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Digital Technology & Electronics
Binary Numbers
http://coddledegg.blogspot.com/2007/05/10-types-of-people.html
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Computers use binary numbers, and
therefore use binary digits in place of
decimal digits.
The word bit is a shortening of the words
"Binary digIT.“
While decimal digits have 10 possible
values ranging from 0 to 9, bits have only
two possible
values:
0 and 1.
http://equinoxstudios.files.wordpress.com/2006/08/equinox_binary_ppl_wall_matrix_1024.png
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vs. Binary
In the decimal system each
digit is associated with a
power of 10. The ‘ones’
digit is associated with 100,
its ‘tens’ digit with 101, the
‘hundreds’ with 102, and the
‘thousands’ digit with 103
and so on.
Example: 4192 = 4 x 103 +
1 x 102 + 9 x 101 + 2x100
So digits of a decimal
number are just the
coefficients of various
powers of 10. These
coefficients are limited to
the digits from 0 to 9.
In binary each bit holds the
value of increasing powers of
2 and the coefficients are
limited to 0 or 1.
 Example: Binary 1011 =
1 x 23 + 0 x 22 + 1 x 21 + 1
x 20 = 8 + 0 + 2 + 1 = 11
5=1 x 22 + 0 x 21 + 1 x 20
 The notation for binary
representation is X2 with X
being the decimal number.
Examples: 52=101
15 = 1x23 + 1x22+1x21 +1x20
152=1111

Binary
http://www.avatarist.com/avatars/Movies/The-Matrix/Binary-Matrix.gif
Decimal
Binary
http://boingboing.net/images/binarymanicure.jpg

http://www.istockphoto.com/file_thumbview_approve/176060/2/istockphoto_176060-binary.jpg
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Some more examples
10 = 1010 13 = 1101
11 = 1011 14 = 1110
12 = 1100 15 = 1111
These are all examples
of four-bit words since
they have four digits.
52=101 is a three bit
word. By adding a zero
in front it can turn it
into a four-bit word.
52=0101
162 = 10000 is a five
bit word.

How many numbers can
be represented by a fivebit word? Define the
range?
25=32 (0-31)
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

In Binary form, the first
non-zero digit is the most
significant bit (MSB).
The MSB is the digit that
most determines the
value of the number.
The last digit is the least
significant bit (LSB)
Bits and bytes.
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
Bits are rarely seen alone in computers.
They are almost always bundled together
into 8-bit collections, and these collections
are called bytes.
With 8 bits in a byte, you can represent
256 values ranging from 0 to 255, as
shown here:
0 = 00000000
1 = 00000001
2 = 00000010
...
254 = 11111110
255 = 11111111
http://www.ollnet.com/school/faculty/DonRoque/binary_path.gif
http://www.ontrackdatarecovery.co.uk/images/newsletters/CD_binaryData.jpg
Bits and bytes continued

CD uses 2 bytes, or 16 bits, per sample.
That gives each sample a range from 0 to
65,535, like this:
0 = 0000000000000000
1 = 0000000000000001
2 = 0000000000000010
...
65534 = 1111111111111110
65535 = 1111111111111111
http://ibsncorp.com/images/clipart/cd_binary.jpg

Analog
Analog Signals – continuous signals varying
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

between two extreme values in a way that
is proportional to the physical mechanism
that created the signal.
As a technology, analog is the process of
taking an audio or video signal and
translating it into electronic pulses.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Digital
http://www.privateline.com/manual/fig3_8.gif
Digital Signal – coded form of a signal that
takes the discrete values 0 or 1 only.
 As a technology, digital breaks the signal
into a binary format where the audio or
video data is represented by a series of
"1"s and "0"s.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Analog History
Thomas Edison is credited

with creating the first device
for recording and playing
back sounds in 1877.
In Edison's original
phonograph, a diaphragm
directly controlled a needle,
and the needle scratched an
analog signal onto a tinfoil
cylinder.
Phonograph
http://www.americaslibrary.gov/assets/aa/edison/aa_edison_phonograph_2_e.jpg
In the Beginning: Etching Tin
An Analog Wave
http://www.privateline.com/PCS/images/Sineani.gif

Below an analog wave representing
the vibrations created by your voice
saying the word ‘hello’ is shown.
 What this graph is showing
is, essentially, the position
of the microphone's
diaphragm (Y axis) over
time (X axis). The
vibrations are very quick
(1000 cycles per second)
Image from www.howstuffworks.com
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Analog
vs.
Advantages

HDTV

Inexpensive, well
established
Rich sound quality
Disadvantages
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Size limitations
Less clarity
Limited error correction
Digital
Advantages



(output is known)
Disadvantages

Less rich sound quality
(less natural sounding)

http://simplykathy.com/wp-content/uploads/2008/08/analog-to-digital-transition1.jpg
Clarity
More information
Easy correct errors
Fairly expensive
http://markun.cs.shinshu-u.ac.jp/learn/osi/e_zub-
Analog to Digital and Back Again.

With digital recording technology (CDs,
DVDs, etc), the goals are to create a
recording with:
high fidelity (similarity between the original
signal and the reproduced signal)
 perfect reproduction (the recording sounds
the same every single time you play it).
Digital recording converts the analog wave into a
stream of numbers and records the numbers
instead of the wave. The conversion is done by
an analog-to-digital converter (ADC).

http://askbobrankin.com/analog-to-digital-tv.jpg
Analog to Digital and Back Again.
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
To play back the music, the stream of numbers
is converted back to an analog wave by a
digital-to-analog converter (DAC).
The analog wave produced by the DAC is
amplified and fed to the speakers to produce
the sound.
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Analog to Digital Conversion

http://media.digikey.com/photos/Maxim%20Photos/175-40-DIP.jpg

Shown is a typical wave where each tick on
the horizontal axis represents onethousandth of a second.
To covert this wave to digital it must be
sampled with an Analog to Digital
convertor (ADC)
In sampling two variables are controlled:
http://communication.howstuffworks.com/analog-digital3.htm
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Sampling rate - Controls the number of
samples taken per second.
The sampling precision - Controls how many
different gradations (quantization levels) are
possible when sampling.
http://communication.howstuffworks.com/analog-digital3.htm
Analog to Digital Conversion

Sampling Rate shown is 1000/sec and
the sampling precision is 10.
The green rectangles
represent samples.
Every 1/1000 of a second,
the ADC looks at the
wave and picks the closest
number between 0 and 9.
The number chosen is shown at the bottom. These numbers
are a digital representation of the original wave.
7 8 9 5 3 4 0 3 7 5 Translated to Binary
0111 1000 1001 0101 0011 0100 0000 0011 0111 0101
Digital to Analog Conversion

To be played the wave must be converted
from its digital form back to an analog form
using a Digital to Analog convertor (DAC)
•The reproduced
wave has a low
fidelity as much of
the original detail was
lost in the conversion.
•This is due to
sampling error.
•Sampling error is
The converted signal from
reduced by increasing
digital back to analog is shown both the sampling
in blue with the original in red. rate and the precision
http://static.howstuffworks.com/gif/digital-converter-box-1.jpg
Aliasing
Poor fidelity in the reconstructed
analog wave.
 Caused by low sampling frequency
compared to the information signal
frequency.

Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Analog

Digital Refinements
Increasing the sample rate and the
precision by a factor of 2 (20 gradations at
a rate of 2,000 samples per second)
Doubling the sample
and precision again
(40 gradations at a
rate of 4,000 samples
per second)
Typically, audio signals are
sampled at a rate of 8000 times
per second 1/8000 or 125µs
Source: http://communication.howstuffworks.com/analog-digital3.htm
AD Conversion Example

Consider the following
potential divider circuit.

If the variable resistor
(arrow) is adjusted from
the bottom to the top at
a constant rate in 4 ms.

The output voltage vs.
time graph would be the
analog signal shown.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
AD Conversion Example
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To covert to digital the signal must
be sampled.
Sampling rate – numbers of times
per second the signal is measured.
Observe for very short period of
time, do not know how the signal
behaves in between the samples.
If we sampled the signal every
1ms, a pulse amplitude modulated
signal (PAM signal) would be
generated with the results shown in
the table.
Note the sampling time is very
short (2.0 μs or 1.0 μs) relative to
the time between intervals and that
is why the sampling time is denoted
with vertical lines of minimal
thickness.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
AD Conversion Example
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The PAM signal sampled values now
need to be converted to binary.
If two-bit words are used, there are
22=4 words available.
Therefore the 0-8V can be divided
into four levels (0-2, 2-4, 4-6, 6-8)
and assign a two bit word to each
level as shown. Each level has a
lower boundary and upper boundary.
So the lower boundary voltage in
each level was divided by 2
(2=highest voltage divided by # of
words) These give corresponding
decimal numbers of 0,1,2,3V which
are then converted to binary.
The results show a loss of
information in the digitization of
data.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008

AD Conversion Example
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To improve the replication we can
increase the sampling rate to 0.5 ms
and use three bit words (23=8 words)
This is shown in the new table.
Note that the lower boundary voltage
was directly converted to binary as
the highest voltage was the same as
the number of words.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Quantization – process of dividing the range of
analog signals (highest value – lowest value) into a
set of levels
• These levels are known as quantization levels.
• The quantization levels are determined by the
length of the binary word used.
Quantization Error (q)
Where,
M m
q
m = minimum value
2n
M = maximum value
n-bit words to digitize
2n = number of
quantization levels.
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Physics for the IB Diploma 5th Edition (Tsokos) 2008
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Physics for the IB Diploma 5th Edition (Tsokos) 2008
Two analog signal values that differ
by less than the quantization error
are assigned the same binary
number.
Calculate the quantization error for
table 1.2 and table 1.3 of the
previous example.
Table 1.2 = 2V Table 1.3 = 1V
Source: Physics for the IB Diploma 5th Edition (Tsokos) 2008
Bit Rate
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
Numbers of bits that can be transmitted
per second.
Bit rate is equal to the sampling rate times
the numbers of bits per sample.
Bit rate = f x n

τ is the time duration (s) of each bit.
Bit rate = 1/ τ
Bit rate determines the bandwidth that will allow a
given digital signal to pass through a
communication line (cable).

A small bandwidth will distort the pulses so the
reconstructed analog signal will be very distorted.
Shannon-Nyquist Sampling Theorem
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
Mathematical Theorem that states the
sampling frequency must be at least twice
as large as the largest frequency in the
information signal.
Nyquist Frequency – sampling frequency
that is twice as large as the largest
frequency of the signal.
Problem: A signal has frequencies ranging from 400
Hz to 5.4 kHz. What is the minimum sampling
frequency that will result in a high fidelity
reproduction?
Digital Signals: Transmission
The Block Diagram shows the process
of converting an analog signal into a
decimal signal and then back into an
analog signal
 Transmission and Reception
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Digital Signals: Transmission
1.
Sample and Hold block.

Here the analog signal is sampled according to the
sampling frequency assigned (sampling frequency
block) to the device and temporarily stored.
2.
ADC Block

Each sample is converted into an n-bit binary
code. (Assume n=8 for simplification) The output
is a set of 8 bits making up one sampled signal.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
http://www.interactivesupercomputing.com/starpexpress/042007/eNewsletter_spring07v3_files/Serial_Animation.gif
Digital Signals: Transmission
3.
4.

_Animation.gif
http://www.interactivesupercomputing.com/starpexpress/042007/eNewsletter_spring07v3_files/Task_Parallel

Parallel to Serial block.
The resulting 8 bits are registered in the serial-toparallel convertor and then each bit is transmitted
one by one along a single line.
Serial to Parallel block
The bits arrive one by one in the serial to parallel
convertor and are registered. When all 8 bits have
arrived, they are assembled into a single 8 bit
word. The eight bits are then simultaneously send
to the DAC
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Digital Signals: Transmission
5.

DAC block.
The DAC coverts the digital signal into an analog
signal and transmits the analog signal.
‘Clock’ blocks

These devices control the process of transmitting
the bits in the parallel/serial and serial/parallel
devices. All of the bits must be sent before the
next set of bits arrive.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Time Division Multiplexing (TDM)

Method used to transmit many digital
signals along the same channel at the
same time.
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
Possible due to the actual sampling time being
so much less than the time between samples
Multiplexer – device used to mix individual
signals to feed them along a single
transmission line.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Advantages of Digital Signals
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Regenerated perfectly (eliminate noise and
distortion) even over large distances.
Relatively inexpensive and available
Errors can be eliminated with error
correction codes.
Signal can be encrypted (scramble the
bits).
Signals can be stored, processed, and
controlled with computers.
Signals can be compressed
Time division multiplexing
Signals can be stored easily on CDs, DVDs,
etc.
A-D Source: Physics for the IB Diploma 5th Edition (Tsokos) 2008
http://www.rob-fra.com/webImages/Cd_animated.gif
CD’s and DVD’s

Data is stored digitally
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The original analog code has been converted to
digital and then stored in binary code (a series of
ones and zeros).
A CD has multiple tracks
The tracks consist of a sequence of pits of varying
length formed in a reflecting information layer.
 The area between the
pits is called ‘lands’.
 The edge of the pit
corresponds to a
binary 1

Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
A CD’s pits and bumps
Physics for the IB Diploma 5th Edition (Tsokos) 2008


Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

The pits are laid out in a path that spirals
out from the center.
The tracks are spaced very close together
~1600 nm.
The pits are with very small dimensions
with the width comparable with the
wavelength of green light.
www.physics.byu.edu/faculty/rees/106/PPT/Class26.ppt
Reading a CD


The laser beam shines on a metallic layer
through a clear plastic coating
As the disk rotates, the laser reflects off the sequence of
bumps and lower areas into a photodetector



The photodetector (photo diode) converts the fluctuating
reflected light intensity into an electrical string of zeros and ones
When the laser beam hits a rising or falling
bump edge, part of the beam reflects from
the top of the bump and part from the lower
adjacent area
Light reflecting from the top and bottom of
the pit is a half-wavelength out of phase, so
the intensity drops
www.physics.byu.edu/faculty/rees/106/PPT/Class26.ppt
Reading a CD
The bump edges are read
as ones
 The flat bump tops and
intervening flat plains are
read as zeros.
Physics of …
 A laser beam has some
finite width.
 When it is incident near
the edge of a pit, some
rays will be reflected from
the pit and the others will
reflect off the land.



Since laser light is
coherent, these
reflections will
interfere.
The light reflecting
from the pit will
travel a distance of
2d farther.
Reading a CD




If the extra difference is
equal to half a
wavelength then the
interference will be
destructive.
Therefore, destructive
interference occurs if
In this case the light
detector would receive a
reflection of zero intensity
This is a binary 1.


The other factor that
affects the laser is
the polycarbonate
coating over the
reflective service.
Its index of
refraction must be
considered.
Physics of Reading a CD
Example: Calculate the standard
depth of a CD pit. A laser with
a wavelength of 780 nm is
used along with a
polycarbonate coating with an
index of refraction of 1.55.
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
CD Storage
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

With CDs fidelity is an important goal, so the
sampling rate is 44,100 samples (words) per
second. With this the output of the DAC so closely
matches the original waveform that the sound is
essentially "perfect" to most human ears.
Calculate the storage capacity of a CD if sampling
is 44,100 words per second, information consists
of 32-bit words (2 channels of 16 pit samples) and
length of 74 minutes.
or
780 Mbytes
CD Source: Physics for the IB Diploma 5th Edition (Tsokos) 2008
DVD’s (digital versatile disk)
DVD’s use shorter wavelength lasers
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
This allows the track separation, pit depth and
minimum pit length to be smaller
Therefore, the DVD can store about 7 times
more information than a CD or about 4.7 Gigabytes.
http://www.beoworld.org/assets/thumbnails/dvdyess.jpg

http://www.grcoatley.mcc.education.nsw.gov.au/ipt_website/images/dvd.jpg
Blu-ray discs
http://uk.gizmodo.com/blu-ray%20disc%20logo.jpg
Name – blu stands for the blue laser
used. Blue lasers have a shorter
wavelength (405 nm) than the red laser used with
DVDs resulting in smaller pit depths and closer
spacing of tracks.
 Different construction with much thinner coating
between the data
and the laser
eliminates several
issues with DVDs.
Results greater storage

Single sided 27 GB
Double sided 54 GB
Source: http://electronics.howstuffworks.com/blu-ray1.htm
LPs


(Records, Vinyl, 45s, Old school…)
Same idea as Edison’s tin cylinder.
Uses a flat rotating disk with a needle that
picks up the scratchings of the record and
converts them into an electronic signal that
is amplified and sent to the speakers.
 Disadvantages
Limited storage capacity
Easily damaged (scratched)
http://www.lp-records.com/turntable-animated.gif
Cassettes
http://coleman300.com/images/animated_gifs/cassette_ani1.gif



Use magnetic recording to store
data in analog form
Sequential devices – must scan
the tape until you reach the track you want.
Plastic tape of ferric oxide
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.


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
Permanently magnetized when exposed to a magnetic
field.
Analog signal converted to electronic signal creates its
own varying magnetic field.
Tape exposed to this a copy of this varying field is
produced.
Advantages: Low price, availability, erasable and reusable
Disadvantages: Sequential nature, limited storage, easily
damaged.
Floppy Disks
Uses magnetic recording where data is
arranged in concentric circles.
 Direct access as data on an outside circle
could be accessed without going through
all the intermediate data.
History

http://farm1.static.flickr.com/28/67613265_558e81a471.jpg?v=0



8 in floppy disk invented by Alan Shugart at
IBM in 1967.
5 ¼ inch floppy disk – 360 kilobytes
3 ½ inch floppy (non flexible) – 1.44 Mbytes.
Source: www.howstuffworks.com
Hard disks



Made of disks of glass or aluminum arranged
on a spindle.
Surface of disks are covered with a
magnetizeable material (cobalt)
Data Storage




Surface divided into tiny regions
Each region is a set 0 or 1 binary digitalized data.
Data is in sectors and tracks with the tracks
being concentric circles and sector is a part of
the track.
Knowing the address the data can be accessed
immediately.
http://www.cksinfo.com/clipart/electronics/computers/drives/hard-drive-internal-large.png
http://docsrv.caldera.com:507/en/HANDBOOK/graphics/harddisk.gif
Digital Storage
Advantages

Large capacity
Fast access

Fast data retrieval

Reliable Storage





http://www.microsoft.com/library/media/1033/windowsxp/using/digitalphotography/images/Prep_03.jpg
Data can be copied or erased easily
Data can be encrypted
Data can be processed and manipulated with
a computer
Easy transport of data physically and
electronically.
What is a digital image?

Essentially, a long string of 1s and 0s that
represent all the tiny colored dots - or pixels
- that collectively make up the image.
2 options for getting a digital image


Scan an image (photo) - record
the pattern of light as a series
of pixel values.
Take a digital picture – samples
the original light that bounces
off your subject, immediately
breaking that light pattern
down into a series of pixel
values
http://www.photo.epson.co.uk/images/Technology/CCD%20scanning_1.jpg

A Charged Coupled Device (CCD) is
the image processor of most digital
cameras
 Another device sometimes used is a
complementary metal oxide
semiconductor (CMOS) device.
 Both convert light into electrons
 Simply, a 2-D array of thousands or
millions of tiny solar cells.

A CMOS sensor
http://www.axis.com/edu/axis/images/ccd.gif
Digital Camera
http://upload.wikimedia.org/wikipedia/commons/a/a1/CCD.jpg
CCD Device History
Invented at Bell Labs in 1969.
 Advantages


High resolution images
Digital form easily processed and
manipulated.
 Much faster capture than conventional
film
 Obtain images of very faint objects.
 Revolutionized astronomy first and then
trickled down into consumer products.

