ENGINEERING PHYSICS-II LABORATORY MANUAL I. B.TECH, II SEMESTER – ALL BRANCHES
NAME:
REGD. NO: BRANCH:
GAYTRI VIDYA PARISHAD COLLEGE OF ENGINEERING FOR WOMEN MADHURAWADA, VISAKHAPATNAM.
Certificate
Certified record of practical work done by Ms………………………………………………........ of first B.Tech, Second Semester, ……………………… Branch bearing registered number…………………… in the Engineering Physics laboratories of Gayatri Vidya Parishad College of Engineering for Women, Madhurawada, Visakhapatnam during the academic year 2012-13. No. of experiments done and certified:
Lecturer in charge
Date
Examiners:
1.
2.
INDEX
1. Thermistor characteristics (Theory part) 1 Thermistor characteristics (Experimental part) 2 – 3 2. Band gap of semiconductor (Theory part) 4 – 5 Band gap of semiconductor (Experimental part) 5 – 7 3. Resistor – Capacitor time constant (Theory part) 8 – 10 Resistor – Capacitor time constant (Experimental part) 10 – 13 4. Zener diode – V – I characteristics (Theory part) 14 – 17 Zener diode – V – I characteristics (Experimental part) 18 – 19 5. Stewart and Gee Apparatus (Theory part) 20 – 21 Stewart and Gee Apparatus (Experimental part) 22 – 23
APPENDIX 6. BREAD BOARD 24 – 25 7. Circuit symbols of components 25 8. Kirchhoff’s laws 26 9. Resistor colour code 27 10. Least square fit method 28 11. Abbreviations 28 12. Tan θ table for selected values 28
DATA SHEETS
S.NO. DATE NAME OF THE EXPERIMENT PAGE NO. REMARKS
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T
THERMISTOR THEORY:
The name thermistor comes from thermally
sensitive resistor. They are basically
semiconducting materials and are of two distinct
classes:
1. METAL OXIDES: They are made from fine
powders that are compressed and sintered at high
temperature. MnR2ROR3R (manganese oxide), Ni O
(nickel oxide), Co OR3R (cobalt oxide), CuR2ROR3R
(copper oxide), FeR2ROR3R (iron oxide), TiOR3R
(titanium oxide) UR2ROR3R (uranium oxide) etc, are
the few examples. They are suitable for
temperatures 200-700 K. If the temperature is
higher than this range then AlR2ROR3R, Be O, Mg O,
ZrOR2R YR2ROR3R and DyR2ROR3R (Dy :dysprosium) are
used.
2. SINGLE CRYSTAL SEMICONDUCTORS:
They are usually Germanium and Silicon doped
with 10P
16P to 10P
17P dopant atoms/cmP
3P. Ge
thermistors are suitable for cryogenic range
1-100 K. Si thermistors are suitable for 100-250
K. After 250 K the Silicon thermistors will
become PTC (positive temperature coefficient)
from NTC.
The resistivity and the conductivity of the
thermistor are related to the concentration of
electrons and holes n and p of the semiconductor
though the relation,
( ) ………………... (1)
The concentrations n and p are strongly
dependent on temperature T in Kelvin.
Where ERaR is called activation energy which is
related to the energy band gap of that
semiconductor. Hence, As temperature
increases, the resistance R(T) changes according
to the relation,
( [
]) ……………. (2)
Where RROR is the resistance of the thermistor at
absolute temperature TRoR. Here TRoR is usually the
reference room temperature. B is a characteristic
temperature that lies between 2000K to 5000K.
The temperature coefficient of resistance is
defined as the ratio of fractional change in
resistance (
) to the infinitesimal change in
temperature .
……….. (3)
The typical value of is about 0.05/K. It is
almost 10 times more sensitive compared with
ordinary metals. Thermistors are available from
1KΩ to 1MΩ.
Advantages:
They are low cost, compact and highly
temperature sensitive devices. Hence are more
useful than conventional thermometric devices.
Using eq. (2) at some constant reference
temperature, say TROR= 300K, the resistance will
be
(
)
Where, (
)
To make the expression to look like a linear
relation to determine the values of A and B
constants, take natural logarithm on both sides of
the above expression,
…………………….. (4)
The exponential curve now became linear. If we
plot the variable
, we will get A and
B constants from the intercept and slope of the
straight line.
DESIGN OF EXPERIMENT:
PRINCIPLE: If we measure the resistance R of
thermistor at various temperatures (T), we can
plot the
graph and obtain the
values of A and B.
How to vary the temperature T?
Using an electric heater we can change the
temperature roughly from 30 to 60 .
How to measure the resistance R?
Using Wheatstone’s bridge.
Wheatstone’s bridge principle:
The circuit shown here is a Wheatstone’s bridge
and it consists of four resistors RR1R, RR2R, RR3R and
RR4R, a
galvanometer
(G) and a Battery
(V). Suppose the
resistance RR4R be
unknown. The
voltage applied
to this circuit by
the battery is
only to set up
some current and
its magnitude has no importance, i.e. whether or
2V or 5V it does not matter at all. Wheatstone
bridge gets balanced, i.e. the Galvanometer
shows a zero deflection when,
G
R3 R4
V
R1 R2
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T
T
Or
If the resistances RR1R and RR2R are equal, then the
bridge will be balanced, i.e. the null deflection in
Galvanometer, when RR4R = RR3R. If we choose R R3R
as a variable resistor, like a decade resistance
box, the unknown resistance RR4R will be equal to
the resistance maintained in the box.
Measurement of resistance of thermistor:
Here in this experiment we employ a
1KΩ (at room temperature) thermistor. We form
a Wheatstone’s bridge with two fixed value
resistors each of 1KΩ resistance along with a
variable decade resistance box. Two arms of the
bridge are formed by 1KΩ resistors and the other
two arms, one with thermistor and the other with
decade resistance box. The sensitivity of
measurement of resistance will be better when
all the four resistors here are of same
(comparable) magnitude, hence the remaining
R’s are 1KΩ each.
Applications of thermistors:
1. They are used as temperature sensing
elements in microwave ovens, heaters
and also in some electronic
thermometers.
2. Used as sensor in cryogenic liquid
storage flasks.
3. Used as compensator for providing
thermal stability to transistor based
circuits.
4. Used in fire alarms, Infrared detectors as
sensor.
THERMISTOR EXPERIMENT Aim:
1. To study the variation of resistance of a thermistor with temperature.
2. To find the temperature (thermoelectric) coefficient of resistance (α) of the thermistor.
3. To determine A and B coefficients.
Apparatus: Thermistor (1 KΩ), electric heater (max 70P
0P C), 1.5 Volt battery or a D.C. power supply, mercury
or benzene thermometer (0 – 110 ), test tube containing insulating oil (edible oil / castor oil),
resistors (1kΩ - 2 No.s), Galvanometer (30 – 0 – 30), resistance box (1 to 1000Ω range),
connecting wires.
Formulae:
(
)
Procedure:
1. Construct the bridge according to the
circuit diagram (UMaintain at least 1000 Ω
Uresistance in the Resistance box before
connecting the circuit, i.e. remove the 1000Ω
plug key).
2. The 1 KΩ resistors are already connected
on the back panel of the board. Hence no
need to connect them again.
3. If a variable D.C. source is given instead
of a battery, set the voltage to 1.5 or 2 Volt
with the help of a multimeter.
4. The bridge gets balanced (Galvanometer
shows “0” deflection) when the resistance of
thermistor gets equal to that of the resistance
box. Remove the plug keys of resistance box
and find out the null point resistance.
T
G
RB RT
1.5 V
R1=1KΩ R2=1KΩ
Electric heater
Test tube with Coconut oil
Ther
mo
met
er
T
G
1.5 V
R2=1KΩ R1=1KΩ
RB
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T
5. Start heating the thermistor by turning on the heater switch on the board.
6. Measure the resistance of thermistor for every two degrees centigrade rise in temperature. Note
the readings up to 60P
0P C in steps of 2P
0P C.
7. At each temperature bridge is not balanced initially and it shows some deflection. It can be made
zero by adjusting the resistances in the variable resistance box. Tabulate the readings.
8. Remove the power supply or battery, soon after you complete the experiment. If you forget
doing this, it will cause the galvanometer to deflect more causing damage to its restore spring.
Graph:
A graph is plotted by taking R versus T (K). This graph gives the value of α.
Another graph is plotted between ln R and (1/T(K)). The slope of this graph gives B and its
intercept on y (ln R) axis gives ln A from which A can be calculated. UBut it is not possible to find
out the intercept from the graph.U It can be done with the help of least square fit method as
described in the Appendix.
Use this method to compute both slope (B) and intercept (ln A) of the straight line. Here assume X as
(1/T) and Y as lnR. The intercept C gives the value of ln A and the slope will give B (in K). From the
intercept find out the value of A (in Ω).
Precautions:
1. Temperature of the thermistor should be
less than 70P
0P C.
2. Thermistor must be immersed completely
inside the hot oil bath.
3. Readings of thermometer must be noted
without parallax.
4. Connections should be made properly
without any loose contact.
5. Resistance must be varied quickly in the
resistance box to get the null point within
the 2P
0PC intervals.
6. Battery must be disconnected immediately
after completion of the experiment.
Viva-Voce Questions:
1. Where do you find applications of
thermistor? Name a few of them. They are useful in temperature sensing and
controlling equipments. Ex. Microwave
ovens, Infrared heat sensors, Liquefied gas
temperature sensors in cryogenics.
