Download Biophysics department, Faculty of sciences, Cairo university [1

Biophysics department, Faculty of sciences, Cairo university
[1]
Measurement of Young's modulus of a bone-equivalent
material
Aim:
1- To measure Young’s modulus of a bone- equivalent material
2- To determine an unknown pan mass.
Theory:
Applying stress on a solid material results in what so called strain.
Stress: is the force applied per unit area of the material
F / A ; it has the unit of pressure N / m 2 , it
is also the restoring force caused due to the deformation divided by the area to which the force is
applied, as the applied force is opposed by an equivalent internal force.
Strain: is the deformation that takes place due to the stress. It is caused by the relative displacement
of particles making up the object. For tensile stress, strain is measured as the change in length
divided by the original length
L / L , therefore, strain is dimensionless.
The ability of the object to reestablish its original shape and length after releasing the applied stress
is called elasticity. Elastic deformation means non-permanent deformation. The restoring forces are
called elastic forces.
The measured strain is directly proportional to the applied stress (Hooke's law). The proportionality
constant is called elastic modulus, for tensile strength it is called Young’s modulus
F
L
E
A
L
1
E  :
Biophysics department, Faculty of sciences, Cairo university
For small stresses, the object can reestablish its original shape and volume. Upon increasing the
stress and at certain value, the object will no longer be able to reestablish its original shape and
volume after releasing the applied stress. This certain stress is called the elastic limit, beyond which,
the object will be deformed permanently which we call plastic deformation. Internal forces
appearing in the body in the course of plastic deformation are also termed plastic forces. At the end,
the object will break down, and this value of stress is called the fracture point.
The stress range beginning at zero stress and ending at the elastic limit is called the elastic region,
during which Hooke’s law is satisfied. The elastic modulus of an object can be defined as
the slope of its stress–strain curve in the elastic deformation region. As such, a stiffer material will
have a higher elastic modulus and better elastic properties.
Tension
(F/A)
C
Stress
B
A
Strain
Figure (1)
Bone as well as some other materials shows an elastic behavior. Bone consists of quite different
materials: Collagen and Bone mineral plus water. Collagen is flexible like rubber, while bone
mineral is very fragile.
Young's modulus for compact bone is greater than that for trabecular bone.
figure (2)
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Biophysics department, Faculty of sciences, Cairo university
In this experiment, loads are suspended from a horizontal bar made of bone equivalent material to
determine Young's modulus
E  of bone, if a weight W  mg is suspended, the depression
Y  of the bar (see Figure (2)) can be given by:
1 4 L3
Y
W
E bd 3
Where
(2)
L , b  and d  are the length, the breadth and the thickness of the bar respectively.
So, we can re-write:
1 4 L3 g
Y
m
3
E bd
Therefore, the Young's modulus
(3)
E  can be calculated as:
m 4 L3 g
E
Y bd 3
As
(m / Y )
(4)
is the reciprocal of the slope of the relation between the depression
and the suspended mass
Y  on the y-axis
m on the x-axis, we can re-write:
1 4 L3 g
E
Slope bd 3
The slope can be calculated from the plot of the depression
and
Y  against the suspended mass m ,
L , b  and d  can be measured by a ruler.
For your knowledge
Bone as well as some other materials shows an elastic behavior. Bone consists of two quite different
materials plus water:
(1) Collagen: It is the major organic fraction, which is about 40% of the weight of solid bone and
60% of its volume.
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Biophysics department, Faculty of sciences, Cairo university
(2) Bone minerals: It is the inorganic component of bone, which is about 60% of the weight of the
bone and 40% of its volume.
Collagen is flexible like rubber, while bone mineral is very fragile. Bone mineral crystals are
believed to be made of calcium hydroxyapatite [Ca10(PO4)6(OH)2] and are rod shaped with diameters
of 20 -70 Ǻ and length of 50 -100 Ǻ.
There are two types of bone:
(1) Solid or compact bone which has constant density throughout the life (ρ = 1.9 g/cm3).
(2) Spongy or trabecular bone.
Young's modulus for compact bone is equal to 1.8 x 1010 N/m2, while it is only 8 x 107 N/m2 for
trabecular bone.
