Why cell phones don''t use a [Li-ion + SuperCap] Combination ? - Page 1
Forum section: electronics / beginners

author: opablo
author_rank: Regular Contributor
timestamp: July 18, 2012, 12:59:16 pm
content: Maybe out there is a SuperCap geek that knows enough to help me understand that questionClearly SuperCaps (currently) are of no interest of the cell phone manufacturers but why ?Maybe they can break on a phone drop ? or get on fire because of fabrication failures ?.. maybe the self-discharge rate is too high ? is it the old mighty $$$ ? why ? why ?I know they have lower capacity/volume than Li-ion but... imagine a 50/50 combination with supercaps... you could be able to charge 50% super fast... and at the same time not rely on them for the whole timeI'm an micro Controller geek (almost all my projects include micro controllers and I just saw a really atractive ebay product to use in flat packs... search there for "1PCS 4F 5.5V Coin V FITN Super Capacitor Memory Backup!"What should stop me from buying 4 of those and power my arduino toys out of it ?

author: T4P
author_rank: Super Contributor
timestamp: July 18, 2012, 01:06:40 pm
content: I guess you didn't do the maths.there's very little energy density to be had out of a supercap and to have a 50/50 we would need to go back to the 80's ( in terms of size ) to even get decent 900mAH ratingsTrust me, 4farads is hardly anything

author: KTP
author_rank: Frequent Contributor
timestamp: July 18, 2012, 01:13:33 pm
content: For those who don't want to bother doing the math.A 10 Farad supercap will drop by 0.5V with a 100mA current draw in 50 seconds. (I = CdV/dt)So talk really fast!

author: digsys
author_rank: Unknown
timestamp: July 18, 2012, 01:23:32 pm
content: PLUS, with a supercap, one momentary short and the entire energy is dumped instantly !!They can be quite dangerous.

author: KTP
author_rank: Frequent Contributor
timestamp: July 18, 2012, 01:52:19 pm
content: Quote from: digsys on July 18, 2012, 01:23:32 pmWell...try shorting a LiPo battery...wait, let me get out of the room first.

author: Psi
author_rank: Super Contributor
timestamp: July 18, 2012, 02:06:56 pm
content: It's because size is a critical factor with cellphones, there just isn't any spare.If you put a supercap in you'd have to remove something else and the supercap would still be so small it wouldn't do much.You'd need a supercap 20 times the size of the li-ion battery.

author: opablo
author_rank: Regular Contributor
timestamp: July 18, 2012, 02:30:59 pm
content: It's not that my math is wrong; it's that I don''t know the math requiredthat's why I posted in the beginners forum.another guy said the same thing in other words: "You'd need a supercap 20 times the size of the li-ion battery."I didn't know how far off they were density-wise. It's clearly more than I though.Maybe someone actually knows the math involved to actually know the real density ratio?Is it converting the V, F to Jules, divide by volume for both the cap and the battery and getting the ratio of that ?I will actually try to learn and do the math to see what I end up with. For fun (?)Quote from: DaveXRT on July 18, 2012, 01:06:40 pm

author: KTP
author_rank: Frequent Contributor
timestamp: July 18, 2012, 02:59:40 pm
content: I gave you one simple formula in the 2nd reply. Current in a capacitor (in amps) is equal to the capacitance value (in farads) multiplied by the change in voltage (in volts) with respect to time (in seconds).Or I = C * dV/dtI like this formula because it is easy to use with simple values like current draw to immediately see how much the capacitor voltage will drop over a certain time period.But for total energy stored in a capacitor charged to a certain voltage, down to the last drop:W = 0.5 * C * V^2 in Joules (not Jules..though I did go out on a date with her and she had a lot of energy)For our 10 Farad capacitor charged to 3V, it would have a maximum energy stored of 45JA typical AA battery, capable of supplying 0.5 amps for four hours at 1.5V has a maximum energy of about 3 watt hours, or 10,800 watt seconds, or 10,800JSo there you go...10F cap gives you 45J, a AA battery gives you 10,800J

author: Short Circuit
author_rank: Frequent Contributor
timestamp: July 18, 2012, 03:08:58 pm
content: Even if charge per volume were acceptable; phones must use microUSB for charging these days, no way to get more than ~2amps through that connection...

author: Flávio V
author_rank: Regular Contributor
timestamp: July 19, 2012, 09:38:44 am
content: The main reasons have already been mentioned, but i will add some more..To reach 3.4V the super cap must be duple cell or coin type...you may think than it is fine but it isn't...A small super caps have VERY LOW ripple current(100mA for a 100mF philips one, 800mA maxwell 1F 2.7V one)The other problem of super caps is their energy volume,but if you don't mind than your "iphone","Lumix" whatever...has come ioxus 5kF 2.5V super caps then there is no problem...A type of super cap than would work better would be the hybrid super caps, ioxus seems than is the only than makes then, but they have some drawbacks, like lower operating voltage of 2.3V and limited lifecyles(+-20k) the positive part is than has near 2-3x more capacity than regular super caps.And to use almost all the energy from the super cap it would be needed to have a step-up and a step-down switched power supply built in the phone.Also, and to finish, super caps don't discharge instantly, they can take more than 30 seconds to discharge because they store a loot of energy and have some resistance...I have thought to use super caps in my fluke 97 to not use batteries but even if i use 150F caps(a bit oversized for the battery mount) it would only have charge(ideal) for 1 minute...at 5W power max consumption..