Simple battery charger

[tnnelectro]

Hi everyone,

to charge the 12V - 1,2Ah lead acid battery I found this scheme very simple and unpretentious ... what do you think?





with the V1 trimmer I adjusted the charge voltage to 13.7V as indicated on the battery ....

3 questions:

- does the battery have to be disconnected when I adjust the charge voltage?

- the L2 led lights up when the battery is charged or discharged?

- is a 15V power supply for LM317 okay?

thank you.

 

[Harald Kapp]

I adjusted the charge voltage to 13.7V as indicated on the battery ....

That should be ~14.4 V. 13.7 V for the battery + 0.7 V for the voltage drop across D1

does the battery have to be disconnected when I adjust the charge voltage?

Yes. Otherwise the battery may influence your measurement, depending on its charge state. Specifically when the battery charge is low and D1 is conducting.

the L2 led lights up when the battery is charged or discharged?

When charged, but the indication is rather imprecise. When LED L2 will light up depends on the color of the LED (different colors require different voltages).

s a 15V power supply for LM317 okay?

At 1.5 A the dropout voltage (V_out - V_in) can be up to 2.5 V. At 14.4 V (see above) output voltage that is equivalent to 16.9 V input voltage. A 15 V power supply is insufficient.

Also note: 2.5 V × 1.5 A = 3.75 W. You'll need a heatsink to dissipate this power. Otherwise the thermal limiter will be triggered and reduce the output voltage to limit power dissipation.

 

[tnnelectro]

Thanks for the reply...

then seen the zenner diode D2 which is 11V, the led L2 turns off

when the battery drops below this voltage?

 

[Harald Kapp]

Below 11 V + V_LED. at the output of the regulator.
With respect to the battery voltage the voltage drop across D1 and R3 have to be taken into account, too.
As the current will drop at the end of the charge cycle, let's ignore the voltage drop across R3, that leaves D1 with ~ 0.7 V.
The "threshold" for the indicator LED is then 11 V + V_LED - V_D1 = ~ 11.9 V. (A red LED typically has a forward voltage of ~ 1.6 V, therefore the indicator would turn off below ~12.6 V.)
It will not be a sharp on/off transition, the LED will slowly fade as the voltage across the battery goes down.

Have a look at e.g. this more sophisticated circuit. Includes a short explanation of the operation.

 

[bertus]

Hello,

And if the voltage is 14.4 Volts, as Harald recommended, the current through the led will be (14.4 - 12.6) Volts / 15 Ohms = 120 mA, wich will likely burn the led.
You would get a better indication of the battery status when you move D2 to point 1 of the output connector.


In the origenal schematic it only sees the regulator.

Bertus

 

[Harald Kapp]

The circuit is probably meant to be used with a green LED (to indicate end of charge). The forward voltage of a green LED is ~ 2.2 V, that would reduce the current to approx. 33 mA (13.7 V from the regulator - 11 V from the zener diode - 2.2 V for the green LED). That would be o.k.
The circuit is imho a cheap one that may work, but can be improved significantly.

 

[tnnelectro]

thanks for the answers ...
but I have a doubt that the zenner diode D2 must not be connected after the diode D1?

 

[Harald Kapp]

I have a doubt that the zenner diode D2 must not be connected after the diode D1?

You then need to correct the voltage as you "measure" directly across the battery
Assuming 2.2 V for the LED, in this position it will light up at V_bat = 11 V + 2.2 V = 13.2 V, but you want to charge the battery to 13.7 V ...

 

[tnnelectro]

ok I did a test I brought the battery to about 8V and then I put it in charge .. but the L2 led is on ...?

now the battery is about 12.4V and the led is still on ....?

 

[Harald Kapp]

Which circuit? The one from post #1 or from post #5?
Also measure the actual voltage at the terminals of M04.
Also check the polarity of the zener diode.

 

[tnnelectro]

circuit post#1...

 

[Harald Kapp]

In that case the charge current develops a high voltage drop across R3 that is enough to turn on the LED. You wouldn't have this effect with your modification in post #5, but you'll have to change the zener voltage. Or simply add a normal silicon diode in series to the LED to add the 0.7 V you are now missing since D1 is out of the indicator circuit
As I stated before: this circuit is not particularly well designed.

 

[tnnelectro]

ok well so if I replace the zenner with one of 11.7V, how many volts of the battery should the led turn off?

 

[tnnelectro]

could this then be more efficient then ...?

 

[bertus]

Hello,

Did you have a look at the thread on AAC that wendy gave you?
SLA Battery Charger
There are some good ideas from SgtWookie in there.

Bertus