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'[EE]: NiCd Battery Charging (New Topic!)'
2001\05\24@012531 by Josh Koffman

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Greetings. Just as a heads up, this is a new question, and has nothing
to do with NiMH battery packs, robots, or PIC controlled chargers. That
said, here is my problem :)

I recently completed a prototype of a hand held tester (yes it has a PIC
in it). For my power source I hacked up a 9 volt NiCd battery and used 5
of the 6 cells inside. I wanted to use all 6, but space limitations made
me eliminate one. Anyway, my original plan was to use a commercially
available battery charger, and just build a custom cable to connect it
to my device. Now I'm faced with a problem...I no longer have a complete
9V battery. I figure that each cell in my 5 cell pack will see about an
extra  0.3V when charging. They are in series, so current should remain
the same correct? Will this extra voltage cause problems? I suppose I
could rig the remaining cell into the charging cable somehow, then I'd
have a complete pack again. My worry is that the exterior cell will
never have any load on it, so it will always remain charged (except for
internal discharge). Therefore there would be 5 cells with similar
charge states, and one with a higher charge state. I don't know how this
would effect the charging cycle. Any ideas?

Josh
spam_OUTjoshyTakeThisOuTspammb.sympatico.ca

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2001\05\24@024402 by Spehro Pefhany

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At 12:21 AM 5/25/01 -0500, you wrote:
>Greetings. Just as a heads up, this is a new question, and has nothing

>charge states, and one with a higher charge state. I don't know how this
>would effect the charging cycle. Any ideas?

As a first approximation, you can put a couple of 1N400x diodes in
series with the charger. They will drop around 0.6~0.7V each, or about
the same in total as one cell.

Best regards,

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2001\05\24@052437 by Roman Black

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Maybe you could put a couple of diodes in the
wire to the commercial battery charger. That will
drop 1.2v to 1.4v, about perfect to compensate
the missing cell. ??
-Roman




Josh Koffman wrote:
{Quote hidden}

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2001\05\24@063601 by Vasile Surducan

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Clean design, Roman will love it because is very close on his one resistor
charger. ( BTW I charge battery the same, but only mine, not others... )
However I want to point some things and explain why after a while your
logger show a discharge.
Could be a malfunction or a desired option, regarding the maximum output
swing of your 7805 current source generator:
7805 need about 4V across if is standard, and about 1.25 to 2.5V if is
low drop, 0.6V on D1, 3.6V to 4.5V on the cell and 0.3V to 1V on T1.

4+0.6+4.2+0.6 = 9.4V  for medium above values

The max voltage on sensing resistor R1 is 12V-9V= approx 3V.
These 3V must ensure current polarisation for the reference pin of the
7805. So, a current limitation will occure if battery is charged, which
sould be a good thing.
Do another aquisition using a 15V for supply and see what's happent.
However for 9V battery charging you need to increase input acordingly if
the cells are series.
Cooool micromouse !

Cheers, Vasile





On Thu, 24 May 2001, Roman Black wrote:

{Quote hidden}

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2001\05\24@065520 by Alan B. Pearce

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>Cooool micromouse !

even cooler debugging technique for the mouse!!!

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2001\05\24@135922 by Josh Koffman

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Vasile, I think you replied to the wrong message :) My question had
nothing to do with a 7805, and the Micromouse was on someone else's page
:)

Just so you know :)

Josh
EraseMEjoshyspam_OUTspamTakeThisOuTmb.sympatico.ca

Vasile Surducan wrote:
{Quote hidden}

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2001\05\25@041858 by Peter L. Peres

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Imho place two 1N4007 diodes in series with the charger + lead. This
should fake a battery for the simpler chargers.

Peter

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