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Transcript
Department of
Applied Sciences
LAB MANUAL
Basic Electrical & Electronics Lab (BTEE 102)
B.Tech 1st Year – All branches [2012-13]t
of
V KCT COLLEGE OF ENGG & TECH, FATEHGARH
Punjab Technical University
LAB MANUAL
KCT College of Engg. & Tech.
Department- AS
INDEX
Sr no
Name of Experiment
1
To experimentally verify the ohm’s law
2
To verify Kirchhoff's Voltage law in a series circuit and parallel circuit
3
To find voltage, current relationship and power factor of a given R-L circuit
4
To find out the line voltage, phase voltage relationship, line current and phase
Current relationship in case of star-connected and delta connected, 3-phase
balanced load.
5
6
To measure resonance in AC circuits
To study Hysteresis property of the given magnetic material
7
Full-Wave Rectification
8
To verify the working of LVDT
9
Semiconductor (or crystal) diode characteristics.
10
To study the truth table of all the gates
11
12
To study the input and output characteristics of a PNP/ NPN transistors in
common base configuration
To perform open-circuit and short-circuit test on a transformer
13
To study the speed control of a dc shunt motor
14
To connect, start and reverse the direction of rotation of a 3-phase induction
motor.
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EXPERIMENT-1
Objective: To find voltage, current relationship and power factor of a given R-L circuit.
Apparatus Required: Variac or 1-phase auto-transformer-1, Single-phase ac load [or lamps
and choke coils], Moving-iron voltmeter [0-250 V]-1, Moving iron ammeter [0-5 A]-1,
Dynamometer .~-pe wattmeter (250 V, 5 A)-1, Double pole, single throw switch (DPST
Switch)-1, Connecting ieads.
Theory: Current flowing through an ac circuit is given as
I= V/Z
where V is the ac supply voltage (voltage applied to the circuit) and Z is the impedance of
the circuit in ohms.
Power factor of an ac circuit is given as P
Power factor, cos ~ = VI
Where P is power of the given load circuit in watts, V is the voltage applied to the
circuit in volts and I is the current in amperes flowing through the circuit. Connection
Diagram:
Procedure: Variac, ammeter, wattmeter, voltmeter, load, (lamps in series with choke coils)
are connected, through double pole single throw switch, to single phase
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supply mains as shown in fig. 1. The readings of ammeter, voltmeter and
wattmeter are noted for var~ous settings of variac.
Observations:
S.No. Voltmeter Reading, Ammeter Reading,
V in Volts
1.
2.
3.
4.
Wattmeter
Power Factor of
I
Reading,
the Circuit,
in Amperes
P in Watts
Cos ~ = P
VI
Conclusion: 1. Current I increases directly in proportion to applied voltage V.
2. Power factor of the circuit is same through out for a given load.
Note: If required, the readings of the instruments can be recorded with the different loads
(by varying the number of lamps and chokes connected in series).
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EXPERIMENT-2
Objective: To find out the line voltage, phase voltage relationship, line current and phase
Current relationship in case of star-connected and delta connected, 3-phase balanced
load.
Apparatus Required: TPIC switch (500 V, 15 A)-1, Moving iron type voltmeter (0500 V)-2, Moving iron ammeter (0-10 A)-2, 3-phase balanced load (say a 3-phase, 440 V,
50 Hz, 7.5 kW induction motor), Board containing six open terminals-1 and Connecting
leads.
Theory: In star-connections
Line voltage, VL =~ p h as e vo lt age =~ V p
Line current, IL = Phase current, Ip In deltaconnections
Line voltage, VL = Phase voltage, Vp
Line current, IL =~ phase current I p
Connection Diagram:
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Procedure: The connections of the terminal box of the induction motor to the stator phase
windings of the motor are shown in fig. 3.3 (a). First of all the connections in terminal box
are made, as shown in fig. 3.3 (b). The motor is started and put on load and readings of both
voltmeters and ammeters are noted for different loads on the motor.
