Battery - Reduce output amperage but maintain voltage

[GaryDamp]

I have a device that operates at 12V and 2 amps. I would like to use a battery, like a car battery, and shunt the amperage so that I achieve the output needed by the device. Can you suggest a circuit that would accomplish this?

Thanks  

[johnsold]

Your question does not make sense. Your device will have an equivalent resistance of 12 V/2 A = 6 ohms. The battery produces a voltage that is approximately 12 V, depending on its state of charge. Its output current will be determined by its actual voltage divided by the resistance of the load.

Real batteries have an internal resistance which appears to be in series with an ideal voltage source. But for a car battery that resistance is typically a small fraction of an ohm and will not be significant until you draw several tens of amperes or more.

Are you looking for a voltage regulator to maintain exactly 12.0 V from a lead acid battery? Something else?

Lynn

[GaryDamp]

Hi Lynn

Yes, I am seeking a constant 12V from a lead acid battery.

Thanks    

[johnsold]

That is not easy. The problem is that the battery voltage can vary from about 14.4 V while charging down to <11 V while discharging, especially under heavy load or when nearly completely discharged.

Standard linear regulators can only produce output voltages lower than their input voltages.  This is true for both series and shunt regulators. Some switching regulators (buck/boost types) can be designed to either decrease or increase the votlage from input to output.  The design of such circuits is not trivial. The inductors may need to be custom designs, depending on the exact specifications.

If you can limit operation to times when the battery voltage is above ~12.6 V, you could build a low-dropout regulator. Unfortunately a 12 V lead-acid battery will drop to 12.6 V very quickly unless the discharge rate is very low. At 2 A it would be below 12.6 V in minutes.  It would likely use a power transistor and an op amp along with a voltage reference. If you need current limiting, additional circuitry would be required.  

Lynn

[GaryDamp]

Thanks Lynn

I understand the issues concerning the battery drawing down below 12V and the 2A pull from the device itself.  While this project now dosen't appear practical I might work if:

a) two 12V batteries in series were placed in series,

b) a low drop-out regulator was introduced (and probably the additional circuitry cause a steady 12V), and

c) understood the draw characteristics of the device - that is, how much current is drawn as the device goes through its cycle. The device is be a motion detecting video camera that is wireless - that is network addressible - has its own ip address. It may be that it has a very low draw when there is no motion.

Do you think the above might be doable?

Thanks for your comments.

[Shane-C]

Lynn,

Placing the batteries in series will give you ~24 volts.  You will still run into all the same issues regulating this to 12V, though with the propery regulation circuitry, you should be fine.  

Regards,

Shane C

[GaryDamp]

Thanks, all!

[johnsold]

Shane,

I started a post saying that but got interupted.

Gary,

An LM350 with a few resistors and capacitors should work well with two batteries in series.

Lynn

[stargateok]

I have a similar question, but reading these questions and answers didn't help.  I also have equipment that now uses a "switching" 12v 2amp wall adapter.  I am trying to get completely off-grid and I don't want ANY AC usage.  How can I run my camera and DVR of of my battery bank(36 Sealed 100AH 10 amp batteries) some are hooked 12v parallel and some are hooked 24v series.  My concern is the "switching" aspect of the wall adapter, can the equipment run OK without it and do I need to drop the 10amps to 2 amps?

[crossrulz]

Don't worry about the "switching" power supply.  That is just the architecture of the regulator used in the adapter.  All you care about is that it puts out 12V.  So as long as your battery bank puts out the 12V and can supply at least 2A, you will be fine.