# 12v 500mA DC out of battery

Discussion in 'General Electronics Chat' started by naveed, Apr 26, 2012.

1. ### naveed Thread Starter Active Member

Mar 12, 2008
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I want to run my modem on battery. It takes 12v 500mA DC.

I have a 6v 4.0Ah led acid battery, I can afford two batteries if the circuit becomes simple that way.

Please help me how to get 12v 500mA out of 12v battery? i.e., how to limit the current.

2. ### K7GUH Member

Jan 28, 2011
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You might not have to do anything special. The 500 ma specifies how much current the device will draw. If it is operating normally, it will not take more than 500 ma. You may wish to insert a fuse in the power lead to the modem. This will blow if the device draws more than 500 ma.

3. ### mcgyvr AAC Fanatic!

Oct 15, 2009
5,394
1,194
Why?
How long do you "expect" this device to run on the battery?...It won't be long at all.. Not to mention a batteries voltage goes down as its being used.. So you might start at 12V with 2 batteries but it will quickly go below that and there is a good chance your modem will turn off long before your batteries are even close to "used up"

4. ### naveed Thread Starter Active Member

Mar 12, 2008
42
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Once I tried charging my phone directly with 6v battery. And the charging IC of my phone blown off.

I think something should be done to limit current.

5. ### strantor AAC Fanatic!

Oct 3, 2010
5,011
3,094
Then put 24Ω of resistance in series to limit the current to 500mA

6. ### naveed Thread Starter Active Member

Mar 12, 2008
42
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Because I am a Pakistani. Here light comes for 1 hr after 5 hrs of loadshedding. And I want to do some freelance projects(on my laptop).

Anyway then should I use 3 batteries? i.e., 18v. Is there an easy circuit to make 12v 500mA out of 18v?

7. ### naveed Thread Starter Active Member

Mar 12, 2008
42
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Are you kidding??

I mean according to what I know is that resistance in series don't limit current but voltage.

8. ### t06afre AAC Fanatic!

May 11, 2009
5,936
1,229
If your modem use 12 volt 500mA. You can use two 6 volt batteries in series. The modem will draw the current it need and nothing more. However put a fuse close to the batteries to limit the current if something goes wrong. Use say around 1A fuse. How long you can depend on this setup before the batteries goes out. I do not know.

9. ### strantor AAC Fanatic!

Oct 3, 2010
5,011
3,094
I was making a humorous comment about the fact that you think the device needs current limiting, but otherwise, no, I wasn't kidding - resistance in series limits current, not voltage. Depending on the consumption of the modem, there would be a voltage drop across the resistor, hence less voltage at the modem, but the amount is unpredictable.

10. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
512
You could use three batteries for 18V and then step down with a cheap buck regulator like LM2575-12. You should not have to use any current limiting at all, the load will draw only what it needs. The 2575 is a 1A (max) reg so it can easily supply 0.5A.

11. ### WBahn Moderator

Mar 31, 2012
24,454
7,653
Does the 500mA number come from the modem's documentation, or is it just the current listed on the power supply's label?

The modem will draw different amounts from a 12V source depending on its immediate needs and this amount should never exceed the amount listed in the documentation (unless something is wrong). The power supply will deliver whatever current is needed by the load up to the amount listed on its label (usually a bit more and sometimes quite a bit more).

Do you have a multimeter with ammeter that covers half an amp? If so, then put it in series between your existing power supply and the modem and get a decent feel for home much current the modem actually draws while you are using it as heavily as you can. Also note the current when you aren't using it at all and it is just sitting there idle. The peak current is probably noticeably below the 500mA and the idle current is probably well below that.

If you put two 6V-4Ah batteries in series, then you basically get a 12V-4Ah battery. At 500mA, you would expect about 8 hours of service, however the battery voltage might drop below what the modem needs to function well before then. On the other hand, the actual average current draw is almost certainly well below 500mA, so you might get as much as a day or so before that happens. I'm gonna give a WAG and say that you might realistically expect perhaps 12 hours of usable service.

Perhaps the best solution would be to buy a DC-DC converter module. In the RC world these are called BECs (battery eliminator circuits) and are very common and fairly cheap. You can get one for your needs for \$20 to \$40 (here in the U.S.). I would recommend looking for one that can deliver 12V at at least 1A and that can work with input voltages of, say, 4V to 16V.

With this, you could take a rechargable lead acid battery and connect it to a battery charger that is plugged into the wall. Use a strong enough charger so that it can replace all the charge used from the battery during the power-out periods during the brief power-on periods and make sure that the battery is rated for that charge rate. At that point, you basically have built a UPS that you can just leave connected to the wall and to the modem and shouldn't have to touch it for long periods of time.