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FORUMS Canon Cameras, Lenses & Accessories Canon EOS Digital Cameras 
Thread started 31 Oct 2008 (Friday) 03:03
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40D at ISO 3200 - why is it not on by default

 
David ­ Ransley
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Oct 31, 2008 03:03 |  #1

Any ideas why canon leaves the ISO 3200 feature off and hidden away on the 40D? Does it generate to much heat or reduce battery lifetime if left on and not used? Or what?

Shot taken: F4, 1/125. ISO 3200, Noise reduced with Canon Digital Photo Professional 3.4.1.1.


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Juan ­ Zas
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Oct 31, 2008 06:19 |  #2

Because is not a native ISO, it´s just an electronic game ... but you can select its activation to be always on by Cf-n ...


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apersson850
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Oct 31, 2008 06:23 as a reply to  @ Juan Zas's post |  #3

If you didn't understand that explanation, then I'll expand it a bit. This is still a bit simplified, but anyway.

The ISO levels 100-1600 are achieved by changing the amplification when reading the data from the CMOS sensor.
When you use expansion modes, in this case H, which is equivalent to ISO 3200, the camera actually exposes according to ISO 1600, then just multiply by two. This reduces the range from the sensor, effectively making the 14-bit sensor in the 40D a 13-bit instead.
So there's a quality loss from a different reason, and that's why Canon makes you think once more, before you can use it.


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Oct 31, 2008 08:09 as a reply to  @ apersson850's post |  #4

ahhh makes perfect sense now




  
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gjl711
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Oct 31, 2008 08:30 |  #5

apersson850 wrote in post #6596653 (external link)
...
When you use expansion modes, in this case H, which is equivalent to ISO 3200, the camera actually exposes according to ISO 1600, then just multiply by two. This reduces the range from the sensor, effectively making the 14-bit sensor in the 40D a 13-bit instead.....

Hmm.. this is not how I understand the expanded modes. M y understanding has always been that the H mode is basically dialing in an exposure compensation value of -1 then when the raw file is processed the exposure is pushed by +1 thus you gain a full stop in speed. You can accomplish the same thing by shooting EC-1 and in DPP add exposure +1 and will get the exact same pic.


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Oct 31, 2008 11:26 |  #6

gjl711 wrote in post #6597180 (external link)
Hmm.. this is not how I understand the expanded modes. M y understanding has always been that the H mode is basically dialing in an exposure compensation value of -1 then when the raw file is processed the exposure is pushed by +1 thus you gain a full stop in speed. You can accomplish the same thing by shooting EC-1 and in DPP add exposure +1 and will get the exact same pic.

Originally Posted by apersson850

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...
When you use expansion modes, in this case H, which is equivalent to ISO 3200, the camera actually exposes according to ISO 1600, then just multiply by two. This reduces the range from the sensor, effectively making the 14-bit sensor in the 40D a 13-bit instead.....



That's the same thing as what Anders said. An exposure compensation of -1 is the same as exposing at ISO 1600 and pushing +1 stop is the same as mutiplying by 2.

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gjl711
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Oct 31, 2008 11:34 |  #7

So why would you loose 1/2 of the data? Something is not making sense.


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number ­ six
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Oct 31, 2008 13:36 |  #8

gjl711 wrote in post #6598166 (external link)
So why would you loose 1/2 of the data? Something is not making sense.

You don't. You lose 1/14 of the dynamic range. A shift of one bit is the same as multiplying ISO by 2.

-js


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David ­ Ransley
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Oct 31, 2008 14:05 |  #9

Thanks, it clarifies the concept and I will keep this in mind when moving into the expanded range. That may also be the reason why Canon shows H in the LCD and not 3200 anymore. It is not real 3200 :-) The 50D obviously uses the same method to go from 3200 to its expanded mode.


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gjl711
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Oct 31, 2008 17:33 |  #10

number six wrote in post #6598953 (external link)
You don't. You lose 1/14 of the dynamic range. A shift of one bit is the same as multiplying ISO by 2.

-js

But loosing 1 bit does not loose 1/14, it looses 1/2 What you guys are saying is that you go from 2 to the 14th to 2 to the 13th or 1/2 of the data. Isn't the data just remapped to a different range?


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Oct 31, 2008 17:59 |  #11

gjl711 wrote in post #6600230 (external link)
But loosing 1 bit does not loose 1/14, it looses 1/2 What you guys are saying is that you go from 2 to the 14th to 2 to the 13th or 1/2 of the data. Isn't the data just remapped to a different range?

Arithmetically it is 1/2 of the data...2^13 vs 2^14...but photographically it is taking 14 levels of range to 13 levels...the top and bottom are both the same, but you lose an intermdiate value, and that level is mapped (rounded) up or down to the next range instead. You're both right!


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40D at ISO 3200 - why is it not on by default
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