9V BATTERY CHARGER CIRCUIT

[cobra149]

Hi, found this circuit on the net. I have attached the schematic and a drawing of the pcb layout. Just need to know if it will correctly charge a 9v battery safely, and if the pcb I have drawn is correct.

Thanks in advance for any help.

View attachment 41995

 

[audioguru2]

Of course the battery connects to the output, not the input.
The circuit limits the charging to (9.1V - 0.7V for the diode)/15mA, plus or minus 5%.
But which kind of battery? A 7.2V Ni-Cad or an 8.4V Ni-MH?

A 7.2V Ni-Cad has 6 cells and charges to 8.4V so 8.4V (it might be 8.0V) might be too low.
An 8.4V Ni-MH has 7 cells and charges to 9.8V so 8.4V (it might be 8.0V) is much too low.

At only 15mA a 200mAh battery takes more than 13 hours to fully charge.

EDIT: I originally forgot about the diode.

 

[cobra149]

Thanks. The 9v batteries I use are 280ma 8.4v NiMh. So would I use a 10v zener reference  instead ?.
Also What resistor would I use to give me 28ma c/10 in this circuit.

Thanks.

 

[audioguru2]

The 9v batteries I use are 280ma 8.4v NiMh. So would I use a 10v zener reference  instead?

Your batteries ARE NOT 280mA. They are 280mAh which is not their maximum current, it is the current they can provide for one hour. Their maximum current might be a few Amps for less than 5 minutes.

If you use a 10V zener diode then the diode at the output of the charger will reduce it to 9.3V which is not fully charged but is OK.

Also What resistor would I use to give me 28ma c/10 in this circuit?

The datasheet for the LM317 says it produces 1.25V (plus or minus 5%) between its output and the ADJ pin. So for 28mA use Ohm's Law:
1.25V/28mA= 45 ohms. Use 47 ohms which is the nearest standard value. The heating in the resistor is (1.25V squared)/47 ohms= 0.03W. A 1/4W resistor will remain cold.

Do not discharge less than about 7.7V. Remove the battery from the charger after it charges for 12 to 14 hours.