Bone material is as strong as granite in compression and 25 times stronger than granite under
tension. Healthy compact bone is able to withstand a compressive stress of about 170 N/m 2 before it
fractures. The midshaft of the femur has a cross-section area of about 3.3 cm2; it would support a
force of about 5.7 x 104 N (6 tons).
Generally, bones are not as strong under tension as they are in compression; a tension stress of about
120 N/mm2 will cause a bone to break. However, bone is stronger under tension than many common
materials such as porcelain, oak wood and concrete.
Procedure:
L , b  and d  in cm.
1.
Measure
2.
Take the zero reading (without the hook suspended from the bar).
3.
Apply various loads on the hook. Then record the reading of the meter and calculate the
depression (Y) in cm in each case. Increasing the load by equal amounts (100 grams each). Take
six or seven readings both with increasing and with decreasing loads.
Note: Remove the load from the hook after taking each reading.
4.
Tabulate the results.
5.
Draw the relation between the applied load (m) and the corresponding mean depression (Y).
6.
From the straight line obtained, determine the slope of equation (3) and the intercept (Figure 3).
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Biophysics department, Faculty of sciences, Cairo university
7.
From the slope determine the Young's modulus (E) as:
1 4 L3 g
Slope 
E bd 3
Therefore,
1 4 L3 g
E
Slope bd 3
7.
The straight line intersects the negative mass axis at a value corresponding to the mass of the
pan (hook), mo.
Y (cm)
Slope = Y/m
m (g)
mo
Figure (3)
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Biophysics department, Faculty of sciences, Cairo university
Results:
Zero Reading =
Mass (m)
Depression (y)
in (g)
in (cm)
50
100
150
200
250
L=
cm
b=
cm
d=
cm
Slope = Y/m =
cm/g
1 4 L3 g
E
Slope bd 3
E=
dyne/cm2
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Biophysics department, Faculty of sciences, Cairo university
[2]
Cathode ray tube (Oscilloscope)
Aim:
1.
Determine of the amplitude, the periodic time and the frequency of the given waves.
2.
Sketch the given waves.
Theory:
The cathode ray tube is a vacuum tube containing a gas at a pressure of 0.01 mmHg. It contains an
electron gun at its narrow end. The inner side of its wide end is coated with a fluorescent material
such as zinc supplied representing the screen. The electron gun (cathode) emits a beam of electrons
which is controlled and focused to a bright spot on the fluorescent screen.
A part of the kinetic energy of the electrons is transferred to the fluorescent material which gives
light. The color of the light depends on the kind of the coating (fluorescent material) and on the
energy of the falling electrons. A bright spot appears on the screen and determines the position of the
falling electrons on the screen.
The tube contains deflecting system near its middle. The deflecting system may consist of two pairs
of metal plates which produce two electric fields or of two pairs of coils which produce two
magnetic fields. These fields deflect the beam of electrons.
The inner side of the wide end of the tube is coated with a layer of emulsion of carbon which is
connected to the cathode for carrying electrons back to the cathode without the accumulation of a
charge on the screen.
The cathode ray oscilloscope is a cathode ray tube having a deflecting system consisting of two
pairs of metal plates (x1, x2) and (y1, y2) which produce two electric fields perpendicular to each
other and at the same time perpendicular to the direction of motion of electrons.
If the two vertical plates (x1, x2) are connected to alternating source, they produce an electric field in
a horizontal plane. The direction of the electric field moves from right to left and vice versa.
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Biophysics department, Faculty of sciences, Cairo university
If the frequency of the current is more than 16 Hz, the eye sees a straight line due to the persistence
of vision.
If the horizontal plates (y1, y2) are only connected to an alternating source, they produce an electric
field in a vertical plane. The direction of the field varies up and down.
