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FORUMS Cameras, Lenses & Accessories Canon Digital Cameras 
Thread started 13 Feb 2008 (Wednesday) 17:57
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-=1D Classic Lovers !!!

 
PM01
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Jan 10, 2009 01:29 as a reply to  @ post 7044173 |  #976

That should do the trick!




  
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Vascilli
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Jan 10, 2009 01:31 |  #977

Well then the experiments will begin, once I acquire my 1D (16th birthday present from myself) and a few dead batteries.


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PM01
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Jan 10, 2009 01:36 as a reply to  @ Vascilli's post |  #978

Excellent! Please do keep up posted!




  
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Vascilli
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Jan 10, 2009 01:37 |  #979

It may not be for a while. I should have the 1D in a month though.


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Vascilli
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Jan 10, 2009 02:36 |  #980

Okay I did some messed up math, I think I may have gotten something wrong, but if I used a 1750mAh LiPoly pack with 14.8V, and the 1D takes 12V, then you have a 486.111.. Amp battery or something like that, which would calculate to about a 175 Ohm resistor needed. Sounds right or am I totally wrong?


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Headcase650
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Jan 11, 2009 10:04 |  #981

Lipoly batteries can be very dangerous and I wouldn't experiment with them. I use to fly RC plains and if you punctured one on impact they would explode in a big way, or charged them wrong or backwards they would explode into a fire ball. These batteries are made to carry a lot of power in a lightweight package and dump it rather quickly, perfect for RC flight, They even need to be charged on a flat hard concrete surface so that if they do burn up nothing gets damaged. The bad thing is that once you short one out, even if by accident it stats the chemical reaction and its gonna explode. There is no stopping it. Lets say you drop your 1D and crack or puncture the lipo battery, battery bursts into flames, magnesium camera catches fire. Its just a bad idea all round.


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Headcase650
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Jan 11, 2009 10:16 as a reply to  @ Vascilli's post |  #982

Check out these fire balls.
http://www.youtube.com​/watch?v=4OsBc8RqSKU (external link)
http://www.youtube.com​/watch?v=SQ0SNESIkWk&N​R=1 (external link)
http://www.youtube.com​/watch?v=GbhXf_jhwgE (external link)


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TopGear1Ds
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Jan 11, 2009 11:37 |  #983

Headcase650 wrote in post #7052083 (external link)
Check out these fire balls.

:shock:.. Holy crap!


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canonloader
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Jan 11, 2009 11:58 |  #984

Bet they don't want you on an airplane with that in your pocket. ;)


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René ­ Damkot
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Jan 12, 2009 13:27 |  #985

Vascilli wrote in post #7044384 (external link)
Okay I did some messed up math, I think I may have gotten something wrong, but if I used a 1750mAh LiPoly pack with 14.8V, and the 1D takes 12V, then you have a 486.111.. Amp battery or something like that, which would calculate to about a 175 Ohm resistor needed. Sounds right or am I totally wrong?

Totally wrong.

the 1750mAh figure is a measure of capacity: The battery can deliver 1 mA for 1750 hours, or 1.75A for an hour, or ... you get the point.

A resistor will give a voltage drop depending on the current running through it (U = R * I). Not what you'd want in this case: You want a current independant voltage drop.

I think a simple diode (external link) would be a better choice: 0.7V drop each. So 4 diodes will give a 2.8V drop.
Totally unsure about how much current the camera draws at a maximum, thus also about what diodes to use. Maybe a 1N4001 is fine, maybe not.

Also, obviously, the warning about the Lithium batteries is a good one...


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PM01
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Jan 12, 2009 14:11 as a reply to  @ René Damkot's post |  #986

7812 voltage regulator.

http://www.geocities.c​om …nvalley/2072/3p​involt.htm (external link)

Nice, simple and quick.




  
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PM01
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Jan 12, 2009 14:13 as a reply to  @ PM01's post |  #987

Also available here.

http://www.radioshack.​com …dex.jsp?product​Id=2062600 (external link)




  
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René ­ Damkot
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Jan 12, 2009 14:49 |  #988

Only if the camera doesn't draw too much current!

Otherwise you'd need to use a heatsink.


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Vascilli
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Jan 12, 2009 18:44 |  #989

René Damkot wrote in post #7060364 (external link)
Totally wrong.

the 1750mAh figure is a measure of capacity: The battery can deliver 1 mA for 1750 hours, or 1.75A for an hour, or ... you get the point.

A resistor will give a voltage drop depending on the current running through it (U = R * I). Not what you'd want in this case: You want a current independant voltage drop.

I think a simple diode (external link) would be a better choice: 0.7V drop each. So 4 diodes will give a 2.8V drop.
Totally unsure about how much current the camera draws at a maximum, thus also about what diodes to use. Maybe a 1N4001 is fine, maybe not.

Also, obviously, the warning about the Lithium batteries is a good one...

I knew I did something wrong. I got totally different results every single time I did it. :confused:

And given the risks of LiPoly, Li-ion it is. (Or maybe just a bunch of Eneloops)


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mrerico
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Jan 13, 2009 01:58 |  #990

I love how they laugh in the second video :)




  
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