2. Explain the principle of Wheatstone’s
bridge.
In the bridge circuit, the potential at the two
nodes across which the galvanometer is
connected will be same when the four
resistors RR1R to RR4R satisfy the relation
3. After obtaining the data from this
experiment, you will have the values of A
and B coefficients. Can you determine
the temperature of your body? I will
provide you only a thermistor and a
multimeter. If yes, describe the method.
If No, justify your answer. Yes, it is possible. Suppose that you want
to measure your body temperature. Just
keep it in tight contact with your body
(cover it tightly with skin). Use the
multimeter to measure the resistance of this
thermistor. After few seconds of contact
with body, thermistor attains constant
resistance. With the known A and B
coefficients, we can measure the body
temperature by substituting in
(
)
(
)
T1 T2
T in K
R1
R2
in K-1
Slope = B
ln R
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REFERENCES:
1. Physics of semiconductor devices, S. M. Sze, 3 P
rdP ed, John Wiley publications, chapter 14, sensors,
Thermal sensors, p.744-746.
2. The art of Electronics, Paul Horowitz, 2 P
ndP ed, Cambridge university press, chapter 15,
Measurement transducers, Thermistors, p.992-993.
3. Electronic devices and circuit theory, R. Boylestad, 7 P
thP ed, Prentice hall publications, Art. 20.11
Thermistors, p.837-838.
4. Electronic sensor circuits and projects, Forrest Mims – III, Master publishing, p.13, 46-47.
5. Advanced level physics, Nelkon and Parker, 3P
rdP Ed, Wheatstone’s bridge, p.829-834.
BAND GAP OF SEMICONDUCTOR USING PN JUNCTION DIODE
THEORY: PN junction diode is an example for extrinsic
semiconductor. It can be biased in both
forward and reverse directions. The current
that flow through the diode when its junction
is biased with a voltage V will be
(
)
With
.
Where,
V = Applied voltage across junction
IRsR = Reverse saturation current, a constant
dependent on temperature of junction
η = A constant equal to 1 for Ge (high
rated currents) and 2 for Si (low rated
currents)
VRTR = Volt equivalent of temperature
=
, T = Temperature of junction in
Kelvin
A = Area of cross – section of junction
e = Elementary charge = C
DRp(n)R= Diffusion constant for holes
(electrons)
for holes and
for electrons
= Mobility of holes
LRp(n)R = Diffusion length for holes
(electrons)
pRnoR = Equilibrium concentration of holes
(p) in the n – type material
=
nRpoR = Equilibrium concentration of
electrons (n) in p – type material
=
nRiR = Intrinsic carrier concentration (/cmP
3P)
nRiRP
2P =
B = A constant independent of T
ERGR = Energy band gap of semiconductor
(in Joule)
NRA R= Acceptor ion concentration (/cmP
3P)
NRD R= Donor ion concentration (/cmP
3P)
The term IRsR is highly temperature dependent.
The expression for it can be written as
(
)
(
)
(
)
(
)
(
)
(
)(
)
(
)(
)(
)
(
)
Experimentally it was observed that the
mobility term in the bracket varies as .
Hence,
………………………… (1)
is a constant whose magnitude is in nano or
pico ampere.
Under reverse biased condition applied
voltage V will be negative and hence the
expression for current through diode will be,
(
)
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Diode will have only the reverse saturation
current flowing through it. The negative sign
indicates that the current is flowing in opposite
direction to that of forward bias. Hence the
current IRDR through diode in reverse bias will
be
(
)…………. (2)
Applying natural logarithms on both sides
implies,
[ (
)]
(
) ……………. (3)
This is the equation of the straight line with
ln(IRDR) as ordinate(y – axis) and 1/T as abscissa
(x – axis). ln(IR0R) is the y intercept of the graph.
If we plot 1/T versus ln(IRDR) graph, its slope
with x – axis gives the value of (–
). By
knowing the Boltzmann constant kRBR we can
evaluate the energy band gap of the
semiconductor, similarly we can estimate the
value of Boltzmann constant if we know the
energy band gap of the given semiconductor.
Applications:
1. We can use this to make a diode
thermometer.
DESIGN OF EXPERIMENT:
PRINCIPLE: If we measure the reverse
saturation current through the diode by
varying its temperature, we can plot the
graph and obtain slope (–
).
Which diode is suitable for this?
OA79, Germanium diode, used as envelope
detector in amplitude demodulation circuits.
Why this OA 79? Why not any other?
Because reverse current variation is more in
the case of Germanium than with silicon.
Hence for a small temperature range of
variation (30P
0P to 60P
0PC), it is better to choose
Ge diode than any other silicon diodes. If we
want to do this experiment with silicon diodes,
we must have an electric heater capable of
giving temperatures up to 150P
0PC.
How to vary the temperature T?
Using an electric heater we can change the
temperature roughly from 30 to 60 .
How to measure the reverse current?
Using a moving coil micro ammeter.
Biasing the diode:
Use a constant voltage D.C. power supply or a
battery to bias it in reverse direction. The
voltage applied must be very low, 2 Volt. In
case of an ideal diode the reverse current does
not vary with applied reverse voltage. But in
practical diode case, it increases with increase
in reverse voltage. This is due to the increase
of leakage currents across the junction with
applied voltage. At room temperature, the
reverse current may be small and different for
same type of diodes, but it follows the
equation (2). The values of IRoR may vary from
diode to diode.
Description of heater:
The heater contains an electric heating
element attached to a stainless steel container
holding some cold water. A test tube
containing oil is immersed in the water bath.
Oil is an insulator of electricity and hence it
is used for heating the diode. This also
provides uniform heating of diode. The diode
with properly insulated connecting wires is
immersed in the oil bath. Thermometer is also
kept inside the oil bath to measure its
temperature. We cannot directly insert the
diode inside the water bath as tap water
contains lots of minerals dissolved in it and
acts like conductor. This will short circuit the
diode.
Useful data:
From the data sheet of the OA 79 diode:
Material of the diode is Germanium.
Maximum surrounding temperature is
60P
0PC.
Maximum allowed reverse current
through the diode is 60µA.
BAND GAP OF SEMICONDUCTOR EXPERIMENT
Aim: To determine the energy band gap of the
material of the semiconductor by studying the
variation of reverse saturation current through
given PN junction diode with temperature.
Apparatus: OA 79 Ge diode, heater (max 60 P
0P C),
thermometer, test tube containing insulating
oil (edible oil or castor oil), power supply (2V
D.C.), connecting wires, micro ammeter
(0 - 50 µA) and a voltmeter or multimeter.
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VD
ID
T
Stainless steel container water bath
Test tube containing oil
A
2 V
+ +
_ _
+
Electric Heater
Formula:
Reverse current through diode is given by
(
)
Where, ERGR is the energy band gap of the
material of the semi conductor diode, T is the
absolute temperature of the diode junction and
kRBR = 1.38 x 10P
-23 PJ/K is Boltzmann constant.
Circuit diagram:
Caution: Set the applied reverse bias voltage
at 2 Volt. Do not increase this value more. Do
not heat the diode beyond 60P
0PC.
Procedure:
1. Build the circuit as shown in the circuit
diagram.
2. Observe the initial temperature of the
thermometer. If it is high (>30P
0P C) then
replace the water in the heater jar with
some cold water and try to reduce the
temperature below 30P
0P C.
3. Apply the reverse voltage (2 Volt) by
adjusting the potentiometer (if a battery is
given, then there is no need of doing this
adjustment).
4. Switch on the heater. Note down the
reverse current in the micro ammeter for
say, every 2P
0PC rise, in temperature of the
diode (if micro ammeter is not available,
you can use a multimeter in D.C. current
mode under 200 µA ranges).
5. Tabulate the readings.
6. Complete the calculations relevant to the
tabular form and get the answer for slope.
7. Plot a graph between lnI and 1/T to obtain
its slope.
8. Calculate the ERGR from both slopes obtained
from graph and table.
Precautions:
1. Readings of thermometer must be noted
without any parallax error.
2. Reverse bias voltage must be regulated at 2
Volt throughout the experiment.
3. Diode should be completely immersed inside
the oil bath.
GRAPH:
Plot a graph by taking the values of ln I vs
1/T. Find out the slope of the curve. Do not
consider the origin of this graph.
Usually we start at 300K and go up to 333K,
hence 1/T varies roughly from to . So start at 2.98
and go up to 3.34 by choosing the scale
On 1/T axis as
Usually IRDR varies from 2 µA to 60 µA. So ln
IRDR varies roughly from to – 13.2. So start at
– 9.7 and go up to – 13.2 by choosing the
scale
On ln I axis as
Slope (ERGR/kRBR) can be calculated from both
straight line data fit as well as from the
graph.
Viva-Voce questions:
1. Distinguish intrinsic and extrinsic
semiconductor.
If the semiconductor material consists of no
impurities (dopants), then it will be intrinsic
(pure) semiconductor. If it contains dopants
acceptor type [p-type] – III group elements
or Donor type [n-type] – IV group elements
then it will be an extrinsic semiconductor.
2. What are the band gaps of Silicon and
Germanium?
For silicon; (eV = electron volt)
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For Germanium,
T is the temperature of the sample in Kelvin.
At 300K, ERG R= 0.72 eV for Ge; ERGR= 1.1 eV
for Si.
