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I want to replace a battery, which has a load of LEDs and a temperature sensor, with a super capacitor.

How can I calculate the specifications of the super capacitor?

Thanks!

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  • \$\begingroup\$ Battery capacity in generaly stated in Ah , Your supercaps are in Farads which is Ampsecond per Volt . So this means that your supercaps wont operate the Leds for as long as a battery does .If this is acceptable to yo then you can have something that has essentialy no maintainence . \$\endgroup\$ – Autistic Jan 8 '16 at 21:38
  • \$\begingroup\$ You have a good answer here! electronics.stackexchange.com/questions/156277/… \$\endgroup\$ – WeGoToMars Mar 8 '16 at 20:40
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You need four pieces of information

1) How much current do you need (in amps)?

2) How long do you want to run your circuit from the cap (in seconds)?

3) What it the largest voltage (normally the regular operating voltage) which will be applied? and

4) What is the lowest voltage out of the capacitor which will still allow the circuit to function?

Let's call 1) i, let's call 2) $\delta$t, 3) is Vmax, and 3) minus 4) is $\delta$V.

The relationship between time, voltage and current in a capacitor with a value of C in Farads is $${\frac{i}{C} = \frac{dV}{dt}}$$ For a constant i, which is a reasonable first approximation in this case, this becomes $${\frac{i}{C} = \frac{{\Delta}V}{{\Delta}t}}$$, or$${C = \frac{i{\Delta}t}{{\Delta}V}}$$ and the capacitor must have a voltage rating of Vmax.

To walk you through this, lets say you need .25 amps for 5 minutes (600 seconds). Let's say your battery puts out 3.7 volts when fresh, and your circuit will work down to a battery voltage of 3.0 volts. Then $$C = {\frac{(.25)(600)}{(3.7 - 3.0)}}$$ $$C = {\frac{150}{0.7}}$$

and you need a 214 Farad supercap with a voltage rating of 3.7 volts minimum.

EDIT - As Brian Drummond has pointed out, supercaps often have a high internal resistance. If you want to take this into consideration (and you had better do so), you need to quantify the resistance R of the cap at the current level which you are using. Then the capacity calculation remains the same, but if the capacitor voltage rating is Vcap, instead of being Vmax, $$Vcap = Vmax + (iR)$$ In the above example, if the ESR of the supercap is 20 ohms, $$Vcap = 3.7 + (.25)(20)$$ and $$Vcap = 8.7$$ In this case you would definitely need to find a different supercap.

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Simple. What current do you need and for how long? Multiply those two to get the charge you need.(amps * seconds = coulombs).

So if you need 1 milliamp for an hour, that's 0.001 amps * 3600 seconds = 3.6 coulombs.

Now decide the voltage drop you can tolerate. If it's 0.5V (say, your circuit works from 3 down to 2.5V), divide the charge by that voltage (dC = dQ/dV). So the capacitance you need is 3.6 / 0.5 = 3.6 * 2 = 7.2 Farads for this example.

Now plug in your own numbers.

EDIT : Be aware that some supercapacitors have a relatively high ESR, or series resistance, perhaps 10 or 50 ohms. At 1 mA, that would reduce the available voltage by 10 to 50mV, but at higher currents this gets more problematic.

Check the datasheet for your choice of capacitor, and if necessary, allow for this voltage in the voltage drop calculation above.

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