2. 14.2.1 Define capacitance. 14.2.2 Describe the structure of a charge-coupled device (CCD). 14.2.3 Explain how incident light causes charge to build up within a pixel. 14.2.4 Outline how the image on a CCD is digitized. Topic 14: Digital technology 14.2 Data capture and digital imaging

3. Define capacitance.  A capacitor is an electronic component that stores charge (and thus, energy in an electric field).  In its simplest form, a capacitor consists of two metal plates separated by a non-conductor called a dielectric. The dielectric prevents charge from conducting from one plate to another, so that the capacitor can maintain its charge.  If the positive and negative terminals of a battery are connected to the capacitor, one plate will take on a positive charge q, and the other will take on a negative charge -q, and the two charges will be the same in magnitude. Topic 14: Digital technology 14.2 Data capture and digital imaging p l a t e d i e l e c t r i c plate + - Plate Dielectric +q+q -q-q capacitor Oreo

4. Define capacitance.  Essentially, the emf of the battery “sucks” electrons off of the positive plate and “pushes” them onto the negative plate until the attraction between the opposite charges on the plates prevents further transfer of charge.  Experiments show that the bigger the voltage V of the battery, the more charge q is deposited on the capacitor’s plates. Thus q  V, or charge is proportional to voltage.  The proportionality constant C is known as the capacitance. Thus Topic 14: Digital technology 14.2 Data capture and digital imaging capacitance q = CV ( or V = q/C or C = q/V ) FYI  From the last equation we see that capacitance C is measured in coulombs per volt, which is called a Farad (F). Thus 1 C V -1 = 1 F.

5. Define capacitance. Topic 14: Digital technology 14.2 Data capture and digital imaging capacitance q = CV ( or V = q/C or C = q/V ) FYI  The schematic symbol for a capacitor is.  The symbol for the capacitor is not on the physics data booklet. Neither is the formula for capacitance.  In fact, the PDB has no formulas for Topic 14. EXAMPLE: A 250  F parallel plate capacitor is connected to a 6.2 V dc power supply. What charge will accumulate on each of its plates? SOLUTION:  Use the first form: q = CV = 250  10 -6 (6.2) = 0.0016 C. + - +q+q -q-q C V

6. Define capacitance. Topic 14: Digital technology 14.2 Data capture and digital imaging capacitance q = CV ( or V = q/C or C = q/V ) PRACTICE: A 1.3 pF capacitor has a charge equivalent to that of 8500 electrons. What is the potential difference across its plates? SOLUTION: NOTE: p = pico = 10 -12.  Use the second form: V = q/C = 8500(1.6  10 -19 )/1.3  10 -12 = 1.0  10 -3 V (1.0 mV). FYI  Clearly, the smaller the capacitor, or the more electrons are stored in it, the higher its p.d. will be.

7. Describe structure of charge-coupled device CCD.  If within a capacitor we place a photosens- itive material, charge will build up within the capacitor (because of the photoelectric effect) proportional to the intensity of the light (number of photons).  Electrodes placed across the photosensitive capacitor will then have a p.d. develop across them according to V = q/C.  We call a single such unit a picture element or pixel.  A typical charge-coupled device (CCD), which is used in a camera, has thousands of these units set up in an array. Topic 14: Digital technology 14.2 Data capture and digital imaging photosensitive material capacitor plate pixel V

8. Describe structure of charge-coupled device CCD.  This picture represents a three-by-four rectangular array of 12 pixels.  Since each pixel has two plates, it must have two electrodes. Thus there are 24 electrodes in this (very small) array.  A back-lit array is more sensitive. Topic 14: Digital technology 14.2 Data capture and digital imaging front-lit array FYI  In a front-lit array light comes in on the same side as the electrodes. Obviously some light is blocked by the electrodes. back-lit array

9. Describe structure of charge-coupled device CCD. Topic 14: Digital technology 14.2 Data capture and digital imaging EXAMPLE: Two CCDs are shown here. One is for a camera, and one is for a scanner. Which one is which, and how do you know? Which CCD is probably less expensive? SOLUTION:  The first is for a camera since it captures a whole rectangle of light in an instant. Having more pixels makes it more expensive.  The second is for a scanner since it is just a row of pixels. It only needs to scan line by line as an original is passed over it. It uses fewer pixels and is thus cheaper.