3. How do you test the diode for its polarity
using a multimeter?
There will be a Usymbol of diodeU on the
multimeter’s mode changing dial. Turn the
dial to diode testing mode. Connect the two
leads of the multimeter to the two leads of
the diode. If the multimeter shows infinite
resistance (it shows a “1 ” Or “OL” means
out of range, very large), then it is reverse
biased and the terminal of diode that is
connected to positive (red probe) of
multimeter will be the UcathodeU of the diode
and the other one will obviously be the
anode. Similarly, if the meter shows some
finite resistance like few hundred (150, 540
etc), then it is forward biased, i.e. the
terminal of diode that is connected to positive
(red probe) of multimeter will be the UAnodeU
of the diode and the other one will be the
Cathode. During this process, multimeter
applies some known voltage across its leads
and measures its resistance.
4. If I reveal the material of the diode used,
can you estimate the Boltzmann constant
from this experiment? If yes, describe how
do you do it, if no, say why?
(Think and answer)
5. Why do we observe small current (of the
order of Micro amp) in this experiment?
What are responsible for this small
current?
Because reverse current is due to the
minority carries only. As their number is
very small the current is also small.
6. In which biasing of diode are you doing
this experiment?
Reverse bias.
7. Can you determine the band gap by
changing the bias of the diode? If yes,
describe how you do it. If no, explain
why? (think and answer)
8. If I give you a silicon diode and the same
experimental set up (micro ammeter 0-
50range), can you find out its band gap?
Justify your answer.
No, the reverse current variation is very
small of the order of few nano amperes per
degree centigrade and hence it not possible
to observe the variation in reverse current
with the micro ammeter for a temperature
range of 30-60P
0P C
9. What is the magnitude of reverse current
in silicon at moderate temperatures?
Few tens of nano amperes.
10. Can you make a diode thermometer
using this setup? If yes, say how? If no,
say why?
Yes, once if we know the value of IR0R
(antilog of intercept of lnI vs 1/T graph)
from the experiment, we can measure the
T. Just bring the diode in contact with the
body whose temperature is to be measured
and measure the reverse current (IRDR)
accurately. As we know the IR0R and IRDR we
can determine the T in Kelvin for that body
using the relation (
).
References:
1. Electronic devices and circuits, Millman and Halkias, McGraw hill student edition
p.126-132.
2. Semiconductor device physics and technology, SM Sze, M K Lee, 3P
rdP Ed, John wiley,
P.107.
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RESISTOR – CAPACITOR CIRCUIT – TIME CONSTANT
THEORY:
Resistor and capacitor combination circuits
have got great importance in the field of both
electrical and electronics engineering. This is
the fundamental circuit to understand the
working of many complex electronic circuits.
Consider a resistor of resistance R and
a capacitor of capacitance C. They can be
biased with an external D.C. source to start the
process of charging.
Suppose that switch S is closed.
Apply Kirchhoff voltage law (KVL) to the
above circuit of Charging.
Where, the sub scripts R and C denote the
voltages across the resistor and capacitor. If
we assume a current of i(t), a function of time,
then from Ohm’s law,
.
If the instantaneous charge on the plates of the
capacitor is , then
, Where, C
is the capacitance of the capacitor.
Substituting them in above eqn. implies,
;
But, is equal to the rate of change of
charge in the circuit, i.e.
Differentiate this expression with respect to
time t.
Voltage V across battery does not change with
time,
Put a trial solution,
Going back to differential equation,
(Assuming α≠0)
This implies
(
)
Voltage across capacitor will be,
(
)
, when capacitor charges to maximum
value qR0R, the voltage across it will be VO.
Hence,
(
)
Current will be,
(
)
(
);
(
)
Hence,
At t = 0, is , the maximum current
flowing through the circuit. Using,
(
)
Where, iR0R is the current in the circuit at t=0,
iR0RR represents the initial voltage across the
resistor, i.e. V;
( (
))
(
)
(
)
S
V
R
C VC
VR
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DISCHARGING OF CAPACITOR: Similarly for discharging of capacitor, KVL
gives
Proceeding in the same manner as above, we
can show that voltage across the capacitor
during discharging will be
(
)
GRAPHS:
APPLICATIONS:
1. In electronic circuits, capacitor is a
crucial element. Its behavior in
electronic circuits can be better
understood from this experiment.
2. We can use this circuit to produce time
delay in D.C. operated circuits.
3. Capacitor – Load resistor combination
is used in filter circuits.
4. To set or reset the Digital IC’s at the
time of start or during their operation.
DESIGN OF EXPERIMENT:
PRINCIPLE: To study the charging process,
we maintain a constant D.C. voltage across the
serial combination of Resistor and Capacitor
to study how the voltage changes across the
capacitor with time by noting down the
voltage at capacitor in regular intervals of
time. After plotting the graph of voltage vs.
time we can compute the time constant for the
circuit from charging process.
To study the discharging circuit, we first fully
charge the capacitor to the maximum value
and then discharge it through resistor. Once
again the voltage across the capacitor in this
process is to be noted in regular intervals of
time and graph is to be plotted.
Which capacitor is suitable for this?
Any electrolytic capacitor with fairly large
capacitance. (2200 µF, 3300 µF, 4700 µF etc)
Why large capacitance?
Because we want to observe this process by
means of a manually countable time period,
say few tens or hundreds of seconds. If it is a
very small capacitance, then it will be difficult
to observe the variation of voltage with time, it
drops rapidly.
What must be the voltage rating of the
capacitor?
In our lab we often use 12 to 22 Volt D.C.
power supplies. Standard capacitors are
available from 12V, 16V, 25V, 35V etc. So it
is preferred to use at least 25V rating for
capacitor. If it has more voltage rating then it
Voltage across capacitor with time during CHARGING
C A P A C I T O R
Time
τ = Time constant τ = Time constant Voltage across capacitor and Resistor with time
during DISCHARGING
S
V
R
C VC
VR
TIME
VO
LTA
GE
V
O
V
t
VC(t)
VR(t)
VC(t) + VR(t) = V at any ‘t’
Capacitor
Resistor
CHARGING PROCESS Vs. TIME
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can also be used safely. But a 12V or 16 V
rating for capacitor may not be safe always.
Which value of resistor we have to choose?
As we wish to study this process slowly, it is
better to use a resistor R that gives an RC
product of approximately 100 to 400 seconds.
If you are using a 3300 µF, 25V rating
capacitor, it is better to use at least 50KΩ
resistor which gives
.
If we use less value of resistor than this, it will
charge or discharge very quickly and it
becomes difficult to note down the readings.
What should be the minimum wattage
(power rating) for resistor?
In the above example, if we use an 18 volt
supply, then the maximum current that flows
in the circuit will be
. So,
Hence any wattage carbon resistor is suitable
(1/8P
thP Watt, 1/4P
thP Watt or even ½ Watt is OK)
How to measure the Voltage?
Using a multimeter in D.C. voltage mode.
Biasing the capacitor:
As we are using electrolytic capacitor beware
of the polarity. The positive of capacitor must
be connected towards the high potential
(positive) of the D.C. source.
R – C CIRCUIT – TIME CONSTANT EXPERIMENT
Aim: 1. To study the charging and discharging
processes of Resistor and Capacitor
series (R-C) circuit.
2. To determine the time constant of the
circuit from the charging and
discharging curves.
Apparatus:
Resistors, electrolytic capacitors, D.C.
power supply, Multimeter, bread board,
connecting wires and stop watch.
Formulae:
Across capacitor,
[
]……. While charging
[
]……….. While discharging
Circuit diagram:
CHARGING PROCESS
DISCHARGING PROCESS
Procedure:
Set the multimeter in 20 Volt D.C. voltage
mode
Charging process:
1. Connect the circuit on the bread board as
shown in the above figure (charging).
2. Calculate the theoretical value of the time
constant using the above formula.
3. Switch on the circuit. Join the nodes A and
B to the multimeter leads. Join the two
leads and Uhold them togetherU.
4. Switch on the stop watch and separate the
probes of multimeter simultaneously. Start
counting of time.
5. Note the initial reading of voltage across
capacitor as zero volts.
6. While joining the ends of the capacitor with
a wire, it gets discharged and the voltage
across it becomes zero. As the multimeter
S
V
R
C VC
VR
A
B
A: connect this point to +ve (RED) probe of
multimeter
B: connect this point to – ve (BLACK) probe
of multimeter
S
V
R
C VC
VR
A
B
A: connect this point to +ve (RED) probe of
multimeter
B: connect this point to – ve (BLACK) probe
of multimeter
ENGINEERING PHYSICS LABORATORY – II
11 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
probes are connected to the ends of the
capacitor, joining the leads of the
multimeter discharges the capacitor to zero
volts. You must hold the two leads of
multimeter together Uuntil Uyou switch on the
stop watch.
7. Note down the voltages in regular intervals
of time (say 10 or 20 sec) and tabulate
them. it can be decided on the basis of
time constant, if time constant is about 220
sec, then we can go in steps of 20 sec; if it
is about 100 sec or less, we can go in steps
of 5 or 10 sec.
8. Take the readings until the capacitor
charges to maximum voltage (of power
supply).
Set the multimeter in 20 Volt D.C. voltage
mode
Discharging process:
1. Replace the resistor in the charging circuit
with another high value resistor.
2. Disconnect the wire joined to positive of
power supply from the resistor.
