capacitors in series

it is going to work, sorry for the post. One of the capacitor is rated 1000uF and the other 2200uF, I am just thinking that one might take more voltage then the other. v=Q/C proportionally, so 1000uF will have twice more voltage.

If the capacitors are equal, the applied voltage will split equally across them, and you can use two 50V caps in a 35V circuit. If the caps aren't equal, the voltage will be higher across the smaller cap. (The charge is the same on both caps because they're in series, and voltage = charge / capacitance V=Q/C). So if you had two capacitors with a 25V rating, and the ratio between their values was greater than 25/10, or

-- john

I did the circuit like said, I get 35 volt at the bridge, but when I add the capacitors, I get 50v (49.8v), I was sure I would be getting 35 volts across both...

ken

I see you figured it out for yourself. I'd just like to inject a note of caution about the electrolytics. It's very common for large caps to be 20 or 30% off nominal value, often higher. If the 2200 is that much higher and the 1000 isn't, the latter might have end up with more than

-- john

For electrolytics as DC filters, the voltage split may depend more on the relative leakage current of the two caps than on their capacitance values. It's probably safer to use a single capacitor with a voltage rating higher than 50V for your power supply.

-- john

Something to consider: The tolerance of electrolytic caps is typically very large, maybe even

Second, leakage and ESR, and probably a lot of other things will also contribute to voltage drop.

So, the actual capacitance ratio only controls the AC voltage division across the capacitors. If the capacitors are filtering rectified AC and thus dropping a significant DC Voltage component then the Capacitance ratio cannot be used to predict the (DC) voltage drop across the individual capacitors. For this the other effects become dominant. For this reason, series connections of electrolytics in power filters is probably not advised.

In situations where it is done by paralleling the caps with resistors controls the (DC) voltage division.

Sorry, but I never went down this road so I don't know the criteria for selecting the resistors other than the basic voltage division. I would think that the resistor current would have to be selected as a ratio of the ripple current but that is only a guess.

Hope this helps ....

You mean you got 35V without the caps ? That's because it was just rectified ac.

The caps will charge up to the peak of the recified ac's votage which will inded be ~ 50V in this cae.

Graham

Have you considered the ripple voltage? Assuming you are using 60 Hz AC, it will be about I*.0083/C where I is the current you draw from the supply in amperes, and C is the value of your filter capacitor, in farads. For example, assume you draw 1/2 amp from your ~50 volt supply, and have a 2200 uF 63 volt cap. Your ripple voltage will be .5*.0083/.0022 or about 1.89 volts. And rather than using the nominal 2200 uF figure, you should look at the the +/- spec on the cap and use - spec to compute the lowest value the cap might be.

There's a whole lot more to consider than the above when designng a power supply. This site may help:

Ed

are they the same capacitance?

Bye. Jasen

Here is one way to calculate parallel "ballast" or "bleeder" resistors for a DC application of 2 series caps:

First you need to know the approximate worst case leakage of the capacitors in question.

Then take the following worst case conditions:

- The maximum voltage rating of one of the caps

- Assuming one cap has maximum leakage and the other has zero leakage

i.e. The capacitor with the leakage with drag the mid rail voltage away from its nominal half rail, creating a greater voltage across the cap with no leakage. You don't want to have more than the maximum capacitor voltage across the non-leaky cap.

Once you assume these worse case conditions then the circuit is easy to analyse.

Each resistor is: R=(CapLeakRes * (MaxCapVolt-(Vrail/2)) / (Vrail-MaxCapVolt)) * 2

The leakage will be voltage dependant, but this simple formula is a good starting point.

Dave :)

On Sun, 2 Jul 2006 19:14:49 -0400, in message , "Ken O" scribed:

Assuming you are using a DC voltmeter: you are measuring average voltage. assuming you are measuring the output of a rectifier: an unfiltered rectifier will have a lower average voltage than a filtered rectifier.

Now, about ratings. When designing circuits, it is always a good idea to use a process called "derating." Derating is the use of components with a rating comfortably in excess of the actual running conditions of the circuit. In this case, with a filtered DC output of 50V, good derating is the use of a capacitor rated at 75-100VDC.

yes you are right, I just calculated that I needed at least 2700uF and approximately 75v, but I'll go to 100v to make sure. That means that the capacitors I used are know blown up. ..

Ken

Yes, two 25V caps in series will give 50V operational overall but cap total cap will be 1/2. I'll also add a 470K resistor across each one to ensure that each one share the voltage equally. Steve Balstone stephen balstone