Now the TPIC switch is made off and the connections in terminal board are changed,
as illustrated in fig. 3.3 (c) for delta connections. The motor is started and put on
load and the readings of both ammeters and voltmeters are noted for different loads
on the motor.
Connection S. No. ~ Voltmeter
s
Star
Delta
Reading Vl,
1
2
3
l
Voltmeter
Ammeter
Reading V2,
Reading Al,
Line Voltage Phase Voltage Line Current
in Volt5, VI
Ammeter V~
~
Vp
IP
in Volts, Vp in Amperes, IL Reading A2,
Phase Current
in Amperes, Ip
23
Results : In star connections, line voltage V comes out to be ~ times phase voltage
Vp and line current IL comes out to be equal ~o phase current Ip. In delta connections,
lines voltage V L comes out ,;o be equal to phase voltage V p and line current IL comes
out to be times ~ phase current Ip
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Experiment-3
Objective: To perform open-circuit and short-circuit test on a transformer and
determine the following:
(a.) the transformation ratio.
(b) the transformer efficiency at 25%, 50%. 75%, 100%, l50% load at pf of 0.8 lagging
and to pplot the characteristic
Apparatus Required: Single phase transformer (preferably 1 kVA with one winding
rated at 230 V)-1, V ariac-1, Low range voltmeter-1, Low range ammeter-1, Low range
low power factor wattrneter-1, W~attmeter, ammeter and voltmeter of normal ranges
(depending upon the rating of transformer under test) one each, DPIC switch-1 and
connecting leads.
Theory: Open-circuit test is performed to determine the iron loss and the
transformation ratio. Since during this test no current flows in the open-circuited
secondary, the current in the primary is merely that necessary to magnetise the core at
normal voltage. Moreover, the magnetising current is very sma11 fractioii of the fullload current (usually 3 to 10 per cent of full-load current) and may be neglected as far
as copper loss is concerned consequently, this test gives this core loss alone practically.
The transformation ratio, K-_ Secondary turns _ Secondary voltage on opencircuit Primary turns Voltage applied to primary
=V2 / V1
The short-circuit test is performed to determine the full-load copper loss. In this test
~ince the terminals of the low voltage winding are short-circuited, therefore, the
trans:ormer becomes equivaTent to a coil having animpe~arice -equal--t-o the
impedance of both the windings. The applied voltage is very small as it is adjusted
to cause flow of rated current in the windings, so the flux linking with the core is
very small and, therefore, iron losses are negligible. The power drawn from the
supply, therefore, represents full-load copper loss.
Transformer efficiency at any given load is given as
_
Output ______________ x
100 Output + iron loss + copper loss at given load
where Output = Voltmeter reading in open-circuit test, Vl
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x ammeter reading during short-circuit test, I5 x pf
Iron loss, Pi = Power input in open-circuit test with rated voltage = P o and
Copper loss, P~ = Power input in short-circuit test
Connection Diagrams: See figures 8.20 and 8.21
Procedure: 1. (ipen-circuit Test. The transformer is connected, as shown in fig. 8.20
on the low voltage side. The ammeter A is a low range ammeter and W is a low
range and low power factor wattmeter. The high voltage side is kept opencircuited. AC supply at rated voltage is switched on to the transformer through DPIC
switch. High range voltmeter V2 is connected across the secondary of the transformer.
Readings of voltmeters V l and V2, ammeter A and wattmeter W are noted and
tabulated as below.
2. Short-circuit test. The transformer is connected as shown in fig. 8.21, the low
voltage winding is short-circuited by connection having negligible resistance and
good contacts. The voltage applied to the primary is increased in steps so that
ammeter A carries 0.25, 0.5, 0.75 time rated current, rated current, and 1.5 times
rated current.
The efficiency of the transformer is calculated at different loads as mentioned
above and a curve is plotted between efficiency and load current.
Observations:
(i) Open-circuit test:
Primary voltmeter reading, V l = ........
Secondary voltmeter reading, V 2 = .......
