12V to 5V

Discussion in 'Electronic Design' started by Constant Meiring, Jun 11, 2006.

1. Constant MeiringGuest

Hi there.

I'm pretty stupid concerning electronics, but I want to play around a
bit.

I have two 6Volt UPS batteries that can give 4A/Hour (It's written on
the battery). When I connect these two in series I get 12V. I probably
will still have 4A/Hour (I don't know please correct me if I'm wrong).
So now I have 12V. How would I get that 12V to 5V so I have a 12V and a
5V output from the batteries.

Another thing, how does positive and negative DC work? The two UPS
batteries when connected together and measured with a mulitmeter (+
with + and - with -), it says 11.something volts, but when I connect it
(- with + and - with +) I get -14.something. Why is the negative
voltage higher than the positive. How does this work?

Constant Meiring

2. mcGuest

I have two 6Volt UPS batteries that can give 4A/Hour (It's written on
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.

3. John FieldsGuest

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.
---

4. redbellyGuest

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

Mark

5. redbellyGuest

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

6. mcGuest

Those results are puzzling. Connected in parallel, the two 6-volt
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.

7. redbellyGuest

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

Mark

8. Constant MeiringGuest

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?

10. redbellyGuest

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 http://www.digikey.com . 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

11. John FieldsGuest

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

13. PeteSGuest

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:

http://www.fairchildsemi.com/ds/LM/LM7805.pdf

(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