3. Join this wire to the ground. (If there is any
charge that is left on the positive plate of
the capacitor that will be discharged
through the resistor).
4. Join the positive probe of multimeter
(attached to the positive of capacitor) with
positive of the power supply and Uhold itU.
(As we want to study the discharge process,
the capacitor must have sufficient charge.
So charge it directly by joining the positive
lead of capacitor to the positive of power
supply).
5. Note down the voltage on the capacitor at
this “zero” time.
6. Switch on the stop watch and
simultaneously UremoveU the positive probe
of multimeter from the positive of power
supply. (this will allow the capacitor to
decay through resistor)
7. Note down the values of the voltage across
capacitor in the same intervals of time as
above until some minimum value of
voltage appear across it. (Say about two
volts)
Graph: Plot a graph by taking the voltage across capacitor versus time both for charging and
discharging cases.
Choose the time axis scale as 1 division = 20 sec and the voltage axis scale as 1division = 1 volt. Use
separate graph sheets to plot these graphs.
From each graph calculate the time constant of the circuit. On charging graph, the time taken
for the capacitor to charge to 63% of the maximum value is its time constant. Similarly on
discharging graph, time taken by the capacitor to discharge to 37% of its maximum value is its time
constant.
Precautions:
1. Do not short the ends of the power supply. This will damage your power supply.
2. Connections on the breadboard must be tight. Avoid loose contacts.
3. Start counting of time immediately after closing the switch wire while charging as well as
discharging the capacitor.
4. While charging the capacitor wait for sufficiently long time to reach the saturation value. Do not
stop the process at the middle.
Viva – voce questions:
1. Will you get a different answer to the
time constant of RC circuit if we replace the
power supply from say 12 Volt to 18 Volt?
Justify your answer.
As the time constant does not depend on the
voltage of the power supply, the answer
remains same as RC.
2. While discharging the capacitor, suppose
that I forgot to switch on the stop watch.
Some time lapsed. Then I have started
taking readings of voltage across capacitor
versus time. Will it give me correct answer
for τ? Justify.
Yes. We will get correct answer. Where ever
we start counting of time, there the voltage
across capacitor will be the new maximum
voltage V. From there the voltage starts falling
exponentially. Measure the time elapsed
between the instant at which voltage is V and
the instant where the voltage became 0.37 V.
ENGINEERING PHYSICS LABORATORY – II
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It gives the time constant.
3. During discharging of capacitor if the
power supply is turned off, will you get
correct answer to the time constant? Justify
your answer.
For discharging process we do not need any
power supply. Hence nothing happens. (Power
supply is initially required to charge the
capacitor. Once after charging the capacitor, it
is not required)
4. Can you find out the value of an
unknown resistor/capacitor (either C or R
will be given to you, but not both) from this
experiment? If yes, describe how you get it.
If No, say why it is not possible?
Yes. Build the charging or discharging circuit
with the given components. Study the voltage
variation in small intervals of time. (Because
we initially do not know whether to take 5 sec
time intervals or 10, 20 seconds). If the
voltage falls or increases rapidly, then take the
smallest time interval possible. If the voltage
is not varying much then accordingly you can
decide whether to take 10 sec, 20 sec or even
30 sec time intervals. Measure the time
constant using the voltage – time graph. This
gives R.C product value. As one of them is
known, other can be estimated.
5. I have started studying the discharging
of a capacitor (R=56KΩ and C= 3300 F,
Voltage source is 12V). My aim is only to
find out its time constant. I have done the
experiment and noted the voltage versus
time up to 900 seconds. Is it an efficient way
of doing the experiment?
RC= 184.8 sec, hence we can stop at roughly
around 300 sec range to know the time
constant experimentally from graph. So it is
waste of time to take the readings up to 900
sec. If the variation in voltage reduces, say
during one or two minutes time if the voltage
drops by 1 volt, then we can stop.
6. Find out the voltage VROR across the
resistor after a very long time. What will be
the voltage across C after very long time?
After long time (say few tens of minutes), the
voltage across the capacitor will be equal to
the voltage of the D.C. voltage source. If we
measure the voltage VRRR across R, after long
time it will be zero. (Use charging equation)
7. In certain logic integrated circuits (IC’s),
it is necessary to SET or RESET the output
of the circuit. It is something like the reset
button of your digital stopwatch which
makes the output time Zero. It is required
to send a high voltage (something like 12V)
pulse for a short duration of time and later
it should go down to zero (0 Volt). Can
you produce it by using this circuit? If Yes,
where to take the output? If yes, suggest
where to take the output voltage if I need a
low voltage (0 Volt) initially that should go
quickly to a high voltage (12Volt).
(Think! And answer!)
8. If the answer to above question is Yes,
what value of resistance would you suggest
in achieving this action, Smaller or larger?
Justify your answer.
We should take large value of resistor and
small value of capacitor. For a given RC,
initially the voltage appears entirely across R
at time t = 0. If R is small, then this will
produce an unwanted spike (sudden extraction
of current from the power supply). This is
undesirable. Hence if we take large R and
small C, these spikes would not exist.
9. Transistors are devices which can be used
as electronic switches. You will learn about
the action of transistors in later courses. For
the time being suppose that the Base of a
transistor is a control (something like tap head
which controls the water flow) which can turn
On or Off the transistor depending on the
voltage applied at Base. Suppose that I want to
watch a programme on a television after some
known time delay, say after 10 minutes
(10 60=600 sec). My transistor can work as a
switch that can turn the T.V. On or Off
depending on the voltage applied at its Base.
Can you design an RC network that gives me a
voltage of 10 volt after 600 sec? You can
choose the resistor of your choice, but
capacitor is 3300 F and Voltage source is
15Volt. Decide where to take the output of this
RC circuit. (Think and answer)
10. During discharging of capacitor if we
plot , where, V is the voltage
across capacitor at time t, what will be the
shape of the graph? If we take the slope of
the graph at any point what does it give?
V VC
VR
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It will be a straight line with negative slope
During discharging process,
(
)
V = voltage of power supply
(
)
[
] [ (
)]
[
] [
]
Y = m X
Hence if we plot [
] vs. time t
during discharging process, we will get a
straight line passing through origin with
slope -1/RC.
11. During charging, if we plot
(
) vs. time t, what will be the
shape of the graph? (V(t)=voltage across C,
V = voltage of D.C. source)
It will be a straight line passing through origin
with slope (-1/RC). For charging process,
(
)
V = voltage of power supply
(
)
(
)
(
)
(
)
Y = m X
References:
1. Fundamentals of physics, Robert Resnick, David Halliday, Walker, 7P
thP edition, Art. 28.4, RC
circuits, p.882-887.
2. University Physics, Young, Art. 26.4, RC circuits, p.896-900.
ENGINEERING PHYSICS LABORATORY – II
14 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
ZENER DIODE VOLT – AMPERE CHARACTERISTICS
THEORY:
Semiconductors are basically of two types.
Intrinsic semiconductors: These are in their
purest form, without any impurities (Dopants).
Extrinsic semiconductors: These are
impurity added (Doped) intrinsic
semiconductors. Doping is a process of adding
impurity atoms to the pure semiconductors.
The reason for this doping is only to increase
the conductivity of the semiconductors. By
adding Group III elements Boron, Aluminum,
Gallium, Indium (Trivalent impurity) to the
pure semiconductors, it becomes P – type. By
adding Group V elements Nitrogen,
phosphorus, Arsenic, Antimony, Bismuth
(pentavalent impurity) it becomes N – type. P
– type has excess of holes as majority carriers
and N – type has excess of electrons as
majority carriers.
Diode is a semiconductor based electronic
component. It is formed by joining a p – type
section of semiconductor with n – type
section. It has anode (p – type) and cathode
(n – type). It is a polar device, i.e. its operation
will depend on the direction of connection
(biasing).
The above symbol represents an ordinary
P – N junction diode. A denotes the positive
(high potential end) Anode and K denotes the
negative (low potential end) of the diode.
Diode acts like a mechanical check valve,
that conducts (allows flow of liquid) only
when the Anode is at relatively high potential
with respect to the cathode. Suppose that A is
at 10 Volt potential and K is at 9.3 Volt
potential. Then the diode will conduct (closed
switch or Forward Bias) a current from anode
to cathode in the direction of arrow shown in
diode symbol. If the potentials are reversed,
i.e. A at 9.3V and K at 10V, it does not
conduct, acts like infinite resistance (open
switch or Reverse bias).
Forward Bias: Anode of the diode will be at a
relatively high potential than that of cathode.
Reverse bias: Cathode of the diode will be at
a relatively high potential than that of Anode.
Zener and Avalanche diodes are
heavily doped p-n junction
diodes. It is their Ucircuit symbolU.
The doping levels (amounts of added
impurities) are considerably different from
those normally found in a rectifier (PN) diode.
This diode preferably used in REVERSE
BIAS.
A rectifier diode cannot be used in the
breakdown region as it makes permanent
damage to the junction. However, zener and
avalanche diodes are designed to use in the
breakdown region. These diodes are used for
voltage reference and voltage regulator
circuits. There are two mechanisms that cause
a reverse-biased p-n junction to break down:
the UZener effectU and Uavalanche breakdownU.
Either of the two may occur independently, or
they may both occur simultaneously. Diode
junctions that break down below 5 V are
caused by the UZener effectU. Junctions that
experience break down above 5 V are caused
by Uavalanche breakdownU. Junctions that
break down around 5 V are usually caused by
a combination of the two effects.