10. Describe structure of charge-coupled device CCD. Topic 14: Digital technology 14.2 Data capture and digital imaging EXAMPLE: The CCD shown here is an array of 512 by 1024 pixels. The dimensions of the array are given. What are the dimensions of each pixel, and how many pixels are there altogether? SOLUTION:  Match the dimensions: 0.018 m / 512 pixels = 0.000035 m pixel -1. 0.036 m /1024 pixels = 0.000035 m pixel -1.  Thus each pixel is 0.000035 m by 0.000035 m.  The array has a total of 512  1024 = 524288 pixels.  This is 2 19 pixels. 1.8 cm 3.6 cm

11. Explain how incident light causes charge to build up within a pixel.  The photoelectric effect causes charge to build up on a pixel. Photons striking the photosensitive material of the pixel can free electrons.  The longer the exposure or the brighter the light (the more intense it is) the more electrons are released.  More electrons means more charge, and more charge in the capacitor means a bigger p.d. across the capacitor, according to V = q/C. Topic 14: Digital technology 14.2 Data capture and digital imaging Photosensitive material FYI  The photosensitive material is usually silicon.

12. Outline how the image on a CCD is digitized.  To digitize a picture whose image is stored on a CCD two values are needed for each pixel. (1) The value of the voltage at the pixel which represents the intensity. (2) The position of the pixel in the CCD array. Topic 14: Digital technology 14.2 Data capture and digital imaging EXAMPLE: A CCD array consists of 512  1024 pixels, each of which can distinguish between 256 shades of gray. How much memory in MB is needed (a) to store gray-scale values of all the pixels? SOLUTION: Use 1 byte = 8 bits.  Since 256 = 2 8 values, we need 8 bits for each pixel’s gray-scale value.  Thus (2 19 pixels)(8 bits / pixel)(1 B/8 bits) = 524288 B = 0.52 MB.

13. Outline how the image on a CCD is digitized.  To digitize a picture whose image is stored on a CCD two values are needed for each pixel. (1) The value of the voltage at the pixel which represents the intensity. (2) The position of the pixel in the CCD array. Topic 14: Digital technology 14.2 Data capture and digital imaging EXAMPLE: A CCD array consists of 512  1024 pixels, each of which can distinguish between 256 shades of gray. How much memory in MB is needed (b) to store the positions of all the pixels? SOLUTION: Use 1 byte = 8 bits.  There are 512  1024 = 524288 = 2 19 pixels, so we need 19 bits for each pixel position (address).  Thus (2 19 pixels)(19 bits / pixel)(1 B/8 bits) = 1245184 B = 1.25 MB.

14. Outline how the image on a CCD is digitized.  To digitize a picture whose image is stored on a CCD two values are needed for each pixel. (1) The value of the voltage at the pixel which represents the intensity. (2) The position of the pixel in the CCD array. Topic 14: Digital technology 14.2 Data capture and digital imaging EXAMPLE: A CCD array consists of 512  1024 pixels, each of which can distinguish between 256 shades of gray. How much memory in MB is needed (c) to store one picture? SOLUTION:  Just total the previous two answers. 1.25 MB + 0.52 MB = 1.76 MB.

15. A typical 256-gray-scale photograph. Note that each pixel has a single gray shade.

16. 14.2.5 Define quantum efficiency of a pixel. 14.2.6 Define magnification. 14.2.7 State that two points on an object may be just resolved on a CCD if the images of the points are at least two pixels apart. 14.2.8 Discuss the effects of quantum efficiency, magnification and resolution on the quality of the processed image. 14.2.9 Describe a range of practical uses of a CCD, and list some advantages compared with the use of film. 14.2.10 Outline how the image stored on a CCD is retrieved. 14.2.11 Solve problems involving the use of CCDs. Topic 14: Digital technology 14.2 Data capture and digital imaging

17. Define quantum efficiency of a pixel.  The quantum efficiency QE of a CCD is the ratio of the number N e of photoelectrons emitted in a pixel to the number N  of incident photons.  The ideal QE (or sensitivity) for a CCD would be 1, i.e.: for each photon that strikes the pixel, one electron is freed up to contribute to the charge (and thus the voltage).  In reality, the QE of a CCD depends on the frequency of the incident photons, and whether the CCD is back-lit or front-lit. Topic 14: Digital technology 14.2 Data capture and digital imaging quantum efficiency QE QE = N e /N  http://ccd.com/ccd101.html

18. Define magnification.  The linear magnification M of the lensing system that projects an image on the CCD for recording is given by the ratio of the length of the projected image L image and the length of the actual object L object :  True magnification can come only from the lenses in the camera, whereas “pseudo” magnification can come digital- ly, by making the pixel images each larger than the actual pixel size during display.  Digital magnification produces visually grainy results, whereas lens magnification (true) gives clear images. Topic 14: Digital technology 14.2 Data capture and digital imaging linear magnification M M = L image /L object

19. Define magnification. Topic 14: Digital technology 14.2 Data capture and digital imaging PRACTICE: Decide whether M > 1 or M < 1 for each of the following. Justify your response. (a) You snap a picture with your cell phone camera of a friend. (b) An electron microscope records the image of a record groove. SOLUTION: Use M = L image /L object. (a) Your friend is certainly bigger than the picture you recorded in your camera. Thus L image < L object so that M < 1. (b) A record groove is smaller than the picture of it recorded by an electron microscope. Thus L image > L object so that M > 1.

