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Digital Imaging with Charge-coupled devices (CCDs)

Digital Imaging with Charge-coupled devices (CCDs). Capacitor. Q = CV. Capacitor. 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. Example.

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Digital Imaging with Charge-coupled devices (CCDs)

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  1. Digital Imaging with Charge-coupled devices (CCDs)

  2. Capacitor

  3. Q = CV Capacitor

  4. 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.

  5. 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

  6. Charged-coupled device (CCD)

  7. 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.

  8. 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.

  9. 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)

  10. Charged-coupled device (CCD) • The number of electrons released, and the voltage created across the pixel is proportional to the intensity of light.

  11. 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)

  12. Charge-coupled device (CCD) • The charges are the moved horizontally, where the voltage is amplified, measured, and and passed to a digital converter.

  13. 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.

  14. 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.

  15. CCD uses - Telescopes

  16. CCD uses – Digital Cameras

  17. CCD uses – Endoscopes

  18. 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.

  19. 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.

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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

  26. Resolution • Two points are resolved if their images are more than two pixel lengths apart.

  27. Finished!

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