A zener diode is produced by moderately
doping the p-type semiconductor and heavily
doping the n-type material (see Fig below).
Observe that the depletion region extends
more deeply into the p-type region.
Under the influence of a high-intensity electric
field, large numbers of bound electrons within
the depletion region will break their covalent
bonds to become free. This is ionization by an
Depletion region
Heavily doped
N – Side
Moderately doped P – Side
Denotes atoms/ions
A REVERSE BIASED ZENER DIODE
Bubbles ( ) denote holes and black dots ( ) denote electrons
ENGINEERING PHYSICS LABORATORY – II
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electric field. When ionization occurs, the
increase in the number of free electrons in the
depletion region converts it from being
practically an insulator, to being a conductor.
As a result, a large reverse current may flow
through the junction. The actual electric field
intensity required for the Zener effect to occur
is approximately 3 X 10P
7P Volt/meter. From
basic circuit theory we recall that the electric
field intensity E is given by
V = E d
where
E = electric field intensity (volts per meter)
V = potential difference (volts)
d = distance (meters)
In terms of the p-n junction depicted in above
Fig. we note that the applied reverse voltage is
V and the depletion region width is the
distance d. The narrower the depletion region,
the smaller the required reverse bias to cause
Zener breakdown. A small reverse bias can
produce a sufficiently strong electric field in a
narrow depletion region. By controlling the
doping levels, manufacturers can control the
magnitudes of the reverse biases required for
Zener breakdown to occur. Only certain
standard zener diode voltages are available.
These range from 2.4 to 5.1 V. With lightly
doped p-type material, the depletion region
may be too wide for the electric field intensity
to become sufficient for Zener breakdown to
occur. In these cases, the breakdown of the
reverse-biased junction is caused by avalanche
breakdown (see Fig below).
The depletion region is wider because it
extends more deeply into the p region.
Reverse saturation current is a current flow
across a reverse-biased p-n junction due to
minority carriers. Even though the electric
field strength is not large enough to ionize the
atoms in the depletion region, it may
accelerate the minority carriers sufficiently to
allow them to cause ionization by collision.
The specifics may be outlined as follows:
1. The depletion region is too wide to allow
an electric field intensity of at least 3 X10P
7P
V/m.
2. The minority carriers are accelerated by the
applied electric field.
3. The minority carriers gain kinetic energy.
4. The minority carriers collide with atoms in
the depletion region.
5. The valence electrons of the atoms receive
enough energy from the collisions to
become free (conduction band) electrons.
6. As a result, the number of free electrons in
the depletion region increases to support a
large reverse current. This avalanche of
carriers is also termed as “carrier
multiplication" since one minority carrier
may ultimately cause many free electrons.
The V- I characteristic curve for a zener diode
will be similar to rectifier diode in forward
bias condition. Its behavior in reverse bias is
different from rectifier diode.
Important points from V – I characteristics:
1. Cut – in Voltage (VRγR): During forward
bias of the diode, if we slowly vary the
voltage across the diode, there will be no
observable current up to a characteristic
voltage known as Cut – In or Break – in
voltage or Knee voltage. The minimum
forward voltage to be applied to the diode
to make it just conducting is called its Cut –
in voltage. For Silicon diodes, this cut – in
voltage will be approximately 0.6 to 0.7
Depletion region
Heavily doped
N – Side
Moderately doped P – Side
The black dotted electrons on the P-side are minority carriers that are “Feeling” forward bias and travelling with high speed, colliding with ions of depletion region causing them to release electrons. Their number increases drastically and an avalanche (flood) of electrons are released (Avalanche breakdown)
MECHANISM OF AVALANCHE
BREAKDOWN
MECHANISM OF AVALANCHE BREAKDOWN
ENGINEERING PHYSICS LABORATORY – II
16 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
Volt. For Germanium diodes it will be
approximately 0.2 to 0.3 Volt.
2. Break – down voltage (VRZR): During
reverse biasing of diode, initially there will
be no current through the diode.
(Exception: if we use a micro ammeter, we
can observe some small current, a milli –
ammeter does not show any current ) As we
increase the magnitude of reverse voltage,
there will be a characteristic voltage for the
diode at which it starts conducting
infinitely. Sudden raise of current will be
observed at this point leaving the voltage
across diode almost constant. This voltage
is called the Break – down voltage. For
voltages less than 5V zener break down is
dominant and for voltages greater than 5V,
Avalanche breakdown is dominant.
3. Dynamic Resistance (RRFR and RRZR): During
forward or reverse biasing of diode there
are points at which the current through
diode increase rapidly. At these points the
variation of current with voltage is non –
linear, reflecting that these devices are non
– Ohmic. For Ohmic devices, that obey
Ohm’s law, the resistance does not change
with applied voltage and hence they have
some fixed value of resistance. But here in
the case of diode, the resistance changes
with applied voltage. So we define the ratio
of differential change in Voltage across the
diode with the corresponding differential
change in current through it as the
UDynamic ResistanceU.
4. Material of the diode: Depending on
the cut – in and break – down voltages as
described above, we can decide the make of
the diode.
APPLICATIONS:
1. As voltage regulators for both line
regulation and load regulation in D.C.
power supplies.
2. Used in generating reference voltages
for transistor based and integrated
circuits.
DESIGN OF EXPERIMENT:
PRINCIPLE: To study the V – I
characteristics of the zener diode, we must
measure the current through the diode by
applying various voltages to the diode in both
forward and reverse biases. This can be done
with a variable voltage D.C. source and a
milli – ammeter.
What is the D.C. source?
A variable D.C. power supply with zero
minimum voltage to at least 15 to 20 V
maximum voltage. Its power rating must be
sufficient to draw at least 100 mA current at
these voltages. In our lab we are going to use a
0 – 20 V variable D.C. source with 1Ampere
maximum current.
How to choose the diode?
The zener break – down voltage should not
exceed the maximum voltage supplied by the
D.C. source. As a rule of thumb, the difference
between maximum voltage of the source and
the break – down voltage of the diode must be
greater than at least 5V. If the D.C. source has
maximum voltage of 15 Volt, we can use
zener diodes of break – down voltages up to
10V. The power rating of the diode is
specified by the manufacturer. If we want
mA
RS
REVERSE BIAS V VD
mA
RS
FORWARD BIAS V VD
VR
VR
ENGINEERING PHYSICS LABORATORY – II
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more current through the diode, we must use
high power rated diodes. In our lab we use
either half watt or one watt rated zener
diodes. Their voltage ratings usually vary from
5V to 13V.
How to recognize its polarity?
There will be a ring (band) on the cathode side
it will be the negative of diode and obviously
the other one will be positive of diode. If the
band is not visible, you can test it with a
multi-meter.
How to test a diode with a multi-meter?
There will be a diode symbol on the
multi-meter dial knob. Turn it to the diode
testing mode. Join the positive (red probe) of
multi-meter to one end of the diode and the
negative (black probe) to the other end of
diode. If the meter shows a low resistance of
say few hundred, it means that the diode is
forward biased, i.e. the leg of diode connected
to positive (red probe) is it’s positive and vice
– versa. If the multi-meter shows an infinite
resistance, it means that it is in reverse bias,
i.e. the leg of diode connected to the positive
(red probe) of multimeter is its cathode
(negative) and vice – versa.
How to check whether a diode is working or
spoiled?
To check whether the diode is working or
spoiled use the multimeter test as described
above. If the diode shows very low resistance
in both directions, it is spoiled. If it shows
high resistance only in one direction, it is in
good condition.
What is the function of series resistance
RRSR?
RRSR is used for limiting (controlling) the
current through diode. The value of this
resistor can be decided by the power rating of
the diode.
How to measure the current?
In our lab we have milli – ammeters of 0 – 50
and 0-100 range. We can also use the digital
multimeter (DMM) in current measuring
mode.
Fixing the values of components:
Apply KVL to the forward bias circuit.
During forward bias VRDR =0.7 V approximately
for silicon diode. If the power rating of Zener
is P, maximum current it can hold without
being destroyed is iRmaxR, then,
Or
This will tell us the maximum current the
diode can withstand when a voltage of VRDR is
applied to it. The resistor must be capable of
controlling the current to this threshold value.
As a rule of thumb, we restrict our self
to a threshold current value which is much
lower than the value predicted by the above
expression for iRmaxR. If the predicted value is
say 90 mA, then we restrict to ¼ of this value,
say 20 to 25 mA. Take this value as iRmaxR.
This is to ensure the durability of the diode. If
the applied maximum voltage by the D.C.
source is say 20 V, then
(In forward bias)
Suppose that the zener is a half watt rated.
Then,
. Hence it
can withstand 700 mA. But our
milli – ammeter has only 50 or 100 mA range,
it is better to restrict up to 30 mA in forward
bias. So, IRmaxR is 0.03Amp. Hence,
or
.
The nearest standard resistance value is 680Ω.
The power rating of the resistor can be
calculated using .
Suppose that the maximum current goes up to
30 mA in the resistor, then,
Nearest standard wattage is 1 Watt. If we use a
1KΩ resistor in place of 680Ω, the current
drops and even a 1KΩ half watt resistor can
withstand the maximum current. So, when we
wish to reduce the resistance we must increase
its power rating and vice versa.
For reverse bias replace VRDR with the break
down voltage of the zener diode, say 5.6V, ½
watt rating, then,
.
