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# Batteries Charger & PSU - for Cameras

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Hi Guys,

I was wondering, how can this project charge four NI-MH batteries from only 5V in series with a diode, and how can a resistor in series with a voltage source supply "constant current"?

The AA size NI-MH batteries that I have charging right now at 150mA, which is less than their recommended current of 210mA, measure 1.4V each, and they are not fully charged yet.

So 4 X 1.4V = 5,6V plus 0.7V for the backwards-protection-diode = 6.3V or more to reach a full charge. Additional voltage must also be added for a voltage drop across the current limiting resistors. From a 7805 5V regulator?
The voltage across the batteries increases while being charged, therefore the current will also vary when charged through a simple resistor.
These problems are solved if the 7805 and its current-limiting resistors are replaced with another LM317 set as a current regulator:

1) Connect a current-setting resistor between its output and its reference pins, and take the output current from the reference pin.
2) Calculate the value of the current-setting-resistor: 1.25 divided by the desired current in amps (10 ohms gives 125mA). 1/2W is fine.

Caution: A fully-charged battery that is still charging gets hot and may be damaged! Frequently check them, snce if you start charging batteries that are not completely discharged then you won't know how long to charge them for.

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The regulator is connected as a constant current source not a voltage regulator!

Ante ::)

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Sorry you are right now I see what you mean :o :o :o!
The regulator is supposed to be connected as a constant current source!

Ante ::)

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Although the 7805 can be a current regulator, do you see that it can't be used here, since it would need at least 7V across it (plus the 6.3V or more for the batteries)?. The transformer voltage is too low for that. If the transformer voltage was increased then the heatsinks must be bigger.
An LM317 will current regulate with only 3.2V across it.

Since the PSU may be charging batteries and powering a camera at the same time, then the transformer must be rated for 1.5A.

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Yes I see, 9Volt transformer, rectifier, and cap gives around 13Volts with zero load. One regulator, resistor, diode and the shunt in the A-meter

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I thank Audioguru for his notes. I very sorry but I missed circuit. I forget to draw a wire between 100 pot. - diode connect and GND of 7905. I'll modify my project for electronics-lab, too.
I agree with Audioguru. In his opinion sec. voltage of transformer (9V) too low for four batteries. He's right, however I still charge simultaneously only 2 batteries. So it was enough. I didn't think and didn't calculate this 9V is little. I think more voltage (15V) won't be bed besause cooling is soluble and before charging I descharge batteries until about 1V, so shouldn't be overcharging during given charging time.
Please write if somebody dosn't agree with me!

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Hi Sandor

Everyone make mistakes, no big deal. Let

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Dsander,
Thank-you for admitting that you missed drawing an important wire on your schematic. So then the 7805 will be a current regulator.
I am glad that you realise that the supply voltage is too low for a 7805 current regulator to charge 4 batteries. If you use a 12V AC transformer then you will get about 15V DC supplying the regulators. The regulators will get mighty hot when supplying a low voltage (charging only 1 battery and/or supplying 800mA to a 2.9V camera), so when you update your project please mention heatsinks.
By the way, with the batteries being fed a constant current then you don't need the pot nor the ammeter. The current will be the same for 1 battery as for 4. Just calculate the resistor and solder it in.

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Hi Audioguru, Ante !

Thanks for your notes. My project was updated before I have read your advice, so I couldn't mention nessesary of regulator heatsink and the contest have already finished. :(
I think it is not frequent occurrance using of digitalis camera with PSU and battery charging simultaneously, but it can occur.
Therfore I applied pot and ammeter because I charge not only NiMH batteries but older NiCd bat., too with lower charging current.

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D. Sandor,
Thanks for correcting your schematic on the projects page, which is here:
http://www.electronics-lab.com/projects/power/035/index.html
But you have also changed the transformer to 15V AC, which will give more than 19V DC. So now there will be more heating, but it has enough voltage to charge 8 batteries. 12V AC would halve been fine to charge 4 batteries, with less heating., or the original 9V AC with an LM317 current-regulator for even less heating.
I can see your point about using a meter instead of having a battery-select switch or calibrated dial on a pot. It gives indication that charging is occuring! Most chargers use an LED.

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