CCD Devices
Contents:
•Capacitance
•How they work
•Resolution
•Magnification
Concept 0 - Capacitance
C=q
V
C = capacitance (Farads)
q = Charge on the capacitor (C)
V = voltage across the capacitor (V)
demo big cap
++++++++++++++++++++++++++++++++++++++ ++++++
- -------------------------------------------------------------
A CCD pixel has a capacitance of 1.7x10-12 F. What is
the voltage across it if it has been charged 6.0x104
electron charges? (1 e = 1.602E-19)
V = q/C = (6.0E11*1.602E-19/1.7E-16) = 5.7 mV
TOC
Whiteboards:
Capacitance
1|2|3
TOC
What is the charge on a 250 microfarad
capacitor if it has been charged to 12 V?
q = CV = (250E-6 F)(12 V) = 3E-3 Coulombs
.0030 C
W
What is the capacitance of a CCD pixel if it
has .014 V across it when it has a charge of
2.13x10-15 C?
C = q/V = (2.13E-15 C)/(.014 V) = 1.52x10-13 F
1.52x10-13 F
W
Example: A 3.1x10-10 m2 CCD pixel has a
capacitance of 2.3 pF (x10-12) If after being
exposed to light, it has picked up a charge of
3.45x10-14 C, what is the voltage across it? How
many electrons were displaced from the pixel?
If photons displaced these electrons, then what
is the photon flux in photons/m2 incident on the
CCD?
(e = 1.602E-19 C/electron)
V = q/C = (3.45E-14C )/(2.3E-12 F) = .015 V
# electrons = (3.45E-14C )/(1.602E-19 C/electron) = 2.154E5 electrons
photons/m2 = (2.154E5 electrons)/(3.1E-10 m2) = 6.95E14 photons/m2
A sunny day = about 1000 W/m2 ≈ 3x1021 photons/sec (550 nm)
(a force of 3.3x10-6 N/m2
0.015 V
2.2E5 electrons
6.9E14 photons/m2
W
How a CCD Works (Charge-Coupled Device
)
•Photons charge electrodes
•Voltages from pixels is
shifted off the chip
•The voltage is converted to
digital
www.microscopy.fsu.edu/primer/digitalimaging/concepts/ccdanatomy.html
TOC
How a CCD Works
CCDs only register light,
color is achieved using
filters
www.microscopy.fsu.edu/primer/digitalimaging/concepts/ccdanatomy.html
TOC
jpeg
compression
photon
memory card
Resolution, sensor size and magnification
Nikon D3x: (CMOS)
Sensor size = 35.9x24.0 mm = 861.6 mm2
6048x4032 resolution = 24 Mega pixels
Nikon D3x
Human eye: (rods and cones)
Sensor size = 1100 mm2
11,000x11,000 = 121 Mega Pixels
Magnification:
image/object ratio
TOC
Resolution, sensor size and magnification
Nikon D3x: (CMOS)
Sensor size = 35.9x24.0 mm = 861.6 mm2
6048x4032 resolution = 24 Mega pixels
(Let’s assume a quantum efficiency of 70%)
A) What is the area of each pixel?
B) If light with a photon flux of 4.5x1019 photons/m2/s is incident on
the sensor for 1/250th of a second, what is the resulting charge on the
pixel?
C) If the voltage across the pixel is 2.4 mV, what is the capacitance of
the pixel?
D) If I am 183 cm tall, and my image on the sensor is 19.2 mm tall,
what is the “magnification” in this case?
TOC
Nikon D3x: (CMOS)
Sensor size = 35.9x24.0 mm = 861.6 mm2
6048x4032 resolution = 24 Mega pixels
(Let’s assume a quantum efficiency of 70%)
A) What is the area of each pixel?
total area = (35.9E-3m)x(24.0E-3) = 0.0008616 m2
#pixels = 6048x4032 = 24,385,536 pixels
area/pixels = (0.0008616 m2)/(24,385,536 pixels) = 3.53324E-11 m2/pixel
B) If light with a photon flux of 4.5x1019 photons/m2/s is incident on the sensor for
1/250th of a second, what is the resulting charge on the pixel?
Photons per pixel resulting in charge =
(0.70)(4.5x1019 photons/m2/s)(1 s/250)(3.53324E-11 m2/pixel) = 4451884.921
e charges per pixel
So this is (4451884.921)(1.602E-19) =7.13192E-13 C
C) If the voltage across the pixel is 2.4 mV, what is the capacitance of the pixel?
C = q/V = (7.13192E-13 C)/(2.4E-3 V) = 2.97163E-10 F = 297 pF
D) If I am 183 cm tall, and my image on the sensor is 19.2 mm tall, what is the
“magnification” in this case?
image/object = (1.92 cm)/(183 cm) = .0105x (it’s a smallifier)
TOC
Whiteboards:
Random CCD Crap
1|2|3|4
TOC
My camera has a 25.4 mm x 58.4 mm CCD
sensor with 4.0 Mega pixels. What is the area of
each pixel in square meters?
Area per pixel = (25.4E-3 m)(58.4E-3 m)/(4.0E6) = 3.7084E-10 m2
3.7084E-10 m2
W
A pixel builds up a potential of 5.2 mV. How
many photons hit it if it has a capacitance of 13
pF, and a quantum efficiency of 75%?
q = CV = (13E-12F)(5.2E-3) = 6.76E-14 C
# electrons = 421,972
#electrons/#photons = .75, #photons = (421,972)/.75 = 562,630
560,000 photons
W
A single pixel with an area of 3.7x10-10 m2 is hit
with 562,630 photons. What is the light
intensity in photons/m2?
photons/m2 = (562,630)/(3.7x10-10 m2) = 1.5E+15
1.5x1015 photons/m2
W