6.1
Materials Science
Laboratory Skills
Experiment – A Trying Out Action
Step 1 / Understand the principles involved in the procedure of the experimentStep 2 / Your skill in arranging and handling different measuring instruments
Step 3 / Determination of errors in various measurements and planning/suggesting how these contributions may be made of the same order so as to make the error in the final result small
Step 4 / A comprehensive view of the whole experiment with sufficient number of observations, correct calculations, proper graphs and result
Problem Solving - An Effective Approach Aid
Hint 1 / Read the problem carefullyHint 2 / Understand what is given and what is required
- Write down the given data, with units, using the symbols
- What is the unknown and what is its symbol
- What is the connection between the unknown and the data
Hint 3 / Watch the units
- You may need to make conversions
Hint 4 / Think about your answer
- Whether it makes sense
- Is the sign correct
- Are the units appropriate
Hint 5 / Reading the graph
- Note the units in which the variables are expressed
- Note the variables whether positive or negative
- A graph describes something that is happening and you should be able to say what it is
6.1(a)Band Gap Determination using Post Office Box
Aim
To find the band gap of the material of the given thermistor using post office box.
Apparatus Required
Thermistor, thermometer, post office box, power supply, galvanometer, insulating coil and glass beakers.
Principle and formulae
(1)Wheatstone’s Principle for balancing a network
Of the four resistances, if three resistances are known and one is unknown, the unknown resistance can be calculated.
(2)The band gap for semiconductors is given by,
Eg = 2k
where k = Boltzmann constant = 1.38 10 – 23 J /K
RT = Resistance at T K
Procedure
- The connections are given as in the Fig. 6.1(a).1. Ten ohm resistances are taken in P and Q.
- Then the resistance in R is adjusted by pressing the tap key, until the deflection in the galvanometer crosses zero reading of the galvanometer, say from left to right.
- After finding an approximate resistance for this, two resistances in R, which differ by 1 ohm, are to be found out such that the deflections in the galvanometer for these resistances will be on either side of zero reading of galvanometer.
- We know RT = .This means that the resistance of the thermistor lies between R1 and (R1+1). Then keeping the resistance in Q the same, the resistance in P is changed to 100 ohm.
- Again two resistances, which differ by one ohm are found out such that the deflections in the galvanometer are on the either side of zero. Therefore the actual resistance of thermistor will be between and .
Table 6.1(a).1 To find the resistance of the thermistor at different temperatures
Temp. of thermistorT = t+273 / / Resistance in P / Resistance in Q / Resistance in R / Resistance of the thermistor RT =R / 2.303 log10 RT
K / K-1 / ohm / ohm / Ohm / ohm / ohm
Fig 6.1(a).1 Post Office Box - Circuit diagramFig 6.1(a).2 Model Graph
Observation
From graph, slope = (dy / dx) = ……
Calculation
Band gap, Eg = 2k(dy / dx) =…..
- Then the resistance in P is made 1000 ohms keeping same 10 ohms in Q. Again, two resistances R and (R+1) are found out such that the deflection in galvanometer changes its direction. Then the correct resistance.
= (or)
= R+1 = 0.01R (or) 0.01(R+1)
- Thus, the resistance of the thermistor is found out accurately to two decimals, at room temperature. The lower value may be assumed to be RT (0.01R).
- Then the thermistor is heated, by keeping it immersed in insulating oil. For every 10 K rise in temperature, the resistance of the thermistor is found out, (i.e) RT’s are found out. The reading is entered in the tabular column.
Graph
A graph is drawn between in X axis and 2.303 log RT in Y axis where T is the temperature in K and RT is the resistance of the thermistor at TK. The graph will be as shown in the Fig.6.1(a).2.
Band gap (Eg)=2k slope of the graph = 2k ()
where K = Boltzman’s constant.
Result
The band gap of the material of the thermistor = ………eV.
6.1(b) Resistivity Determination for a Semiconductor Wafer
using Four Probe Method
Aim
To determine the energy band gap of a semiconductor (Germanium) using four probe method.
Apparatus Required
Probes arrangement (it should have four probes, coated with zinc at the tips). The probes should be equally spaced and must be in good electrical contact with the sample), Sample (Germanium or silicon crystal chip with non-conducting base), Oven (for the variation of temperature of the crystal from room temperature to about 200°C), A constant current generator (open circuit voltage about 20V, current range 0 to 10mA), Milli-voltmeter (range from 100mV to 3V), Power supply for oven Thermometer.
Formula
The energy band gap, Eg., of semi-conductor is given by
where kB is Boltzmann constant equal to 8.6 × 10 – 5 eV / kelvin , and is the resistivity of the semi-conductor crystal given by
;
Here, s is distance between probes and W is the thickness of semi-conducting crystal. V and I are the voltage and current across and through the crystal chip.
