Super caps formula needed.

[Dark Prognosis]

I am looking for the formula for replacing a battery of a known size with a super cap.

Battery currently in the circuit is 1.2vdc @ 40mah what would the cap need to be to replace that or any battery specs I plug in?

[IanB]

Well, a super cap is not a battery. One does not replace the other.

However, you have not told us about your battery. What chemistry is it? What physical size is it? What capacity does it have (in mAh)? What does it look like? Is it a primary (disposable) battery, or a secondary (rechargeable) battery? What is its purpose and application in the circuit?

All batteries are DC by definition. No common battery has a nominal voltage of 1.4 V (except zinc air hearing aid batteries). NiCd and NiMH are labeled as 1.2 V, while alkaline are labeled as 1.5 V. But no battery actually has the voltage on its label in real life so this voltage is not really useful for design purposes.

[Dark Prognosis]

It is a 1.2vdc 40mah and all I want is a capacitor that will deliver 1.2vdc at 40mah.  Charging of the cap is already done for sake of simplicity (basically the battery is a rechargeable type (chemical makeup is a Ni-Mh) and I wish to replace it with a capacitor).

Corrected my typo above.

[Rufus]

It is a 1.2vdc 40mah and all I want is a capacitor that will deliver 1.2vdc at 40mah.

Q = CV = IT

You need to decide on V which is how much the voltage on the capacitor is allowed to change.

Then lol at the price when you have worked it out.

[IanB]

OK, it's a 40 mAh NiMH cell. The nominal voltage is 1.2 V (forget the DC bit, it's irrelevant).

A NiMH cell is about 1.4 V when fully charged, and let's say we can allow it to drop down to 1.0 V under load when discharged.

As Rufus said, the charge on a capacitor is given by

Q = CV.

In this case, we need the change in charge between 1.4 V and 1.0 V, so

dQ = C dV

Firstly, 40 mAh in coulombs is

 40 mAh x 3600 s/h  = 144 000 mAs = 144 As.

We plug that into our formula:

144 = C (1.4 - 1.0)

=> C = 144 / (1.4 - 1.0) = 360 F

So you need a supercapacitor with a capacity of 360 farads rated to about 1.5 V.

[Dark Prognosis]

Thanks to you both.

This battery is from a solar cell rock from the local dollar store (Dollar Tree) and this battery is 5mm thick and 1cm round that has a tab on the + and - that is soldered in (grrrrrr).  The solar cell pumps out 2.45vdc and 24.5ma in direct sun light so the battery will be fully charged in about 98 mins of direct sun.

The circuit is rather ingenious and simple at the same time as it powers a single white LED (fully charged the LED will go for about 10 hours) which means it is a joule thief type circuit as it has a coil and one 4 legged type ULTRA thin (about 3 notebook paper in thickness) transistor type deal (so bloody small I can't read the number on it) but unlike the normal solar powered stuff this circuit uses no CdS cell to tell it when the light source is removed it just knows when the power coming in has decreased below a specified point and simply turns on the LED.

Sad about all of this is that the circuit died (One Hung Lo at their greatest) the moment I stuck it in the sun for a few hours.  The other 5 I have work fine but the wife would kill me if I touch any of those to tear them down.

[IanB]

Does Dollar Tree still sell them, or are they from a while ago? For a dollar apiece I could pick some up to play with.

The simplest modification would be to put a bigger battery in there (if it would fit) such as an AAA cell.

The trouble with feeding 25 mA into a 40 mAh cell is that overcharging the cell will kill it in short order. I wonder if the circuit has any kind of charge limiting arrangement?

[Dark Prognosis]

Dollar Tree still sells them as it was last month's hot item of the week (about 3 weeks ago) and they have a lot in one of mine but not the other but the other never carries much of anything so sells out fast.

Let me get the wife to loan me her phone and I will snap a pic or two of the circuit board and post back.

[IanB]

I'm going for an evening stroll to have a look...

[Dark Prognosis]

Just got the phone so best I have figured out how to do with it and the back of the board still has the hot glue on it.

[IanB]

Here's what I got for my dollar.

The battery fits into a battery holder with the spring terminals and the writing on the IC is YX8018. The solar cell is 1" x 1.25" (2.5 cm x 3.5 cm). I'll have to see what kind of output it produces when the sun comes out.

[Psi]

A slightly easier way to calculate it is to use the formula   "current(A) / capacity(F) = voltage drop per second"

So for the 360F cap previous calculated and 40mA constant discharge the voltage drop per second is

0.04 / 360 = 0.0001111V

Put that into volts per hour   60 * 60 * 0.0001111V  = 0.40V

So if the initial capacitor voltage is 1.4V, then after 40mA load for 1 hour the voltage will be 1.0V.

