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

2. Capacitor

3. Q = CV

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. 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 m 2 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 m 2 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 m 2 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 10 8 )/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 m 2 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 10 5

23. Example The area of a pixel on a CCD is 8.0 x 10 -10 m 2 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 10 5 The number of absorbed photons is therefore = 0.70 x 4.9 x 10 5 = 3.4 x 10 5

24. Example The area of a pixel on a CCD is 8.0 x 10 -10 m 2 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 10 5 = 3.4 x 10 5 The charge corresponding to this number of electrons is 3.4 x 10 5 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 m 2 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 10 5 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!