Procedure
- Connect one pair of probes to direct current source through milliammeter and other pair to millivoltmeter.
- Switch on the constant current source and adjust current I, to a described value, say 2 mA.
- Connect the oven power supply and start heating.
- Measure the inner probe voltage V, for various temperatures.
Graph
Plot a graph in and log10 as shown in Fig.6.1(b).2. Find the slope of the curve . So the energy band gap of semiconductor (Germanium) is given by
Table 6.1(b).1 To determine the resistivity of the semi-conductor for various temperatures:
Current (I) = …………mA
S.No. / Temperature / Voltage (V) / Resistivity (ohm. cm) / 10 – 3 / T / Log10
in°C / in K / (Volts) / (K)
Observations:
Distance between probes(s)= ……………………..mm
Thickness of the crystal chip (W) = ……………………mm
current (I) = ………………..mA
Fig 6.1(b).1 Four Probe Setup
Fig 6.1(b).2 Model Graph
Result
Energy band gap for semiconductor (Germanium) is Eg =….eV
Source of error and precautions
- The resistivity of the material should be uniform in the area of measurement.
- The surface on which the probes rest should be flat with no surface leakage.
- The diameter of the contact between the metallic probes and the semiconductor crystal chip should be small compared to the distance between the probes.
6.2 Determination of Hall Coefficient and carrier type
for a Semi-conducting Material
Aim
To determine the hall coefficient of the given n type or p-type semiconductor
Apparatus Required
Hall probe (n type or p type), Hall effect setup, Electromagnet, constant current power supply, gauss meter etc.,
Formulae
i)Hall coefficient (RH) = × 10 8 cm3 C – 1
where VH= Hall voltage (volt)
t = Thickness of the sample (cm)
I= Current (ampere)
H= Magnetic filed (Gauss)
ii)Carrier density ( n ) = cm – 3
whereRH= Hall coefficient (cm3 C – 1 )
q= Charge of the electron or hole (C)
iii)Carrier mobility () = RH cm2V – 1 s – 1
whereRH= Hall coefficient (cm3C – 1 )
= Conductivity (C V – 1 s – 1 cm – 1 )
Principle
Hall effect: When a current carrying conductor is placed in a transverse magnetic field, a potential difference is developed across the conductor in a direction perpendicular to both the current and the magnetic field.
Table 6.2.1 Measurement of Hall coefficient
Current in the Hall effect setup = ------mA
Current in the constant current power supply (A) / Magnetic field (H)(Gauss) / Hall Voltage (VH)
(volts) / Hall coefficient (RH)
cm3 C – 1
Observations and Calculations
(1)Thickness of the sample = t = cm
(2)Resistivity of the sample = = V C – 1 s cm
(3)Conductivity of the sample = = CV – 1 s – 1 cm – 1
(4)The hall coefficient of the sample = RH = × 10 8
= ------
(5)The carrier density of the sample = n =
= ------
(6)The carrier mobility of the sample= RH
= ------
Fig 6.2.1 Hall Effect Setup
Procedure
1. Connect the widthwise contacts of the hall probe to the terminals marked as ‘voltage’ (i.e. potential difference should be measured along the width) and lengthwise contacts to the terminals marked (i.e. current should be measured along the length) as shown in fig.
2.Switch on the Hall Effect setup and adjust the current say 0.2 mA.
3.Switch over the display in the Hall Effect setup to the voltage side.
4.Now place the probe in the magnetic field as shown in fig and switch on the electromagnetic power supply and adjust the current to any desired value. Rotate the Hall probe until it become perpendicular to magnetic field. Hall voltage will be maximum in this adjustment.
5.Measure the hall voltage and tabulate the readings.
6.Measure the Hall voltage for different magnetic fields and tabulate the readings.
7.Measure the magnetic field using Gauss meter
8.From the data, calculate the Hall coefficient, carrier mobility and current density.
Result
1. The Hall coefficient of the given semi conducting material =
2. The carrier density =
3. The carrier mobility =
6.3. To study V-I Characteristics of a Light Dependent Resistor (LDR)
Aim
To measure the photoconductive nature and the dark resistance of the given light dependent resistor (LDR) and to plot the characteristics of the LDR.
Apparatus Required
LDR, Resistor (1 k), ammeter (0 – 10 mA), voltmeter (0 – 10 V), light source, regulated power supply.
Formula
By ohm’s law, (or) ohm
where R is the resistance of the LDR (i.e) the resistance when the LDR is closed. V and I represents the corresponding voltage and current respectively.
Principle
The photoconductive device is based on the decrease in the resistance of certain semiconductor materials when they are exposed to both infrared and visible radiation.
The photoconductivity is the result of carrier excitation due to light absorption and the figure of merit depends on the light absorption efficiency. The increase in conductivity is due to an increase in the number of mobile charge carriers in the material.
