Electronics > Beginners

Capacitor as power supply circuit

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mrpsychotic:
I have seen those heavily discounted maxwell ultracapacitors on electronic goldmine lately, and with the holidays around the corner I plan on having my hands on one soon.

The two possibilities i might get are 1 3000F 2.7V capacitor, or 2 2600F 2.5V capacitors which can make 5200F 2.5V or 1300F 5V.

(A side question: capacitors cannot really be chained in a long series like resistors can without running into problems, right? Isn't there a problem with the energy not being distributed evenly?)

My main question is what circuit or ic I can use to turn this large capacitance into a useful power supply. Because of its discharge curve, If I wanted to run something that needed 2 volts, it would stop running although there is still plenty of energy. The other issue is the inherent low voltage of ultracapacitors in general. I hear this is because the activated carbon used to achieve such high capacitance has a low breakdown voltage and there isn't much they can do about it. If I wanted to create a power supply that delivered a constant 5 volts from a capactior with lower voltage, how would I go about doing this?Thanks!

Simon:
why do you want to use a super capacitor as a power supply ? why not just use a power supply ?

I'm not certain but beware with large capacitors, they could draw a large amount of current on charge-up and blow things (once a diodes data-sheet would tell you the maximum capacitance it can charge)

Feanor:
Boost converter is the answer. You can use a switch mode boost converter to get your required 5V out of the super capacitor even as it discharges.

Simon is right, be careful when charging or discharging these things! A small accident like shorting out your 9V battery terminals will not cause much damage but a 2300F capacitor is another story. Use a series resistor to control the current at least.

Generally the higher the capacitance the lower the rated voltage is, it all comes down to the formula for capacitance of two parallel plates with a dielectric between them. A is the area of the plates, Er is the relative permittivity and pi is well, that's pi.

C = (Er* A)/(4*pi*d)

The d is the distance between the plates of the capacitor. So if we make this distance really small we get a large capacitance. But if this distance is very small then the voltage required to breakdown the dielectric between the plates is also very small, because they are so close together. We also want to choose a dielectric with a high relative permittivity so we can not always choose one with a high breakdown voltage.

Murphy:
Ultracaps are a great replacement for batteries in some applications because they can be charged(and discharged) extremely quickly, their ESR is very small. Hundreds to thousands of farads can supply significant power for a reasonable amount of time and then be zapped back to full in seconds.

Yeah you need a boost converter, like the ones in LED flashlights that suck a AA all the way down.

Not sure on the load balancing, usually it's not a big deal but ultracaps could be different. I know the humongous banks that Maxwell sells have some sort of active circuitry attached to them that looks like it's for load balancing.

Simon:
the most efficient way of charging the caps would be a switchmode converter with an automatically increasing voltage, probably you could make it current controlled up to a certain max voltage so you don't dissipate heat in the resistors and have control over the speed of charge and current demand