Silicon chip with a surface covered
with light sensitive elements – pixels.
 Each pixel releases electrons when
struck by light – photoelectric effect
 Key characteristics

http://www.shortcourses.com/sensors/sensors1-0.html
Number of electrons released when light
is incident on a pixel is proportional to
the intensity of the light.
 So charge and the potential difference
across a pixel are also proportional to
the intensity of light on that pixel.

http://www.axis.com/edu/axis/images/ccd.gif
CCD
CCD Theory


http://www.mssl.ucl.ac.uk/~gbr/2b65_fig/epic_pn_ccd.gif

A pixel is essentially a
 Energy of a single
small capacitor.
photon of light of
The electrons released
frequency f is
have some charge Q.
So a potential difference
develops at the ends of  h = 6.63 x 10-34 Js
the pixel where
(Planck Constant)


http://www.creativepro.com/files/story_images/070604_fg1.gif
C is the capacitance of
the pixel.
This potential difference
can be measured.

From the wave eqn.
λ = wavelength
So,
http://kepler.nasa.gov/sci/techdemo/images/CCD01.gif
CCD Image Capture

Exposure

Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.

Open shutter –
exposes CCD surface
to light for a set time.
Causes charge and
voltage to build up in
each pixel.
Next, a potential
difference is applied to
each row of pixels

This forces the charge
stored in each pixel to
move to the row below
(charge coupled)
CCD
http://solar.physics.montana.edu/nuggets/2000/001201/ccd.png


Starting from the bottom
row, the charge of each
pixel is moved vertically
down into the register.
Then one by one the
charge is moved
horizontally where the
voltage is amplified,
measured, and passed
through an ADC until the
entire row is processed.

Processing



The processing computer
stores the voltage in each pixel
and the location of each pixel.
Process is repeated for the
next row until the entire image
is processed.
Since charge and thus voltage
is proportional to the light
intensity the image can be
recreated on the computer
screen.
Process for a B&W photo.
Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
CCD Image Capture
Building a CCD
http://www.axis.com/edu/axis/images/ccd.gif
Color Image

For a color image, pixels are
arranged in groups of four.



Demosaicing
Algorithm


2 with green filters, 1 red and 1
blue (Bayer filter)
The intensity of light in pixels of
the same color (example: green)
are recorded as before
A computer program
(demosaicing algorithm) is used
to find the intensity of green light
in each pixel by interpolation
based on the intensity in
neighboring green pixels.
The same process is repeated for
red and blue.
Thus, the intensity of green, red,
blue lights are known and a
colored image can be found by
combining all of these.



High end cameras
Use three separate
sensors with a beam
splitter that splits the
light to separate
beams that hit each
sensor.
Expensive and bulky
CCD Image Characteristics
Quantum efficiency – ratio
Example Problem:
of the number of emitted
The area of a pixel in a
CCD is 8.0 x 10-10m2 and
electrons to the number of
its capacitance is 38 pF.
incident photons.
Light of intensity 2.1 x 10 70-80 % CCD
3 W/m2 and wavelength
 4% best quality photographic
4.8 x 10-7 m is incident on
films
the collection area of the
 1% human eye
CCD for 120 ms.
Calculate the potential
 Why, CCDs are used to
difference established at
measure the apparent
the ends of a pixel,
brightness of stars.
assuming that 70% of the
 Higher quantum efficiency
incident photons cause
means the image takes less the emission of electrons.
http://www.olympusmicro.com/primer/digitalimaging/concepts/quantumefficiency.html
time to form in low
intensity incident light.
CCD Image Characteristics
Magnification –
ratio of the length
of the image as it
is formed on the
CCD to the actual
length of the
object.


Example Question:
A digital camera is used to take
a photograph of the eye of
an insect. The area of the
real eye is 1.4 x 10-6 m2 and
the area of the image eye is
9.5 x 10-6 m2. Calculate the
magnification.
Determined by the
properties of the
lens that focuses
the light.
http://k53.pbase.com/o4/34/564334/1/64338540.Xd7b5P9k.20060730damsel02b.jpg
CCD Image Characteristics
Resolution – amount of


detail an image can
capture and is
measured in pixels.
Two points are
resolved if their
images are more than
two pixel lengths
apart.
Higher resolution
images are clearer as
they include more
detail.
Photo courtesy Morguefile

Example Question
The magnification
produced by a 3.0
megapixel digital
camera with a
collecting area of 12
mm2 is 1.5.
Determine if this
camera can resolve
two points a distance
3.2 x 10-3 mm apart.
CCD Medical Uses
Endoscopy – a thin tube is inserted into
the patient to make observation of
internal images. CCDs allow real-time
digital images of this.
http://lmc.lorma.org/images/video_endoscopy.jpg
http://images.goshdawnit.com/uploaded_images/upper-endoscopy-770149.jpg
http://www.fairchildimaging.com/main/images/applications/xray_hand.jpg
CCD Medical Uses
X-rays – special CCDs have been
developed that can detect X-rays.
Detect x-rays much more sensitive
require shorter exposure time, but
are still very expensive.
http://www.anl.gov/techtransfer/images/CCD_X-ray_detector_72dpi.jpg
http://www.z-beamlet.sandia.gov/images/photcount.gif
CCD Source: Physics for the IB Diploma
5th
Edition (Tsokos) 2008
Digital World
Function
Digital
Analog
Complexity
Detailed list of
instructions (computer
code) needed for the
conversion from an
input to a digital
signal and then back
to an output
Simple – direct
conversion from
pressure variations of
a sound to electrical
variations of the
signal
Quality
High fidelity requires
high frequency
sampling & sampling
(quantum) levels
Fidelity is virtually
indistinguishable from
input but vulnerable
to damage or
corruption
Reproducibility
Optical reading can
ensure same result
each time
Playback often affects
quality of future
playback (wear)
Source: Kirk, Tim, Physics for the IB Diploma, Oxford University Press 2007
Digital World
Function
Digital
Analog
Retrieval Speed
Very quick retrieval
though large amount
of info take slightly
longer (video),
different sections of
data can be retrieved
without additional
delay
Slower retrieval due
to mechanical
process. Accessing
different sections of
data require additional
time (fast forwarding
a tape)
Portability
Advances in
miniaturization allows
large amount of
information to be
stored on small
devices
Stored material can
be very compact,
typically analog
storage requires more
space than digital
Manipulation
Easily performed
without significant
corruption
All types increase
chance of corruption.
Source: Kirk, Tim, Physics for the IB Diploma, Oxford University Press 2007
Digital World - Implications
Potential store almost infinite
amounts of information
 Access to this information
 Lower cost of storage than other
means
What implications does this ever
increasing data storage capacity hold
for a society – morally & ethically,
socially, economically, and
environmentally

