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'Off topic: a question about Ni-cads'
1996\12\28@174915 by Xaq

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This is not exactly a pic question, but it is about the battery pack I am
using to power a pic.

I recently bought four new C size ni-cad batteries (1.2V 2200Ah).  I wired
them up in series (making 4.8V). I read on the side of the battery "Charge
for 15 hours at 220mA."  Now here is my question- what is an easy way to
keep the current at 220mA.  Sure, you could put a 21ohm resistor in series
with the pack, that is if the battery pack had no internal resistance.  But
the amount of current drawn by the pack changes depending on how charged it
is.  Am I worrying to much about details?

Well, here is why I ask.  I charged this pack up over 15 hours, and then
connected my pic.  The pic ran for about 5 minutes and then stopped.
Voltage on the batteries was 4.1V and falling.   I measured the current
drawn by the pic circuit.  It was only pulling about 60mA.  Why didn't the
batteries last MUCH longer?!?

I am either not charging the pack correctly or I have a bad cell.  If I
have a bad cell, is there any easy way to test this? They were supposed to
be new batteries, they were expensive ($7 bucks each) .

Thanks for any advise,

Zach

1996\12\29@175926 by Reginald Neale

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Zach said:
{Quote hidden}

Unless all four cells are at the same state of charge initially, charging
them in series can lead to problems.

Measure the voltage of each cell under load. If there are dramatic
differences, (20%) you should charge them individually before using them in
series. Simplest way is to break the series, use a 12VDC supply and
independent dropping resistors of appropriate wattage for each cell.

Reg Neale

1996\12\29@181041 by Bob Blick

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>keep the current at 220mA.  Sure, you could put a 21ohm resistor in series

Remember that Ohm's Law deals with the voltage dropped through the resistor,
not with either the supply voltage or the voltage at the load. For get 220mA
through 21 ohms, you need 4.62 volts more than your battery pack, or about
10 volts. It sounds to me like you were not supplying it with enough
voltage. Also consider that the charging voltage will average about 1.35
volts per cell. Yet another thing to consider is that a brand spanking new
nicad battery should be charged to double its capacity, so give it 20 hours
at 220 mA.

Cheers, Bob

1996\12\29@190436 by Lee Jones

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> I recently bought four new C size ni-cad batteries (1.2V 2200Ah).

I assume you mean 2200 mAH.  If you can buy 2200 AH capacity in
a C size battery package, then someone's made a breakthrough! :-)


> I wired them up in series (making 4.8V). I read on the side of
> the battery "Charge for 15 hours at 220mA."  Now here is my
> question- what is an easy way to keep the current at 220mA.
> Sure, you could put a 21ohm resistor in series...

A constant current source driven by a 10V or higher supply.

Or build a battery charger based on an active control element
such as the BQ200x series or a PIC.

Or just spend $50-100+ on a battery charger in the R/C car
section of your local (or mail order) hobby shop.  Most of
the "real" R/C cars (i.e. not the ones from Toys'R'Us) use
6 of these cells connected in series as a 7.2V pack.

You can also get basic battery charging & maintenance info
from books on R/C cars.  "Radio Control Car Action" magazine
also has some information and lots of advertisements (but is
aimed at a non-technical 16-18 year old crowd).


As an aside, the 220mA charge current is the C/10 (capacity
divided by 10) rate.  Nearly any ni-cad can be charged using
this method for 14-16 hours.

Most modern cells can also handle much higher rates up to 2C
or 3C.  Using a charge current of 4-5 Amps, you must monitor
the pack.  The Hitec CG-325 charger I have has programmable
voltages & currents.  It's a peak detection charger.  It stops
charging when the battery pack voltage peaks & then sags very
slightly.  Run from a 12V battery, it can fully charge a dead
7.2V 1800mAH pack in 20-30 minutes.

