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Digital Imaging with Charge-coupled devices (CCDs)
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Capacitor
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CapacitorQ = CV
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Example
• The capacitance of two parallel plates is 4.5pF. Calculate the charge on one plate when a voltage of 8.0 V is applied to the plates.
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Example
• The capacitance of two parallel plates is 4.5pF. Calculate the charge on one plate when a voltage of 8.0 V is applied to the plates.
Q = CV = 4.5 x 10-12 x 8.0 = 3.6 x 10-11 C
= 36 pC
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Charged-coupled device (CCD)
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Charged-coupled device (CCD)
• Silicon chip varying from 20 mm x 20 mm to 60 mm x 60 mm.
• Surface covered in pixels (picture elements) varying from 5 x 10-6 m to 25 x 10-6 m.
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Charged-coupled device (CCD)
• Each pixel releases electrons (by the photoelectric effect) when light is incident on it.
• We may think of each pixel like a small capacitor.
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Charged-coupled device (CCD)
• The electrons released in each pixel constitute a certain amount of charge Q, and so a potential difference V develops at the ends of the pixel (V = Q/C)
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Charged-coupled device (CCD)
• The number of electrons released, and the voltage created across the pixel is proportional to the intensity of light.
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Charge-coupled device (CCD)
• The charges on each row of pixels is pushed down to the next row until they reach the bottom row (the register)
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Charge-coupled device (CCD)
• The charges are the moved horizontally, where the voltage is amplified, measured, and and passed to a digital converter.
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Charge-coupled device (CCD)
• The computer processing this information now knows the position and voltage on each pixel.
• The light intensity is proportional to the voltage so a digital (black and white image) is now stored.
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Charge-coupled device (CCD)
• To store a colored image, the pixels are arranged in groups of four, with 2 green filters, a red and a blue.
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CCD uses - Telescopes
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CCD uses – Digital Cameras
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CCD uses – Endoscopes
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Quantum efficiency
• The ratio of the number of emitted electrons to the number of incident photons.
• About 70% for a CCD (4% for photographic film and 1% for the human eye.
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Example
• The area of a pixel on a CCD is 8.0 x 10-10 m2 and its capacitance is 38 pF. Light of intensity 2.1 x 10-3 W.m-2 and wavelength 4.8 x 10-7 m is incident on the collecting area of the CCD for 120 ms. Calculate the p.d. across the pixel, assuming that 70% of the incident photons cause the emission of an electron.
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Example
• The area of a pixel on a CCD is 8.0 x 10-10 m2 and its capacitance is 38 pF. Light of intensity 2.1 x 10-3 W.m-2 and wavelength 4.8 x 10-7 m is incident on the collecting area of the CCD for 120 ms. Calculate the p.d. across the pixel, assuming that 70% of the incident photons cause the emission of an electron.
The energy incident on a pixel
= 2.1 x 10-3 x 8.0 x 10-10 x 120 x 10-3 = 2.0 x 10-13 J
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Example
• The area of a pixel on a CCD is 8.0 x 10-10 m2 and its capacitance is 38 pF. Light of intensity 2.1 x 10-3 W.m-2 and wavelength 4.8 x 10-7 m is incident on the collecting area of the CCD for 120 ms. Calculate the p.d. across the pixel, assuming that 70% of the incident photons cause the emission of an electron.
The energy incident on a pixel
= 2.1 x 10-3 x 8.0 x 10-10 x 120 x 10-3 = 2.0 x 10-13 J
The energy of one photon
= hf = hc/λ = (6.63 x 10-34 x 3.0 x 108)/4.8 x 10-7
= 4.1 x 10-19 J
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Example
• The area of a pixel on a CCD is 8.0 x 10-10 m2 and its capacitance is 38 pF. Light of intensity 2.1 x 10-3 W.m-2 and wavelength 4.8 x 10-7 m is incident on the collecting area of the CCD for 120 ms. Calculate the p.d. across the pixel, assuming that 70% of the incident photons cause the emission of an electron.
The energy incident on a pixel = 2.0 x 10-13 J
The energy of one photon = 4.1 x 10-19 J
The number of incident photons is then
= 2.0 x 10-13 / 4.1 x 10-19 = 4.9 x 105
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Example
• The area of a pixel on a CCD is 8.0 x 10-10 m2 and its capacitance is 38 pF. Light of intensity 2.1 x 10-3 W.m-2 and wavelength 4.8 x 10-7 m is incident on the collecting area of the CCD for 120 ms. Calculate the p.d. across the pixel, assuming that 70% of the incident photons cause the emission of an electron.
The number of incident photons is then
= 2.0 x 10-13 / 4.1 x 10-19 = 4.9 x 105
The number of absorbed photons is therefore
= 0.70 x 4.9 x 105 = 3.4 x 105
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Example
• The area of a pixel on a CCD is 8.0 x 10-10 m2 and its capacitance is 38 pF. Light of intensity 2.1 x 10-3 W.m-2 and wavelength 4.8 x 10-7 m is incident on the collecting area of the CCD for 120 ms. Calculate the p.d. across the pixel, assuming that 70% of the incident photons cause the emission of an electron.
The number of absorbed photons is therefore
= 0.70 x 4.9 x 105 = 3.4 x 105
The charge corresponding to this number of electrons is
3.4 x 105 x 1.6 x 10-19 = 5.4 x 10-14 C
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Example
• The area of a pixel on a CCD is 8.0 x 10-10 m2 and its capacitance is 38 pF. Light of intensity 2.1 x 10-3 W.m-2 and wavelength 4.8 x 10-7 m is incident on the collecting area of the CCD for 120 ms. Calculate the p.d. across the pixel, assuming that 70% of the incident photons cause the emission of an electron.
The charge corresponding to this number of electrons is
3.4 x 105 x 1.6 x 10-19 = 5.4 x 10-14 C
The p.d. is thus V = Q/C = 5.4 x 10-14/38 x 10-12 = 1.4 mV
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Resolution
• Two points are resolved if their images are more than two pixel lengths apart.
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Finished!