Communication Challenges
Attenuation – loss of the signal’s power due
to losing energy to the surroundings.
 Corrected by using repeaters that boost
the signal strength along the path
Noise – addition of any random electrical
signals to the original electrical signal.
Leads to distortion.
Digital Signals less susceptible to both


Detector systems only need to make out a 1 or 0
so larger variation in amplitude easier to make out
even with attenuation
Noise correction is possible through circuits that
recreate the original signal known as regenerator
or shaper circuits.
http://www.kxcad.n6et/autodesk/3ds_max/Autodesk_3ds_Max_9_Reference/graphics/il_vlight_attenuated.jpg
http://www.comm-spec.com/images/articles/repeater-diagram.jpg
http://pendrey.net.au/Portals/0/TeachingAids/Noise/Noisy%20Signal.PNG
Communication Options - WIRED
http://www.acclima.com/images/2wire.png
Wire pairs – 2 wires connecting a
sender and a receiver


Advantages


Example: Microphone, amplifier, speaker
or intercom
Simple and cheap
Disadvantages
http://markun.cs.shinshu-u.ac.jp/learn/osi/e_zub-4-2.gif
Susceptible to noise and interference
 Cannot transmit very high frequency
signals

Communication Options - WIRED
http://www.rogershelp.com/assets/digital/basic/basic2.gif
Coaxial cables – central wire
surrounded by second wire in the
form of an copper tube or mesh with
an insulator separating the wires.
Carry frequencies upto 1 GHz, but higher
frequencies will attenuate more. A 100
MHz signal will need repeaters every ½
km. Upper limit is 140 Mbit/s
Examples: TV signals from antennas to
receivers, standard underground telephone
links.


http://www.printercode.com/Images/Coaxial%20Cable.jpg
Communication Options - WIRED
Coaxial cables – central wire
surrounded by second wire in the
form of an copper tube or mesh with
an insulator separating the wires.


Advantages – Simple and straightforward.
Less susceptible to noise than wire pair
Disadvantage – Noise is still a problem
http://www.printercode.com/Images/Coaxial%20Cable.jpg
Communication Options - WIRED
Fiber Optics– Laser light used to send
signals along optical cables.
Same frequency limit as cables ~ 1 GHz
 Less attenuation than coaxial cable with
repeaters spaced 10s or even 100s of
kilometers apart


Example: Long-distance
telecommunication
Communication Options - WIRED
Fiber Optics– Laser light used to send
signals along optical cables.
 Advantages: Over coaxial
Higher transmission capacity
 No electromagnetic interference
 Smaller, cheaper, quieter and more
secure

http://ddp13fiberoptics.files.wordpress.com/2008/09/fiberoptics1.jpg

Disadvantages:
More difficult to repair
 Regenerators more complex

Communication Options - WIRELESS
Radio Waves– Electromagnetic waves
that travel at the speed of light in a
vacuum can be used to transmit
signals. Obey c = fλ
 Advantages/Disadvantages




High frequency end transfer large amounts of data.
Ground based (terrestrial) radio broadcasting while
expensive is cheaper than satellite communication
Noise depends on modulation process (AM or FM)
Attenuation depends on the spectrum used
Radio Frequencies
Range
Frequency
Wavelength
Utilization
Very low frequency (VLF)
3 - 30 kHz
100 - 10 km
Long range communication
Low frequency (LF)
30 – 300 kHz
10 – 1 km
Long distance communication,
radio broadcasting
Medium frequency (MF)
300 – 3000 kHz
1000 – 100 m
AM radio broadcasting
High frequency (HF)
3 – 30 MHz
100 – 10 m
International broadcasting,
radio controlled devices,
amateur radio, long distance
ship communication
Very high frequency (VHF)
30 – 300 MHz
10 – 1 m
FM radio, TV, aircraft
communication, emergency
services communication, radio
astronomy
Ultra high frequency (UHF)
300 – 3000 MHz
100 – 10 cm
Satellite communication, TV,
mobile phones, microwave links,
radar
Super high frequency (SHF)
3 – 30 GHz
10 – 1 cm
Mobile phones, satellite
communication, microwave links
Extra high frequency (EHF)
30 – 300 GHz
10 – 0.1 cm
Radar, radio astronomy
Microwaves – radio waves > 1 GHz
Infrared (IR) waves – very short range communication, e.g. television remote
Comparison of Channels of
Communication
Channel
Carrier
Frequency
Bandwidth
Average
Distance
Between
repeaters
Specific
Attenuation
(dB/km)
Copper wires 20 kHz
20 kHz
10 km
10
Wire Pairs
10 MHz
500 kHz
5 km
25
Coaxial
Cable
2 MHz (telephone) 500 MHz
1 GHz (TV)
10 km
100 m
6
200
Microwaves
5 GHz
100 MHz
50 km
Distancedependent
0.2 THz
10 GHz
80 km
0.20
in free space
Optic Fibers
Geostationary (geosynchronous)
Satellites – maintains the
same position relative to a point on the
earth’s surface
 Possible by placing the satellite in orbit
high above the equator.
 Distance is such that the period of the
satellite is that of the Earth’s 24 hrs
allows the satellite to appear fixed.
(Distance ~ 3.6 x 104 km or 5.6 radiuses of
the Earth from the earth’s surface)
http://www.makingthemodernworld.org.uk/learning_modules/geography/07.TU.01/illustrations/07.IL.04.01.gif
http://www.centennialofflight.gov/essay/Dictionary/meteorology/DI63G2.jpg
Satellite Communications
Satellite Communications
http://spaceplace.nasa.gov/en/kids/goes/goes_poes_orbits.shtml
Geostationary Satellites – maintains the same
position relative to a point on the earth’s
surface
 Used to relay information between sender
and receiver
 Frequencies used are >1GHz (SHF range or
above) to deliver large-bandwidth signals at
relatively low power and amplitude
 Uplink frequency is different than the
Up-link Frequency
Down-link Frequency Bandwidth
downlink
frequency.


