30 second recharging Rechargeable Battery

[jaxbird]

It's entirely impossible to have a horizontal voltage trend like that...

No it's using internal regulator, so it would be flat. However it appears to be linear (!) from the pictures posted.

[miguelvp]

It's entirely impossible to have a horizontal voltage trend like that...

Not if the cap is at a higher voltage (3.8V) and rectified to the 1.5V output like he claims he does. He mentioned in the video @1:38 that when the voltage drops to 2.2 volts it will damage the cap, so the logic inside shuts it off when the cap is at 2.3V. I'm not saying it's all in the open and will work.

Also, $10K is hardly a scammer's target, I think he wholeheartedly believes what he is doing but made a couple of mistakes. I think the whole Graphene is BS and a cover up for his mistake.

If you look here in page 5:
http://www.mouser.com/pdfdocs/LICpresentationforstorage20131110.PDF

His specs point to: LIC2540R 3R8277, but that has 25mm in diameter.

A AA battery has the following dimensions: 49.2–50.5 mm length and 13.5–14.5 mm in diameter.

So he can only use LIC1235R 3R8406 for his AA battery but that only has 40F at 17.77mAh capacitance. when he keeps on quoting the 270F (at 120mAh and he doesn't mention that part)

Page 7 has the charge time stated at 30 seconds, also shows capacitance drop of 20% in 10 years. Also it doesn't like hot spaces and it's most happy at 60 degrees Celsius or lower temperature.

But there are many problems with the kickstarter even if it was just that one mistake:

He is offering AAA batteries too (44.5 mm in length and 10.5 mm) in diameter and there is no supercap of that size by that manufacturer, and of course it will hold less than the 40 Farads

He also offers C sized batteries  (50 millimeters length and 26.2 millimeters diameter) Now he can use the 270F supercap.

And also offers D sized batteries (61.5 millimeters length and  33.2 millimeters diameter).

But AA and AAA wont meet his claims so he went for the Graphene escape.

The chart, doesn't look fake since the cap might jump back to 2.3V after being shutoff, but he is connecting the data points in a ramp.

So he blatantly lied about the 270F cap when he shows in the picture the 40F one that will fit in the casing.
His toy demo well, he is just proving that it doesn't discharge in 30 seconds but that doesn't mean it has the same density as a AA battery, his C battery might have the density of a AA battery maybe but can it deliver 6000 mAh that rechargable C batteries can hold? Nope it's only 120 mAh and if you look in his chart, that's where it cuts off.

So his battery only holds 1/50th of a fully charged NiMH rechargeable C size battery with a 30 second charge. So if it takes 25 minutes to fully charge the NiMH C size battery there is no gain really, the charge rate is about the same. Sure you can use it quicker but then you have to make sure the device is under 60 degrees celsius if you want the cap to live it's 10 years at which point it will only hold barely 100 mAh of energy.

Scam? I don't think so, I think is just plain being naive. But there might be a market for a quick device that only needs to operate for a short amount of time.

Phone batteries hold what 2600 mAh and that will last say 12 hours being optimistic? sure you can charge your phone in 30 seconds but it will only give you about an half an hour of battery life (about 1/22th of the real battery)

Hmm took too long to post so jaxbird beat me to the regulator thing

[miguelvp]

Also to match the energy needed by a cellphone that uses a small 1200 mAh battery he will need 10 of the big super caps in parallel with diode balancing to be able to charge a phone in 30 seconds, that's a pretty big battery pack. And that will hold the same energy as the 1.5V AA battery he claims he has.

[Rasz]

So I skipped a coffee and signed up.

skip another coffee and join my 'cat videos and brooklyn bridge' project

[GeorgeOfTheJungle]

Nope it's only 120 mAh and if you look in his chart, that's where it cuts off.

But the chart says 1.2 Ah not mAh, no?

[miguelvp]

He also doesn't explain the constant load value used on that chart.

[mikerj]

Wonder what the quiescent current of the regulator is?  Given the very low capacity it won't take an awful lot to have the 'battery' drain itself quite quickly.

[bktemp]

Using CMOS/mosfet regulators the quiescent current can be in the nA range. There are many ultra low quiescent current LDOs up to 6V input voltage.

[tom66]

I've seen many switchers with Iq<10nA but that's not for an output that is switching. If the output is switching Iq can be in the 100uA's level, which will drain a supercap quickly, but a li-ion 18650 will last some time.