Input power on no load, P o = Wattmeter reading, W= .........
V_
Transformation ratio, K = V l
...................
(ii) Short-circuit test: Rated primary current,
Rated kVA X 1,000
I
-
_ ...........
f Rated primary voltage
Power factor, cos ~ = 0.8 (lag)
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Ammeter Reading
Wattmeter Reading
in Amperes, IS
in Watts, PS
Transformer Efficiency,
V~ Is cos~
~~ =
_
x
Vlls cos~+P o+Ps
100
Vlls x0.8+P
V~Is ax +P
0.8s x
1.
2.
3.
4.
5.
100
Result: 1.During short-circuit test, power input, P S varies as the square of the input
current, I5 i.e. copper loss varies as the square of the load current.
2. The curve plotted between the transformer efficiency and load current [fig. 4)
shows that the efficiency becomes maximum when copper loss equals iron loss.
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EXPERIMENT-4
Objective: To study the speed control of a dc shunt motor and to draw the speed variation
with respect to change of (a) field current (field control) and (b) armature current (armature
control).
Apparatus Required: 250 V dc shunt motor of capacity, say, 2 kW-1, PMMC
ammeter (0-2.5 A)-1 and (0-10 A)-1, Rheostats-high resistance, low current-1 andlow
resistance, high current-1, Tachometer-1, DPIC switch-1 and Connecting leads.
Theory: The speed of a dc shunt motor is given as
Hence the speed of a dc shunt motor can be changed by either changing the shunt fie7d
azn~xit h~~ (by inserting an external resistance in the field circuit) or changing the back
emf, Eh, (i.e. V- IQ Ro) by inserting an external resistance in the armature circuit.
Connection Diagram:
Procedure: The connections are made, as shown in fig. 10, and the rheostats are put in
zero positions. The dc supply is switched on by closing the DPIC switch.
Field control: The reading of ammeter A1 is noted and the speed of the motor is
measured by tacliometer. Now the field rheostat setting is changed in steps
(increasing the resistance)
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and ammeter Al readings are noted and the motor speed is measured and
recorded. Now the field rheostat setting is brought back to zero.
Armature Control: Resistance is increased in steps in the armature circuit for
armature resistance control. The ammeter A z readings are noted and the motor speeds
are measured
and recorded.
Armature control rheostat is brought back to zero and the supply is switched off through
DPIC switch.
Observations:
S.No.
Field Control
Armature Control
Ammeter A1 Read-
Speed of Motor
Ammeter A~ Read-
Speed of Motor
ings in Amperes, Ish
in RPM
ings in Amperes, Io
in RPM
1
2
3
4
5
Result: l. With the increase in resistance in the field circuit, the field current
decreases i.e. field becomes weaker and, therefore, speed increases.
2. With the increase in resistance of the armature circuit, voltage drop in
armature increases i.e. back emf Eh decreases and, therefore, speed
decreases.
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EXPERIMENT-5
Objective: To connect, start and reverse the direction of rotation of a 3-phase induction
motor.
Apparatus Required: Three phase induction motor-1, Star-delta starter-1, TPIC
switch1, Tools such as screw driver, plier, test pen etc. and connecting leads.
Theory: Motor is connected to the 3-phase ac supply mains through star-delta
starter and TPIC switch, as shown in fig. 11 (a). The direction of rotation of a 3-phase
induction motor can be reversed by interchanging any two terminals at the TPIC
switch, as shown in fig.l l(b)
Procedure: The connections of a 3-phase induction motor are made to the stardelta starter and to the TPIC switch, as shown in fig. 11 (a). The TPIC switch is
closed and the motor is started !-y taking the lever of the starter to the start (star)
position and then with a jerk to the run position (or delta connections). The
direction of the rotation of the motor is observed. Say, it is in clock-wise direction.
Now the motor is stopped by pushing the stop button and supply to the motor is
r'emoved by opening the TPIC switch. The two leads of the motor are interchanged to
the T1~I~C\switch, as shown in fig. 11 (b). TPIC switch is closed and the motor is started
again. Tl,~e direction of rotation of the motor is observed.