It is a customary to connect the two vertical plates (x1, x2) to a source which gives a smooth potential
difference. This source is a special electronic valve. The circuit is known as a sweep circuit. The
voltage on the vertical plates' increases gradually until reaching a maximum value, then it vanishes
suddenly and this is periodically repeated. As a result of this alternating voltage, the beam spot start
its motion from left to right in a straight line across the screen at constant speed and then jumps
quickly back to repeat the motion from left to right. When an alternating voltage is desired to be
investigated, this voltage is connected to the two horizontal plates (y1, y2). At the same time the
saw-tooth voltage is connected to the two plates (x1, x2). The luminous spot draws a graph of the
voltage under investigation.
Figure (1)
By varying the frequency of the saw-tooth voltage until it matches the frequency of the investigated
voltage, the fluorescent screen will show a stationary voltage and the fluorescent screen will show a
stationary graph. Therefore, knowing the frequency of the tooth voltage, the frequency and the
Periodic time of the investigated voltage can be known.
Frequency (f): It is the number of complete cycles per unit time.
Periodic time (T): It is the time taken to complete one cycle of the wave.
Amplitude (A): It is the maximum displacement of the wave.
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Biophysics department, Faculty of sciences, Cairo university
Procedure:
1.
Connect the generator to the oscilloscope to produce different waves (sine, square and
triangular).
2.
Draw the obtained sine waveform from the screen of the oscilloscope.
3.
From the graph determine the amplitude (A), the periodic time (T) and the frequency (f) of the
wave using the following formulae:
Amplitude (A) = (Vpp/2) x (volt/div.)
(Volt)
Periodic time (T) = Distance between two successive crests x (time/div.)
Frequency (f) = 1/T
4.
(Hz)
Repeat steps (2 and 3) for the square wave and the triangular wave.
9
(Second).
10
wave
Square
wave
Triangular
Sine wave
shape
Wave
(divisions)
↕y
(volt/div)
(volts)
Vpp
(Volt)
= Vpp/2
Amplitude
(divisions)
x
(time/div)
(s)
Time (T)
Periodic
(Hz)
or
(s-1)
= 1/T
Frequency
Biophysics department, Faculty of sciences, Cairo university
Results:
Useful link: http://www.kentchemistry.com/links/AtomicStructure/flash/CathodeRayTube.swf
Biophysics department, Faculty of sciences, Cairo university
[3]
Measurement of the Radius of Sucrose Molecule
Aim:
1.
To determine the absolute and the relative viscosity coefficients of the different sucrose
concentrations.
2.
To determine the radius of sucrose molecule.
3.
To find the concentration of unknown sucrose solution.
Theory:
Viscosity: is the frictional resistance offered by a liquid against the displacement of its own
molecules.
The viscosity coefficient of a liquid is generally measured by observing the time required for a
definite volume of the liquid to flow through a standard capillary tube under a known pressure
difference (Figure 1). The device used to measure the time of flow is called Ostwald’s viscometer.
A
B
Figure (1). Ostwald’s viscometer.
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Biophysics department, Faculty of sciences, Cairo university
Poiseuille’s law:
The law governing the laminar flow of viscous fluids through capillary tubes is derived by
Poiseuille and the coefficient of viscosity (η) is given by:
η = ( P r4 t) / 8 VL
(1)
Where ΔP: is the pressure difference at the ends of the capillary tube.
t: is the time of flow of the liquid from the upper mark A to the lower mark B.
r: is the radius of the capillary tube.
L: is the length of the capillary tube.
V: is the volume of the liquid of viscosity (η).
Units of η:
Since

Force.of .vis cos ity / A
F .S

Velocity.gradient .of .the.liquid
A.v
Where A is the area in m2. Therefore the unit of viscosity (η) will be:
N.m.sec/ m2.m= N.s/m² or dyne.s/cm²
= kg.m.m.s /s².m².m = kg/m.s
or
gm/cm.s = Poise
If the time of flow of equal volumes of two different liquids through the same
capillary are measured under the same experimental conditions, therefore the ratio between their
viscosities (η1 and η2), (the relative viscosity coefficient, r) is given by:
ηr = η1/η2 = [( P1 r4 t1) / 8 VL] / [( P2 r4 t2) / 8 VL ]
ηr = P1 t1 / P2 t2
(2)
P is the average difference in pressure.