Procedure
- The connections are given in as shown in Fig. 6.3.1.
- The light source is switched on and made to fall on the LDR.
- The corresponding voltmeter and ammeter readings are noted.
- The procedure is repeated by keeping the light source at different distances from the LDR.
- A graph is plotted between resistance and distance of LDR from the light source.
- The LDR is closed and the corresponding voltmeter and ammeter readings are noted. The value of the dark resistance can be calculated by Ohm’s law.
Result
- The characteristics of LDR were studied and plotted.
- The dark resistance of the given LDR = …….. ohm
Fig. 6.3.1 Circuit diagram Fig.6.3.2 Model graph
Observation
Voltmeter reading when the LDR is closed = …… V
Ammeter reading when the LDR is closed = ……. A
Dark resistance = = ……. ohm
Table 6.3.1 To determine the resistances of LDR at different distances
S.No / Distance(cm) / Voltmeter reading
(V) volt / Ammeter reading
(I) mA / RR
k
6.4 Determination of Energy Loss in a Magnetic Material – B-H Curve
Aim
(i)To trace the B-H loop (hysteresis) of a ferrite specimen (transformer core) and
(ii)To determine the energy loss of the given specimen.
Apparatus Required
Magnetizing coil, CRO, given sample of ferrite etc.,
Principle
The primary winding on the specimen, when fed to low a.c. voltage (50 Hz), produces a magnetic field H of the specimen. The a.c. magnetic field induces a voltage in the secondary coil. The voltage across the resistance R1, connected in series with the primary is proportional to the magnetic field and is given to the horizontal input of CRO. The induced voltage in the secondary coil, which is proportional to dB/dt (flux density), is applied to the passive integrating circuit. The output of the integrator is now fed to the vertical input of the CRO. Because of application of a voltage proportional to H to the horizontal axis and a voltage proportional to B to the vertical axis, the loop is formed as shown in figure.
Formula
Energy loss = Area of loop Unit: Joules / cycle / unit vol.
where N1= number of turns in the Primary
N2= number of turns in the Secondary
R1= Resistance between D to A or D to B or D to C
R2= Resistance between upper S and V (to be measured by the student on B-H unit)
C2= Capacitance
A = Area of cross section = w t (m)
L = Length of the specimen =2 (length + breadth) (m)
Table 6.4.1 To find width of the transformer core (w)
LC = …cm
S.No. / MSR / VSC / TR = MSR + VSC LCUnit / (cm) / (div) / (cm)
Mean (w) = ………………10 – 2 m
Table 6.4.2 To find thickness of the transformer core (t)
LC =… cm
S.No. / MSR / VSC / TR = MSR + VSC LCUnit / (cm) / (div) / (cm)
Mean (t) = ………………10 – 2 m
Observations
N1= Number of turns in the Primary= 200 turns
N2= Number of turns in the Secondary = 400 turns
R2= Resistance between upper S and V= 4.7 kilo-ohm
C2= Capacitance= 4.7µF
A = Area of cross section (w t) =… m2
L = Length of the specimen = 2(length + breadth) = … m
w = Width of the transformer core=… m
t = Thickness of the specimen=… m
R1= Resistance between D to A or D to B or D to C=
Sv= Vertical sensitivity of CRO=
SH= Horizontal sensitivity of CRO=
Fig 6.4.1 Experimental Arrangement
Fig 6.4.2 Hysteresis Loop Fig 6.4.3 Top view of BH curve Unit
Table 6.4.3 To find breadth of the transformer core (w)
LC = …cm
S.No. / MSR / VSC / TR = MSR + VSC X LCUnit / (cm) / (div) / (cm)
Mean (b) = ………………10 – 2 m
Table 6.4.4 To find length of the transformer core (l)
LC =… cm
S.No. / MSR / VSC / TR = MSR + VSC LCUnit / (cm) / (div) / (cm)
Mean (l) = ………………10 – 2 m
Procedure
1.Choose appropriate resistance values by connecting terminal D to either A,B or C.
2.Connect the primary terminals of the specimen to P,P and secondary to S, S terminals.
3.Calibrate the CRO.
4.Adjust the CRO to work on external mode. The time is switched off. Adjust horizontal and vertical position controls such that the spot is at the centre of the CRO screen.
5.Connect terminal marked GND to the ground of the CRO.
6.Connect terminal H to the horizontal input of the CRO.
7.Connect terminal V to the vertical input of the CRO.
8.Switch ON the power supply of the unit. The hysteresis loop is formed.
9.Adjust the horizontal and vertical gains such that the loop occupies maximum area on the screen of the CRO. Once this adjustment is made, do not disturb the gain controls.
10.Trace the loop on a translucent graph paper. Estimate the area of loop.
11.Remove the connections from CRO without disturbing the horizontal and vertical gain controls.