The push button is pushed and the TPIC switch is made off.
Observations and Results: The direction of rotation of the motor in second case
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is found opposite to that in first case.
EXPERIMENT NO 6
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Procedure : lnsert IC 7432 on the IC base of the logic experimental trainer board.
Give + 5 V input to pin 14 and graund pin 7. Give inputs A and B to pins 1 and 2 using
binary switches. Connect output pin 3 to LED. ~witch on the trainer. Observe LED
output. Try with different combinations of t A, B. Prepare truth tabie. Repeat for other
gates.
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EXPERIMENT NO 7
Title: Semiconductor (or crystal) diode characteristics.
Objective:
1. Trace the circuit meant to draw the diode-characteristics;
2. Measure the current through the diode for a particular value of forward voltage;
3. Plot the forward characteristics of a germanium and a silicon diode;
4. Compare the forward characteristic of a Ge diode with that of a Si diode;
5. Calculate the forward static and dynamic resisrance of the diode at a particular
operating point.
APPARATUS REQUIRED: Experimentai board, regulated power supply,
milliammeter, electronic multimeter.
CIRCUIT DIAGRAM: The circuit diagram is given in Fig.
Brief Theory: A diode cr5~duets in forward bias (i.e. when its anode is at higher poi~ntiai
than its cathode). It does not conduct in reverse bias. When the diode is forward-biased, the
barrier potential at junction reduce~.
The majority carriers then diffuse across the junction. This causes current to flow
through the diode. In-reverse bias, the barrier -potential increases, and almost no current
can flow through the diode.
The external battery is connected so that its positive terminal goes to the anode and
its negative terminal goes to cathode. The diode is then forward biased, The amount of
forward bias can be varied by changing the externally applied voltage. As shown in
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Fig. E.4.1.1, the external vo7tage applied across the diode can be varied by the
potentiometer Ri. A series resistor (say, 1 ks2) is connected in the circuit so `that
excessive current does not flow through the diode. VVe can note down different
values of the current through the diode for various values of the v~lta,ge across it. A
nlat between this voltage and current gives the clzode forward characteristics
At a given operating point we can determine the static rcsastance (Ra) and dynamic
resistance (ra) of the diode T'rom its characteristic. The static resistance is defined as the
ratio of the dc voltage to dc current, i.e.
Rd =V/I
The dynamic resistance is the ratio of a small change in voltage to a- small ~hange in current, i.e.
PROCEDURE:
1 Find the type number of the diodes connected in the experirnental board.
2. Trace the circuit and identify different components used in the cii•cuit.
Read the value of the resistor using the colour code.
3. Connect the milliammeter and voltmeter of suitable ranges, say, (i, to
25 mA for ammeter and 0 to 1.5 V for voltmeter.
4. Switch on the pawer supply. With the help of the potentiometer R t,
increase the voltage slowly.
5. Note the milliammeter and voltmeter readings for each setting of t~e potentiometer.
Tabulate the observations.
6. Draw the graph between voltage and current.
7 At a suitable operating point, caiculate the static and dynamic resistance of tlte
diode, as
illustrated in Fig. E.4.L2.
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Experiment no 8
Objectives:
To experimentally verify the ohm’s law.
Apparatus:
DC Power Supply
DC current source
Few Resistors
Wheatstone bridge
Multimeter
THEORY:
Ohm’s Law:
The voltage across an element is directly proportional to the current through it. The
ohm’s law can be written mathematically as:
V  IR
where R = Resistance
V = voltage across the resistance R
I = Current through the resistance R
PROCEDURE:
A. Ohm’s Law:
1.
2.
3.
4.
Connect the circuit as shown in figure 2.
Set the DC voltage supply to 10 Volts.
Set the resistance R to 100 ohms.
Measure the voltage across the resistor and the current through the resistor and write
the results in Table 1.
