I’m curious about when digital gain comes into play. Initially, I thought that if I had an object with an intensity of 10 in a pixel with values ranging from 0 to 255 (uint8), increasing the digital gain to 5 would make the pixel value 50. However, in reality, values smaller than 50 are produced. I’m wondering how digital gain works and at which stage it operates.
Furthermore, I need to perform imaging in very low-light conditions. Since I need to capture video at 14fps, I can’t increase exposure further. Pixel intensity ranges from 2 to 3 in uint8 format and flickers slightly. The analog gain is at its maximum. I believe uint8 format is insufficient to capture the flickering of this signal. Therefore, I would like to obtain raw values of 10 bits. Is there a way to achieve this? My sensor is IMX708.
Hello JerryChang.
Thank you for leaving a response.
The intensity of light reflected from the target itself varies slightly due to changes in ambient lighting. I need to capture changes in light over time with a camera, but it’s challenging to capture subtle changes because the sensitivity is too low. Even if the brightness of a real object changes from 2 to 2.1 (for example), uint8 won’t capture this change, and it will still be indicated as 2. This is why I want to acquire the original signal in 10 bits, not just 8 bits.
I’m also curious about when the ISP digital gain is applied. Is it multiplied onto the signal before it’s converted to uint8, or is it applied after? If it’s applied earlier, I hope that by representing the lost details of the signal in 8 bits, even if only slightly, it might enhance the overall fidelity of the image.