Welcome to OnlineConversion.com Forums

 [ Home ] [ Forum Home ] Register FAQ Calendar Search Today's Posts Mark Forums Read

 Convert and Calculate Post any conversion related questions and discussions here. If you're having trouble converting something, this is where you should post.* Guest Posting is allowed.

#1
08-18-2007, 10:44 AM
 Unregistered Guest Posts: n/a

I am considering building an electric car. It may not be possable, but I would like to use Super-capacitors instead of batteries. I need to know how to convert farads in to volts, and how to rate farads as amp hours. Can you help me?

Thanks!
#2
08-22-2007, 12:49 AM
 Robert Fogt Administrator Join Date: Dec 2005 Location: Seattle, WA Posts: 3,416 Rep Power: 12

No direct conversion from farad to volts or amp*hours. Here are the formulas.

1 farad = 1 coulomb / volt
1 coulomb = 1 amp/second
#3
09-17-2008, 09:30 PM
 Unregistered Guest Posts: n/a

While true, that answer is a bit disingenuous.

I think this is how it works, but I may be wrong.

So you divide the Capacitance by the cell voltage
(Not pack voltage, since the cells are in series and
the same current goes through each cell)
The Maxwell packs I am looing at are 2.7 volts.

X/2.7 is the number of coulombs the pack can supply.

Now, Amps is Coulombs per second, so you would
get X/2.7/60 = X/162 Amp Hours (hours at one amp)

That gets you to Farads/162 = Amp-Hours for this cell type.
Or 162 * Amp-Hours = Farads

This means to get a 12A-H pack for my electric bike I need
nearly 2000 Farads. For 32 volts with 16V-500F packs
I need 8 at \$770 each.

Looks like if my math is right, my bike will be batteries, not
ultra-capacitors.
#4
09-18-2008, 03:30 PM
 JohnS Double Ultimate Supreme Member Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,233 Rep Power: 18

Quote:
 Originally Posted by Unregistered While true, that answer is a bit disingenuous. I think this is how it works, but I may be wrong. X Farads = X(Coulomb/Volt) So you divide the Capacitance by the cell voltage (Not pack voltage, since the cells are in series and the same current goes through each cell) The Maxwell packs I am looing at are 2.7 volts. X/2.7 is the number of coulombs the pack can supply. Now, Amps is Coulombs per second, so you would get X/2.7/60 = X/162 Amp Hours (hours at one amp) That gets you to Farads/162 = Amp-Hours for this cell type. Or 162 * Amp-Hours = Farads This means to get a 12A-H pack for my electric bike I need nearly 2000 Farads. For 32 volts with 16V-500F packs I need 8 at \$770 each. Looks like if my math is right, my bike will be batteries, not ultra-capacitors.
X farads at 2.7 V is 2.7 X coulombs. However, there are 3600 s in an hour, so 2.7*X coulombs is 2.7*X/3600 Ah. Still worse, the capacitor voltage will decrease linearly as it discharges. A battery maintains a fairly constant voltage until it is almost discharged, then plummets to zero.

Real comparability would require looking at the total stored energy in joules. For the capacitor, 0.5*C*VČ, for the battery V*Amp hours*3600. You need 2700 farads, rated at 32 volts. If you connect 16 V packs in series, you'll need 5400 F in each pack. I recommend batteries. Supercapacitors aren't really super.
#5
09-11-2010, 01:41 PM
 Unregistered Guest Posts: n/a

The problem with using capacitors is that as they deplete the voltage also drops as apposed to batteries were the voltage stays the same till they are "dead"
#6
11-19-2010, 05:07 PM
 Unregistered Guest Posts: n/a

I feel the best way to solve the problum would be to use both batterys and caps. The caps can be conected so they would "release" energy at a lowr.p.m. such as starting off and then would power back up and could be used to once again give a "shot' of powewr in hill erias. The thing you should put the most concideration into is the type of battery and may i suggest lithium. Very light wait to the power, jewls, amps voltage for this type of battery is superior. But be prepaired to pay for the best batterys also. i may not be able to spell well but i know what i am talking about. just think about it and make your own disition. i just had to throw it out there for you because thats what i would do and i have experiance with electrical vehicals and bikes as well. Utilize any and all possable options. Just don't over engineere. over redundance can be the death of a project. "KISS" is the best way.
Keep
It
Simple
Stupid
KISS besides if it simple when you have a problum some time and you will with any first production you can always solve the problum with a couple of jumpers to get you home.
#7
11-29-2010, 08:33 AM
 Unregistered Guest Posts: n/a

