Battery pack with indicator

I am thinking of creating an external battery pack for a camera which internal circuit works at 7.2v.

When shopping for cells I found this:

http://www.tmart.com/2pcs-Ultra-Fire-18650-3.7V-5000mAH-Lithium-Battery-Yellow_p191451.html

Two of this in series would offer 7.4v and 5000mAh, so it would give juice for quite a while to power the camera. Anybody has experience with these cells? Do these cells need circuit protection?

Also, I would like to add a custom battery fuel gauge indicator similar the one the macbooks pro use:

http://icdn6.digitaltrends.com/image/battery-indicator-macbook-pro-650x0.png

But I don't know where to start on creating something like that because I can only find already made ones that are too big or bulky.

Thanks

 

Thanks gregfox. That is exactly what I was fearing, too good to be true. So what brands are more reliable for Li-Ion cells?

As usual the big challenge is space, which is approximately 100mmx26mmx26mm. It could roughly fit two 26500 or C size cells placed one on top of the other perfectly or 8 AA cells.

I am trying to get at least 7.4v (it would be better if I could get to 12v) and at least 3000mAh.

I have been browsing so many battery websites and I can not seem to find the perfect fit. I found this but I suspect the quality might be poor as well:

http://efestpower.com/Product/1652072426.html

 

Thank you so much for the Hobby King recommendation! They have a great selection and great prices.

Unfortunately today I realized I made a mistake, the camera needs 12v instead of 7.4v. Since the space is limited this changes everything dramatically.

I could use a step up from 7.4v to 12v but it would reduce the capacity.

So again I am hitting another wall... Now I need to fit ideally 4x Li-Ion cells to get 14.4v and regulate it to 12v with at least 2000mAh. Would a step down from 14.4v to 12v give me more amperage?

 

Thank you so much gregfox!

So if I step up from 7.4v to 12v then there will be a loss in capacity? right? Is there a math to calculate the amperes that get lost? by doing that is it going to generate heat?

And if the input in the camera is 12v-20v, do I still need to step down or regulate the voltage to 12v? would this process generate heat as well?

 

Thanks guys!

I do not know if I am doing something wrong but if I follow the math:

12v / 7.4v x 3000mAh = 4864.86

then 4864.86 / 0.8 = 6081 mAh???

Am I doing it wrong? Or is it instead:

7.4v / 12v x 3000mAh = 1850

1850 / 0.8 = 2312.5 mAh

The result is the loss or the final amperage?

 

Thanks steve for the explanation.

I apologize for the confusion but this is exactly what I needed:

Two 3.7v 3000mAh Li-Ion cells in series obtain 7.4v 3000mAh and stepped up to 12v will run at 1875mAh. I hope I am correct now

 

Thanks guys for all your help, and so sorry I keep using the wrong terminology. I also forgot to multiply 1875 by 0.8 to obtain the 1500mAh.

gregfox, I am just looking for all the options. Either 7.4v battery pack stepped up to 12v or a 14.8v stepped down to 12v. Whatever is more efficient, takes less space and lasts longer.

Maybe I should have been more clear from the beginning. What I am trying to create is a pistol grip that would provide external power to the camera and serve as a LANC controller at the same time. This is the camera:

http://www.blackmagicdesign.com/products/blackmagicpocketcinemacamera

It uses rechargeable EN-EL20 7.4v Lithium Ion batteries. Even though the camera uses these 7.4v batteries, the external DC input is rated 12v-20v. And I believe the camera draws between 800mAh to 1000mAh, I have to confirm (I hope I am not saying something wrong again). That is why I am trying to reach 12v.

My knowledge in electronics is very basic as you can tell, I am more of an enthusiast designer. I am doing this for fun and to see if it would be possible. So the big challenge is to figure out if it would all fit inside of the grip.