There are lots of other chargers currently available with
this kind of performance.
                                               Lee Jones

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1996\12\29@192554 by David Schmidt

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Find yourself just about any data book that has information on the
LM317 adjustable voltage regulator.  In there you should find a diagram
and equations for a constant current regulator circuit.
(attempted ascii art below)

 You can do this with two parts, the LM317 and a single
resistor.  Should total less than $2.00

I think the equation for finding the resistor value is  1.2V/A = R
where A is your desired charging current (220mA = 0.22A, so R = 5.45ohms)

                  in  -------------  out    R
raw dc in + ----------|             | -----\/\/\/----> (+) to batt
filtered              |  LM317      |              |
                     |-------------|              |
                        adj |                     |
                            ----------------------|

       (-) ------------------------------------------> (-) to batt
Dave

1996\12\30@013618 by carl watley

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Hi Zach,

>I recently bought four new C size ni-cad batteries (1.2V 2200Ah).  I
>wired
>them up in series (making 4.8V). I read on the side of the battery
>"Charge
>for 15 hours at 220mA."  Now here is my question- what is an easy way
>to
>keep the current at 220mA.  Sure, you could put a 21ohm resistor in
>series

I like to use a light bulb. When the bulb dosn't glow the battery is
charged.
A five volt bulb might work nicely.

>with the pack, that is if the battery pack had no internal resistance.
> But
>the amount of current drawn by the pack changes depending on how
>charged it
>is.  Am I worrying to much about details?


Yes, You are only in danger of ruining the battery with too much current.


>Well, here is why I ask.  I charged this pack up over 15 hours, and
>then
>connected my pic.  The pic ran for about 5 minutes and then stopped.
>Voltage on the batteries was 4.1V and falling.   I measured the
>current
>drawn by the pic circuit.  It was only pulling about 60mA.  Why didn't
>the
>batteries last MUCH longer?!?
>

Good question, sounds like ether the pack was not suffintly charged or
you do have a
problem.

Carl

1996\12\30@072513 by Cheng Tan

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Try to discharge the batteries to about 0.9V and then charge it up again.
Your resistor will also drain out the current of the batteries as the
internal resistance will increase with charge decrease.

You may have memory effect in your batteries or they are unmatched (i.e.
different internal resistance). That will not allow you to get maximum
power output.


'Off topic: a question about Ni-cads'
1997\01\01@131641 by bill
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> I recently bought four new C size ni-cad batteries (1.2V 2200Ah).  I wired
> them up in series (making 4.8V). I read on the side of the battery "Charge
> for 15 hours at 220mA."  Now here is my question- what is an easy way to
> keep the current at 220mA.  Sure, you could put a 21ohm resistor in series
> with the pack, that is if the battery pack had no internal resistance.  But
> the amount of current drawn by the pack changes depending on how charged it
> is.  Am I worrying to much about details?

Zach,

If your charging current is changing significantly during the
charging process, then you must be using a voltage that is very
close to that of the pack with a small resistance.  You could build a
current regulator circuit to produce a constant charging current,
but there is a much simpler solution.  Start with a charging voltage
that is significantly higher than the battery pack.  Subtract the
nominal voltage of a charged cell and calculate the resistance that
is required to get 220ma at that voltage.  For example, if you are
charging the cells individually (which is generally a better
arrangement than charging them in series) and using a 12V supply,
then use a resistor that is somewhere around 49 ohms.  This will
yield your 220ma accross 10.75V (12V minus the approx 1.25V of a
cell)  What?  You don't have any 49 ohm resistors lying around?  47
or 51 will do fine. Battery charging doesn't require great
precision. If you charge the whole pack in series using that same 12V
supply, then you will need about 32 ohms to get 220ma accross 7V (12V
minus the approx. 5V of the pack).  You can use Ohm's Law to
calculate the appropriate resistance for any combination of supply
voltage and cells.  Also make sure your resistors have a high enough
wattage for the power that they will have to dissipate.  The 49 ohm
resistor in the 12V 1cell arrangement will dissipate nearly 2.4
watts.  You might also want to be sure the resistor and power supply
can survive a short circuit so that your charger doesn't die or
catch fire if you have a shorted cell or accidentally short it out.

---

                                      Peace,
                                      William Kitchen

                                      .....billKILLspamspam@spam@iglobal.net

The future is ours to create.

1997\01\03@094752 by Lauw Lim Un Tung

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hi,
try this :                    ``````````
          10 V DC ---------- I  7805  I -----l
                             ``````````      l
                                 l           Z
                                 l           Z 22 Ohm
                                 l           Z
                                 l-----------T
                                             l
                                             l
                                           `````
                                            ```    4x NiCd
                                           ````` 1.2 V C size
                                            ```
                                             l
                                             l
                                            gnd
_______________________________

       Lauw Lim Un Tung
       Pecindilan 5/14
       Surabaya 60273
         INDONESIA
phone  : 62-31-318895
e-mail : tungspamKILLspampeter.petra.ac.id
_______________________________

On Sun, 29 Dec 1996, Bob Blick wrote:

{Quote hidden}

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