So, restrict only to 25 mA.
or
, nearest standard value
is 680Ω. So it is better to use 680Ω or more in
both forward and reverse biasing of the circuit.
ENGINEERING PHYSICS LABORATORY – II
18 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
ZENER DIODE V – I CHARACTERISTICS EXPERIMENT
Aim: 1. To study the volt – ampere
characteristics of the given zener
diode.
2. To determine the Cut – in, Break –
down and dynamic resistances of the
diode from the characteristic curves.
Apparatus: Zener diode (½W), Resistors (1 KΩ, ½W),
Variable voltage D.C. power supply,
milli–ammeter (0 – 50 or 100), Multimeter,
bread board, connecting wires.
Procedure:
FORWARD BIAS:
Circuit:
1. Construct the circuit on bread board
according to the circuit diagram for
forward bias. Zener diode has a black
band on it. It shows the cathode of diode.
2. Vary the potentiometer (knob on the
power supply) and apply various voltages
to the diode in steps of U0.1 VoltU. Note the
current in milli ampere as shown by the
ammeter.
3. Initially, there will be no current through
the diode up to a characteristic voltage,
known as Cut-In voltage. Note values of
current until this characteristic voltage as
zero. Observe carefully for this voltage
and note it down.
4. From here onwards note down the
voltage that you observe across the
diode as function of current through
diode in steps of 2 mA by varying the
potentiometer.
REVERSE BIAS:
Circuit:
1. Connect the circuit in reverse bias as
shown in the circuit diagram, i.e. just
reverse the ends of the diode in the
forward bias circuit.
2. Vary the potentiometer (knob on the
power supply) and apply various voltages
to the diode in steps of U1 voltU starting from
zero.
3. Note the value of current in milli ampere
as shown by the ammeter (initially you
wouldn’t get any current, note them as
zero).
4. At a characteristic voltage known as
UBreak downU voltage, you will get a
sudden raise in the current through the
diode. Observe this carefully and note
down the value.
5. From here onwards note down the
voltage that you observe across the
diode as function of current through
diode in steps of 2 mA by varying the
potentiometer.
Graph: Plot a graph by taking the current through diode versus voltage across diode both for
forward and reverse biases. Split the graph into four quadrants.
Choose the scale on voltage axis (horizontal) as 1 division = 0.1Volt on the positive side and 1
division = 1 volt on the negative side.
Choose the scale on current axis (vertical) as 1division = 1 mA on both positive and negative sides.
From each curve on first and third quadrants, calculate the slope of the graph near cut – in and
break – down points. These slopes will give the dynamic conductances of diode. Inverse of
conductance gives the dynamic resistance of the diode.
mA
RS
REVERSE BIAS V VD
mA
RS
FORWARD BIAS V VD
ENGINEERING PHYSICS LABORATORY – II
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Precautions:
1. Do not short the ends of the power supply. This will damage your power supply.
2. Connections on the breadboard must be tight. Avoid loose contacts.
3. Check the polarity of diode carefully.
4. Do not connect the diode without current limiting resistor. This will Uburn outU the diode in any
bias.
Viva – voce questions:
1. What is the basic application of a zener
diode?
It is used as a voltage regulator.
2. I have a silicon made zener diode with
VRZR=5.2V connected in reverse bias with
a series resistor of 100Ω and a variable
D.C. source of 0-20V. Estimate the
maximum current flowing in the circuit.
Determine the current in the circuit if
the diode direction is reversed. Suggest
the minimum power ratings for both
zener diode and resistor in both
connections.
In reverse bias,
For diode,
For resistor,
.
In forward bias, for Silicon VRZ R= 0.7 V
For diode,
For resistor,
.
3. Design a voltage regulating circuit
which drives a cell phone charging unit
with required output at 5.6 Volt, 300mA
with input voltage of 10 Volt D.C.
IRmaxR for the zener is 300mA, i.e. 0.3A.
Zener voltage rating is 5.6V, for diode,
For resistor,
Voltage drop across it will be 10–5.6= 4.4V
; This circuit will
work with a load (cell phone) of resistance
greater than 18.66 Ω. If the load resistance is
further reduced, the circuit will not work.
4. What do you mean by dynamic
resistance?
It is the resistance offered by the diode due
to the changes occurred in input voltage.
Static resistance of a diode refers to a fixed
resistance at a fixed voltage. But dynamic
resistance is some kind of average
resistance offered by the diode when the
input voltage changes between two closely
separated voltage levels.
5. Which bias would you suggest for
operating the zener diode to exploit to
maximum extent?
Reverse bias only.
6. Can we use an ordinary PN junction
diode to regulate the voltage instead of a
Zener diode? Justify your answer.
Yes, Junction diode offers a forward drop
of about 0.7 Volt (for silicon). Hence by
using a combination of forward biased
diodes we can achieve voltage regulation
even in forward bias. A serial combination
of two forward biased silicon diodes will
provide a forward drop of 1.4V.
7. What is the basic difference between
Zener break down and avalanche
breakdown?
Zener break down is due to the breaking of
bonds in the depletion region because of
applied external reverse voltage.
Avalanche breakdown is due to the rupture
of bonds in depletion region by the
collisions of minority carriers that are
accelerated by the applied reverse voltage.
As the temperature increases, the minority
carrier concentration also increases, giving
more chance for avalanche breakdown.
RS
10V 5.6
ENGINEERING PHYSICS LABORATORY – II
20 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
References:
1. Electronic devices and circuits – Discrete and integrated, Stephen Fleeman, Prentice hall, Art.
2-8, zener and avalanche diodes, p.32-36 (taken verbatim).
2. Electronic devices, 9P
thP Ed, Thomas L Floyd, Prentice hall, unit-3, special purpose diodes,
p.113-126.
3. Electronic devices and circuit theory, R. Boylestad, 7 P
thP ed, Prentice hall publications, Art,
semiconductor diode p.10.
STEWART AND GEE APPARATUS
MAGNETIC FIELD ALONG THE AXIS OF CIRCULAR CURRENT CONDUCTOR
THEORY:
Biot – Savart’s Law:
Consider a current carrying conductor (of
arbitrary orientation) as shown in figure. It
carries a current of I. The magnetic field at
a point P at distance from an element of
the conductor will be given by
| |
|
|
| |
Here θ is the angle between the radius vector r
and the length element ds.
, is the free
space permeability constant.
Direction of magnetic field is in the direction
of the cross product of ds and r, given by right
hand screw rule.
Magnetic Field on the Axis of a Circular
Current Loop:
Consider a circular wire loop of radius R
located in the yz plane and carrying a steady
current I, as shown in Figure. We are going to
calculate the magnetic field at an axial point P
at a distance x from the center of the loop.
Consider element of the wire. Using Biot
savart’s law, the field at P due to this will
be
The angle between the ds and r is 90P
0P. So,
And its direction is indicated in the figure.
Also from the figure the angle between vector
r and the y – axis (smaller angle side) is . So,
makes an angle with the x – direction.
Its components along X and Y – directions are
and respectively. When we consider
the entire elements of the loop, their y –
components will cancel with each other
due to the circular symmetry of the coil and
only the x – components survive. So, the total
field at P due to all elements will be,
∮ ∮
Throughout the loop the values of θ and r
remains unchanged and hence can be taken
outside the integral.
∮
∮ is the circumference of the coil.
From the figure,
√
Therefore,
ENGINEERING PHYSICS LABORATORY – II
21 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
(√ )
√
If the coil contains n number of turns, then the
field gets multiplied by that factor.
The direction of this is always either
parallel or anti – parallel to the axis of the coil.
The field B at the centre of the coil can be
obtained by putting x = 0,
At ,
( )
√
Hence, the field B falls to
√ times of its
maximum value BRoR at the center. We can
use this point to calculate the radius of the
coil, without measuring it physically with a
scale, from the experiment.
DESIGN OF EXPERIMENT:
PRINCIPLE:
The variation in B along the axis of the
circular coil can be studied experimentally
with the help of a Tangent Galvanometer.
How to measure B?
The coil will produce a magnetic field . There is a huge EARTH magnet that will
produce another field , known as the
horizontal component of Earth’s magnetic
field. The plane containing the axis of the
hypothetical Earth bar magnet is called the
MAGNETIC MERIDIAN.
If we place the plane of the coil in the
magnetic meridian, then there will be two
mutually perpendicular magnetic fields, one in
the North – South direction (Earth) and the
other in the East – West direction (coil). If we
use a magnetic compass near the coil (which is
already set in magnetic meridian), it will
experience a torque due to the action of the
two magnetic fields and will settle ultimately
in the resultant direction of the two fields.
H = 0.38 Oersted or 0.38 X 10 P
- 4P Tesla
By measuring θ we can estimate the
experimental value of using the above
relation.
What is the coil?
A circular frame holding the coil of variable
number of turns is mounted vertically on a
platform. The platform can be adjusted to
make it horizontal with the help of two
leveling screws. The set up has 2 turns, 50
turns and 500 turns of coil for experimenting.
We use only the 50 turn coil.
How to set up the current in the coil?
Using a fixed voltage D.C. source.
How to measure the current?
Using an ammeter of 0 – 3 Amp range.
How to vary the current the circuit?
By using a 20 Ω Rheostat.
Why to adjust the current?
As we measure the magnetic field as a
function of angle, it is necessary to restrict our
self to some fixed range (30P
0P-60P
0P). Hence it is
required to adjust the current to get the desired
value of deflection θ.
