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### Topic: Nice, I just bought myself 2 Super Capacitors for \$18 bucks + P&P (Read 3965 times)previous topic - next topic

#### cjdelphi

##### Oct 23, 2012, 04:07 pm
http://www.ebay.com.au/itm/160892379881?ssPageName=STRK:MEWNX:IT&_trksid=p3984.m1439.l2649

Quote

150F 2.7V super capacitors
Tolerance -20%~+80%
Voltage 2.7V
Surge voltage 2.85V
Nominal impedance: AC 10 megaOhms, DC 14 megaOhms
Working and storage temperature range -40~+60?
Lifecycles: Standard charge-discharge mode > 100000 cycles,?C/C, ?30%, ESR ? 4 times specified ESR
Dimensions: ?D: 25.0±1.0mm, L: 50.0±2 .0mm, H: 6.8±0. 5 mm, P: 10.0±0.2mm

I'm going to wire them up in Series (just over 5v) and power an Arduino Board

But how long is 2 150F cap's in Series giving a little over 5v going to power a standard Arduino Uno?

#### fungus

#1
##### Oct 23, 2012, 04:58 pm

But how long is 2 150F cap's in Series giving a little over 5v going to power a standard Arduino Uno?

Maybe Grumpy can calculate that for us...

I'm wondering where you're going to find enough electrons to charge it up.
No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

#### fungus

#2
##### Oct 23, 2012, 05:00 pm

But how long is 2 150F cap's in Series giving a little over 5v going to power a standard Arduino Uno?

Maybe Grumpy can calculate that for us...

I'm wondering where you're going to find enough electrons to charge it up.

PS: Seeing as how you've got two and they only cost \$9, are you going to connect one up backwards to mains electricity and film the explosion/crater?

No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

#### cjdelphi

#3
##### Oct 23, 2012, 05:19 pm
nah it's going to power some atmel processor on my arm or sleve powering a mini compiter.. i get a highest lightest power plug ask the place i'm at to charge my phone i dump it directly into the cap via a some kind of regulator i guess limit the current to 1/2 amp and after a minute or so say Thanks i charged it enough to make my call.

awww what a nice man saving money by being quick - nah lady he can recharge his entire phone with what he just took

#4
##### Oct 23, 2012, 05:21 pm
This seems to be pretty high impedance:

"v=Nominal impedance: AC 10 megaOhms, DC 14 megaOhms"

A better indication of what these parts can do is their ESR.

See these articles, which indicate a low ESR cap is needed.
http://cds.linear.com/docs/LT%20Journal/LTJournal-V20N3-06-df-LTC3625-Jim_Drew.pdf
http://cds.linear.com/docs/Design%20Note/dn166f.pdf

This link to a 150F 2.7V cap at digikey is for a part with 14mOhm (milli-ohm)
which indicates how fast you charge up the cap and also how fast it can discharge.

http://www.maxwell.com/products/ultracapacitors/docs/datasheet_hc_series_1013793.pdf
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

#### cjdelphi

#5
##### Oct 23, 2012, 05:43 pm
http://www.ebay.com.au/itm/160892379881?redirect=mobile

#### cjdelphi

#6
##### Oct 23, 2012, 05:48 pm
i think these are significant lower or he may as well sell a double A battery instead... i'll try 7 amps at 2.8v for short bursts and let you know how i go charging them... should be some wicked kick action for a solenoid!

#### Tom Carpenter

#7
##### Oct 23, 2012, 07:52 pm
I would be my guess that person who wrote the eBay listing confused mOhm with MOhm, and that those Z figures are in milliohms not megaohms.
~Tom~

#### JoeN

#8
##### Oct 23, 2012, 09:04 pm
Forgive my total lack of basic electrical knowledge, but doesn't wiring these in series actually make the voltage through the whole circuit including the capacitors themselves 5.4V and exceed the specification of the two capacitors?

I will never ask you to do anything that I wouldn't do myself.

#### Tom Carpenter

#9
##### Oct 23, 2012, 09:15 pm
No, that would be in parrallel.