5. Determine the value of the resistance using Ohm’s law R=V/I and record in the Table
1.
6. Repeat step 2 to 5 for the other resistors (1000 ohms, 10 K ohms).
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1 K ohms
R
Vs=10V
Figure 1: Ohm's Law
TABLE 1
Resistor (Nominal Value)
100 
1 K
10 K 
Ohm-meter Reading
R=V/I
Percent Deviation from Nominal Value
Percent Deviation = (Nominal Value – Ohm-meter Reading) / (Nominal Value)
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EXPERIMENT NO 9
Kirchhoff’s Laws
Objective:


To verify Kirchhoff's Voltage law in a series circuit.
To verify Kirchhoff's Current law in a parallel circuit.
Series Circuit
R1
VS = 10V
Kirchhoff's Voltage law:
= IR1 + IR2
junction
R2
Parallel Circuit
VS = 10V
VS = VR1 + VR2
R1
R2
Kirchhoff's Current law: IT
in a loop
at a
Part A - Series Circuit:
1. Calculate the total resistance, circuit current and voltage drop across each resistor if
wired in a series circuit as shown in the figure.
Total Resistance:
Circuit Current:
Voltage Drop across the 470 Ohm Resistor, R1:
Voltage Drop across the 1K Ohm Resistor, R2 :
Power-supply Voltage using Kirchhoff's Voltage law :
(should be equal to the sum of the voltage drops)
2. Obtain a 470 Ohms and a 1K Ohms resistors and make sure you got the correct ones.
Measure and record their actual resistances.
R1 = ___________________
R2 = ___________________
3. Using the Breadboard, connect the resistors in SERIES as shown in the figure below:
R1 is the 470 Ohm resistor and R2 is the 1K Ohm resistor.
R1
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4. Measure the total circuit resistance.
Total Resistance:
5. Connect the Power-supply to the SERIES resistors as shown in the figure below.
6. Set the Power-supply to 10 Volts.
7. Measure the voltage drop across each resistor and the Power-supply voltage.

Voltage Drop across the 470 Ohm Resistor:

Voltage Drop across the 1K Ohm Resistor:

Power-supply Voltage (should be equal to the sum of the voltage drops):
8. Verify that all the measured values tally (matches) with the calculated values.
Part B - Parallel Circuit:
1. Using the Breadboard, connect the PARALLEL circuit. R1 is the 470 Ohm resistor
and R2 is the 1K Ohm resistor.
2. Calculate the current flow through each of the resistors, and the total circuit current.
To calculate current through each of these resistors use ohm’s law: I = V/R
Current through:
IR1 =
IR2 =
Total Circuit Current using Kirchhoff's Current law :
Total Circuit Current using Ohm’s law :
IT =
IT = VS / RP = ?
1/RP = 1 / R1 + 1 / R2
RP =
( R1 + R2 ) / ( R1 x R2 )
R1
R2
RP
R1 is the 470 Ohm resistor and R2 is the 1K Ohm resistor.
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RP = ?
3. Measure the current flow through each of the resistors, and the total circuit current.
Current through:
IR1 =
Total Circuit Current:
IT =
IR2 =
9. Verify that all the measured values tally (matches) with the calculated values.
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EXPERIMENT 10
Title: to measure resonance in AC circuits
INTRODUCTION
In this experiment the impedance Z, inductance L and capacitance C in alternating
current circuits will be studied.
The parameters of the circuit will be varied to produce the condition called resonance.
The inductive and capacitive reactance are defined as follows:
Inductive Reactance = XL = 2fL
Capacitive Reactance = XC = 1 .
2fC
XL
The impedance in a series
AC circuit is found by
adding the individual
reactances and resistance
as vectors as shown in
Figure 1.
Z
XC
XL  XC
R
Fig. 1
Voltages are all equal to
the current I, times the
individual or combined
reactances. They can be
calculated from a diagram
which has the same form
as that shown in Figure 2.