(GHz)
(MHz)
Keeps (GHz)
transmitting towers
from swamping
6 towers.
4
500
receiving
14 feedback or resonance
11
Prevents
(wave500
30
20
1500
interference)
http://www.radiosatellite.org/sirius_xm_orbits.gif
http://spaceplace.nasa.gov/en/kids/goes/goes_poes_orbits.shtml
Satellite Communications
Polar Satellite – low altitude orbit that
passes over the poles
(typically a few
hundred kilometers from the earth).
Orbital period 1-2 hours
 For extended communication the
satellite must be tracked and only a
limited window with which to
communicate.
 Uses: monitoring the weather, remote
sensing military surveillance (spying)

http://cimss.ssec.wisc.edu/satmet/modules/sat_basics/images/orbits.jpg
Satellite Communication
Function
Tracking
GEOSTATIONARY
No tracking
Constant position with Earth
Fixed communication link
POLAR
Must be tracked
Moves relative to the earth
Only temporary communication
Orbital Height Large distance~ 3.6 x 104 km
above the earth
Weaker signals received,
more attenuation, greater
time delay
Coverage
Large footprint ~ 40% of
earth.
Not able to communicate
with other side of Earth or
polar regions.
Smaller distance ~ few hundred
km above the earth
Stronger signal, less power
required
Uses
Weather, mapping, military
Communication
Restricted coverage
Communication possible with
most of the Earth at some time
Closer distance allows more
detailed monitoring
Geostationary Satellite Problem





If the distance of a geostationary satellite
from the center of the earth is 42,000 km.
How much of the earth can the satellite
see? Assume the radius of the earth is
6400 km.
Calculate the time between the broadcast
of the signal at A and point B receiving it
via the satellite.
This can be shown to correlate to 42% of the
entire earth’s surface.
How many satellites would be required to cover
the entire earth?
Is there any areas of the earth that cannot be
covered by a geostationary satellite?
Satellite
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Geostationary Satellite Problem
Satellite

If the distance of a geostationary satellite
from the center of the earth is 42,000 km.
How much of the earth can the satellite
see? Assume the radius of the earth is
6400 km.
6400
𝑅
−1
cos
= 81°
cos 𝜃 =
42000
𝑑



This can be shown to correlate to 42% of
the entire earth’s surface.
How many satellites would be required to
cover the entire earth? 3
Is there any areas of the earth that cannot
be covered by a geostationary satellite?
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Geostationary Satellite Problem


If the distance of a geostationary satellite
from the center of the earth is 42,000 km.
How much of the earth can the satellite
see? Assume the radius of the earth is
6400 km.
Calculate the time between the broadcast
of the signal at A and point B receiving it
via the satellite.
𝑥=
𝑑2 − 𝑅2
AB & BA is 2x
c=d/t t=d/c
2 𝑑2 − 𝑅2
𝑡=
𝑐
2 (4.2𝑥107 )2 −(6.4𝑥106 )2
𝑡=
3𝑥108
𝑡 = 0.28 s
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Polar Satellite Problem
A satellite is in a polar orbit at a height of
500 km from the Earth’s surface. The
satellite completes one orbit in 95 minutes
and has a speed of 7.6 x 103 m/s
Physics for the IB Diploma 5th Edition (Tsokos) 2008

Calculate the angle by which the earth has
rotated during one revolution of the satellite.
Assume that at time t=0 the satellite is directly
overhead the observer at O.
Estimate the time for which the satellite is
visible to an observer on the equator
Show that the satellite will pass any one point
on the earth’s surface twice in the course of
one day.
Physics for the IB Diploma 5th Edition (Tsokos) 2008


Polar Satellite Problem
A satellite is in a polar orbit at a height of
500 km from the Earth’s surface. The
satellite completes one orbit in 95 minutes
and has a speed of 7.6 x 103 m/s
Physics for the IB Diploma 5th Edition (Tsokos) 2008

Calculate the angle by which the earth has
rotated during one revolution of the satellite.
Assume that at time t=0 the satellite is directly
overhead the observer at O.
Estimate the time for which the satellite is
visible to an observer on the equator
Show that the satellite will pass any one point
on the earth’s surface twice in the course of
one day.
Physics for the IB Diploma 5th Edition (Tsokos) 2008


Physics for the IB Diploma 5th Edition (Tsokos) 2008
Polar Satellite Problem
A satellite is in a polar orbit at a height of
500 km from the Earth’s surface. The
satellite completes one orbit in 95 minutes
and has a speed of 7.6 x 103 m/s
Calculate the angle by which the earth has
rotated during one revolution of the satellite.
Assume that at time t=0 the satellite is directly
overhead the observer at O.
t=95 min
T = 24 h, θ = 360◦

95 𝑚𝑖𝑛
1ℎ
θ = 360° ×
×
= 24°
24 ℎ
60𝑚𝑖𝑛
Polar Satellite Problem
A satellite is in a polar orbit at a height of
500 km from the Earth’s surface. The
satellite completes one orbit in 95 minutes
and has a speed of 7.6 x 103 m/s

Physics for the IB Diploma 5th Edition (Tsokos) 2008
Estimate the time for which the satellite is
visible to an observer on the equator
 From the diagram
𝑅
𝑑
𝑐𝑜𝑠θ =
𝑡=
𝑅+ℎ
𝑣
6400
θ = cos −1
= 22°
5.3 × 10
6900
2𝜃 = 7.6 × 103
 Arclength AB is
𝐴𝐵 = 2𝜋 𝑅 + ℎ
360°
2(22)
𝐴𝐵 = 2𝜋 6400 + 500
= 5300km
360°
Polar Satellite Problem
Estimate the time for which the satellite is visible to
an observer on the equator
 From the diagram
𝑅
𝑐𝑜𝑠θ =
𝑅+ℎ
6400
θ = cos −1
= 22°
6900
2𝜃
 Arclength AB is
𝐴𝐵 = 2𝜋 𝑅 + ℎ
360°
2(22)
𝐴𝐵 = 2𝜋 6400 + 500
= 5300km
360°
𝑑
5.3 × 105 𝑚
𝑡= =
𝑣 7.6 × 103 𝑚/𝑠
Physics for the IB Diploma 5th Edition (Tsokos) 2008
𝑡 = 697𝑠 ≈ 12𝑚𝑖𝑛
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Polar Satellite Problem
A satellite is in a polar orbit at a
height of 500 km from the
Earth’s surface. The satellite
completes one orbit in 95
minutes and has a speed of
7.6 x 103 m/s
 Show that the satellite will
pass any one point on the
earth’s surface twice in the
course of one day.
 Satellite is above observer
O at t=0s.
 The earth rotates by 24◦
by time the satellite
completes one rotation.
Observer no longer visible.
Possible when observer
rotates 180◦
 Takes observer 12 hrs.
 Satellite completes
Physics for the IB Diploma 5th Edition (Tsokos) 2008
12h ×



𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛
95 𝑚𝑖𝑛
×
60 𝑚𝑖𝑛
1ℎ
= 7.58 𝑟𝑒𝑣
≈7.5 rev is a half integral.
Satellite is again directly
overhead the observer
Meets twice a day at the
opposite diametric points
on the equator
http://images.google.com/imgres?imgurl=http://celestrak.com/events/ISS-ASAT15.gif&imgrefurl=http://celestrak.com/events/asat.asp&usg=__mawfhz6G926CRH1AqWsdcEq40FA=&h=1015&w=1383&sz=170&hl=en&start=99&um=1&tbnid=is3VXlskkbaUAM:&tbnh=110&tbnw=150&prev=/images%3Fq%3Dpolar%2Borbit%2Bsatellite%2Banimated%2Bgif%26ndsp%3D20%26hl%3Den%26safe%3Dactive%26rlz%3D1T4ADBR_enUS244US244%26sa%3DN%26start%3D80%26um%3D1
View of ISS Orbit (green) and Debris Ring (red) from Chinese ASAT Test
Mobile Phone System
Mobile phone system developed by
Bell Labs in 1947.
 First mobile phone system established
in Nordic countries in the early 1980s.
 Main parts of mobile phone system

Mobile phone
 Base Stations
 Cellular exchange

Cell Phones
http://www.cordsplus.com/phoneinfo/images/system.jpg
Mobile Phone System


Phone turned on – sends a radio signal to the
nearest base station.
Base station – transmits and receives radio signals.



Range varies from 0.5 km to 30 km.
Connected via cables to the cellular exchange.
Arranged in approximately a hexagonal array.


Covers a geographic area without gaps.
Select a frequency for each call.


Other calls can use
the same frequency
through the time
division multiplexing
technique.
Also used to send
text messages,
photos an, internet access.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Mobile Phone System
http://www.cordsplus.com/phoneinfo/i
mages/system.jpg

Cellular Exchange – controls the operation of many
base stations.



Offers entry into the public switched telephone network
(PSTN).
Allocates a range of frequencies for each cell.
If the mobile phone moves during the call, the cellular
exchange will automatically reroute the phone call to the
base station at the center of the new cell.
http://www.kottke.org/plus/misc/images/iphone-parallels.jpg
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Mobile Phone Source Source: Physics for the IB Diploma 5 th Edition (Tsokos) 2008
Digital Electronics
Purpose – control electricity.
 Two general types
History

Copyright © by Electronic Kourseware Interactive

Switching – on (high) or
off (low) (digital)
Copyright © by Electronic Kourseware Interactive

Regulating – varying
between a minimum and
maximum (analog)
Copyright © by Electronic Kourseware Interactive
Digital Electronics


In a digital circuit there are
only two outputs – HI (1) or
LOW (0)
These output conditions for
all possible inputs conditions
of a circuit can be
represented in a truth
table


A simple example is shown
However, Combining switches or
gates very complex functions can
be performed.
Logic Gates

These combinations are known as
logic gates. There are six basic logic
gates.
Yes
 NOT
 AND
 OR
 NAND
 NOR

Copyright © by Electronic Kourseware Interactive
http://www.cs.nyu.edu/courses/fall99/V22.0436001/gates.png
Logic Gates

AND Logic Gate –


When both inputs are high the output is
high otherwise the output is low.
OR Logic Gate –

Copyright © by Electronic Kourseware Interactive
When both inputs or either input is high
the output is high.
Copyright © by Electronic Kourseware Interactive
Logic Gates

NAND Gate –

An and gate with an inverter (solid
round circle) resulting in the opposite
result of an AND gate.
Copyright © by Electronic Kourseware Interactive
Copyright © by Electronic Kourseware Interactive
http://upload.wikimedia.org/wikipedia/commons/2/2b/Seven_segment_display-animated.gif
Logic Gates


Output often
shown with a
logic indicator
Light Emitting
Diode (LED)


HI – LED On
LOW – LED Off
http://img116.imageshack.us/img116/3852/gates6df.gif
Copyright © by Electronic Kourseware Interactive
Boolean Algebra

Mathematical system used to represent the
operation of digital logic circuit
Copyright © by Electronic Kourseware Interactive


The six basic gates are the basic operations
in Boolean Algebra.
Flow charts are created using these
operators in combinations for desired
output and then circuits are created out of
components that match these functions.
Integrated Circuits (ICs)



Initially discreet (individual) elements were
used to build these logic gates.
Now integrated circuits – whole circuits on
a single chip - are used.
ICs are classified
by their packaging
and family.

IC Packages

Copyright © by Electronic Kourseware Interactive
Dual in Line (DIP), round, and flat- pack.
Copyright © by Electronic Kourseware Interactive
Integrated Circuits (ICs)

IC Families
Transistors
TTL (Transistor-Transistor Logic)
 CMOS (Complementary Metal Oxide
Semiconductor)
 Linear – analog devices

IC Revolution
Copyright © by Electronic Kourseware Interactive
Trivia: Why a Breadboard?
Logic & IC Source: Electronic Kourseware Interactive 2000
Operational Amplifier (op-amp)
Highly versatile integrated circuit
 Main Features

Very high gain in the output voltage (106)
 Very high input resistance – draws
minimal current from the input signal
 Very low output resistance – any load can
be driven no matter how low its
resistance.
 Differential amplifier – produces an output
signal that is proportional to the
difference between two input signals.

Op-Amps
http://www.hi-fi-insight.com/wp-content/uploads/2007/01/Burr_Brown_OPA627AP_Op_Amp.jpg

1.
2.
3.
4.
5.
Designed with 8 pins
our focus is 5 of these.
Inverting input : input will be
changed in sign (V-)
Non-inverting input : no
change in sign (V+)
Output : output signal of the
op-amp V0
Positive voltage supply : +V
Negative voltage supply : -V
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Block diagram of op-amp
Op-Amps
Physics for the IB Diploma 5th Edition (Tsokos) 2008


The output voltage (V0) is directly
proportional to the difference between the
two input voltages, V+ and V-
G0 is the open loop gain ~106 for DC and
low frequency signals.


