I have a Gell Lead-Acid battery, 12V 32AH that will only charge to about 11.3V.
Apparently there is a bad cell in it. Can I keeping the battery but in a different capacity? And if so would I recharge it differently?

I was thinking of putting a voltage regulator on it and using it to power a bedside light(2 series strings of white LED's in parallel...tot 20 LED's) and normaly runs off of 3 C Cells. Would this work? Do I need to look out for anything in particular?

 

It will probably make a better doorstop. The remaining cells will never charge properly.

 

Well, for something really low-current like LED lighting, it might work.

It really depends on how the cell failed. The SLA's I have usually develop a partial short in one of the cells. It's not a "perfect" short, but a "moderate resistance" short, and that resistance can vary quite a bit. The partial short prevents the bad cell from charging up. It also limits the current flow to/from the other cells. That's why a battery that develops a bad cell suddenly seems to have no power at all; if the external resistance is low (like an engine starter, perhaps a scooter motor or uninterruptible power supply) the majority of the battery's power is expended on the internal shorted cell.

It'll take longer to charge such a defective battery than normal, due to the resistive short in the bad cell.

White LEDs typically have a fairly high Vf (forward voltage), somewhere between 3.3v and 4v at their rated current. You control LEDs by current, not voltage. A typical linear regulator (such as an LM317) could be used, but they drop about 1.8v across themselves. So, you could maybe run a maximum of 2 white LEDs in a string that are current-limited by an LM317.

You can also make a current limiter circuit out of a transistor, a couple of resistors and either a standard red LED or a couple of silicon diodes like 1N914, 1N4148, 1N4004, etc. You would need one current limiter circuit per series string of LEDs.

You could also simply use resistors to limit the current flow. However, with the cell already being weak, the light output will vary quite significantly from when freshly charged to half-way discharged.

 

Actually I may be wrong with the LED's in this Lamp. Each LED has it's own resistor. It's possible that the configuration is 10 LED's in parallel that are in turn in series with a second set of 10 LED's in parallel. This means that the current drawn would be 200mA + 200mA, or 400mA. The wallwart I am using is rated for 4.5V and 800mA. Maybe I need to have 6 or 7 parallel sets of 2 series LED's and 1 resistor, all in parallel. this would draw only 120ma - 140mA. If I did the math right I could run the second lighting option for 3hrs/night for 95 days, thus saving untold amounts of energy on my electric bill!!!!!! (Ha Ha).