Since, P = ρ g h /2, then:
r = (ρ1 g h t1 / 2) / (ρ2 g h t2 /2) = (ρ1 t1) / (ρ2 t2)
Where
ρ1 and ρ2: are the densities of the liquid (1) and liquid (2) respectively.
t1 and t2: are the time of flow of the liquid (1) and liquid (2), respectively.
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(3)
Biophysics department, Faculty of sciences, Cairo university
If η2 is the absolute viscosity of known liquid (water), then the absolute viscosity of the unknown
liquid can be determined.
For laminar flow through capillary tube, the coefficient of viscosity is given by equation (1)
[η = ( P r4 t) / 8 VL] and was modified by Einstein to:
η = η1 (1 + 2.5 )
(4)
Where:
η = is the absolute viscosity coefficient of solution.
η1: is the absolute viscosity coefficient of the solvent.
: is the volume fraction of the solute which is equal to Volume of the solute / volume of the
solution.
Equation (4) is valid only for the following assumptions:
1. The solute molecules are rigid and spherical.
2. No mutual interaction between the solute molecules.
3. The solute molecules are larger than the molecules of the solvent.
To measure viscosity coefficient of a solution in Ostwald’s viscometer the solute molecules
should be smaller than the radius of the capillary tube.
Equation (4) is then modified to:
η/η1 = 1 + 2.5 
Then,
ηr = 1 + 2.5 
ηr – 1 = 2.5 
(5)
Since,  = Volume of the solute/volume of the solution and from the assumptions of Einstein`s
equation, the volume of the solute (sucrose) can be given by:
4
Volume of the solute =
3
 r3N
(6)
Where N is the number of sucrose molecules and r is the radius of sucrose molecule.
Now, to calculate the volume of the solution:
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Biophysics department, Faculty of sciences, Cairo university
The concentration (C) of a solution is given by
C = mass of the solute / volume of the solution
So, Volume of the solution = mass of the solute / concentration of the solution
Volume of the solution = m / C
(7)
m= Mwt.N / Nav
By substituting by the value of m in equation (7), then:
Volume of the solution = N. Mwt / NA. C
=(
4
3
 r³ C) NA / Mwt
(8)
(9)
Substituting for  from equation (9) in equation (5) we get:
ηr – 1 = 2.5 [(
ηr – 1 = [2.5 (
4
3
4
3
 r³ C) NA / Mwt ]
 r³ ) C NA ] / Mwt
(ηr – 1 )/C = [2.5 (
4
3
 r³) NA] / Mwt
By substituting for Nav , the entire constant together, we get:
(ηr – 1 )/C = [6.3 x 1024 r³] / Mwt
(10)
Since the molecular structure of sucrose is C12 H22 O11 , therefore its molecular weight will be:
Mwt = (12 x 12) + (22 x 1) + (11 x 16) = 342
and (ηr – 1 )/C is the slope of the straight line obtained from the practical work. So, equation (10)
becomes:
Slope = (6.3 x 1024 r3) / 342
And
r = [(slope x 342) / 6.3 x 1024]1/3
(11)
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Biophysics department, Faculty of sciences, Cairo university
Procedure:
1.
Prepare 5 sucrose solutions with different concentrations.
2.
Use clean and dry viscometer.
3.
Introduce certain volume of distilled water in the wide tube of the viscometer.
4.
Force up the water through the capillary by suction with a syringe through a rubber tube
attached to the end of the viscometer until the water fills the bulbs and rises slightly above the
mark A, then remove the syringe.
5.
Allow the water to flow back through the capillary tube and a stopwatch is started when the
water passes the upper mark (A) and is stopped when the liquid passes the lower mark (B).
The time of flow of the water (tw) is recorded. Measure tw three times then take the average tw
average.
6.
Repeat the above steps for the five solutions of different concentrations of sucrose,
find tS average for each solution.
7.
Find the relative viscosity (ηr) from the relation r = ρs ts / ρw tw .
8.
Find the absolute viscosity coefficient (ηs) of each solution from the relation:
ηs = 0.01 x ηr [The absolute viscosity coefficient of water (ηw) is 0.01 poise].
9.
Plot a relation between ηr – 1 and C.
