How to calculate the performance of a battery charger (LR06)

[hic]

How do you measure to integrate variations?

for once, with patience, it can be done with a stopwatch, a voltmeter and an ammeter and noting the values ​​at time intervals noted on a paper.

Otherwise it can be done with a microprocessor programmed for that which measures and calculates.
In addition there are integrated circuits realizing this entirely automatically and used in the commercial wattmeters.
see Maxim-ic.com
http://www.maxim-ic.com/datasheet/index.mvp/id/6454


microchip.com
http://www.microchip.com/
http://www.embeddedstar.com/weblog/2007 ... p-mcp3909/

and many others :
http://www.futurlec.com/News/Analog/Energy.html
http://www.google.fr/search?num=100&hl= ... =&aql=&oq=


http://www.microchip.com/stellent/idcpl ... odeId=2889
http://www.google.com/powermeter/about/index.html

hi dedeleco

Manually, It is more accurate to note the time of variations
than to do a survey.
Whenever a variation appears, you take it chronologically.

[Christophe]

In my opinion it is the energy contained in the battery divided by consumed EDF energy.

It is !

On the other hand in terms of unity, I have for the first mAh and for the second Joules ...
How do I recover equivalent units ... ??

Thank you

Oh, that was the question? Ben as we say others: just spend all the same energy: Wh. In passing, to be rigorous, mAh are not an energy in the strict sense! Ampere is not a unit of energy.

W * duration in seconds = Joules
W * time in hour = Wh

So we have the equivalence 1 Wh = 3600 Joules

mAh * Voltage = mWh = 1Wh / 1000

Practical example:

a battery of 2200 mAH under 1.2V assuming 1.2 constant (false) will supply an energy of 2200 * 1.2 = 2640 mWh = 2.64 Wh is 2.64 * 3600 = 9504 joules.

If you had to recharge this battery with a charger which consumes 2.3 W during 6h is 2.3 * 6 = 13.8 Wh = 49680 joules then the yield would be of 2.64 / 13.8 = 9504 / 49680 = 19.13%

For the precision that you want it is necessary to necessarily integrate the drop of current of the discharge of the battery (thus to have the curve of discharge and to make the integral one), in 1er approximation you can make an average: (initial tension - tension final) / 2 assuming that the fall is linear (it is not, but better than no correction)

The yield of my example would fall to 2.2 / 13.8 = 15.94% if we assumed 1.2 at the beginning and 0.8V at the end of the discharge according to this method.

In the strict sense it would be necessary to do the same method listed EdF ...

How is it that you have joules EdF side, normally we are in Wh or kWh no? What do you use as material?

[Gaston]

In my opinion it is the energy contained in the battery divided by consumed EDF energy.

It is !

As Citro said, it depends on the definition of "contained".

If it is the energy recovered during the discharge of the battery, it is the efficiency of the set {battery + charger} which is measured.
This yield therefore also depends on the battery used.

If it is the energy injected into the battery during charging (less simple to measure), it is the performance of the charger alone.