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Old 27th May 2008, 11:09 PM   (permalink)
Default Voltage Conversion

I am currently working on building a battery for my Nikon D200 DSLR Camera so that it can basically be powered for very long periods of time to do extended (think months) time lapse photography.

I have consulted another photographer doing this in the Arctic so have my design all layed out except for one part. I need to attach the camera to a 12V power source. What this other photographer has done is simply solder two leads to a AA battery tray in the camera, run these leads into a voltage regulator and then the voltage regulator to the 12V battery.

What I need help doing is creating or finding a voltage regulator to conver the 12V car battery to power the camera instead of the 6 AA batteries that would normally be in the trade.

The other option is to run an inverter off of the battery and then into an AC adapter but there is a lot of conversion in there and isn't as efficient (I think).

I should be able to switch out 12V batteries on a regular basis for this application, or I may be attaching a solar panel to trickle charge during future projects.

Thanks for any help.

Best,
Drew
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Old 28th May 2008, 04:01 AM   (permalink)
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Two obvious choices are:

Use a linear regulator which is simple and has about 75% efficiency, which is what the other photographer did.

Use a switching regulator to convert the 12V to the required 9V which is more complex, but can have around 85-90% efficiency.

It depends upon whether the added battery life is worth the extra complexity.

Running a 110V inverter into an AC adapter is likely more inefficient then either of the above.
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Old 28th May 2008, 05:37 PM   (permalink)
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What current does it draw?

Checkout the black regulator, cheap, efficient and simple.
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Old 29th May 2008, 03:09 AM   (permalink)
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Crutschow,

What is the best way to go about obtaining/making one of these regulators? I should say that these things have to withstand some pretty nasty conditions though they will be housed in a waterproof case with the camera.

Hero999,

I am not exactly sure the current it draws. Looking through the manual it can be powered by the following...

One or two EN-EL3 Batteries or six LR6 alkaline, HR-6 NIMH, FR-6 lithium, or ZR-6 nickel-maganese AA batteries.

the EN-EL3 battery is a 7.4V/1500mAh battery.

Unfortunately, thats all I can find at the moment... I will have the camera in hand next week and should be able to learn more.

Thanks again for all the help.

Drew
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Old 29th May 2008, 06:07 AM   (permalink)
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Quote:
Originally Posted by Drew Fulton View Post
I am not exactly sure the current it draws. Looking through the manual it can be powered by the following...

One or two EN-EL3 Batteries or six LR6 alkaline, HR-6 NIMH, FR-6 lithium, or ZR-6 nickel-maganese AA batteries.

the EN-EL3 battery is a 7.4V/1500mAh battery.
Does it say how long it will run on any of those batteries?
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Old 29th May 2008, 07:23 PM   (permalink)
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I would suggest you measure the curren by connecting a multimeter in series with it. You need to know the current requirements in order to build a power supply for it.
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Old 30th May 2008, 01:19 PM   (permalink)
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Sounds good. I will have the camera in hand in the middle of next week and will see what I can learn. I'll check back then. Thanks for the help.

Drew
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Old 3rd June 2008, 10:22 PM   (permalink)
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I finally got the camera in hand and can provide some more information on the power.

On the bottom of the battery grip it reads 9V ---- 2.5A.

The battery itself reads 7.4V, 1500mAh. The camera can run on 1 or 2 batteries.

The tray I will be modifying holds 6 AA batteries has no information.

So how can I connect a 12 V battery to this camera without killing it. I should preface this that I have very little experience with electronics but I can follow directions well.

Thanks everybody.

Drew
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Old 4th June 2008, 12:44 AM   (permalink)
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Quote:
Originally Posted by Drew Fulton View Post
On the bottom of the battery grip it reads 9V ---- 2.5A.

The battery itself reads 7.4V, 1500mAh. The camera can run on 1 or 2 batteries.
I don't think the camera takes 2.5A except perhaps in short bursts. You really need to measure the current or else how long the camera runs on a set of batteries in your intended application. Otherwise it's mostly a guess as to how to build the supply.
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Old 4th June 2008, 12:53 AM   (permalink)
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What's the best way to go about measuring the current? I've got all day tomorrow to run the test.

Thanks,
Drew
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Old 4th June 2008, 08:12 AM   (permalink)
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Quote:
Originally Posted by Drew Fulton View Post
What's the best way to go about measuring the current? I've got all day tomorrow to run the test.

Thanks,
Drew

Hi Drew,
Here's the way I would do it, take the battery from the camera and put a red wire from the positive of the battery to the positive of the battery input on the camera. Connect a black wire from the battery to 1 lead of a dmm then put another wire from the other dmm lead to the negitive input on the camera. Have the dmm set to 10A current and take a pic without the flash then 1 with the flash. these 2 reading will give you the current used in taking a normal pic and 1 with using the flash.

I hope I'm right in the above explanation and if i'm wrong some other member will chime in and correct me I'm sure.

Cheers Bryan
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Old 4th June 2008, 04:45 PM   (permalink)
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Alright everybody, here we go. Camera is hooked up to a digital multimeter and here are the results...

When the camera is just sitting, nothing happens (calling it "sleep" for now), it reads at 0.00.
When the meter is on, it reads 0.30.
When the shutter fires, it spikes to somewhere between 1.5 and 2.0.

Typically, the camera will sit "sleeping" for a while, the meter is activated a second or two before the shutter fires, then the shutter fires, then the meter remains activated about 4 seconds before it returns to the "sleep" mode.

So, thats what I got. Any other measurements I need to take in order to get this going?

Thanks again for all the help.

Best,
Drew
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Old 4th June 2008, 06:01 PM   (permalink)
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Okay. From your results you could use a simple 3-terminal linear regulator to generate the voltage, but the normal bias current draw of typical regulators will drain your battery somewhat over the long period of time you want to operate. For example, the LM317 has a minimum bias draw of 5mA (through the voltage set resistors). This will use about 3.6 ampere-hours per month from the battery. Since a normal 12V car battery is about 50 ampere-hours that may be okay.

If this is acceptable to you than I suggest a simple LM317 regulator with resistors selected to provide the desired 9V. If you look at an LM117/317 data sheet it will show how to calculate the required resistor values for 9V and connect it up.

Since the camera instantaneous current draw is high, you will want a large capacitor on the regulator output to supply the transient. I suggest at least several hundred microfarads (electrolytic type).

If you think you need a lower bias-current regulator, than you may need to build one using a zener diode, a low power op amp and a power MOSFET pass transistor. That could keep the bias current well below a milliamp.

Last edited by crutschow; 4th June 2008 at 06:05 PM.
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Old 4th June 2008, 06:47 PM   (permalink)
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Crutschow,

I am not worried about the minor bias current as in any instance where I will be set up for very long periods of time (aka over a month) then I will either be able to rotate batteries or hook up a solar panel. However, for what I am doing right now, even the solar panel isn't needed.

So it sounds like a LM317 will work perfectly for me. Now here is where my ignorance comes into play. I don't even know how or where to start. Would you mind treating me like a total novice and pointing me in the right direction?

Thanks,
Drew
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Old 4th June 2008, 07:00 PM   (permalink)
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Here's the data sheet for the LM317. Take a look and see if you can make any sense of it especially the sample circuit.

I'm a little busy at the moment, but I can work on it a little later if you need additional help.
LM117 Adj. + Regulator.pdf
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