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Battery pack with indicator

gregfox

Mar 25, 2013
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Fun is a good thing!
Here’s the rub Just because the input is 12V-20V doesn’t mean the camera is seeing that voltage. Most likely the input is from a “brick” that puts out a nominal voltage of under 20VDc, and a on-board regulator drops it down to ~7V or so. It would be my guess that 2 cells (7.4V fully charged 8.4V) would do the trick. I don’t see any need to drop or boost the voltage, but without a schematic, what do I know.
So all the is remaining for you to do is find the proper batteries, and add the low voltage circuit (less the regulator) and your set. You could order the parts from Digikey.com.
Hope that helps.
Buy the way did you mean to say I believe the camera draws between 800mA to 1000mA, and not 800mAh to 1000mAh?
Also, try re-posting your camera as the link seems to be broken.
 

CoDi

Dec 30, 2013
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Thanks gregfox,

Maybe I should contact the manufacturer and ask for more detailed specs.

So sorry about the pictures, I think I figured it out:

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Buy the way did you mean to say I believe the camera draws between 800mA to 1000mA, and not 800mAh to 1000mAh?

Sorry, half the time I don not know if I am using the right terms :confused:
 
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gregfox

Mar 25, 2013
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Cool!
It's always good to contact the manufacturer. Sometimes they are tight in giving out info, but try.
 

CoDi

Dec 30, 2013
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They got back to me and provided the LANC protocol code and told me that the nominal input voltage is 12V, that it might work slightly below that but they can’t guarantee it. They didn't say anything about an internal regulator, but I imagine there might be one stepping it down to ~7v.

Also, the maximum input voltage limit is +18V. When using an external power source, it is very important to make sure you do not exceed the 18V limit or you will risk damaging the camera.

And finally, the typical power consumption is approximately between 6.8W and 7W.

I will keep my research on cells. At least now I have more information to move on :)
 

gregfox

Mar 25, 2013
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Good deal, It seems like the 4 cells with a step down is a good way to go, if you have the room. Keep looking for a proper sized battery pack, or you could make one yourself,

Ohms law for current is I = P/E : I = 7/12 : I = .583 AMP. So if you need 2 hours of operation you would think you need a a battery .583 X 2 hours = 1.166 mAh but wait ....
The way the battery capacity may be calculated differs with manufacturers.
A Li-ion battery (more likely a Li-Polymer battery) will have a full charge at 4.2V and a nominal voltage of 3.7. most manufactures will give an end of life of 2.75V, and if you go much below that voltage, you risk damaging or destroying the battery. I prefer to remain above 3V. So if the manufacturer is claiming lets say 1000 mAh of the battery, they may be claiming it at some lower voltage, therefor I would give myself a cushion and go with a larger capacity battery, say by 20% or so to be safe. Of course the best way is to test the battery under a constant current load, but that's another story.
Oh yes, if the battery has a protective circuit it will self protect itself from over or under voltages.
 
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CoDi

Dec 30, 2013
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Thank you so much for the information gregfox! I am learning a lot!

But I have the feeling that internally the voltage is 7v. Because for some reason the included batteries which are 7.4v 1000mAh only last one hour if you are lucky.

Doing the math again, 7watts/7volts= 1AMP, that would give one hour of operation, I hope I am not saying anything wrong.

The external DC is rated at 12v to make it more compatible with professional external batteries, but it might be stepped down internally to 7v, I am guessing...

Good to know about the life of the batteries! thank again for everything!
 

gregfox

Mar 25, 2013
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I would say your right, and as I said in post #22 The 12V is being dropped down to ~7V. Are you going ahead with this and put together a low battery indicator? Don't give up, it's how you learn.
 

CoDi

Dec 30, 2013
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I really want to move forward but unless I can find a battery pack that's at least 3000mAh it's not worth it. I might have to rework the design to fit bigger cells or think of something else.

I am really curious anyways on figuring out the battery indicator. I found a lot already made online and I suppose I could modify one to fit it in my design. But, how do they work? You have to calibrate the circuit for the current draw? How does it calculate the mAh?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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I am really curious anyways on figuring out the battery indicator. I found a lot already made online and I suppose I could modify one to fit it in my design. But, how do they work?

There's an easy way and a hard way.

The easy way is to measure the voltage on the battery.

The hard way is to monitor the current going in and out of the battery and use this to determine how much charge has gone in and how much has come out. The difference is the charge remaining (it's actually even harder than that)

You have to calibrate the circuit for the current draw?

Not sure what you're asking.

How does it calculate the mAh?

Essentially this is the integral of the discharge current over time. It is calculated by integrating the discharge current over time...

It can be estimated by using the charge currents as a predictor of charge available.

Or if you're asking the simpler question, the mAh rating of a battery is the sum of all cells in parallel. Cells in series do not change it.

So if you have 6 cells each with a claimed capacity of 2000mAh and they were arranged as 2 sets of three cells in parallel, with these sets in series, the capacity of the battery is 3 * 2000 = 6000mAh.

In practice the actual capacity may not be what is claimed, and the actual capacity may vary depending on discharge rate, and temperature
 

CoDi

Dec 30, 2013
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Thanks steve.

The easy way is to measure the voltage on the battery.

I figure measuring the voltage doesn't really give you an accurate prediction of how much longer do you have left of battery life since lithium cells drop down their voltage very quickly at the end of their life, right?

Essentially this is the integral of the discharge current over time. It is calculated by integrating the discharge current over time...

It can be estimated by using the charge currents as a predictor of charge available.

It really sounds very complicated, I am lost already :(

If I have 3x 3.7v 3000mAh cells in series to obtain 11.1v 3000mAh and stepped up to 12v, the battery indicator circuit would look at each cell independently? or does it look at the global current? Would stepping up from 11.1v to 12v influence at all?

When I mentioned earlier to calibrate the circuit, I meant to say to calibrate the circuit on the battery indicator to tell it what voltage would be considered low on the cells, if 3v or 2.75v...
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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I figure measuring the voltage doesn't really give you an accurate prediction of how much longer do you have left of battery life since lithium cells drop down their voltage very quickly at the end of their life, right?

Not quite.

Lithium batteries have a voltage (off load) that quite accurately reflects the state of charge. True, it's not linear, but you know that when a cell reached (say) 2.8V that it has (say) 10% of its capacity left. What its capacity is is another question.

If you google you can find graphs of voltage vs SOC, often at various discharge rates (which means you may be able to get away with measuring the voltage under load.
 

GeoNOregon

Jan 30, 2014
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I'm lost in most of the electronics discussion, but I am up to date on batteries. Earlier, someone mentioned LiPoly batteries and failed to point out they are 3.2v rather than 3.7, the voltage of Li-Ion batteries.

I'm a bit confused why you're talking about stepping up/down the voltage. The camera uses a battery pack of 7.4v, which means it has two Li-Ion batteries of 3.7 each.

You stated, "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".

It seems the simplest thing, to me, would be to build a new battery pack using higher capacity batteries than the 1000mAh OEM batteries.

My go to source for batteries is Tenergy, (all-battery.com is Tenergy's direct sales site), I've been using their batteries for a number of years and while their tech support is hit & miss, their batteries are good quality, (and great prices).

They have a 4200mAh 26650 Li-Ion with buttons or tabs that should give you some good capacity.

If I was thinking about your project, I'd build a replacement battery pack to replace the OEM 1000mAh battery pack and then figure out how to tie in, (in parallel), an additional 4200mAh battery pack in your pistol grip handle to give you a total capacity of 8400mAh.

Just my suggestion...

Good luck.
 
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