If you wire two capacitors in series, the voltage is shared between then proportionally to their capacitance (Q = VC where Q is the charge on the capacitor, which would be the same amount on each). If they both have the same capacitance, then the voltage (in this case 5v) would be shared between them equally, so each would see only 2.5V. The maximum allowed would be 5.4V, not taking into account any small difference is capacitance. 5V is less than this, however 5V nominal could have voltage spikes or noise which may expose the capacitors to more than they are rated for.
~Tom~

#### DVDdoug

#10
##### Oct 23, 2012, 10:28 pm
In the real world, I wouldn't expect two capacitors to charge-up equally.   There will be differences in capacitance,  series resistance, and leakage resistance.  Once the capacitors are fully-charged, the  leakage resistance creates a votage divider for the applied charging voltage, and if the resistance isn't equal, you don't have equal voltage.

To do this properly, I think you should charge each capacitor separately to 2.5V (or 2.7V), and then connect them in series.

#### cjdelphi

#11
##### Oct 24, 2012, 04:51 am

No, that would be in parrallel.

If you wire two capacitors in series, the voltage is shared between then proportionally to their capacitance (Q = VC where Q is the charge on the capacitor, which would be the same amount on each). If they both have the same capacitance, then the voltage (in this case 5v) would be shared between them equally, so each would see only 2.5V. The maximum allowed would be 5.4V, not taking into account any small difference is capacitance. 5V is less than this, however 5V nominal could have voltage spikes or noise which may expose the capacitors to more than they are rated for.

I could use a ZenerDiode to clamp it as closely possible to 5v without going over, a linear regulator would not cut it, too much of a low drop out, a switching regulator would be pushing it too, I think a Zener 1 watt would be the best way to go... or simply place a power diode, let the voltage drop take care of the extra voltage that "could" take place...

#### cjdelphi

#12
##### Oct 24, 2012, 04:54 am

Forgive my total lack of basic electrical knowledge, but doesn't wiring these in series actually make the voltage through the whole circuit including the capacitors themselves 5.4V and exceed the specification of the two capacitors?

If you supply a regulated voltage of 5v to TWO caps in series, each cap should display 2.5v, the voltage in = 5v, you can't get any higher than that, if the Cap's were fully charged that's different then you'd have to discharge the caps to 5v, then i'd use it...

But I'll take a couple of precautions when using with atmega, and I think a Diode (.7v drop should do it)

#### tmd3

#13
##### Oct 24, 2012, 07:02 amLast Edit: Oct 24, 2012, 07:04 am by tmd3 Reason: 1
For a capacitor, the rate of change of the voltage is equal to the current divided by the capacitance:
` dv       i----  =  --- dt       C`
For a constant current, the voltage changes at a constant rate.

Two 150F capacitors in series will have a capacitance of 75F.  For a current of, say, 30 mA, the rate of change of the voltage will be 0.03/75, or 0.4 mV / s.  If the series combination of the capacitors is charged to its limit, 5.4V, and the critical voltage is, say, 4.5V, it'll hit that voltage in (5.4 - 4.5) / 0.0004 = 2250 s, or 37 1/2 minutes.

Taking the positive terminal as the terminal that's, well, more positive, and the direction of current as out of the capacitor, the general formula for the time it takes a capacitor to discharge from an initial voltage Vi to a final voltage Vf is:
`                   i * t ( Vi  - Vf )  =  --------                     C`
and that resolves to:
` ( Vi  - Vf ) * C------------------  =  t         i`
The same formula applies for charging the capacitors - Vf is bigger than Vi, making the voltage term negative, and the current goes into the capacitor, making i negative, so the quotient is positive.  Using the formula, it takes 75 amp-seconds for each volt of charging.  At 1/2 amp, it'll take thirteen and a half minutes to charge the capacitors from a fully discharged state to 5.4V.  It'll take nearly two and a half minutes to get it from 4.5V to 5.4V at 1/2 amp.

But how long is 2 150F cap's in Series giving a little over 5v going to power a standard Arduino Uno?
Translating "a little more than 5V" to, "exactly 5V," and adding the notion that the capacitors start at 5.4V, the answer comes to

( 30  / ( Arduino supply current in amps ) ) seconds.

I don't find any really credible estimates for the supply current of the UNO, but I'll guess that it's about 30 mA, making the time come to a neat 1000 seconds, nearly 17 minutes.