Lead acid battery charger circuit require

[chuckey]

The .51 ohm resistor sets the maximum current, the volt drop across it due to the output current causes the transistor to conduct more and bring the output voltage down. Putting a resistor in series with output reduces the output current too much. You are looking for a voltage drop across the .51 of .7V at your max current, so make it .7/.32 ~ 2 ohms.
Frank

 

[D.A.(Tony)Stewart]

Keep in mind, the ideal SLA charger will have hysteresis or switch to a lower voltage which is temperature compensated for float voltage to prevent self heating and aging after charge current reduces at full charge.

 

[embpic]

if i used 2 ohm resistor instead of .51 ohm then will the current is suitable for battery charging.

 

[BradtheRad]

By replacing the .51 ohm with a 2 ohm, it should do the job.

The 317 will heat up to some extent. If it becomes too hot to keep your finger on it, then you might wish to install a dropping resistor at the incoming wire (upper left) of your schematic. The resistor and the 317 share the power dissipation.

The dropping resistor also provides a degree of safety (in case something malfunctions in the 317 circuit).

The dropping resistor does not need to be as high as 50 ohms. Notice the voltage differential is not 14 V, but only a few volts, between your power supply and the battery.

Suppose your supply is 19 V. The differential might be 8V for a discharged battery (19-11). And it might be 5V for a charged battery (19-14).

Therefore a suitable value for the dropping resistor could be 15 ohms.
No doubt you will need to do some experimenting.

Its power rating should be sufficient to carry .3A. Maybe a watt or two.

 

[embpic]

yes sir but how to calculate the resistor value. because input voltage may change on availabilty of power supply. and mostly i will used input of 24 volts.
and here modified design

 

[chuckey]

1. remove the 50 ohm output resistor.
2. replace the .51 with a 2.2
3. if the input is 24 V, Vbatt 14v, allow 2V for the LM317, so Vdrop = 24 -16 = 8V @.3A therefor the 15 ohm resistor could go as high as 8/.3 ~ 27 ohms.
Frank

 

[BradtheRad]

yes sir but how to calculate the resistor value. because input voltage may change on availabilty of power supply. and mostly i will used input of 24 volts.

A battery can be described as a load which has an opposing voltage. The voltage is changing. This brings a variable into the picture which makes it difficult to calculate a resistor value.

Your power supply voltage is variable. This also makes it difficult to calculate a resistor value.

You will need to experiment, to find the best resistance. Even with a resistor, the 317 should regulate to 14.2 V, and avoid overcharging your battery.

If the supply voltage is high, however, the 317 is likely to overheat. In that case you'll need to put it on a heatsink.

and here modified design
View attachment 106484

Omit the 50 ohm.

Change .51 ohm to 2.

 

[embpic]

Thank you very much sir.
Means i can use this circuit by changing lower resistor (0.51 ohm) and upper left (15 Ohm) to any battery rating.

if the input is 24 V, Vbatt 14v, allow 2V for the LM317, so Vdrop = 24 -16 = 8V @.3A therefor the 15 ohm resistor could go as high as 8/.3 ~ 27 ohms.

what is wattage for this resistor. is it 1/4 w??

 

[BradtheRad]

upper left (15 Ohm) to any battery rating.

what is wattage for this resistor. is it 1/4 w??

As Chuckey points out (post #26), you may need a higher value, perhaps as much as 27 ohms. This is for on possibility that you will use a power supply of 24V.

We'll suppose it carries .3 A.
A 27 ohm resistor will develop 8.1 V across it.
This makes it dissipate 2.4 W.

 

[embpic]

yes sir i got this point.
In this circuit where we can add comparator for showing charging status using led.

 

[BradtheRad]

yes sir i got this point.
In this circuit where we can add comparator for showing charging status using led.

Charging status can be detected when current is flowing. The idea is to find where there is a voltage differential when current is flowing.

You can get this across the 15 ohm resistor.

Notice that there is a voltage differential across the 317 (its input and output pins).
However that will not always mean current is flowing.

 

[colin55]

You don't need both the 50R and the current-limiting resistor in the base-emitter of the transistor.

Change the 0.51R to five 10R 0.25watt resistors in parallel.

And remove the 50R resistor.
The input voltage to the regulator must be higher than 17v

- - - Updated - - -

You don't need both the 50R and the current-limiting resistor in the base-emitter of the transistor.

Remove the 50R resistor.
The input voltage to the regulator must be higher than 17.5v

The 1.2A-Hr battery in the picture you posted is a "maintenance-free" battery and this means the charging voltage rises to about 15v before the cells start to gas.
When charging the battery, the voltage across it rises to about 15.5v and this leaves you very little head-room (and voltage across the regulator) if using a supply less than 17.5v.
300mA is too high for the tiny battery you are using.
Batteries are always rated on 14 hour charging and allow 10% of 1.2 amps for the charging current for 14 hours when the battery is completely dead.
This means the maximum current should be 120mA to prevent gassing.
The current-limiting resistor should be 5R made up of 2 x 10R resistors in parallel.