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Voltage Conversion

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Hi Drew, it's email Tom.

The '2k trimpot' is a variable resistor that can go from about 0 Ohms up to 2000 Ohms. The 'trim' means that it's usually used to calibrate your circuit once and then to be left alone (and resides on the circuit board, thereafter unseen by the user.) The 'pot' is short for potentiometer, which is just a type of variable resistor you turn i.e a knob. For example, the volume control on a radio is just a potentiometer. Almost all knobs you turn are potentiometers. Do a google image search for the various things mentioned and you'll get the idea.

As for the LM317, you already know that the output voltage is determined by the resistors used.
The LM317 is ubiquitous, but you can get regulators that give a predetermined output and thus need no external components. For example the popular regulators LM7805, LM7809, LM7812 etc. will output 5, 9, 12 volts respectively. You see the naming convention.

Look at this online LM317 calculator,

**broken link removed**

If you want to be spot on then simply use your 220Ohm resistor as the first resistor and the trimpot as suggested (small variable resistor) instead of a fixed value second resistor. Using your voltmeter, turn the trimpot until you get your desired output voltage.

Also the camera probably has its own voltage regulation, so you don't have to be spot on. Batteries aren't.

I was going to do something similar (though not month-long) time-lapse. Never followed up on it, but just realised my Canon TC-80N3 Timer remote controller already does never-ending intervalometry. Are you going to use Nikon's equivalent? Anyway, regarding the power, I think you're wise to not go via the AC inverter route.
 
Tom,

Thanks very much. So now I have an idea of what I need. My last question is which resisitors, 1 A, .5A, .25A, etc? Basically, which drawer do I get them out of?

Thanks.
Drew
 
Are those the max. current ratings of the available resistors, or did you mean 1W,0.,5W,0.25W (Watts), in which case they're the power dissipation ratings?

The resistors for the LM317 are just for setting the desired voltage and don't pass a lot of current, so you're safe with the the common 0.25W types.

For resistors, the power dissipated is

P = current*current*resistance.

Current in amps, and resistance in Ohms.

The LM317 will be the thing dissipating the most power, which is why the datasheet should be consulted: google 'LM317 datasheet'. The section 'Absolute maximum values' is of initial interest. The power dissipation is

P=current*voltage drop.

There's nothing to worry about in your case, this is largely academic. The high power spikes only occur very briefly when a photo is taken, otherwise the current drawn is tiny.
 
Hey everybody,

I have now fully assembled my camera with the voltage converter in the middle. Unfortunately, the camera will not fire off multiple images in a burst. After one, sometimes 2 frames, the camera turns off. The power supply doesn't quite hold up enough. I tried increasing the voltage a bit but that didn't work. I am wondering what I need to do to solve this. I am guessing it needs a larger capacitor but would like some more input...

Thanks!

Drew
 
Is the Lm317 definitely putting out the correct voltage? What is it meant to be anyway, 6 x 1.5 = 9volts?

Camera's operating normally with batteries? Everything wired up correctly? the battery compartment of my camera looks a little complicated.

Perhaps the camera really does instantaneously draw more current than the regulator can provide, which is at least 1.5Amps. Maybe the solution would be a regulator with a higher max current rating. To be honest I would be surprised if it really needs much more than 2Amps.

Don't know about a capacitor, regular AA batteries work without anything fancy. You're using a 12volt battery which provides a smooth voltage level.
 
Everything looks good to me and works fine with just a battery. The bottom of the battery pack says 9.5 V and 2.5A. When I was testing, I didn't try to fire any bursts but I never saw anything near 2.5A. It barely got to 2A. I am now wondering if its the frame bursts that gets the currents up to 2.5A.

The way I powered the camera is by soldering wires into the tray that normally holds the AAs. Therefore I don't really have to deal with which contact is which in the compartment.

The trimpot is set perfectly. When I hook everything up with a multimeter in place of the camera, the voltage is set perfectly at 9.5V. I hook the camera in and I can fire frames singly but when I try to fire multiple frames the camera shuts off like you had pulled the battery. Sometimes the shutter even stays open and the camera produces an error.

So do I need to look for a higher rated regulator? Any suggestions?
 
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After one, sometimes 2 frames, the camera turns off. The power supply doesn't quite hold up enough. I tried increasing the voltage a bit but that didn't work. I am wondering what I need to do to solve this. I am guessing it needs a larger capacitor but would like some more input...
Sounds like it's drawing more current than the regulator can supply. You can try increasing that regualtor output capacitance.

How many frames do you need to fire in a row? Since it works for a least one frame, I would multiply the present output capacitance by the number of frames you want, to get the capacitance you need.

There's no particular limit on how large the capacitance can be.
 
One frame is minimum. Ideally I could pull in 5 image bursts with a period in between. For example, 5 frames are fired in order to compensate for high dynamic range (5 frames with slightly different exposures and then later combined), then we wait at least 10 seconds and take another burst. I'll try the capacitor solution and check out the link as well.

Thanks guys.

Drew
 
Alright, I have now installed a larger capacitor, going from a 470uF to a 2200uF. Looks like I can get a 3 frame bracketing burst quite easily which is all the camera can do automatically.

I am quite pleased with the result and would like to thank everyone for all their help. I saved a ton of money doing this on my own rather than purchasing something and I really couldn't have done it without the step by step instructions.

This is a great forum and the willingness to share solutions is refreshing.

Thanks again,
Best,
Drew
 
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