[jaxbird]

It has now achieved funding.

So we will get to follow the progress and see what happens.

My guess is 'capacitor company took all the money but didn't deliver' will be the final update.

[Richard Crowley]

My guess is 'capacitor company took all the money but didn't deliver' will be the final update.

But at least supporters will have the satisfaction that they kept his kids toys in batteries.

[williefleete]

My guess is 'capacitor company took all the money but didn't deliver' will be the final update.

along with the company laughing behind his back and throwing the "homebrew" mix in the trash

[EEVblog]

Like many other projects I think he vastly underestimated the cost and the quality/support people expect.

He vastly overestimated his ability to understand what mAh means 

[Jay_Diddy_B]

Hi,
Check the recent comments on Kickstarter:

https://www.kickstarter.com/projects/shawnpwest/30-second-charging-rechargeable-battery/comments

Sky, is challenging Shawn West's claims.

Regards,

Jay_Diddy_B

[jaxbird]

Sky, is challenging Shawn West's claims.

Shawn seems very defensive in all his comments, like "See, you're trying to start a p*ssing contest. I'm not arguing through comments." and I have addressed it on that blog. And I understand that he is a qualifies engineer but so am I. So, in turn this will just turn in to a "you know what'ing" contest. and
I was wonder when you were actually going to begin with the slander. etc.

You'd think someone inventing a breakthrough in super capacitor technology would be more proud to share performance details with the world. I know I would.

Why doesn't he do a live session demonstrating a few full charge and discharge cycles, with a programmable load, while taking questions? I'm sure that would increase contributions significantly.

Kickstarter is going to get hurt by allowing these kind of projects take money from people.

[miguelvp]

Well that Sky character keeps posting every hour and many times on each hour.

Not sure where his 100 Amps come from but he keeps on linking this thread.

Also some other one posted
1150 mAh over 26 seconds will require over 150A.
Wouldn't that be more 8.3 Amps?

So the challengers sound like nuts in my opinion bringing the name of this blog constantly to the comments.

[Jay_Diddy_B]

Hi,
I estimated a charging current of 73A in the original post of the message. I assumed that the capacitor was 1100F.

It is not easy to predict what the charging current will be if the storage device is not a linear capacitor.

However, there is a well known concept the Conservation of Energy (one of the laws of physics).

If we assume lossless, ideal case, very unlikely:

Energy in = Energy out

That is the conversion efficiency electrical to chemical, chemical to electrical is 100%.

Energy out = 1100 mAh x 1.5V = 1100E-3 x 60 x 60 x 1.5 = 5.9 KJ

to replace 5.9 KJ in 30 seconds requires a power of

5900 / 30 = 196 Watts This is the average power over 30 seconds.

If the terminal voltage is constant at 3.3V

The charging current is 196 / 3.3 = 59 Amps.

If the voltage rises from 1.8V to 3.3V during the charging process the current is required is higher.

Assuming a capacitor:

5900J = 1/2 x C x V² _max - 1/2 x C x V² _min

5900J = 1/2 x C (V² _max - V² _min)

C = 5900J / 1/2 x (V² _max - V² _min)

assume:

V_max = 3.3V

V_min = 1.8V

C = 5900 / 1/2 x (3.3² - 1.8 ²)

= 1542 Farads

to charge from V_min to V_max

I= C dv/dt

= C x (V_max - V_min) / time

= 1542 x (3.3 - 1. / 30s

= 77 Amps


quod erat demonstrandum



Regards,

Jay_Diddy_B

[jaxbird]

Well that Sky character keeps posting every hour and many times on each hour.

Not sure where his 100 Amps come from but he keeps on linking this thread.

Also some other one posted
1150 mAh over 26 seconds will require over 150A.
Wouldn't that be more 8.3 Amps?

So the challengers sound like nuts in my opinion bringing the name of this blog constantly to the comments.

Well, a lot of people with no interest in electronics seem to have problems with the difference between mAh and mA.

But assuming 100% efficiency, 1150mAh over 26 seconds requires ~159.23A charge current.

Edit: but of course Jay_Diddy_B above is more correct, as the charge and discharge voltages are not the same. However, still a very potent charger is needed.