20. State that two points on an object may be just resolved on a CCD if the images of the points are at least two pixels apart.  Consider the images of an A that are being projected on two CCDs: FYI  A bigger image has a better resolution (a).  The resolution depends on how the image happens to fall on the pixels (b). Topic 14: Digital technology 14.2 Data capture and digital imaging (a) (b)

21. State that two points on an object may be just resolved on a CCD if the images of the points are at least two pixels apart.  In each of the two CCDs below two H’s are being projected. The width of the H in (a) is < 2 pixels, the width of the H in (b) is 2 pixels. Topic 14: Digital technology 14.2 Data capture and digital imaging (a) (b) Resolution is not guaranteed Resolution is guaranteed

22. Discuss the effects of quantum efficiency, magnification and resolution on the quality of the processed image.  The higher the quantum efficiency, the closer to 1 : 1 the photon : electron ratio is. This would imply that the CCD can operate under low-light conditions or very fast sample speed.  Optical (as opposed to digital) magnification is desired to prevent graininess.  A CCD with a very large array of very small pixels will ensure a quality resolution. Topic 14: Digital technology 14.2 Data capture and digital imaging FYI  So if you are looking for a good digital camera you want a large CCD array inside.  You want high quantum efficiency.  You want true, optical magnification.  Obviously, this will cost you some money.

23. Describe a range of practical uses of a CCD, and list some advantages compared with the use of film.  Cameras.  Video cameras.  Telescopes.  Medical imaging.  Spectrophotometer. FYI Compared to film, CCDs have the following advantages:  Easier storage of images.  Images can be digitally enhanced and filtered.  Images can be transferred without noise. Topic 14: Digital technology 14.2 Data capture and digital imaging

24. Outline how the image stored on a CCD is retrieved.  One method of accessing the information on the CCD array is using a shift register. Observe the animation below: It repeats, emptying the array. 110 001 010 110 111 110 010 001 110 Topic 14: Digital technology 14.2 Data capture and digital imaging 6 1 2 6 7 6 2 1 6 the shift register

25. Solve problems involving the use of CCDs. Topic 14: Digital technology 14.2 Data capture and digital imaging  Magnification is the ratio of the length of the image to the length of the object.  M = L image /L object.

26. Solve problems involving the use of CCDs. Topic 14: Digital technology 14.2 Data capture and digital imaging  To be resolved, the image points must be at least two pixels apart.  There are 4000000 pixels.  The CCD is 16 cm 2.  Thus A pixel = 16/4000000 = 0.000004 cm 2.  Assuming a square pixel, it has a width w given by w =  A = 0.0020 cm.  M = L image /L object so that 1.5 = L image /0.002  L image = 0.0030 cm.  This is 1.5 pixel of separation. Close to resolution…

27. Solve problems involving the use of CCDs. Topic 14: Digital technology 14.2 Data capture and digital imaging  The charge is proportional to the number of photons that knock loose electrons (the photoelectric effect).  The number of photons is proportional to the intensity and the length of exposure.  Intensity is a property of the incident light.

28. Solve problems involving the use of CCDs. Topic 14: Digital technology 14.2 Data capture and digital imaging  M = L im /L obj =  A im /  A obj. Thus M =  (A im /A obj ) =  ( 1.0  10 -3 /2.5  10 -2 ) = 0.04.  L im = ML obj = 0.040(0.50 mm) = 0.020 mm.  A pixel = 2.3  10 -10 m 2  L pixel = 0.000015 = 0.015 mm.  2L pixel = 0.030 mm > L im.  You need separation to be at least two pixels wide so the images are not resolved.

29. Solve problems involving the use of CCDs. Topic 14: Digital technology 14.2 Data capture and digital imaging  Quantum efficiency is the ratio of the number of electrons freed, to the number of photons striking the photomaterial.

30. Solve problems involving the use of CCDs. Topic 14: Digital technology 14.2 Data capture and digital imaging quantum efficiency QE QE = N e /N   QE = N electrons /N photons. Thus N electrons = (QE)N photons = 0.80(5.5  10 4 ) = 44000 e -. q = eN electrons = (1.6  10 -19 C) 44000 = 7.04  10 -19 C. V = q/C = 7.04  10 -19 /40  10 -12 = 1.8  10 -4 V. capacitance q = CV ( or V = q/C or C = q/V )

31. Solve problems involving the use of CCDs. Topic 14: Digital technology 14.2 Data capture and digital imaging  Each pixel has a voltage proportional to its intensity.  Each voltage is converted to a digital value.  The location of each pixel is stored along with its intensity.  We thus have an image converted to a digital string of bits.  Playback and conversion of this string allows transfer back to a screen.