is in MOST commercial designs I've seen the voltage share resistors have been too high a value.

If you want to do a calc. The leakage current will be given on the data sheet, if not, use a default formula of I = 0.01CV where V is the applied voltage not the voltage rating of the cap.

Typical example: 220uF caps in series with 250V across each
the calc gives 550uA leakage current.
3 to 5 times that is 1.65mA to 2.75 mA - Tom's recommendation of 2mA would be good.
So 2mA with 250V across it means the voltage share resistors in this case should 120K maximum and will dissipate 480 mW. So use 2W metal films.

How many times have you seen this typical example but with 330K or 470K voltage share resistors ??? - That is just not good enough.

Cheers,
Ian
 
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#16
Thanks again for the replies...

Off topic...how much capacitance is generally required for a r/c power supply?
How may mV's of ripple before it becomes audible in a preamp? I imaging 10mV would be audible...but what about 5mV? My last supply simulated to less than 1mV.
I read somewhere that lower value caps sounded "better" smoother and more dynamic?
I'm not sure I agree yet...but I'm open to thoughts/ experience.
Generalization / specifics are both welcome.
 
#19
tomchr said:
For two caps of equal capacitance (C) in series, the total capacitance will be C/2. The total voltage rating will be 2*V, assuming that the charge is evenly shared.

To ensure equal sharing, place a resistor in parallel with each cap. All the resistors should be the same resistance.

Now... In high-voltage circuits, it's commonly advised (if not required) that a bleeder resistor is added in parallel with the supply capacitance. This is to discharge the cap when the power is turned off. What I do in my circuits is to make the charge sharing resistor the bleeder resistor. So I'll have, say, 100 kOhm 3 W across each capacitor and two 400 V 100 uF caps in series. This gives me a total of 50 uF with a working voltage rating of 800 V.

I tend to run about 2 mA through the bleeder resistor. This is enough current to power a high-efficiency LED. So I've taken to include a red LED in series with the bleeder resistor. This serves as an on-board power ON indicator. For debugging it's pretty handy to have a visual indicator of power on (do not stick fingers into amp!)

~Tom
Click to expand...

Any chance you have a sketch of this circuit,,, I just did a similar recap, with two 100uF 450V caps, to get 900v 50uF,,, there are two sets in this ckt, in series, replacing two 80uF 450V cans,,,There is a 150K 2W R across each set...

Regards,
John
 
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