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Can I safely connect two Super capacitors in Series ?

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Apr 1, 2013
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Can I safely connect two Super capacitors in Series ?

My super capacitor rating is 10F 2.5V and I cannot connect it across 5V. So, can I connect two super capacitors in series and connect them across 5V from 7805 ? This way capacitance will reduce to 5F and voltage will increase to 5V as they are in series. Will this harm the capacitors I mean the Super capacitors ? I am using Super Capacitor for embedded application so that it does some task on power failure. When mains failure is detected it has to do a small task before shutting down and hence I want to use Super capacitors.

You have nothing to regulate the voltage across each capacitor. When power is applied the capacitor with the least capacity (they will probably not be the same) will get over-voltage. The capacitor that has the lowest leakage current will also get over-voltage.

You can use this circuit, it makes sure none of the caps charges over 2.5V at any time.

[dismissively waves hand] Relax, guys!
Yup. It'll work - just connect 'em in series.

That being said though, Audioguru is right - and a little attention to balancing the voltages across them will go a long way in terms of reliability. The most common approach is to use "voltage balancing resistors" (e.g. which is simply two *equal valued* resistors in parallel with each cap. This effectively forms a voltage divider that drives the junction of the two capacitors to mid-rail, i.e. 2.5V in your case.

You want the current through these resistors to swamp your leakage current, which of course means you'll trade-off how long the capacitors will hold the rail voltage up as they discharge, since they'll also be pouring current into the resistors across them. If you really can't tolerate any extra discharge current (i.e. you want to use big resistors or omit them entirely) and you're paranoid, shunt the capacitors with 2V7 zener diodes instead and all will be well (do BOTH, and the design will be bulletproof! :)

I suspect with 5F load across a 7805 it will struggle to keep it's output up anyway. For a significant time it will be in current limiting state.

The problem I see with balancing resistors is they would need to be very low value to ensure each capacitor sees an equal voltage, there is no margin for error, if the voltage isn't equally shared, one of the capacitors will go pop! I'm thinking (without doing any math) the resistors may have to be 10 Ohms or less so they will use much of the 7805's capability before even starting to charge the caps. From the first post, I'm guessing the 7805 is also powering the remaining circuits as well.

It's a tricky problem, I think the best solution may be to simply add series diodes across the capacitors, so they work like Zeners but only using the 0.6V - 0.7V Vf stacked up instead. Four 3A diodes in series across each capacitor would probably work. To get an exact voltage it may be necessary to mix normal and Schottky diodes.


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