12V to 5V

What will you be powering with it?

Those results are puzzling. Connected in parallel, the two 6-volt batteries should give 6 volts (or slightly more, up to about 7, if they are fully charged). Why you are getting 11, I don't know. Connected in series they should (and do) give 12 volts or a little more (up to 14, wich is what you observe).

What kind of voltmeter are you using?

As for basic questions, it's time to go to the library and get a basic electricity book.

It's not 4A/Hour, it's 4 Ampere-Hours, and that's called the capacity (C) of the battery.

What that means is that if the battery is fully charged and you take

4 amps out of it for an hour the battery will be discharged. Or if you take 2 amps out of it for two hours, or if you take 1 amp out of it for four hours, or any combination of amperes and hours where the product of the two is 4. Big "but", though, that "but" being that you can only get 4AH out of the battery at C/10 or C/20, depending on the manufacturer. C/10 would be 0.4A for 10 hours, and C/20 would be 0.2A for 20 hours.

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You\'re right, but it\'s still 4AH. ;)

Try replacing the battery in your meter, and measure it again.

Mark

He has the batteries in series for both measurements. He is referring to the METER'S + and - terminals connecting the two batteries' + and -. The readings should have the same absolute value, just differing by a

+/- sign.

Mark

AH!! Sorry.

Then it's a digital multimeter whose DC offset is WAY out of adjustment. A correctly adjusted multimeter should read the same voltage if you swap + and -, except that you get a - in front of the number.

It could just be the battery is low. The readings on a perfectly good DVM can get rather wacko when this happens.

Mark

Ok, thanks with all the meter related help. But I still don't know how to change that 12V to 5V... :)

Is there a simple circuit I can use to do something like that?

The program I used is here:

The circuit posted by HKJ (on 17 June 2006) is typically used for this. You might find the parts at Radio Shack ... if not they can be ordered from places like

. Get 2 or 3 of the 7805's, so you have a spare in case one burns out. (part # 497-1441-5-ND at Digikey).

Mark

Yes; I already posted a solution for you, that being to use a 12V to

5V DC to DC converter connected across both batteries in series. That combination will give you better efficiency and longer battery life than if you use a linear regulator with a 5V output connected to 12V or, possibly, with a low-dropout regulator connected to +6V.

If you want to use a linear regulator you'll need to let us know what the maximum load current will be and then we'll be able to advise you further.

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John Fields
Professional Circuit Designer

Thanks for all of your help, I saw in the Pic above there is an IC. What IC is it?

Thanx constant meiring

(By the way... Radio Shack doesn't exist in South-Africa. Lol)

The IC is a 7805 (which comes in a bewildering array of packaging options for different requirements such as output current and thermal dissipation). As it has a dropout voltage (the minimum difference between input and output) of typically 2V, you could not use it from paralleled 6V batteries. [You need a minimum input of about 7V or so for it to work].

Typical datasheet here:

(now made by Fairchild, National, OnSemi [the old Mot SPS], TI and others).

If, however, you use it from 12V, you will have a large amount of thermal dissipation, as John noted. At very low currents that may be ok (although it's very inefficient, but that may be ok for you too).

As John said - tell us the load and we'll be better equipped to advise you.

Cheers

PeteS