10. From the graph find the slope and then substitute in equation (11) to find the radius (r) of
sucrose molecule
r – 1
Slope = r – 1
C
Concentration (gm/ml)
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Biophysics department, Faculty of sciences, Cairo university
Safety and Precautions:
 Be careful when dealing with any glasses in the lab, it could be easily broken, deal with it
gently.
 The device (viscometer) must be washed very well before use.
 All liquids must be filled to 2/3 of the large bulb, the assumption of the fixed volume is very
important to get best results.
 The liquid must be sucked slowly to avoid the formation of any air bubbles.
 The least time calculated should be that concerning water because the water has the lowest
viscosity among liquids.
 The straight line drawn from the relation between ηr – 1 and C must pass through the origin
because we begin with water which has a concentration of 0 gm/ml of sucrose and the ηr – 1 of
water equal to 0.
 The concentration must be in gm/ml otherwise you must convert it to gm/ml if it is given in
other units.
 If an unknown solution is given to calculate its concentration then, we deal with it like any
other liquid, we calculate its ηr – 1, and then we can get easily its concentration in gm/ml from
the graph.
 Start with water then the liquid with the least concentration and so on.
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Biophysics department, Faculty of sciences, Cairo university
Results:
Solution Concentration
(C) (gm/ml)
Density
t (s)
r = stys/wtw
()
ηs = 0.01 x ηr
( poise)
(gm/cm3)
Water
1
2
3
4
Unknown
r = [(slope x 342) / 6.3 x 1024]1/3
Useful links: http://www.physics.usyd.edu.au/teach_res/jp/fluids/viscosity.pdf
http://www.youtube.com/watch?v=Gs3gfwG9a7k
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r - 1
Biophysics department, Faculty of sciences, Cairo university
[4]
Measurement of Short Length Using Vernier Caliper
Aim:
1. To study the vernier caliper as a tool to measure short lengths (e.g. beans lengths).
2. To know how to handle the data statistically by determining the mean value (Xtheo) of the sample
and its standard deviation (σ).
3. To draw a histogram for the results and to find the mean value (Xexp) from it.
Theory:
The Vernier Caliper is a precision instrument that can be used to measure internal and external
distances extremely accurately. The example shown below is a manual caliper. Measurements are
interpreted from the scale by the user. This is more difficult than using a digital vernier caliper which
has an LCD digital display on which the reading appears. Manually operated vernier calipers are more
trusted than digital ones.
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Biophysics department, Faculty of sciences, Cairo university
How to read a measurement from the scales
Example 1:
A. The main metric scale is read first and this shows that there are 13 whole divisions before the 0
on the hundredths scale. Therefore, the first number is 13.
B. The’ hundredths of mm’ scale is then read. The best way to do this is to count the number of
divisions until you get to the division that lines up (coincide) with the main metric scale. There are
21 divisions on the hundredths scale
C. This 21 is multiplied by 0.02 giving 0.42 as the answer (each division on the hundredths scale is
equivalent to 0.02mm
D. The 13 and the 0.42 are added together to give the final measurement of 13.42 mm (the object
length).
Example 2:
32
22
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Biophysics department, Faculty of sciences, Cairo university
Frequency Distribution:
If the results of measuring the length of a number of beans are n values, e.g. X1, X2, X3, X4, X5,…..,
Xn, then two quantities are defined, namely, the mean value (X) and the standard deviation (σ). The
members of a population can be grouped and tabulated into small successive increments of their
magnitudes. Any one group is called class. The number of members within each class forms a
frequency distribution of the observed values. The frequency and class can be represented by a
graph consisting of adjacent rectangles each of base width equals to the class interval and height
equals to the corresponding frequency of the class. This representation is called the histogram.
Histogram gives information about the frequency distribution of the observed magnitudes (see the
figure in below). From the histogram, the experimental mean value (Xexp) can be found from the half
of the maximum column.
Frequency
Class interval
Definitions and Relations:
The mean value (X): It is the arithmetic average of a group of numbers.
The mean = the sum of the values / the number of values (e.g. the mean of 1, 3, 7, 10 and 13 is 6.8).
