Need to limit current in small space

Hi.

My camera take a small battery that doesn't last too long. I thought if I could put 2 batteries in parallel (because I have a suitable enclosure that happens to fit the bottom of the camera and is even the same colour) I could then cram a couple of diodes and/or resistors in series/parallel with the batteries to limit the amperage to what one battery alone outputs.

Each battery is 7.2~7.4vdc @ 1500ma

So two in parallel would keep the voltage constant but raise the amperage to 3000ma, correct? I want to drop that back to 1500ma which should close to double my battery life without having to manually swap batteries so often.

Any circuits available?

Would this be sufficient (click to view larger image)?:

. . .

Thanks for your time and your help!

I think you misunderstand those amperage ratings. When you have only one battery at, say, 7.2 volts, then ohm's law dictates the actual current drain of the battery. The current drain is not 1500 mA. When you read 1500 mA off the battery, you should look a little closer because what you are actually reading is 1500 mAH. The H stands for hours and this value is in fact the capacity of the battery. You use this number by dividing it by the actual current drain to understand the number of hours it will be able to supply that current for.

Anyways, when you attach two batteries in parallel, then theoretically, you get the capacity of both batteries without changing the voltage. So when you parallel those two 1500 mAH batteries, you would theoretically get 3000 mAH of capacity, which means you do indeed double the life of the batteries. Theoretically, anyways. There are some practical problems that tend to lower this number. Like for example, since the batteries are not chemically exactly identical, it is possible for one battery to discharge partly into the other one over time. I don't have much other experience with putting two batteries in parallel, so I'll let others post on that topic.

I don't think you need to include any additional parts at all if you just want to try it out. There is no need to add any current limiting functions because the load will continue to take what it took before, despite the change in capacity of your battery, as long as the voltage is the same as before.

Oh, and by the way, your diagram actually shows the two batteries connected in SERIES which will double the voltage but not the capacity, and may blow up your camera. Parallel is where you attach + to + of the two batteries and also - to -.

No, those batteries are connected in series not parallel.

You don't need a resistor to limit the current, providing the batteries are both the same capacity, voltage and inittially charge to approximately the same level, just connect them up in parallel (+ to + and - to -) and it'll be fine.

You need to up on ohms law - by your question you obviously don't understand it.

https://en.wikipedia.org/wiki/Ohms_law

Basically your battery has a constant voltage, (the mAh rating is the capacity measured in miliamp hours and is irrelevant here) the camera is a load resistor that draws a current depending on its resistance which is the same regardless of the mAh value of the batteries.

V = IR

For example a:
1 ohm load draws 1 Amps from a 1V supply.
2 ohm load draws 2 Amps from a 2V supply.
2 ohm load draws 1 Amp from a 2V supply.

Suppose your camera draws 1.5A (750mA) from your 7.2V battery:
Resistance of camera = 7.2/0.75 = 9.6 ohms.

Your 1500mAh (1.5Ah) battery will last:
Time = 1.5/0.75 = 2 hours/

Two 1500mAh (1.5Ah) batteries in parallel will last:
Total Battery Capacity = 2*1.5 = 3Ah
Time = 3/0.75 = 4 hours which is twice as long (obviously)

Thanks for your input, guys!

I feel much more confident - I wanted to be sure BEFORE doing anything that could wreck my camera -- better safe than sorry!

Thanks again!

This is getting very confusing. I know I'm weak in decimals and metric and exponents, etc., but I understood 1.5 Amps would be 1500 milliamps, and 750 milliamps would be .75 Amps. Where have I gone wrong?

Hero999 said:

Suppose your camera draws 1.5A (750mA) from your 7.2V battery:

Click to expand...

AllVol said:

This is getting very confusing. I know I'm weak in decimals and metric and exponents, etc., but I understood 1.5 Amps would be 1500 milliamps, and 750 milliamps would be .75 Amps. Where have I gone wrong?

Click to expand...

You haven't, I presume the 1.5A is the total current, and 750mA would be the current from each battery (half of 1.5A) - assuming batteries in parallel.

wubs,
Like RadioRon said, if you put the batteries in parallel, there will be no problems, you will raise the capacity and you will not need to use any kind of current limiting devices.

But use the same batteries brand, because theirs chemical composition will be similar.

There is no need for a Amperage limiter.

The digital camera only use as much mA as it needs.

Two batteries will last twice as long as one.

And wire it up so all 3 batteries (1 internal and 2 external)
are all in parallel and you tripple the time before running out off power.