Charging the batts inside the Cmoy
Jun 27, 2006 at 6:48 PM Thread Starter Post #1 of 14
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Im a begginer DIYer and i wonder how to connect the circuit to the power jack (in Cmoy amp) in order to recharge 2x9V batteries from the wall charger. Could someone post the schematics of such connection?

THX in advance.
 
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Jun 27, 2006 at 7:10 PM Post #3 of 14
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Jun 27, 2006 at 7:26 PM Post #4 of 14
Quote:

Originally Posted by Matez
But if I forget and let the charger charge batts for eg. 20h ,will it damage my batteries?


I've left them on the charger for 3 days straight at charge rate of about .08C with no effects that I have noticed. May have decreased the life of the batteries in the long term but I'd say if it doesn't happen that frequently, then no big deal.
 
Jun 27, 2006 at 8:40 PM Post #5 of 14
3 another newbie questions :

1. Where should i put the power switch? As close as possible from battery '+'?

2. Could you post some example of charger which would be ok for 2x9V battery Cmoy?

3. What type of "recharging circuit" does PA2V2 have?

thx in advance
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Jun 27, 2006 at 9:49 PM Post #6 of 14
1. It doesn't need to be very close. Just use the amount of wire you need to be able to assemble the casing later.

2. Could you post some example of charger which would be ok for 2x9V battery Cmoy? Any AC-DC wallawart with an output of 22volts or more and at least 200-300mA of current.

3. What type of "recharging circuit" does PA2V2 have? The PA2V2 uses a single resistor to set the charge current. it has pros and cons comparing withthe one above. I would go with the one above.


Good luck

Manuel
 
Jun 28, 2006 at 11:34 AM Post #7 of 14
Quote:

Originally Posted by MASantos
3. What type of "recharging circuit" does PA2V2 have? The PA2V2 uses a single resistor to set the charge current. it has pros and cons comparing withthe one above. I would go with the one above.
Good luck

Manuel



Could you tell me more about this pros and cons? Im curious
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Jun 29, 2006 at 1:53 AM Post #10 of 14
Quote:

Originally Posted by Matez
Could you tell me more about this pros and cons? Im curious
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The LM317 charger is actually simpler to implement than a resistor based charger. It will put out a constant current regardless of the supply voltage (within reasonable values of course).

A resistor based charger puts out a declining charge current. I built one for a Pimeta with 2x9.6V (8 cell) parallel Maha Powerex batteries, with a 25.6V supply. With one OR bridge diode drop, it puts about 25V on the resistor. It starts charging at about .14C and outputs about .08C when the battery is fully charged. I think a resistor based charger, carefully configured, may do less damage if it is "over charged" because of the tapering off of the current. The downside is that you have to carefully configure the supply voltage and resistor value to the battery. You might have trouble using, say, a cheap unregulated supply. I use a Tread with the standard LM317 adjustable regulator to get the 25.6V output. I breadboarded the circuit and experimented with a few different resistors to get the charge curve I wanted.
 
Jun 29, 2006 at 6:57 PM Post #11 of 14
And the very last questions
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How do i know when my 250mAh 9V battery is fully charged? Could anyone explain me how to count this final charge rate while using different wall chargers? Eg. on a 14V ,500mAh DC charger...

thx in advance and sry for newbie questions
rolleyes.gif
 
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Jun 29, 2006 at 7:09 PM Post #12 of 14
Quote:

Originally Posted by Matez
And the very last questions
smily_headphones1.gif


How do i know when my 250mAh 9V battery is fully charged? Could anyone explain me how to count this final charge rate while using different wall chargers? Eg. on a 14V ,500mAh DC charger...

thx in advance and sry for newbie questions
rolleyes.gif



If you use a resistor based charger, you can measure the current draw with a meter. It will decline and eventually level off. When it levels off, presumably the battery is fully charged.

If you use a current regulator such as lm317, you could probably observe the same thing, more or less, as a declining voltage drop across the regulator's input/output pins, but I have never actually bothered to do that.

There is no magic light bulb that goes off when the battery is charged. I just leave mine on overnight to get a full charge and I try to configure it for a 10-12 hour charge.

I think there is a general consensus here that you would have to leave your batteries charging for a year or more at 0.1C for it to significantly impact the battery life. Leaving the batteries on a 0.1c charge for even a couple days at a time is not going to destroy them. Unfortunately, the battery datasheets I have seen are very vague on this subject.

In the final analysis, a 100% healthy battery that is discharged does not perform well in your amp. A battery at 85% of original capacity works much better when it is fully charged when needed. Don't obsess, just keep the sucker charged
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Jul 3, 2006 at 9:19 PM Post #13 of 14
If I don't want to run amp while charging battery ,do i need so high Voltage input (15V) ? Will smaller voltage be enough (eg. 9V)? I have 9V ,500mA power supply and I dont want to buy another one just to power 1x9V Cmoy...

once again sorry for noobish questions
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Jul 3, 2006 at 9:27 PM Post #14 of 14
Quote:

Originally Posted by Matez
If I don't want to run amp while charging battery ,do i need so high Voltage input (15V) ? Will smaller voltage be enough (eg. 9V)? I have 9V ,500mA power supply and I dont want to buy another one just to power 1x9V Cmoy...

once again sorry for noobish questions
smily_headphones1.gif



You need to supply about 1.55V per cell to the battery, after any diode drops if you are using an OR bridge or other "flow control" diodes. 9V batts are either 7 or 8 cell. An 8 cell battery needs about 12.4V, a 7 cell about 10.85V.

A 9V supply will not fully charge a 9V battery, regardless of the number of internal cells.
 

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