Example: A 3.1x m 2 CCD pixel has a capacitance of 2.3 pF (x ) If after being exposed to light, it has picked up a charge of 3.45x C, what is the voltage across it? How many electrons were displaced from the pixel? If photons displaced these electrons, then what is the photon flux in photons/m 2 incident on the CCD? V 2.2E5 electrons 6.9E14 photons/m 2 V = q/C = (3.45E-14C )/(2.3E-12 F) =.015 V # electrons = (3.45E-14C )/(1.602E-19 C/electron) = 2.154E5 electrons photons/m 2 = (2.154E5 electrons)/(3.1E-10 m 2 ) = 6.95E14 photons/m 2 A sunny day = about 1000 W/m 2 ≈ 3x10 21 photons/sec (550 nm) (a force of 3.3x10 -6 N/m 2

Resolution, sensor size and magnification Nikon D3x: (CMOS) Sensor size = 35.9x24.0 mm = mm x4032 resolution = 24 Mega pixels (Let’s assume a quantum efficiency of 70%) A) What is the area of each pixel? 3.533E-11 m 2 /pixel

Resolution, sensor size and magnification Nikon D3x: (CMOS) Sensor size = 35.9x24.0 mm = mm x4032 resolution = 24 Mega pixels (Let’s assume a quantum efficiency of 70%) B) If light with a photon flux of 4.5x10 19 photons/m 2 /s is incident on the sensor for 1/250 th of a second, what is the resulting charge on the pixel? 3.533E-11 m 2 /pixel, 7.131E-13 C

Resolution, sensor size and magnification Nikon D3x: (CMOS) Sensor size = 35.9x24.0 mm = mm x4032 resolution = 24 Mega pixels (Let’s assume a quantum efficiency of 70%) C) If the voltage across the pixel is 2.4 mV, what is the capacitance of the pixel? 3.533E-11 m 2 /pixel, 7.131E-13 C, 2.972E-10 F,

Resolution, sensor size and magnification Nikon D3x: (CMOS) Sensor size = 35.9x24.0 mm = mm x4032 resolution = 24 Mega pixels (Let’s assume a quantum efficiency of 70%) D) If I am 183 cm tall, and my image on the sensor is 19.2 mm tall, what is the “magnification” in this case? 3.533E-11 m 2 /pixel, 7.131E-13 C, 2.972E-10 F, x

Nikon D3x: (CMOS) Sensor size = 35.9x24.0 mm = mm x4032 resolution = 24 Mega pixels (Let’s assume a quantum efficiency of 70%) A) What is the area of each pixel? total area = (35.9E-3m)x(24.0E-3) = m 2 #pixels = 6048x4032 = 24,385,536 pixels area/pixels = ( m 2 )/(24,385,536 pixels) = E-11 m 2 /pixel B) If light with a photon flux of 4.5x10 19 photons/m 2 /s is incident on the sensor for 1/250 th of a second, what is the resulting charge on the pixel? Photons per pixel resulting in charge = (0.70)(4.5x10 19 photons/m 2 /s)(1 s/250)( E-11 m 2 /pixel) = e charges per pixel So this is ( )(1.602E-19) = E-13 C C) If the voltage across the pixel is 2.4 mV, what is the capacitance of the pixel? C = q/V = ( E-13 C)/(2.4E-3 V) = E-10 F = 297 pF D) If I am 183 cm tall, and my image on the sensor is 19.2 mm tall, what is the “magnification” in this case? image/object = (1.92 cm)/(183 cm) =.0105x (it’s a smallifier)

My camera has a 25.4 mm x 58.4 mm CCD sensor with 4.0 Mega pixels. What is the area of each pixel in square meters? E-10 m 2 Area per pixel = (25.4E-3 m)(58.4E-3 m)/(4.0E6) = E-10 m 2

A pixel builds up a potential of 5.2 mV. How many photons hit it if it has a capacitance of 13 pF, and a quantum efficiency of 75%? 560,000 photons q = CV = (13E-12F)(5.2E-3) = 6.76E-14 C # electrons = 421,972 #electrons/#photons =.75, #photons = (421,972)/.75 = 562,630

A single pixel with an area of 3.7x m 2 is hit with 562,630 photons. What is the light intensity in photons/m 2 ? 1.5x10 15 photons/m 2 photons/m 2 = (562,630)/(3.7x m 2 ) = 1.5E+15