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Charging circuit 4 AAA NiMh

mach7

Jan 8, 2013
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Hello everyone!

I am trying to charge 4 AAA NiMh rechargeable batteries with this simple circuit. Do you have any suggestions for better results? The resistor is getting really hot during charging. Is that normal? I use a 12V, 2A transformer.

Thank you.
 

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BobK

Jan 5, 2010
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A proper NiMH charger has to detect the end of the charge cycle by either detecting a rapid temparature rise or a voltage drop.

You can use a trickle charger at 0.05C which will take about 24 hours to fully charge the battery, any other simple charger is going to be bad news.

As far as your circuit goes, the resistor is in the wrong place, it should be after the tank capacitor, and a bridge rectifier would be preferred over a single diode.

The resistor must then be chosen to get 0.05C, based on the input voltage and the battery capacity. If you give me those two numbers I can suggest the correct resistor.

This site has a lot of good info about battery charging:

http://batteryuniversity.com/


Bob
 

mach7

Jan 8, 2013
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Thank you very much Bob.

Sure it must detect voltage drop or rapid temperature rise but it not necessary for me, now. There is no way to make a simple charger like this? Only trickle charge will work?

The resistor should be in series after the capacitor?

The input voltage is 12V and the batteries is 800mah each. I have 4 of them in series.

Thanks for the link also.
 

Edwin Fitzpatrick

Dec 26, 2012
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Charging AAA batteries,MACH7

You need to build a constant current and voltage charger with a LM723 I have the best charger circuit for ni-cads,nh-mh and small gel cells. I can send yoy the circuit and a PC board layout. Need to know you address or e-mail. I can scan the print.
Edwin fitzpatrick
 

BobK

Jan 5, 2010
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If you overcharge, you will cause the batteries to vent. NiMH are the most difficult of all the common rechargables to charge correctly. Read what battery university has to say.

A trickle charger is the only safe option for a simple charging circuit.

The only other option is to use a timer, but that is fraught with danger. If the batteries are not fully discharged when you start they will overcharge and vent.

The capacitor across the battery is doing nothing. It is just going to track the battery voltage, not providing any additional charge current. If you put the diode to the capcitor then to the resistor to the battery, the capacitor will charge during the time that the input waveform is less greater and discharge into the battery when it is lower. This will give you about 16V for charging from a 12V supply.

Now, to calculate the resisstor. The charger voltage is 16V. The battery voltage is 4.8V, so the difference, which must show up across the resistor, is 11.2V. The current desired is 800ma / 20 = 40ma. So

V = I R

11.2 = 0.04 R
280 = R

Bob
 

BobK

Jan 5, 2010
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You need to build a constant current and voltage charger with a LM723 I have the best charger circuit for ni-cads,nh-mh and small gel cells. I can send yoy the circuit and a PC board layout. Need to know you address or e-mail. I can scan the print.
Edwin fitzpatrick
\
Why don't you share it with all of us?

Bob
 

mach7

Jan 8, 2013
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Bob, thank you again. I ll study also the battery university articles.
So i ll put all components (diode, capacitor, resistor and batteries) in series?
I thought that batteries don't vent so easy, with some extra hours of charging, am I wrong?

Edwin, thank you for your offer. Sure you can share it here.
 

BobK

Jan 5, 2010
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Here is what the circuit shold look like:
 

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mach7

Jan 8, 2013
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Thanks! So is this circuit safe as long as i check the charging time?
Why you write 12V AC? I use a transformer 220V to 12V, 2A DC.
 

BobK

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It is safe forever as long as you calculate the resistor for charging at 0.05C which is 40 ma.

If you are using a 12V DC supply, then take out the diode and capacitor and recalculate the resistor based on 12V instead of the 16 you would get with the circuit I gave you.

Bob
 
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mach7

Jan 8, 2013
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Really safe? That's good news!
With this circuit, we have 12-4,8=7,2V across the resistor. So R=V/I=7,2/40mA=180 Ohm.
Is this correct?
 

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mach7

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Bob, I have another question.
I am charging 4 batteries of 800 mAh in series.
Should I calculate the 0.05C for only one (40 mA) or for all of them (160mA)?
 

(*steve*)

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If they're all in series, the mAh rating of the battery is the same as any 1 cell (assuming all cells are the same). Only the voltage changes.

So four 800mAh 1.5V cels in series gives you 6V at 800mAh.

Four 800 mAh 1.5V cells in parallel gives you 1.5V at 3200 mAh.

So the answer is that it must remain at 40mA.
 

mach7

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Thanks steve.

How about the Watts of the resistor?
I think it must be, W=V*I=7,2*40mA=0,28 Watt, so I am ok with a 1/2 Watt resistor.
 

(*steve*)

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best to calculate the wattage based on the resistor being shorted to ground. That way if you do something silly, you won't kill it. P = 144/180. I would go with a 1W resistor.
 

mach7

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How many hours should I charge my 4 800mAh batteries at 40 mA?
24 hours is ok?
 

(*steve*)

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You will probably need 28 hours.

I'm basing this on the C/10 charge rate normally being specified for 14 hours.
 

mach7

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Is it safe to charge also at C/10 if I want faster charging?
 
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