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### Topic: Converting Farad to mAh (Read 24672 times)previous topic - next topic

#### fkeel

##### Apr 22, 2012, 08:54 pmLast Edit: Apr 22, 2012, 09:16 pm by fkeel Reason: 1
Lets say I want to know how many mA per hour my supercapacitor can supply and assuming my capacitor is rated 2.7V and has 2F:

Code: [Select]
`Farad = (Ampere per second) / VoltFarad * Volt = Ampere per secondso 2F * 2.7V = 5.4 Ampere per second5.4 Ampere per second = 0.0015 ampere per hour //dividing it by 36000.0015Ah = 1.5mAh `

edit: this part is wrong

but. this assumes constant voltage. Actually there is a linear drop in voltage until the capacitor reaches zero.

So the actual energy it can provide is 1.5mAh/2

Answer: A capacitor with 2F and 2.7 volt can supply 0.75mAh (asuming I can somehow use all the energy to the last bit.)

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Is this correct?
http://embodimentlabs.tumblr.com/
http://paulstrohmeier.info/

#### RuggedCircuits

#1
##### Apr 22, 2012, 09:00 pm
A bit complicated. A 2F 2.7V capacitor can store Q=CV=5.4 Coulombs of charge. Now 1 mAh is 0.001 Columbs per second (0.001A) multiplied by 3600 seconds or 3.6 Coulombs.  So I think the capacitor is equivalent to 5.4/3.6 = 1.5mAh.

Of course, the capacitor voltage is going to go down linearly towards 0V, not like a battery, if you draw a steady 1.5mA from it for 1 hour.

--
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#### fkeel

#2
##### Apr 22, 2012, 09:07 pm
yeah. that's basically my reasoning. however: does the 1.5mAh already take the linear decrease of voltage into account? i.e. at 50% discharge the voltage has dropped to 1.35V, changing the formula we used to calculate those 1.5mAh.

Cause its a linear drop, can we simply divide the 1.5mAh by 2, in order to get the actual available energy? (think of it as a graph. voltage is X and time is Y. the area underneath the voltage is double the size as it would be if it would decrease in a linear manner.)

But somehow that seems both too complicated and too simple at the same time.

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so. back to my original question. does 1.5 already take the voltage drop into account?
http://embodimentlabs.tumblr.com/
http://paulstrohmeier.info/

#### RuggedCircuits

#3
##### Apr 22, 2012, 09:13 pm
Both mAh and Coulombs are measures of charge (hence energy) storage. There is no consideration of voltage, just a measure of how much energy the capacitor can store when fully charged. It doesn't matter how the voltage varies from 2.7V to 0, linear, non-linear, etc.

Think about a 5.4L tank of water. The capacity is 5.4L. If you drain it at 1.5mL per second, it will take 1 hour to fully drain it. Or you can drain it at 3mL per second for half an hour. Or start, stop, start, stop, etc. It doesn't matter -- the capacity of the tank is 5.4L.

Same for the capacitor. The capacity is 5.4 Coulombs of charge. If you drain it at 1.5mA it will take 1 hour to fully drain -- hence you can say it's equivalent to 1.5mAh of capacity. Any other way you do it (3mA, linear, non-linear, etc.) does not change how much charge (hence energy) the capacitor can store.

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#### fkeel

#4
##### Apr 22, 2012, 09:15 pm
cool. thanks for clarifying. I was getting myself all confused.
http://embodimentlabs.tumblr.com/
http://paulstrohmeier.info/

#5

#### fkeel

#6
##### Apr 23, 2012, 12:52 am
dont know what I was thinking. I know that. anyway. thanks for pointing it out, grumpy_mike
http://embodimentlabs.tumblr.com/
http://paulstrohmeier.info/

#### MarkT

#7
##### Apr 23, 2012, 11:22 am
Also you cannot blithely assume an ultra capacitor is a linear device as its about as far from an ideal parallel-plate capacitor as its possible to get!
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### fkeel

#8
##### Apr 23, 2012, 06:39 pm
Yes, MarkT, you are right. Grumpy Mike already pointed that out.

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I was just told that 4 of the capacitors I described at start (so 2.7V & 2F) In parallel would be equal to 75mAh. This information comes from the manufacturer.

Does anyone have any idea how they arrive at that conclusion?

http://embodimentlabs.tumblr.com/
http://paulstrohmeier.info/

#### RuggedCircuits

#9
##### Apr 23, 2012, 06:42 pm
Voltage will go down exponentially under constant resistance load, linearly under constant current load.

I don't know how the manufacturer justifies 75mAh, unless my calculations above are quite wrong. I'd be curious to see if they can justify it.

--
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#### MarkT

#10
##### Apr 23, 2012, 06:48 pm

Yes, MarkT, you are right. Grumpy Mike already pointed that out.

*

I was just told that 4 of the capacitors I described at start (so 2.7V & 2F) In parallel would be equal to 75mAh. This information comes from the manufacturer.

Does anyone have any idea how they arrive at that conclusion?

They multiplied by 3.6 instead of dividing by 3.6 - and didn't check their calculations (marketing department then!)
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### Grumpy_Mike

#11
##### Apr 23, 2012, 06:49 pm
Quote
Voltage will go down exponentially under constant resistance load, linearly under constant current load.

True, I suspect a processor chip is not of these.

#### fkeel

#12
##### Apr 23, 2012, 06:54 pmLast Edit: Apr 23, 2012, 07:23 pm by fkeel Reason: 1
thanks @rugged circuit. I was really confused, as during my research I taught myself that it is both constant and exponential... now it makes sense.

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I will ask them how they got that number. I just wanted to check with you guys first, as I did not want to call them on a mistake which was on my side...

p.

edit: they just corrected themselves, saying its 6mAh
http://embodimentlabs.tumblr.com/
http://paulstrohmeier.info/

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