12.Determine the vertical sensitivity of the CRO by applying a known AC voltage, say 6V (peak to peak). If the spot deflects x cm, for 6V the sensitivity = (6/x10 – 2 volts / metre. Let it be Sv.
13.Determine the horizontal sensitivity of the CRO by applying a known AC voltage, say 6V (peak to peak). Let the horizontal sensitivity be SH volts / metre.
14.The energy loss is computed from the given formula.
Result
Energy loss of the transformer core is given as ______Joules/cycle/unit vol.
6.5. Determination of Paramagnetic Susceptibility – Quincke’s Method
Aim
To measure the susceptibility of paramagnetic solution by Quincke’s tube method.
Apparatus Required
Quincke’s tube, Travelling microscope, sample (FeCl3 solution), electromagnet, Power supply, Gauss meter.
Principle
Based on molecular currents to explain Para and diamagnetic properties magnetic moment to the molecule and such substances are attracted in a magnetic filed are called paramagnetics. The repulsion of diamagnetics is assigned to the induced molecular current and its respective reverse magnetic moment.
The force acting on a substance, of either repulsion or attraction, can be measured with the help of an accurate balance in case of solids or with the measurement of rise in level in narrow capillary in case of liquids.
The force depends on the susceptibility , of the material, i.e., on ratio of intensity of magnetization to magnetizing field I/H. If the force on the substance and field are measured the value of susceptibility can be calculated.
Formula
The susceptibility of the given sample is found by the formula
= kg m– 1 s– 2 gauss– 2
Where is the density of the liquid or solution (kg/m3)
is the density of air (kg/ m3)
g is the acceleration due to gravity (ms-2)
his the height through which the column rises (m)
H is the magnetic field at the centre of pole pieces (Gauss)
Fig. 6.5.1 Quincke’s Setup
Table 6.5.1 To find the rise in the capillary tube of the solution:
Microscopic reading without field (h1) = ….. cm
LC = … cm TR = MSR + (VSC LC)
S.No. / Current (i) / Field (H) / Travelling microscope reading (h2) / Differenceh = h1 ~h2 / h / H2
(m – 1 )
Ampere / Gauss / MSR (cm) / VSC (div) / TR (cm) / 10 – 2 m
Mean h/H2 = …………….
Observation:
= density of the liquid or solution= …..kg/m3
= density of air= … kg/ m3
Calculation:
The magnetic susceptibility of the given solution =
Procedure
- The apparatus consists of U-shaped tube known as Quincke’s tube. One of the limbs of the tube is wide and the other one is narrow.
- The experimental liquid or the solution (FeCl3) is filled in the tube in such a way that the meniscus of the liquid in the narrow limb is at the centre of the magnetic field as shown in the figure.
- The level of the liquid in the narrow tube is read by a traveling microscope when the magnetic field is off (h1).
- The magnetic field is switched on by switching on the electromagnet. Adjust the regulator knob available with the power supply to the electromagnet and fix the current to be 0.3A. The raised level of the column is read with the traveling microscope and noted in the table as (h2).
- The experiment is repeated by varying the field by changing the current insteps of 0.3 A upto the maximum and each reading is noted.
- To determine the magnetic field (H), the hall probe flux meter (Gauss meter) is used.
- The flat portion of the hall probe is placed perpendicular to the magnetic field i.e. between the pole pieces at the center parallel to the poles.
- Switch off the electromagnet power supply. By adjusting, the gauss meter knob and fix the field to be zero.
- Switch on the electromagnet and adjust the current to be 0.3A. Note the field value from the gauss meter. Repeat the same as before till attaining the maximum current and note the reading in the table.
- Calculate the magnetic susceptibility using the above formula.
Result
The magnetic susceptibility of the given sample = …………… kg m– 1 s– 2 gauss– 2
6.6. Dielectric Constant Measurement
Aim
To determine the dielectric constant of the given sample at different temperatures.
Apparatus required
The given sample, capacitance meter, dielectric sample cell, digital temperature indicator etc.
Formula
- The dielectric constant of the sample is given by,
r= C / C0 (No unit)
where C = capacitance of the sample (farad)
C0= Capacitance of the air capacitor having the same area and thickness as the sample (farad)
- The capacitance of air capacitor is given by,
where 0= permittivity of free space
= 8.854 1012 farad / metre
A = area of the plates of the capacitor
(A = r2 : r = radius of the sample)
d = thickness of the sample (or) distance between the plates (m)
Principle
The capacitance of a capacitor increases when it is filled with an insulating medium. The increase in the capacitance depends on the property of the medium, called dielectric constant (). It can be measured using either static or alternating electric fields. The static dielectric constant is measured with static fields or with low frequency ac fields. At higher frequencies, values of dielectric constant become frequency dependent. The dielectric constant varies with temperature also.