Quote:
 Originally Posted by JohnS X farads at 2.7 V is 2.7 X coulombs. However, there are 3600 s in an hour, so 2.7*X coulombs is 2.7*X/3600 Ah. Still worse, the capacitor voltage will decrease linearly as it discharges. A battery maintains a fairly constant voltage until it is almost discharged, then plummets to zero. Real comparability would require looking at the total stored energy in joules. For the capacitor, 0.5*C*VČ, for the battery V*Amp hours*3600. You need 2700 farads, rated at 32 volts. If you connect 16 V packs in series, you'll need 5400 F in each pack. I recommend batteries. Supercapacitors aren't really super.
I like this method. Just remember, for those who are not engineers, these equations can be written in straight english.

Power in a capacitor (as measured in Joules) is equal to the voltage it is charged to "squared" times its capacitance in Farads. If you have millifarads, you must multiply by 1000. if you have uF (microFarads) you must multiply by 1 million (1 000 000). Remember not to use the capacitors rated voltage. That's just the highest charge it can handle before it explodes. They sometimes explode anyways if your charge is too close to the rated voltage, so select a capacitor with an oversized rated voltage for safety (the maximum charge voltage times 2 perhaps).

The power of a fully charged battery (again, measured in Joules) is the battery voltage, times the A-h rating of the battery. If you have only the mA-h rating, you must multiply by 1000. Now multiply by 3600. (the number of seconds in an hour). Now you can compare.

By the way, you can always put a capacitor accross the terminals of a battery in theory, but in practice you could hurt yourself. The battery has a very low internal resistance, as does the capaitor, so the amount of current is equivalent to shorting the battery terminals with a screwdriver. I don't recommend it. I am not sure what the result will be, but with this much current, "somethings gotta give". If you plan to do this, in a real engineering design, you probably know what you're doing so you would use a resistor to dampen the charge speed until the two voltages are equivalent.

Also remember that the capacitor will discharge your battery much faster (two or three weeks?) than if your battery had nothing connected to it (6 months?).
#8
11-29-2010, 10:05 AM
 JohnS Double Ultimate Supreme Member Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,233 Rep Power: 18

Quote:
 Originally Posted by Unregistered I like this method. Just remember, for those who are not engineers, these equations can be written in straight english. Power in a capacitor (as measured in Joules) is equal to the voltage it is charged to "squared" times its capacitance in Farads. If you have millifarads, you must multiply by 1000. if you have uF (microFarads) you must multiply by 1 million (1 000 000). Remember not to use the capacitors rated voltage. That's just the highest charge it can handle before it explodes. They sometimes explode anyways if your charge is too close to the rated voltage, so select a capacitor with an oversized rated voltage for safety (the maximum charge voltage times 2 perhaps). The power of a fully charged battery (again, measured in Joules) is the battery voltage, times the A-h rating of the battery. If you have only the mA-h rating, you must multiply by 1000. Now multiply by 3600. (the number of seconds in an hour). Now you can compare. By the way, you can always put a capacitor accross the terminals of a battery in theory, but in practice you could hurt yourself. The battery has a very low internal resistance, as does the capaitor, so the amount of current is equivalent to shorting the battery terminals with a screwdriver. I don't recommend it. I am not sure what the result will be, but with this much current, "somethings gotta give". If you plan to do this, in a real engineering design, you probably know what you're doing so you would use a resistor to dampen the charge speed until the two voltages are equivalent. Also remember that the capacitor will discharge your battery much faster (two or three weeks?) than if your battery had nothing connected to it (6 months?).
Actually, if you are calculating energy and have millifarads, you must divide by 1000 to convert to farads, and microfarads would require division by 1 million. In later paragraph, you must divide by 1000 to convert milliampere-hours to ampere-hours.

 Thread Tools Display Modes Linear Mode

 Posting Rules You may post new threads You may post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Main Forums     Convert and Calculate     Resources     General Chat

All times are GMT -8. The time now is 10:43 AM.