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Metering and exposure compensation on the D800...

km6xz

St Petersburg, RU
3559 posts

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"RE: Metering and exposure compensation on the D800..."

km6xz Moderator Awarded for his in-depth knowledge in various areas, including Portraits and Urban Photography Nikonian since 22nd Jan 2009
Sat 15-Dec-12 08:30 PM

I think a wrong track is being followed.
Gain does not require higher voltage or require more current to be drawn. Gain comes in two stage types, analog and digital.
The analog amplification occurs right after a pixel is scanned when the accumulated photon count is measured and sent to an analog amplifier the conditions the level and provides isolation(buffer stage) for the optimum dynamic range of the ADC, analog to digital converter.
This takes a snapshot of the analog voltage representing the charge state of the pixel and creates a numerical binary word that represents the analog voltage. The pixel has a dynamic range of the difference in voltage in the pixel with no light, comprising mostly of noise of several types that sets the lowest voltage that a pixel can register as black. The maximum voltage level is the number of photons that be accumulated in the pixel between reads is called "full-Well" value. Any additional photon hitting that pixel will not be registered.
The difference between those two values is the dynamic range of the sensor itself.
Pixels are arranged in rows that are read out by sequentially reading and clearing the values in each pixel in a row when addressed by selecting a row number and a column that uniquely addresses a pixel. Each column has a dedicated ADC that creates the number in binary format representing the voltage level that in an analog of the number of photons detected in the time between the last read of that pixel. All column are converted at once since there are converters attached to each column.
The analog signal needs to be preconditioned before being sent to the analog to digital conversion stage, because the ADC has it own DR and minimum value below which is ambiguous noise. After a number is generated, the signal from then on is in the digital domain and digital gain can be applied by arithmetically multiplying the recorded value by a gain. Any noise present at this point is multiplied as well. There are a number of sources of noise, both analog(amplifier noise, conversion noise, shot noise, 1/f noise). Changing ISO changes a combination of the analog preconditioning and primarily the digital multiplication.
All this is to lay the groundwork for explaining where current drain occurs. The reading of a pixel is a very low current process, due to the very high input impedance of the input amplifier, current varies very little between the lowest idle current of the amplifier and highest voltage output. Minimum idle current is sought because any heat generated increases noise, but not enough idle current and distortion increases in the amplifier. That analog stage can also act as an attenuator so get the low extended ISO values under the natural ISO for the D800 of 100.
To reduce losses, noise and complexity, all the ADC and first amplifiers are etched right on the sensor silicon wafer.

Reading a sensor does not increase or decrease current based on light or well level. The only significant current is the switching losses when addressing each pixel. That is constant regardless of the signal level or gain, and is primarily determined by the clock rate of the addressing. Low pixel count sensors have fewer reads per second so switching losses are less, sensors and related electronics would stay cooler. Same with computer memory, run at low clock speeds, chips run cool, and heat increases as clock speed increases regardless of what data is being stored. The rate a sensor is scanned determines the current draw per pixel and more pixels mean more heat(current)independent of light level or gain. The major current draws are fast addressing, the DSP (digital signal processor, Expeed3 in this case, is a very fast throughput device so generates a lot of heat through switching losses. Digital circuits pull almost no current in a steady state but switching from a logic 1 to 0 or 0 to 1 takes a finite time short period where the state if changing in voltage, ramping up or down, in nanoseconds where losses occur due to capacitance. The faster it runs the more current. What the data is, makes no difference. It is a multicore processor so as more cores are activated for some computation intensive tasks like NR, current and heat can spike upwards but the image data makes no difference, just what processes are being applied to it.

So if battery drain changes with photo content or ISO, it is not due to the amplification or reading the sensor, would be from the additional processing that some images might require after the images has already been recorded.

The big current draws are backlights, AF motor current, VR etc and the DSP, assist light, on-board flash and mirror motors.
Stan
St Petersburg Russia

Visit my Nikonians gallery.

This is a hot, active topic! Metering and exposure compensation on the D800... [View all] , walk43 , Mon 10-Dec-12 03:31 PM
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