Why to restrict only to 30P
0P-60P
0P range?
When using the instrument it is important to
adjust matters so that the deflection is never
outside the range 25° to 65° and preferably it
should be between 30° and 60°. This is
because the value of θ is to be used in the form
tan θ and an effect which can be called 'error
magnification' arises. The matter will be made
clear by considering the following examples:
Suppose the deflection can only be
observed with an accuracy of half a degree.
Let us consider how this possible error will
affect the values of the tangents of deflections
10°. tan 10° 30' = 0.1853 and tan 9° 30' =
0.1673 thus tan 10° 30' - tan 9° 30' = 0.0180.
Now tan 10° 00' = 0.1763.Thus an observation
of θ = 10° ± 0.5° leads to a statement that tan θ
= 0.1763 ± 0.0090. This represents a possible
error of over 5% in tan θ.
N
ENGINEERING PHYSICS LABORATORY – II
22 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
STEWART AND GEE APPARATUS EXPERIMENT
AIM:
To study the variation in magnetic field with
distance along the axis of circular current
carrying conductor.
APPARATUS:
Stewart and Gee type galvanometer, battery
(D.C. Source 2 Volt - 1Amp), commutator,
rheostat, Ammeter and connecting wires.
CIRCUIT DIAGRAM:
FORMULA:
Where
R0R = 4 X 10P
-7P Henry/meter
n = No. of turn in the coil
i = Current flowing through the circuit
x = Distance of the magnetic compass
from the center of the coil
a = Radius of the coil.
If x and a are expressed in centimeter, then the
resultant expression will be
In gauss, the same formula will be,
DESCRIPTION OF EQUIPMENT:
It consists of a circular coil in a vertical plane
fixed to a horizontal frame at its middle point.
The ends of the coil are connected to binding
screws. A magnetic compass is arranged such
that it can slide along the horizontal scale
passing through the center of the coil and is
perpendicular to the plane of the coil. The
magnetic compass consists of a small magnet
and an aluminum pointer is fixed
perpendicular to the small magnet situated at
the center of the compass. The circular scale in
the magnetic compass is divided into four
quadrants to read the angles from 0 to 90
and 90P
0P to 0P
0P. A plane mirror is fixed below
the pointer such that the deflections can be
observed without parallax.
PROCEDURE:
1. The circuit should be connected as shown
in the diagram.
2. Remove the power connection applied to
the circuit.
3. Place the compass exactly at the centre of
the coil.
4. Adjust the arms of the magnetometer until
the pointer of compass becomes parallel to
it. Rotate the compass until the pointer
reads 0P
0P -0P
0P.
5. Suppose that the coil is placed in magnetic
meridian and switch on the power to
circuit. It will show some deflection.
Carefully adjust the rheostat and bring the
deflection to 60P
0P -60P
0P.
6. Interchange the plug keys of the
commutator and reverse the current
direction in the coil. Note down the
deflections of compass.
7. If your coil is exactly in magnetic meridian,
then the readings of compass should not
differ by more than 5°P
Pfrom their previous
values, before interchanging the
commutator. If this is not satisfied, once
again turn off the power and make the
pointer parallel to the magnetometer and
repeat this until you get all four deflections
within 5P
0P variation.
8. Move the compass to 10 cm distance on
both east and west directions on the
magnetometer and obtain the deflections
with both directions of current.
9. If all the eight deflections that you have
obtained in above case lie within 5P
0P, you
can start taking deflections at various
positions. UNow the instrument should not
be disturbed while moving the compass U.
Otherwise repeat the adjustment by
disconnecting the power.
A
C
S.G. coil
Rh (20Ω) 2 VOLT D.C.
0 to 3 Amp
ENGINEERING PHYSICS LABORATORY – II
23 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
Tanθ
Position of compass
East 0 West
Tanθ
East 2R West
√
10. Start at 0 cm position and obtain four
deflections. Vary the position to 2 cm either
East or West and obtain four more
readings. Tabulate them.
11. Proceed in the same way at 2, 4, 6…. cm
on both East and West until the deflection
falls less than 20P
0P. Tabulate the readings.
PRECAUTIONS:
1. The Stewart and Gee apparatus should
not be disturbed after the adjustments.
2. Observations are noted down without
parallax.
3. The ammeter and rheostat should be
kept far away from the deflection
magnetometer
Graph:
A graph is plotted taking distance of the compass from the
center of the coil along X-axis and tan along Y-axis. The
shape of the curve is as shown in the figure and is symmetric
about Y-axis. The magnetic field is found to be maximum at
the center of the coil. The radius of the coil ‘a’ is determined by
measuring its circumference. The current flowing through the
circuit ‘i’ and the number of turns in the coil ‘n’ are noted. The
value of magnetic induction is calculated from the above
formula and is compared with the experimental formula B = H
tan θ.
Viva-Voce questions:
1. What are the magnetic forces acting on
the compass when it is mounted on the
axis of the coil? Mention their directions.
The forces are, due to Earth’s magnetic
field along the geographic north direction
and due to coil along either east or west
direction.
2. What is the direction of the magnetic
field produced by the coil?
Along East or West, i.e. perpendicular to
the plane of the coil.
3. Why do we adjust the maximum
deflection at 60° ?
To restrict the error in the measurement of
θ and hence in the tan θ to less than 5%, we
always adjust the maximum deflection to
60°.
4. State Biot-Savart’s law.
Refer to text.
5. Define magnetic meridian.
It is the plane containing the axis of the
earth’s hypothetical bar magnet.
6. Why the ammeter should be placed far
away from the coil?
If it is sufficiently close to the coil, its horse
shoe magnet will influence the resultant
deflection of the compass which is an
undesirable effect.
7. What is the function of rheostat in this
experiment? To vary the current in the circuit and to
bring the deflection to desired value.
8. Can you determine the radius of the coil
without measuring it with a scale?
Yes, consider the tan θ vs. position graph.
Maximum value of tan θ is obtained at the
centre. Calculate the value of
√ .
Draw a horizontal line intersecting the tan θ
axis at this value. The line intersects the
graph (curve) at two different points. The
graphical distance between these two points
will give the diameter of the coil and half
of it will give the required radius of the
coil.
REFERENCES:
1. Advanced level physics, Nelkon and parker, magnetic fields due to conductors, p.935 2. Fundamentals of physics, Resnick, Halliday, Walker, 7P
thP ed, Example 30.3, p.942 (for fig).
APPENDIX ENGINEERING PHYSICS LABORATORY – II
24 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
BREAD BOARD: The figure shows sets of five holed boxes. Each hole in a five hole box has METAL CONTACT with the remaining four holes in that
box.
HORIZONTAL BUSES:
The series of holes on the top and bottom parts of the bread board are called horizontal buses. For indexing purpose they were named as A, B, C, D, E, F, G
and H in the figure. UIN PRACTICAL BREAD BOARD YOU WILL NOT FIND ANY SUCH NAMING. U
A bus: The five hole pairs are joined to each other by a metal strip on the back side of bread board. If you insert a battery positive lead in any of the holes in A
bus, the other holes will also have the same potential. Similarly the buses B, C, D, E, F, G and H also have the same hole connections. The above said eight
horizontal buses are independent of each other, i.e. A and B do not have any connection, similarly A and E ; B and F etc, are not connected.
USUALLY THE A BUS IS RESERVED FOR POSITIVE OF THE D.C. SUPPLY. SIMILARLY C BUS IS RESERVED FOR GROUND (NEGATIVE OF
D.C. SUPPLY)
A
B
A
B
C C
D D
E
F
E
F
G
H H
G
1 2 3 4 5 6
7…………………………………………………………………………………………………………………………………………………………………….56 57 58 59 60
……………………………………………………………………….
661 62 63 64 65 66 67………………………………………………………………………………………………………………………………………………………………………………. 120
……………………………………………………………………….
VER
TIC
AL
BU
SES V
ERTIC
AL B
USES
APPENDIX ENGINEERING PHYSICS LABORATORY – II
25 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
BATTERY/
D.C.
SOURCE
VARIABLE
VOLTAGE
D.C. SOURCE GROUND ZERO POTENTIAL
LIGHT EMITTING
DIODE (LED)
INDUCTOR
RHEOSTAT
Temperature
Sensitive Resistor
VERTICAL BUSES:
The five holed buses numbered as 1, 2,…29,30…58,59,60,….120 in the fig. are called VERTICAL
BUSES. There is a metal strip on the back side of five holes in each vertical bus. Hence there is no
connection between 1 and 2 buses. This is same for all other vertical buses. Hence if we insert any
component lead in a vertical bus, the remaining four holes will come in contact with the component.
There are two rows of such vertical buses in the middle of the bread board in between the horizontal
buses. Vertical buses are used for inserting the components like resistors, capacitors and IC’s.