The mean value (X) of different values X1, X2, X3,…,Xn, can be calculated as follows;
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Biophysics department, Faculty of sciences, Cairo university
n
X  X  X  .....  X
1
2
3
n
X=
n
 X i2
i 1
=
n
(1)
The Standard Deviation (σ):
The value of σ determines the state of the sample, and the amount of deviation from the mean value.
If the value of σ is very small, thus the sample is said to be equilibrated (i.e. all the values are very
narrow). If the value of σ is large, thus the sample is said to be un-equilibrated (i.e. all the values are
different).
=
 (X i  X) 2
n 1
=
 (ΔX i ) 2
(2)
n 1
Class Interval: Interval contains group of members in desired range.
Frequency: Number of members in each group.
Procedure:
1. Study the vernier caliper as a tool to measure the length of beans.
2. Determine the lengths of 15 beans.
3. Classify the results into at least 5 class intervals.
4. Draw the histogram distribution.
5. From the histogram, determine the mean experimentally (Xexp) and compare it with that calculated
theoretically (Xtheo).
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Biophysics department, Faculty of sciences, Cairo university
Results:
Serial Number (n)
n = 15
Xi (cm)
 Xi =
Xi = (Xi – X)
 (Xi)  0
Range = Max. Length – Min. Length
Class Interval Width = Range/Number of Intervals
22
(Xi)2
 (Xi)2 =
Biophysics department, Faculty of sciences, Cairo university
Class Interval
Frequency
Useful link: http://janggeng.com/vernier-caliper/
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Biophysics department, Faculty of sciences, Cairo university
[5]
Verification of Ohm`s Law
Aim:
To verify Ohm`s law by drawing a graph between the potential difference and the current.
Theory:
Ohm`s law relates the potential difference applied at the terminals of a conductor and the current
flowing through it.
The current produced in a given conductor kept at constant temperature is directly proportional to the
difference of potential across its terminals.
This proportionality stated as Ohm`s law
V α I
V=IR
(1)
Where V is the potential difference in volt (V), I is the current in Ampere (A) and R is the resistance
in Ohm (). Therefore,
R = V/I
(2)
The Ohm is the resistance of a conductor that supports a current of one ampere when a potential
difference of one volt is impressed across its terminals.
The substances which obey Ohm’s law on applying the electric potential across them are called
ohmic materials; such as: Aluminum, silver, gold, while those that don’t obey Ohm's Law are called
non-ohmic materials such as diodes.
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Biophysics department, Faculty of sciences, Cairo university
Procedure:
1.
Measure both resistances Rs and RL and then choose the lowest one as the resistance under
investigation to verify Ohm's law.
2.
Connect the circuit as shown in the Figure.
RL
A
RS
V
E
3.
Switch on the circuit and note the values of the current (I) flowing through the circuit (through
RS) as indicated by the ammeter in Ampere (A) and the potential difference (VRs) across the RS
as indicated by the voltmeter in volt (V).
4.
Repeat step (3) several times with changing the applied voltage and record the results in table.
5.
Plot a graph between the potential difference (VRs) and the current (I) as shown below.
I (Ampere)
I2
I1
V1
V2
VRs (volt)
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Biophysics department, Faculty of sciences, Cairo university
6.
Since the graph is a straight line relation, therefore, Ohm's law is verified.
7.
From the slope of the line (= 1/RS), calculate RS and compare it with the measured value.
Precautions and Safety

Do not touch or use any electronic component if your hand is wet.

Check up the electric circuit with the laboratory demonstrator before switching it on.
.
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Biophysics department, Faculty of sciences, Cairo university
Results:
RL =
RS =
Potential difference, VRs
Current , I
(volt)
(Ampere)
Slope =
RL exp =
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Biophysics department, Faculty of sciences, Cairo university
Digital multimeter
Digital multimeter
Power supply
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Biophysics department, Faculty of sciences, Cairo university
[6]
Characteristics of a P-N Junction Diode
Aim:
1.
To plot the characteristic curve of a given p-n junction diode.
2.
To Find the voltage drop (VF) of a given p-n junction diode.
