Digital Imaging with Charge-coupled devices (CCDs)

Capacitor

CapacitorQ = CV

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

• 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

Charged-coupled device (CCD)

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.

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.

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)

Charged-coupled device (CCD)

• The number of electrons released, and the voltage created across the pixel is proportional to the intensity of light.

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)

Charge-coupled device (CCD)

• The charges are the moved horizontally, where the voltage is amplified, measured, and and passed to a digital converter.

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.

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.

CCD uses - Telescopes

CCD uses – Digital Cameras

CCD uses – Endoscopes

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.

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.

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

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

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

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

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

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

Resolution

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

Finished!