Discussion in 'The Projects Forum' started by haseebhm, Feb 12, 2013.

1. ### haseebhm Thread Starter New Member

Feb 19, 2010
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I have recently made a charging circuit for my 2.5Ah Lead Acid battery using LM317 IC. Please check the circuit diagram at:
http://electronics-diy.com/schematics/893/battery-charger.jpg

My current circuit doesn't have the transistor Q1. In this case, the charging is too slow. However, I want to charge my battery to 80% within 3-4 hours. After this time, the power goes off again and the battery needs to be charged during this time so that my router keeps on running during the power outage time.
I'm an engineer but not electronics engineer and that's why I need help. Would this circuit charge my battery in the specified time ? If yes, I would add the transistor Q1. Any help in this regard will be appreciated

2. ### gerty AAC Fanatic!

Aug 30, 2007
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Do you have any connection at all in place of Q1?
If so where and what is connected

3. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
510
What is the A-hr rating of the battery?

That device can only put out about one Amp max.

4. ### ErnieM AAC Fanatic!

Apr 24, 2011
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1,825
haseebhm stated it's 2.5AH, so a 1A charge is on the high side. AFAIK lead acid prefers a charge rate of AH/20, or here 2.5 / 20 = .125A

5. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
510

Then it won't work as designed.Q1 is how it regulates the charging current. You might post a schematic of what you actually built up.

You don't say what the source is driving the circuit. An LM317 can drive something like 0.5 - 1.5A depending on input-output voltage differential.

FYI: that circuit need some output capacitance, bigger the better. 0.22uF is not nearly enough.

6. ### gerty AAC Fanatic!

Aug 30, 2007
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That's what I was trying to get out of him, but he hasn't been back..

7. ### haseebhm Thread Starter New Member

Feb 19, 2010
12
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I'm sorry for getting back so late, i was stuck. I have modified my circuit already and placed the Q1 transistor. Now my actual circuit is according to this diagram, except that I have used a 1uF capacitor on the output instead of 0.22uF. What I want to ask is:
1. Whether this circuit will charge my battery in 4 hours ?
2. Would it overcharge the battery if connected for longer duration?
3. Further Details: I have connected the battery through a diode to avoid back-flow of current when power is not available. And at the output, I have placed a 12v zener diode regulator with 1ohm resistor in series for my router. This will be the 12v output for my wireless router. So if the power goes off, my router will still be running through the battery and the battery. Am I doing it right ? Its working the way I want it to, I just have to put a heatsink on LM317 now.

8. ### haseebhm Thread Starter New Member

Feb 19, 2010
12
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Actually, I have used a 1uF capacitor at the output. Is it enough ?
At the input, I have 24V through transformer and full-wave rectifier and two 1000uF capacitors for filtering. Do I need to change that too ?

9. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
510
As I said before:

"You don't say what the source is driving the circuit. An LM317 can drive something like 0.5 - 1.5A depending on input-output voltage differential."

The charge time depends on the charging current. What is driving the circuit? How much current can it put out?

10. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
510
No, I would recommend at least 22 uF.

What is the current rating of the transformer?

11. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
510
The R2/R3/R5 resistors set the final full charge voltage. If you put a diode from there to the battery, the voltage at the battery will be too low.

12. ### haseebhm Thread Starter New Member

Feb 19, 2010
12
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Current rating of transformer is 1500mA.

13. ### haseebhm Thread Starter New Member

Feb 19, 2010
12
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Yes but I have adjusted the R5 potentiometer accordingly.

14. ### haseebhm Thread Starter New Member

Feb 19, 2010
12
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As I mentioned earlier, the circuit is driven by 24v rectified DV input. And at the output is the battery and a router which is connected with a zener diode for regulation. Router takes a current of 500mA.

15. ### haseebhm Thread Starter New Member

Feb 19, 2010
12
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I suppose this is a constant current charging circuit and hence the charging time can be known. Am I right ?

16. ### haseebhm Thread Starter New Member

Feb 19, 2010
12
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The transformer current rating is mA