Theory:
There are three types of materials as shown in figure 1:
1.
Insulators in which the band gap between valance band and conduction band is very large, so
their valance electrons are very tightly bound to their atoms .Therefore, insulators cannot
conduct electric current (Figure 1a).
2.
Semiconductors such as silicon or germanium in which the band gap between valance band and
conduction band is very small, somewhere, between those of conductors and insulators (Figure
1b).
3.
Conductors such as metals in which the electrons are freely moving through the crystal lattice
due to the closeness of the valance and conduction bands. So they conduct electric current
(Figure 1c).
Figure (1). Energy band gaps in materials.
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Biophysics department, Faculty of sciences, Cairo university
The ability of a semiconductor to conduct electricity can be changed dramatically by adding small
numbers of different elements to the semiconductor crystal. This process is called doping (Figure 2).
Figure (2). Silicon crystal doped with phosphorus (P) (n-type) and boron (B) (p-type).
There are two types of extrinsic semiconductors:

n-type (negative-type semiconductor) which contains large number of free electrons that move
freely and can contribute to an electric current.

p-type (positive-type semiconductor) which a large number of positive charge carriers called
holes available for conduction.
Imagine that a p-type block of silicon can be placed in perfect contact with an n-type block. Free
electrons from the n-type region will diffuse across the junction to the p-type side where they will
recombine with some of the holes in the p-type material. Similarly, holes will diffuse across the
junction in the opposite direction and recombine (Figure 3).
The recombination of free electrons and holes in the vicinity of the junction leaves a narrow region
on either side of the junction that contains no mobile charges. This narrow region which has been
depleted of mobile charge is called the depletion layer. It extends into both the p-type and n-type
regions (Figure 3).
A potential difference must therefore exist across the depletion layer. The variation of potential with
distance is shown in (Figure (3) bottom).
We notice that there is a large drop in the number of mobile electrons from right to left; this large
drop is called a potential hill
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Biophysics department, Faculty of sciences, Cairo university
Figure (3). Depletion layer and the potential hill at the junction between two Silicon crystals doped
with Boron (p-type) and with Phosphorus (n-type).
The p-n junction is connected in the electric circuit either through forward or reverse bias (Figure 4).
p
+++
+++
+++
n
p
n
+
-+ ---+ - - -+ ---
+
+
(a) Forward bias
-- - -
++
++
++
-
(b) Reverse bias
Figure (4)
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Biophysics department, Faculty of sciences, Cairo university
By forward biasing the diode, it is easier for electrons to cross from the n-side to p-side and for holes
to cross from p-side to the n-side. Thus, the potential barrier gets lowered at the junction, and the
total current which is the sum of the hole and electron currents will increase. The reduced potential
barrier is shown in Figure (4a). Therefore, forward-biasing a diode results in an easy current flow or
low resistance, and hence increased conduction.
If the diode is reversing biased, the potential barrier is effectively increased as shown in Figure (4b),
therefore, poor electric current flows or high resistance and hence a reduced conduction.
For this reason the diode is known to be a unidirectional conductor that allows electric current to
pass only in one direction (when it is forward biased) and used to rectify current as rectifier
(rectification of alternating currents).
A typical current-voltage (I-V) characteristic of a p-n junction diode is shown in Figure (5). The
forward current increases very rapidly with forward bias whereas the reverse current is very small
(digital voltmeter with poor sensitivity read the reverse current as zero). The forward current
increases very rapidly when the forward voltage exceeds the voltage drop (V F). For voltages less
than VF very small current is permitted to pass through the diode.
I (A)
Reverse bias
Forward bias
VF
Figure (5). I-V characteristic curve of a p-n junction diode.
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V (mV)
Biophysics department, Faculty of sciences, Cairo university
Procedure:
1.
Connect the diode in the circuit in forward bias connection as shown.
E
RL
A
+
2.
-
V
Vary the potential difference (V) across the terminals of the diode and for each voltage value
find the corresponding value of the electric current intensity (I), then tabulate the results.
3.
Repeat step (2) by connecting the diode in reverse bias connection by changing the polarity of
the battery or by reversing the diode.
4.
