Electronics Forum / Basics / July 2007

---
You did nothing wrong but believe that whoever wrote that post knew
what they were talking about.  They did not.
---

---
The problem with that method is that the one ohm resistor looks like
such a small impedance compared to that of the clock that the
voltage dropped across it will be largely insignificant and the
clock will continue to run for nearly as long as if the resistor
wasn't in there.

To measure the capacity of the battery you need to know three
things:

1. The specified capacity of the battery.

2. The specified discharge rate.

3. The specified cutoff voltage.

You already know the capacity, 2000mAH, and you can get the
discharge rate and cutoff voltage ratings from the data sheet for
the battery.

What the deal is with the rate of discharge is that if you have a
battery rated for 2000mAH and the rate is 0.1C, then to get the
2000mAH out of it, then you're allowed to take a maximum of 200mA
from the battery for 10 hours before its voltage falls to cutoff.

If you take more than that, then the capacity will fall and the
product of current and time to cutoff will be less than 2000.
Conversely, if you take less the capacity will rise slightly.

So, to do the test you'll need to start with a fully charged battery
and set up some test equipment like this:  (View in Courier)

    +------+----------+
    |+     |          |+
  [DUT]  [LOAD]  [VOLTMETER
    |      |          |
    +------+----------+  

In order to determine the resistance of the load, first determine
the current that can be taken from the battery by multiplying the
capacity times the rate.  In your case, if you have a 2000mAH
battery with a rate of 0.1C, then the current will be 0.2A.

Next, use Ohm's law to determine the resistance:

         E     1.2V
    R = --- = ------ = 6 ohms,
         I     0.2A

where 1.2V is the nominal voltage of your cell (battery).  I don't
know if it is, though, consult the data sheet and plug in the
appropriate value.

The power it will need to dissipate will be:

    P = I²R = 0.04 * 6 = 0.24 watts,

so you can use two standard 12 ohm +/- 5% 1/4 watt resistors in
parallel for the load.

In order to start the test, connect the load and the voltmeter to
the cell and note the time and the voltage.

Thereafter, monitor the voltage and when it falls to the cutoff
voltage note the time.  If it's less than 10 hours after you started
the test, then the capacity isn't what it's supposed to be.

Keep track of voltage and time and you can plot what the discharge
curve looks like.

Hmm.....  I think this is where things actually went wrong.  This will never
work.

What you really need to do is to connect the 1R resistor in "parallel" with
the battery, not in series.  This will put a hefty load on the battery
(about 1.2A when it is full).  The resistor (you said 10W) will still get
warm, so be careful.  The problem with discharging like this is that the
load decreases as the battery voltage falls.  This makes it a calculus
problem to determine the charge that was in the battery.  If you used a
"constant current" load, calculating battery capacity would be as simple as
timing how long it takes.

With that out of the way, I suspect that you can get an approximate capacity
by using the clock as described.  Assuming that it stops when the battery
voltage drops to about .8V AND pretending that the voltage curve is linear.
You are averaging a 1A load over time, so if the clock stops at 1:30, you
could sorta say that the battery had 1500mAh charge.....sorta.  ;-)

"Jim Elbrecht"

**   Not wired in  "series"   in PARALLEL   !!

**  Neat idea.

**  So four separate clocks ?

However, for testing non rechargeable cells like AA alkalines  -  use a 10
to 15 ohm resistor.

Shame they will be buggered at the end of the test  !

......  Phil