As frequency increases the
gain decreases.
Relationship, limited to
output voltages below the
saturation values (±80% V).
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Op Amps
Ideal



Infinite Gain
Inputs draw no
current.
Output limited to
supply voltages
(±15 V)
Source: Kirk, Tim, Physics for the IB Diploma, Oxford University Press 2007
Reality
 Gain of 105-106
 Input resistance 106
Ω, draws a very
small current
 For a ±15 V supply
voltage the outputs
is fixed between ±13
V , when the output
is 13 V it is
considered saturated
Inverting Op-Amp


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
Has a feedback loop with a feedback resistor (RF)
Part of the output is fed back as input to the op-amp.
Since the feedback is to the inverting input, the
signal is 180° out of phase from the original input –
negative feedback.
Gain of the amplifier is reduced.
Results in the gain being stable over a wide range
of voltages and
frequencies, and
independent of the
characteristics of the
op-amp itself.
Physics for the IB Diploma 5th Edition (Tsokos) 2008




Physics for the IB Diploma 5th Edition (Tsokos) 2008
Inverting Op-Amp
Closed Loop Gain (G) –
Derivation
G0 (open loop gain) ~106
V+=0 so V- at E is
Physics for the IB Diploma 5th Edition (Tsokos) 2008
(virtual earth approximation)
Calculate the closed loop
gain of the inverting
so
and
amplifier shown above
were R = 100 k and RF =
since ~ no current flows
1.0 M.
in the op amp
 If the op-amp is connected
, rearranging to a ±15 V power supply.
Calculate the input voltage
G is the Closed for which the op-amp will
Loop Gain
saturate.
Point E is connected to ground



VR=Vin

Non-Inverting Op-Amp



Two resistors shown, R and RF act as a
potential divider dividing the potential
difference V0 in the ratio RF:R.
The voltage at X is also the voltage at Y.
The closed gain loop of non-inverting
amplifier is
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Non-Inverting Op-Amp
Gain – Derivation

 If potential difference at
X is V, then the potential 
difference across R is also
V.
 The current I in R and RF
is the same and the total
resistance of R and RF is
R+RF.
 So, V=IR and V0=I(R+RF)
 Since the currents are the
same,
Physics for the IB Diploma 5th Edition (Tsokos) 2008
The input voltages are
V+=Vin and
Applying the
open loop gain, G0, to
the circuit shown in
dashed lines. V0=G0(V+V-) or
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Since V0=GVin where G
is the closed loop gain
𝐺𝑉𝑖𝑛 = 𝐺𝑂 𝑉𝑖𝑛 −
𝑅
𝐺𝑉
𝑅 + 𝑅𝐹 𝑖𝑛
Non-Inverting Op-Amp
𝐺𝑉𝑖𝑛 = 𝐺𝑂
𝑅
𝑉𝑖𝑛 −
𝐺𝑉
𝑅 + 𝑅𝐹 𝑖𝑛
Physics for the IB Diploma 5th Edition (Tsokos) 2008

Solving for G

Since the term
the 1 term can be dropped

Physics for the IB Diploma 5th Edition (Tsokos) 2008
Calculate the closed loop
gain of the non-inverting
amplifier above when
R=10 and RF=100
The closed loop gain of a
non-inverting amplifier is If the op-amp is connected to
a ±15V power supply, the
output-input characteristic is
Op-Amp Simple Circuits




Comparator – compares
Physics for the IB Diploma 5th Edition (Tsokos) 2008
one voltage to another.
Voltage to be compared
are the inputs
If a ±15V Power supply is
used with a 15 V Saturation
voltage.
In open loop state the output
will saturate when the
absolute value of the
difference between the two
inputs is greater than ±15µV.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Op-Amp Simple Circuits





In the circuit shown R1
and R2 determine the
input voltages.
If these resistors are
replaced with a potentiometer, and
V- is zero, a sinusoidal signal is
sent by the signal generator.
Shown is the output (oscilloscope)
Since in the first 5ms the
sinusoidal V+ is larger than V- by
15µV the signal saturates.
The reverse then occurs resulting
in this stepped output.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Op-Amp Comparator Circuit
Contains a thermistor –resistor
whose resistance decreases with
increasing temperature.
 Buzzer – operates when the
voltage across it is 30V.
Operation –
 Voltage at the inverting input
determined by the voltage at
point X, reference voltage.
 If R1 and R2 are equal then the
reference voltage at X is zero.
 When the thermistor is cold, its
resistance is high. So the
voltage at the non-inverting
input is high and the output
voltage saturates at +15V.





If the temperature
increases, the voltage at
the non-inverting input will
become negative.
If temperature rises
enough, the output
voltage will saturate at 15V.
The potential difference
across the resistor will be
30V.
The buzzer will sound
indicating a warning of the
increased temperature.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Op-Amp Comparator Circuit



In an alternative setup, the buzzers are
replaced with 2 LEDs.
A green LED is lit if the output voltage
saturates positively
A red LED is lit if the output voltage
saturates negatively.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Op Amp Source: Physics for the IB Diploma 5th Edition (Tsokos) 2008
Schmitt Trigger


Used to reshape digital
signals that have been
distorted.
Input-output
characteristic

Zero input voltage the
output is –V0.
As input voltage
increases, the output
remains at –V0 until the
input reaches the
threshold voltage V2
Here the output then
jumps abruptly to the
value +V0





If the input signal decreases,
the output stays at +V0 until
the lower threshold V1 is
reached.
Here the output now jumps
abruptly to –V0.
So the output is determined
by the two threshold voltages
V1 and V2.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Schmitt Trigger

Schmitt trigger –
works as a standard
comparator


In addition the
reference value is
different when the
input is increasing (V2)
from when it is
decreasing (V1)
Example: consider a
Schmitt trigger with values
V1=0.40V, V2=0.75V, and
V0=3.0V with the input
signal the figure shown.
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Schmitt Trigger
Example:
R1=15k, R2=10k,
R3=100k and a reference
voltage of 3.0V with the
thresholds being 0.75V and
0.40V.
Calculating the threshold voltages
Case 1: Input voltage < V+.
The output will be V0
Current at R2 is

Current at R3 is
The voltage across R1 is V++V0
So the current at R1 is
Since I1=I2+I3
Source: Physics for the IB Diploma 5th Edition (Tsokos) 2008
Physics for the IB Diploma 5th Edition (Tsokos) 2008

Solving for V+ gives the
high threshold

Similarly, the low
threshold is
Sources
Primary Source:
Physics for the IB Diploma 5th Edition (Tsokos) 2008
Supplementary Sources:
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Cutnell & Johnson, Wiley Publishing, Physics 5th Ed.
Complete binary mathematics.
http://www.binarymath.info/
convert binary numbers to decimals
http://www.bellaonline.com/articles/art48652.asp
Basic principles of magnetic recording using digital data in HDD.
http://www.usbyte.com/common/HDD.htm#top
storage types
http://www.jegsworks.com/Lessons/lesson6/lesson6-2.htm
Details of DVDs.
http://electronics.howstuffworks.com/dvd3.htm
Details of Blu Ray.
http://electronics.howstuffworks.com/blu-ray1.htm
Optical recording in a CD.
http://physicsworld.com/cws/article/print/1383/1/world-11-10-7-2
Detailed study of dvds and cds comparison. More emphasis on technology of data
capture.
http://www.iti.uni-stuttgart.de/~ghermanv/Lehre/Seminar/material/Presentation4/talk.pdf
For complete know how on CCDs and image capture in a digital camera
www.howstuffworks.com
Digital Electronics
“Digital Magic’ Mr. Circuit II, Electronic Kourseware Interactive, 2000