# Thread: Sealed Lead-Acid Battery

1. ## Sealed Lead-Acid Battery

hi guys, ive just got hold of a sealed lead-acid battery and i got a few questions if ya dont mind...

Battery Description: 12V, 7Ah, Valve Regulated, Sealed Lead-Acid Type, Rechargeable Battery.

Would this battery be able to supply a 12V PC @ 7A draw for at least 10 seconds? If so how long for in total? For charging what will i have to do? Does the voltage input need to be 12V @ Constant (so from my 12V regulator) or does the current need to be @ Constant? If so what will i need to charge this battery? I electrician my dads knows told me that i need this... Schottky diode,MBR2060CT 10A If 60Vrrm (www.rswww.com and lookup 348-6182) i think this is just a standard diode to stop backflow of current (so other car appliances cant drain the batttery?!)

Yeh so basically wot do i need in order for this battery to recharge and the questions above :-)

The diagram that i had orginally figured goes like this.... (if anyone could ammend it so it would work plz do)...

Dean

2. Yes, a 7 AH battery will easily supply 7 amps for 10 seconds. As for how long... well, the 7 Ampere-hour rating is usually based on a light draw, so it'll be derated a bunch if you pull 7 amps. But, it should easily run for 15 minutes and probably much longer. Read the data sheet for the battery.

Charging: basically, you feed it like you do a regular car battery, only the current must be limited to about C/5 (where C = AH capacity). So, for this battery, the current must be less than 1.4 amps. Easiest way to do this is just a resistor, sized at something like R=V/I, or 2.5V/1 A (to be safer),so about 2.5 ohms. So, try a 2.7 to 3.3 ohm resistor, or larger (within reason). Power rating of the resistor: P=E*I, 2.5V*1A=2.5W, so use at least a 5 to 10 watt rated resistor. The 2.5V comes from the assumption that the car voltage will be around 14.4 V when the alternator is charging, and that a discharged/low gelcell will be around 12V.

The diagram in this thread shows how to arrange a diode and the resistor to charge the battery and provide backup. Using Sealed Lead Acid battery with Casetronic C134

3. ## Quality....

Ok.... so can we confirm the diagram below....

Diodes are 348-6182 from www.rswww.com, and the resistor is 3Ohm at 10W rating?

Would some one please be able to post me the ID number for a resistor that i need from the www.rswww.com website please. Much appreciated!

Dean

4. Originally Posted by BassBinDevil
Yes, a 7 AH battery will easily supply 7 amps for 10 seconds. As for how long... well, the 7 Ampere-hour rating is usually based on a light draw, so it'll be derated a bunch if you pull 7 amps. But, it should easily run for 15 minutes and probably much longer. Read the data sheet for the battery.

Charging: basically, you feed it like you do a regular car battery, only the current must be limited to about C/5 (where C = AH capacity). So, for this battery, the current must be less than 1.4 amps. Easiest way to do this is just a resistor, sized at something like R=V/I, or 2.5V/1 A (to be safer),so about 2.5 ohms. So, try a 2.7 to 3.3 ohm resistor, or larger (within reason). Power rating of the resistor: P=E*I, 2.5V*1A=2.5W, so use at least a 5 to 10 watt rated resistor. The 2.5V comes from the assumption that the car voltage will be around 14.4 V when the alternator is charging, and that a discharged/low gelcell will be around 12V.

The diagram in this thread shows how to arrange a diode and the resistor to charge the battery and provide backup. Using Sealed Lead Acid battery with Casetronic C134
could you pls talk me through about where ya got the 2.5v from :-S i dont get that bit! I would of thought it would of been 12v or sommit close!?

Dean

5. Car running = 14.4 volts minus 12 volts yields a 2.4 volt differential.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•