1. Example: A 3.1x10 -10 m 2 CCD pixel has a capacitance of 2.3 pF (x10 - 12 ) If after being exposed to light, it has picked up a charge of 3.45x10 -14 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? 0.015 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

2. Resolution, sensor size and magnification Nikon D3x: (CMOS) Sensor size = 35.9x24.0 mm = 861.6 mm 2 6048x4032 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

3. Resolution, sensor size and magnification Nikon D3x: (CMOS) Sensor size = 35.9x24.0 mm = 861.6 mm 2 6048x4032 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

4. Resolution, sensor size and magnification Nikon D3x: (CMOS) Sensor size = 35.9x24.0 mm = 861.6 mm 2 6048x4032 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,

5. Resolution, sensor size and magnification Nikon D3x: (CMOS) Sensor size = 35.9x24.0 mm = 861.6 mm 2 6048x4032 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, 0.0105x

6. Nikon D3x: (CMOS) Sensor size = 35.9x24.0 mm = 861.6 mm 2 6048x4032 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) = 0.0008616 m 2 #pixels = 6048x4032 = 24,385,536 pixels area/pixels = (0.0008616 m 2 )/(24,385,536 pixels) = 3.53324E-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)(3.53324E-11 m 2 /pixel) = 4451884.921 e charges per pixel So this is (4451884.921)(1.602E-19) =7.13192E-13 C C) If the voltage across the pixel is 2.4 mV, what is the capacitance of the pixel? C = q/V = (7.13192E-13 C)/(2.4E-3 V) = 2.97163E-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)

8. 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? 3.7084E-10 m 2 Area per pixel = (25.4E-3 m)(58.4E-3 m)/(4.0E6) = 3.7084E-10 m 2

9. 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

10. A single pixel with an area of 3.7x10 -10 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)/(3.7x10 -10 m 2 ) = 1.5E+15