[Dark Prognosis]

Here's what I got for my dollar.

The battery fits into a battery holder with the spring terminals and the writing on the IC is YX8018. The solar cell is 1" x 1.25" (2.5 cm x 3.5 cm). I'll have to see what kind of output it produces when the sun comes out.

I have 3 like that from them only they are all in black.  Nice that you don't need a CdS cell in them for them to turn off/on.  At least yours looks bigger all around than what I have from the rock (the pictures above).  Similar tech wise I bet but beefier due to that LED and battery being larger.

[IanB]

Here's a datasheet for the IC.

[Dark Prognosis]

A slightly easier way to calculate it is to use the formula   "current(A) / capacity(F) = voltage drop per second"

So for the 360F cap previous calculated and 40mA constant discharge the voltage drop per second is

0.04 / 360 = 0.0001111V

Put that into volts per hour   60 * 60 * 0.0001111V  = 0.40V

So if the initial capacitor voltage is 1.4V, then after 40mA load for 1 hour the voltage will be 1.0V.

Though the thing lasts about 10 hours, or more, on a full charge.

[IanB]

The inductor is marked green-blue-brown-silver on mine: 560 uH 10%?

[IanB]

Though the thing lasts about 10 hours, or more, on a full charge.

It may only draw 4-5 mA from the battery and run the LED at low brightness.

[Psi]

Yeah, i don't think he's said what current the unit actually draws.
If it lasts 10 hours it's probably drawing 4mA from the 40mA battery .

[Dark Prognosis]

Here's a datasheet for the IC.

ewwww, I can't read that language.  My coil is red gray red silver (2800 uH 10%).

[Dark Prognosis]

Yeah, i don't think he's said what current the unit actually draws.
If it lasts 10 hours it must be drawing 4mA from the 40mA battery

Yeah, that is the one part of the equation I don't 100% know because I have no working circuit to gleam that info from lest the wife murder me for taking a working rock, lol.

This led is pretty darn bright I have to say and when I point the camera at it I see no flicker so it appears to be a continuous on circuit and bright.  The led is a normal sized led too so I figure 3-5ma but I could not find a 3-5mm @3-5ma and white when I went looking last night online.

[Psi]

It's a little difficult to guess the current as we don't know what the min battery voltage is.
If it has a switch-mode boost converter to run a 4V led from 1V it may discharge the battery down to 0.8V rather than 1.0V.
And we're assuming the current draw is constant.

[Dark Prognosis]

It's a little difficult to guess the current as we don't know what the min battery voltage is.
If it has a switch-mode boost converter to run a 4V led from 1V it may discharge the battery down to 0.8V rather than 1.0V.
And we're assuming the current draw is constant.

Good assumption because everything I have tried shows it to be constant.  That circuit board in my picture is all there is to it.  2 wires from the PV in to it and 2 wires out to the LED.

[Dark Prognosis]

Finally read the 4 pin device on mine and it is JD1803 and below that number it says 27b.

Found this just now -> http://www.instructables.com/answers/Where-can-I-but-a-JD1803-chip/

Seems that 4 pin chip of ours is indeed a joule thief and then some and the inductor size is how much current is getting driven to the LED.

[SeanB]

Bought the same type OHL lights, and broke one in assembling the kit, it just fell to pieces while putting the cover on, so it became the teardown model. It has a single AA NiCd cell in it, and little else. I am using the board with the LED as a torch to illuminate the items I look at with a hand magnifier, replacing the original incandescent lamp and 2 cells, just using 1 of the battery holders and the switch, along with an added decoupling capacitor. I did not install the solar cell, just left it not connected, as I was worried of placing a magnifying lens deliberately in the sun. I will have to replace the single AA primary cell every year or so when it runs down.

[IanB]

So, I did some measurements.

The solar cell can produce 3.4 V when open circuit in sunlight, and can produce a short circuit current of a little over 20 mA. It can feed 20 mA into a single NiCd cell and 18 mA into two NiCd cells in series. That would be a maximum power output of about 50 mW.

So for a single dollar I got that solar cell, a white LED, and a boost converter. I have not measured the performance of the boost converter yet, but I will do that later.

Unfortunately, a quick bit of comparison shopping shows that this is not an economical way to buy solar cells. By the time you have bought enough of these to equal the output of a larger cell it would be far more expensive. Even so I guess they make a fun toy.