[PlainName]

would be more proud to share performance details with the world. I know I would

I wouldn't. Been there, done that, got the ad hom attacks and the threats to buy my product specifically to cause a self-injury and then sue me for my last penny in the courts. I was told that my product didn't work, couldn't work, and would be supremely dangerous if it did anyway. (For the record, it worked very well indeed.) All stated by 'experts' who had never seen it and, from their comments, didn't understand what it did anyway. Of course, some of those experts saw it as a threat to their lucrative trade...

This chap's battery thing has more holes than a very holey thing, but that's based on maths and facts. Without those putting the kibosh on it, I would gloss over his comments that you quote because he does have a point, there. Indeed, this whole topic on EEVBlog is here specifically to denounce crowdfunded projects, and even decent projects get roundly slagged off. Think about how this looks to someone outside this elite.

[miguelvp]

Energy out = 1100 mAh x 1.5V = 1100 x 60 x 60 x 1.5 = 5.9 KJ

Wouldnt that be 1.100 x 60 x 60 x 1.5?

but the result is right 5.9KJ, the problem is that the formula is wrong.

maybe you meant 1.100 Wh x 60 x 60 x 1.5?

Edit Nevermind, W = A * V, got it


But yeah, I think I do understand the problem.

1.1 Ah is 3960 Coulombs

Divided by his claimed 270F will require 14.7 volts to charge that capacitor.

I still think he only has a  270F cap that can hold 120mAh or 432 Coulombs

Attached specs of the 30 second original charging supercap:


Edit: but at that diameter, this will only work on his C size battery.

[miguelvp]

One more thing, since he mentions in the video that the output gets shut off at 2.3V (because under 2.2V will damage the supercap)
He only needs to recharge from a precharged cap.

Say his upper range is 3.5V so that will make 1.2V difference might bring those amps down.

Also there is an error here?

5900J = 1/2 x C (V2 max - V2 min)

C = 5900J / 1/2 x (V2 max - V2 min)

should be C = (2 x 5900J) / (V2 max - V2 min)

so C = 11800/(12.25 - 5.29) = 11800/6.96 = 1695 Farads

1695*1.2/30 = 67.8 Amps, not much reduction.

I still think he doesn't have the 1150mAh storage, he only has 1/10th of that.

[Rasz]

One more thing, since he mentions in the video that the output gets shut off at 2.3V (because under 2.2V will damage the supercap)

no no, that was for Lithium Ion Super Capacitor, his magical quantum dots in the kitchen sink capacitor shouldnt have this property (or any other properties for that matter)

[corrado33]

That sky character is quite... persistent and quite frankly, annoying. I'll agree that this project seems like a bunch of BS, but still, no need to continue asking the same questions over and over again.

For any of you who backed it, what's update #7? It's for backers only.

[hamster_nz]

That sky character is quite... persistent and quite frankly, annoying. I'll agree that this project seems like a bunch of BS, but still, no need to continue asking the same questions over and over again.

For any of you who backed it, what's update #7? It's for backers only.

Nothing much - it is just a pretty much standard "please spread the word" update.

[Jay_Diddy_B]

Hi,
Time to do a little SPICE simulation on a charger. In my previous messages I ignored the effects of ESR in the capacitor. I have included them in the model:




The left side of the model is a constant current / constant voltage charging circuit. As shown the voltage is set to 3.8V and the current to 75A.

The right side of the model is 1500F capacitor with an ESR of 10m Ohm. I estimated the ESR at 10m Ohms. The actually ESR is unknown.

The capacitor is given an initial voltage of 2.2V

Here are the results:



When the charging current is applied the terminal voltage of the capacitor jumps by 75A x 10m Ohm = 0.75V to 2.95V. The constant charging current raises the voltage across the capacitor until the maximum voltage of 3.8V is reached. The terminal voltage is now held constant and the current decreases exponentially with CResr time constant.
After about 50 seconds the capacitor is fully charged.

Some energy is dissipated in the ESR. The simulation that the energy dissipated 1.388 kJ or an average power of 27.8 Watts during the 50 seconds. An average power of 27.8 Watts in something the size of an AA battery for 50 seconds is a lot.



If I increase the ESR to 50m Ohms the initial current is limited to 3.8 - 2.2 /0.05 = 32A. The charging current decays exponentially from the initial value. It takes more than 200 seconds for the capacitor charge.



You can see from the simulation how the capacitor ESR impacts charging.

Regards,

Jay_Diddy_B - BSc (Eng) Hons ACGI (Added my credentials for this post in case my education is challenged)