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what am I doing wrong?

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Old 05-09-2006, 06:32 PM
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paul_c
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Default what am I doing wrong?

Guys, I have a 1700 SCR battery pack. It is brand new (replaced one that I didn't properly cycle, so I'm making sure not to make that mistake again). What I did was take my Sirius Super Test to discharge the pack (8 cell, 9.6 volts). When I first discharged it, I got something around 600 mah. No problem (I discharged with the Super Test for an 8-cell pack, at 250 mah). Then, I slow charged it with a wall wart putting out 100 mah (although it said 4.8volts---it is the wall wart that came with my Futaba TX---the kind with an output for both the TX and RX batts). I then discharged the battery, and the mah was higher. So far so good. So I go to cycle it a couple more times, and now the mah's are coming down! Remember, I'm using the wall wart for about 16 to 18 hours to charge. What is happening?
Old 05-10-2006, 01:50 AM
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Default RE: what am I doing wrong?

I think the batt is not getting full. At 100 ma and 12 volt it should take at least 24-30 hrs to fill it. If it's 4.8 volts it will be even longer.
Old 05-10-2006, 04:51 AM
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paul_c
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Default RE: what am I doing wrong?

ORIGINAL: U-dUd

I think the batt is not getting full. At 100 ma and 12 volt it should take at least 24-30 hrs to fill it. If it's 4.8 volts it will be even longer.
U-dUd,

After I posted, I wondered the same thing as you. But I wasn't sure if the fact that the wall wart was putting out 100 mah at 4.8V affected the rate of delivery. As long as I did not destroy a new battery. I'm going to let it charge, like you said, 24-30 hours. Thanks.
Old 05-10-2006, 06:21 AM
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Nogyro
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Default RE: what am I doing wrong?

I'm a bit confused here. Is the 1700 SCR battery pack 9.6 volt? If so, why are you using the 4.8 volt side of the wall-wart?

If it is 9.6 volts, and you're using the 4.8 volt side of the wall-wart, you're only getting 60 mA or so charge current going into the battery.

I've taken wall warts and cut the transmitter jack off and put on a receiver end. The charger normally puts out 50 mA, and when I use it to charge a 5 cell pack, I get 75 mA charge rate.
Old 05-10-2006, 08:47 AM
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Default RE: what am I doing wrong?


ORIGINAL: Nogyro

I'm a bit confused here. Is the 1700 SCR battery pack 9.6 volt? If so, why are you using the 4.8 volt side of the wall-wart?

If it is 9.6 volts, and you're using the 4.8 volt side of the wall-wart, you're only getting 60 mA or so charge current going into the battery.

I've taken wall warts and cut the transmitter jack off and put on a receiver end. The charger normally puts out 50 mA, and when I use it to charge a 5 cell pack, I get 75 mA charge rate.
I used the 4.8V wall wart because that's what I had in my possesion! I have fast chargers (Sirius, etc), but that was my only wall wart. If I sort through my other wall warts (everyone has a pile of them) and I find one that is putting out 9.6V, can I use this? I really don't want to cut off the TX end, as I do use that.
Old 05-10-2006, 12:31 PM
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Default RE: what am I doing wrong?


ORIGINAL: paul_c

I used the 4.8V wall wart because that's what I had in my possesion! I have fast chargers (Sirius, etc), but that was my only wall wart. If I sort through my other wall warts (everyone has a pile of them) and I find one that is putting out 9.6V, can I use this? I really don't want to cut off the TX end, as I do use that.
Yes you can use the transmitter charge lead coming out of the wall wart. You can make up an adapter if you don't want to cut the TX end off and put a receiver end on it.

If you find a generic wall-wart of some kind, just be sure to check the voltage and mA output. If you have an ohm meter you should check the actual charge rate while you're charging the battery. If all else fails, I've seen some cheap wall-warts that put out 150 mA you could get if you don't want to spring for a new charger. That's what I like about my Hobbico Accu-Cyler Elite. I can set it to charge slow or fast.
Old 05-10-2006, 03:12 PM
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Default RE: what am I doing wrong?

I would use the 12 volt side of it if available and figure the ma as rated current., do the math from there. (the tx side)

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