A bus is reserved for +ve of the power supply. C bus is reserved for -ve of the power supply,
this is also known as ground bus. If the circuit is complex and has many more power
supplies, say, a circuit may run with 18 V, 12 V and 9V power supplies with common ground
(-ve), we can use the B, E, F, D, G, H buses for those power points. Sometimes many
connections are made with a single power point. In that case we can join the A and E buses
with a (jumper) wire to use the entire top line as power bus +V RCCR. Similarly we can join C
and G buses for having a long ground bus. If the circuit is much more complex, then we join
two or more bread boards together to provide more space for the extra components. But the
rule of making a circuit is that its layout must be very clear and understandable to any other
person and at the same time it should use minimum space on the
bread board. COMPONENTS AND THEIR CIRCUIT SYMBOLS:
ZENER
DIODE
CAPACITOR (NON-ELECTROLYTIC)
NO POLARITY,
CAN BE USED
IN BOTH WAYS
CAPACITOR (ELECTROLYTIC)
HAS POLARITY
AMMETER
A +
_
_ +
MICRO AND MILLI -
AMMETERS
µA mA + + _ _
GALVANOMETER
G
PN JUNCTION DIODE
RESISTOR
(FIXED RESISTANCE)
POTENTIOMETER
(VARIABLE
RESISTANCE)
APPENDIX ENGINEERING PHYSICS LABORATORY – II
26 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
i1
i2
i3
B
D
10V
C
i
A
KIRCHHOFF’S LAWS (KCL AND KVL):
1. CURRENT RULE OR JUNCTION RULE (KCL): The
algebraic sum of all currents meeting at any junction (node) of a
circuit is zero. The convention of current direction is that the
current is positive if it moves towards the given node or junction
and it will be negative, if it moves away from the junction. Here iR1R
is positive as the current is approaching the node and the other
currents iR2R, iR3 Rare negative as they move away from the node.
2. VOLTAGE RULE OR LOOP RULE (KVL): The algebraic sum
of the voltage drops in any closed loop of the given
circuit is zero. It means, VRADR+VRBAR+VRCBR+VRDC R= 0.
VRADR means the potential at point A with respect to the
point D. So, VRADR= +10. Similarly V RDAR= –10.
Convention: Assume an arbitrary direction in the given
loop, i.e. say ABCD. If you are travelling from A to
B, the potential drop will be V RABR, equal to – VRDR, the
voltage drop across the diode in forward bias. This is
because the voltage at B is less than the voltage at A by a value equal to the forward cut – in
voltage of the diode (VRDR). Similarly from B to C, VRBCR=+ i RRLR (by using Ohm’s law). If you are
travelling from C to B, then it will be – i RRLR. Similarly, VRCDR= +VRCR, the voltage across the
capacitor. If the capacitor is charged, then the positive plate will be at high potential than the
negative plate by the value of applied voltage. Hence, the equation will be, +10 – VRD R– i RRLR+VRC
R= 0. If we travel from D to A along DCBA path, then the equation will be, –VRC R+ i RRLR+R RVRDR –10
= 0. Hence both equations are one and the same.
COLOUR CODES FOR CARBON RESISTORS
APPENDIX ENGINEERING PHYSICS LABORATORY – II
27 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
REFERENCE BOOKS:
1. ELECTRONIC DEVICES 9P
thP Ed; Thomas L. Floyd; Unit -2, Diodes and applications; Prentice
Hall publications. 2. ELECTRONIC DEVICES AND CIRCUITS; Jacob millman and Christos Halkias. Mc. Graw-
hill publications.
BROWN 1
BLACK 0
RED 2
Error: Gold ± 5%
Black - 0 Brown - 1 Red - 2 Orange - 3 Yellow - 4 Green - 5 Blue - 6 Violet - 7 Grey - 8 White - 9
Tolerance: (Error in the mentioned value of resistance) No colour : ± 20% Gold colour : ± 5% Silver colour : ± 10%
The first two colour bands represent the first two digits
Third colour band indicates the number of ZEROs.
Resistance of above resistor will be 10 with two zeros, i.e. 1000 Ω. Gold band indicates 5% error. i.e. ± 50Ω. Resistance will be (1000±50) Ω. If you measure the resistance you will find it lying between 950Ω and 1050Ω
B B R O Y of Great Britain has Very Good Wife
APPENDIX ENGINEERING PHYSICS LABORATORY – II
28 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
LEAST SQUARE FIT
(FITTING THE DATA TO A STRAIGHT LINE)
To fit the given data to a straight line the following process is to be adopted.
Define a parameter called Residue
∑
The standard deviation S of the data point (xRiR, yRiR) from its average value ( , ) will be
∑
To minimize the deviation with respect to the constants m and c to have a best fit,
After solving the equations we get
∑ ∑ ∑
and
∑ ∑ ∑
After solving for m and c gives
And
Where, refers to the average values of all xRi Rand yRiR respectively.
If the given function is a polynomial of the form y = x P
mP, then use natural logarithm to transform it in
to a linear equation containing logarithmic variables and proceed in the same manner as described
above.
θ error in θ
tan (θ) tan
(θ+0.5) tan
(θ- 0.5) % error in
tan θ
10 ±0.5 0.176327 0.185339 0.167343 5.103143
15 ±0.5 0.267949 0.277325 0.258618 3.490766
20 ±0.5 0.36397 0.373885 0.354119 2.715347
25 ±0.5 0.466308 0.476976 0.455726 2.278461
30 ±0.5 0.57735 0.589045 0.565773 2.015435
35 ±0.5 0.700208 0.713293 0.687281 1.857457
40 ±0.5 0.8391 0.854081 0.824336 1.772394
45 ±0.5 1 1.017607 0.982697 1.745506
50 ±0.5 1.191754 1.213097 1.17085 1.77249
55 ±0.5 1.428148 1.455009 1.401948 1.857676
60 ±0.5 1.732051 1.767494 1.697663 2.015844
65 ±0.5 2.144507 2.1943 2.096544 2.279222
70 ±0.5 2.747477 2.823913 2.674621 2.716881
75 ±0.5 3.732051 3.866713 3.605884 3.494454
80 ±0.5 5.671282 5.975764 5.395517 5.115662
∑
∑
ENGINEERING PHYSICS – II LAB ROLL NO:
G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
Expt. No…………… Date: ……………….. BAND GAP OF EXTRINSIC SEMI CONDUCTOR USING PN JUNCTION DIODE
DATA SHEET Temperature In °C (TC)
T (Kelvin) = TC + 273
Current (I) µA
= ln(I)= ( )2
Total number of observations of made
N = Average
= Average
=
=
=
ENGINEERING PHYSICS – II LAB ROLL NO:
G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
Expt. No…………… Date: ……………….. THERMISTOR CHARACTERISTICS
DATA SHEET Temperature
in 0 C Resistance
in Ω Temperature in K
Xi= 1/T (K-1)
Yi= ln R
Total number of observations of made N =
Average = Average
From TABLE:
B =
A = From GRAPH:
ENGINEERING PHYSICS – II LAB ROLL NO:
G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
Expt. No…………… Date: ……………….. RC CIRCUIT – TIME CONSTANT
DATA SHEET CHARGING PROCESS Resistance R = Capacitance C =
RC (Theoretical) = Voltage of D.C. source V =
Time in sec
VOLTAGE ACROSS
CAPACITOR VC (in volt)
0
Time in sec
VOLTAGE ACROSS
CAPACITOR VC (in volt)
FROM GRAPH:
TIME INTERVAL:
Slope of the graph:
Value from graph =
ENGINEERING PHYSICS – II LAB ROLL NO:
G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
Expt. No…………… Date: ……………….. RC CIRCUIT – TIME CONSTANT
DATA SHEET DISCHARGING PROCESS Resistance R = Capacitance C =
RC (Theoretical) = Voltage of D.C. source V =
Time in sec
VOLTAGE ACROSS
CAPACITOR VC (in volt)
0
Time in sec
VOLTAGE ACROSS
CAPACITOR VC (in volt)
FROM GRAPH:
TIME INTERVAL:
Slope of the graph:
Value from graph =
ENGINEERING PHYSICS – II LAB ROLL NO:
G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
Expt. No…………… Date: ……………….. ZENER DIODE DATA SHEET
FORWARD BIAS Current limiting resistor RS=
REVERSE BIAS Current limiting resistor RS=
Voltage (in volt) across
Am
met
er
Rea
ding
(SU
PPLY
)
ZEN
ER
DIO
DE
RES
ISTO
R
V VD VR ID (mA)
Voltage (in volt) across
Am
met
er
Rea
ding
(SU
PPLY
)
ZEN
ER
DIO
DE
RES
ISTO
R
V VD VR ID (mA)
FROM GRAPH: FORWARD BIAS CHARACTERISTICS: CUT – IN VOLTAGE (Vγ) : SLOPE OF V – I GRAPH IN FORWARD
BIAS FORWARD DYNAMIC RESISTANCE
Material of diode may be……………..
REVERSE BIAS CHARACTERISTICS: BREAK – DOWN VOLTAGE OR ZENER VOLTAGE (VZ): SLOPE OF V – I GRAPH IN BREAK –
DOWN REGION ZENER RESISTANCE IN BREAK – DOWN
REGION
ENGINEERING PHYSICS – II LAB ROLL NO:
G.V.P. COLLEGE OF ENGINEERING FOR WOMEN
Expt. No…………… Date: ……………….. STEWART AND GEE APPARATUS
DATA SHEET Current through the coil i = ……….. Ampere Horizontal component of earth’s field H = 0.38 Oersted Circumference of the coil = radius (a) =
Radius of the coil from the Graph =
Sl.
No Distance
x
Deflection magnetometer readings
Θ = tan θ
Bexp=
H tan θ BTh East West
Θ1 Θ2 Θ3 Θ4 ΘE Tan θE Θ5 Θ6 Θ7 Θ8 Θw Tan θW