For the same values of the potential difference (V) across the diode find the corresponding
values of the reverse electric current intensity (I).
5.
Plot the I-V characteristic curve for the given p-n junction diode in both directions.
6.
From the graph find the voltage drop (VF) of the given p-n junction diode.
Precautions and safety:

Connect the diode in forward bias first since in case of reverse biasing the diode there is no
current flow because of the poor sensitivity of the voltmeter.

Perform the entire experiment in dry position to avoid electric shock.

Do not allow high values of currents to pass through the circuit (I > 0.05A), t may cause the
resistance to melt.
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Biophysics department, Faculty of sciences, Cairo university
Results:
Forward bias
Reverse bias
The voltage drop (VF) =
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Biophysics department, Faculty of sciences, Cairo university
[7]
Mapping of Equipotential Lines of an Electric Field
Aim:
To draw the equipotential lines of an electric field.
Materials:
(1) Power supply (9 V).
(2) Voltmeter.
(3) A large Petri dish.
Theory:
Coulomb's Law:
The magnitude of the electrostatic force (F) between two point electric charges q1 and q2 is directly
proportional to the product of the magnitude of each charge and inversely proportional to the square
of the distance between them.
F
1
Where 4π
0
1
4π
0
qq
1 2
r2
 K is the proportionality constant.
c
35
(1)
Biophysics department, Faculty of sciences, Cairo university
Figure (1)
Electric Field
Electric field intensity due to a point charge:
The electric field due to a point charge is the region surrounding that charge and in which its electric
properties appear. Electric field strength is a vector quantity (it has magnitude or intensity and
direction).
The electric field intensity (E) at a point is defined as the force (F) acting on a unit positive charge
(+q) placed at that point, so:
E
F
1 q1

q
4π r 2
2
0
N/C
(2)
Electric field intensity due to two parallel plates:
Electric field intensity (E) due to two oppositely charged parallel plates separated by a small distance
(d) is:
E
V
d
V/m
Where V: is the potential difference applied on the plates.
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(3)
Biophysics department, Faculty of sciences, Cairo university
An electric field is said to be homogeneous if its intensity (E) is the same at all points of that field.
Electric field direction:
The direction of the field is taken to be the direction of the force that is exerted on a positive point
charge. For a point charge, the electric field is radially outward from a positive charge and radially
inward to the negative charge. Figures (2, 3)
Electric field lines:
The electric field lines are imaginary lines drawn so that the tangent to them at any point gives the
direction of the electric field at that point as shown in Figure (3) (solid lines).
Figure (2)
Figure (3)
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Biophysics department, Faculty of sciences, Cairo university
Potential Difference and Equipotential Lines:
Since a free charge moves in an electric field by the action of the electric force, then work (W) is
done by the field in moving charges from one point to another. To move a positive charge from one
point to another against the electric field would require work supplied by an external force.
Potential difference between two points in the electric field is the work done to move a unit charge
from one point to the other.
Potential at a point is the amount of potential energy per unit of charge at that point.
V
W
q
volts
(4)
If a charge is moved along a path at right angles (i.e. perpendicular) to the field lines, there is no
work done (W = 0) since there is no force component along the path. No work means there is no
potential difference from point to point. So, the potential is constant along paths perpendicular to
field lines, such paths are called equipotential lines (dashed lines in Figure3).
Applications:
An electric field set up by charges may be "mapped" by determining the equipotential lines
(equipotential surfaces in three dimensions) that exist in the region around the charges. Potential
difference is easily read by a voltmeter, whereas the measurement of forces would make numerous
experimental problems.
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Biophysics department, Faculty of sciences, Cairo university
Procedure:
1.
Connect the circuit as shown in the Figure (4).
Figure (4). Circuit for mapping the equipotential lines.
2.
Using the probe of the voltmeter find the points having the same potential at different positions.
3.
Draw the obtained points and connect them to obtain the equipotential lines.
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Biophysics department, Faculty of sciences, Cairo university
Results:
Electromotive
force
(X1,Y1)
(X2,Y2)
(X3,Y3)
(Volt)
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(X4,Y4)
(X5,Y5)