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#### Spilled Fluids

• Full Member
• Posts: 138
« Reply #30 on: May 20, 2015, 08:21:01 PM »

....
Sorry I have no direct idea about equivalence of capacitors vs batteries, need to look this up.

Red_Sunset

Check out Batteryuniversity.com ; you will find all the info you want about batteries, capacitors and super caps as well as how they differ in function and construction

#### Red_Sunset

• Hero Member
• Posts: 548
« Reply #31 on: May 20, 2015, 09:13:30 PM »
Paul, Spilled Fluids,

Thanks for your battery reference site. I couldn't find there immediately what I wanted,
Check out this site:  http://bymark1.hubpages.com/hub/Ultracapacitors-as-batteries

Ultracapacitors as batteries  some details copied and shown below.

We can use the following formula to convert Farads to Amp hours
(Vmin + Vmax)/2 * F / 3600 = Ah
Vmin & Vmax is the upper and lower voltage. This is the voltage range that your device has to work in. As capacitors "run down" and need charging, their voltage drops, so we have to allow for that. If we assume that we want to power a 3 volt radio, we can guess that it should work happily between 3.5 volts and 2.5 volts. That's what the first part of the calculation allows for;
In our example, (Vmin + Vmax)/2 would be (3.5v + 2.5v) / 2 equals 3 volts.

In the rest of the calculation, F - Farads and
3600 - coulomb, a unit of electrical measure. One amp hour = 3,600 C
Ah - Amp hours ( 1000 mAh = 1Ah)
So we now have 3 * F / 3600 = Ah
Let's say we can get hold of a 3.5 volt 100 F super-capacitor. Using 3 * 100 / 3600 = Ah,
we could expect it to power a device drawing 0.083 amps, or 83 milliamps, for one hour.
What if this wasn't enough? What if we wanted a 3 volt capacitor to power a 3 volt device which drew 250mA for one hour?
We can use the formula "current in amps * 3600 / voltage = capacitance needed in Farads" to find out how big the capacitor needs to be. In this example
0.25 * 3600 / 3 = 300 F
The capacitor would need to be 3v 300F to power a 3 volt device drawing 250mA for 1 hour. Ideally, you should have a capacitor of slightly more than 3 volts to allow for the voltage drop mentioned above.

#### Paul-R

• Hero Member
• Posts: 2077
« Reply #32 on: May 24, 2015, 03:12:04 PM »
Thanks for all that useful stuff, Red.

But when you say (Vmax + Vmin), you must mean minus not plus. Vmin will be a parameter which reduces the useful overall power capacity of the cap.

#### Red_Sunset

• Hero Member
• Posts: 548
« Reply #33 on: May 24, 2015, 05:07:01 PM »
Thanks for all that useful stuff, Red.

But when you say (Vmax + Vmin), you must mean minus not plus. Vmin will be a parameter which reduces the useful overall power capacity of the cap.

Paul,

All what the you are trying to do here is find the mid-range or average working voltage.  Remember it is a value you choose based on what you need as the min voltage for the equipment you are going to connect to the Cap.
To can also find the same average by,  do a minus, then divide by 2 then add to Vmin.
Sure the higher Vmin is, the smaller your working range becomes.  The same as with a battery.  Although with a battery, this value is usually set by the battery type,  the lowest discharge voltage before the battery becomes irrecoverable damaged.

Red_Sunset