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Old 18th December 2007, 09:17 PM   (permalink)
Question Just fail to re-charge?

I am doing some research on building a charger for my 12v sla (sealed lead-acid) battery. The requirement is simply bing able to start and stop charging automatically. I realize there is good information on the web (for example, http://electronicdesign.com/Articles...ArticleID=1823)

Ok, the regulated power supply I need should be rated at about 15 volts. However, I am thinking if there is some circuit failure (for instance, diode failure) on the power supply and it is not giving out 12-15 volts dc, am I so-called "back-charging"? So in the case my power source becomes 7 volts and not 15volts, I am further draining the SLA battery that sits at 11 volts!!

In that case I know my charger is not doing its job. But will it cost my battery or blow up my other parts of the circuitry, I am wondering? (please give a refresh to my rusty brain.) Thank you so much.
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Old 18th December 2007, 10:49 PM   (permalink)
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Here is a circuit for a PowerSonic SLA battery charger. I drew it up while repairing one. The diode D6 prevents the battery from draining back into the charger. The circuit has an LED (D4) to indicate when it is on high charge and a section to automatically fold back to a trickle charge.


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Old 19th December 2007, 03:30 AM   (permalink)
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Thank you for your circuitry. The one I got from the web requires a relay but yours does not, which is good.

I am not too sure how this circuitry works without testing it.

A normal Lead-Acid battery should be "heavy" if it is beween 12.8 and 13.1 volts. According to your setup, you make the un-loaded output voltage to sit between 13.98 v and 14.98 v (which will be 13.28 and 14.28 v under normal load). Will your setup be ok?

Another question is, how would you suggest me to revise the given circuit, given that I want to add some Auto turn-on and turn-off circuitry to my existing regulated 16 V dc supply. (I think I can remove the diodes and the LM1728 regulator totally)

Thanks.
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Old 19th December 2007, 04:11 PM   (permalink)
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First, the LM7812 (is LM1728 a typo?) is there to be a switchable regulator. It's output is set by R8 to provide a high charging current. As the battery's terminal voltage approaches this voltage, the charging current drops. This is sensed at R5, and the LM7812 is switched to a lower voltage output, set by R7, that is used to maintain a trickle charge to the battery. The voltages at the output terminals were selected from data sheets for SLA batteries and adjusted to maintain a charging currents required by this circuit.

You could replace T1, D1, and C1 with a regulated supply, but it would need to be 18vDC or greater, to overcome the U1's minimum I/O differential and the Vf drop of D6. Removing the LM7812 would defeat the whole point of this circuit. The circuit switches from high charge current to trickle charge. If you want to do something with your existing regulated 16V power supply you will need to find a different approach. Adding a diode to your original circuit's output (as in my D6), and having the relay cut the power to your regulated supply would probably do what you envision.


This was just an offering of the design of a commercial charger I'm currently using.

Ken
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Old 20th December 2007, 02:22 AM   (permalink)
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Quote:
Originally Posted by KMoffett
As the battery's terminal voltage approaches this voltage, the charging current drops. This is sensed at R5....
Much appreciated for your time to explain. I think I am more knowledgeable with battery charger now.

Question:
I can't figure out how D3 and R5 in the diagram function, and Q1 gets no biasing voltage. Is this a slight drawing mistake?

Thank you again.
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Old 20th December 2007, 02:26 AM   (permalink)
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Quote:
Originally Posted by ykffc
can't figure out how D3 and R5 in the diagram function, and Q1 gets no biasing voltage.
Sorry, I've figured it out now. The circuit is ok, my stupidity and rusty mind.
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Old 20th December 2007, 02:48 AM   (permalink)
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It took a while for me to figure out how it worked too.

Ken
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