VL = IXL
VC = IXC
Vtotal = IZ
VL  V C
VR = IR
Fig. 2
As the frequency is varied from low to high, a minimum value of total impedance Z is
1
found when XL = XC, or f =
. The value of Z at this resonance frequency is Z =
2 LC
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R. If the applied voltage is kept constant, then when Z is a minimum, I will be at a
maximum, so both Z and I have the general form as shown in Figure 3.
Z
I
f
f
Fig. 3
I
The width of the curves in the
above is of great importance
in such devices as radio and
TV receivers (we only want
one channel at a time), and is
measured by the ratio of the
width to the center frequency,
as shown in Figure 4.
Imax
Small R
0.707 Imax
Larger R
f1 fo f2
f
Fig. 4
When the peak is narrow, the
circuit is said to have a high Q,
where the quality Q is defined
as: Q =
fo
f2  f1
. A high Q
corresponds to a small value of
the total series resistance (coil
resistance plus any other
resistance). Q can also be
Z for f = f 2
XL  XC = R
Z for f = f o (Z = R)
Z for f = f 1
45o
45o
XC  XL = R
X
shown to be given by Q = L ,
R
where XL = 2foL, with fo
being the resonant frequency.
Figure 5 indicates the
relationship between Z, R, XL
and XC as f varies from f1 to fo
to f2.
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Fig. 5
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EQUIPMENT & MATERIALS
clips
Simpson 420 function generator
Decade capacitor box
100- composition resistor
Banana wires
2 multimeters
Frequency counter
Oscilloscope
Inductor, 0.01 Henry 2 alligator
2 BNC-to-banana adapters
triangle
Coaxial cable, BNC ends
French curve
Plastic
1 sheet of graph paper
EXPERIMENTAL PROCEDURE
A. Capacitive and Inductive Reactance
1. Set up the circuit as shown in Figure 6. Both
multimeters must be dialed to V to act as
AC voltmeters. Set the decade capacitance
box to 0.5 F. Use alligator clips to connect
the 100- resistor in the circuit. Use the
coaxial cable to connect the function
generator’s TTL output to the frequency
counter’s A input, and use the counter to set
the frequencies accurately by using the
“A Input”, setting the counter for a 1 second
gate time, and a frequency setting of
<10MHz. Set the function generator to
maximum amplitude.
V
V
100 
5.0 X 107 F
Fig. 6
2. Take voltage readings across the function generator and resistor for each frequency
setting. Frequency settings may be made from 2,000 Hz to 10,000 Hz in 2,000-Hz
steps.
3. From Fig. 2, the voltage across the function generator V fg (= Vtotal) is the vector sum
of the voltage across the resistor VR and the voltage across the capacitor V C. From
the right-angle triangle, Vfg2 = VR2 + VC2. Use this equation and the observed
values of voltage to calculate VC.
4. Ohm’s law for the resistor is VR = IR. Calculate the current through the resistor.
5. Ohm’s law for the capacitor is VC = IXC. Because the resistor and capacitor are in
series, they experience the same current. Calculate the measured reactance X C.
6. Theoretically, the reactance of a capacitor is X C = 1/2fC. Calculate this value and
determine the percent difference between this and the measured value. A reasonable
agreement between these two values of X C validates the vector calculation of V C and
the theoretical calculation of XC. Notice that the equation Vfg = VR + VC,
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appropriate for a DC circuit, does not match your data for this circuit.
7. Repeat the above procedure with the 10-mH inductor in the circuit instead of the
capacitor. For an inductor, Vfg2 = VR2 + VL2 and VL = IXL. Theoretically, the
reactance of an inductor is XL = 2fL.
B. RLC Series Resonance
1. Set up the apparatus as in Figure 1, by
connecting the output of the function generator
in series to the Hewlett-Packard multimeter (set
to mA to make it an ammeter), the 10 mH
inductor, the decade capacitance box set to
0.10 F and the 100- resistor held by
alligator clips. Turn on the ammeter and press
the AC/DC button (beside the Range button) to
produce a ~ sign in the display. This indicates
that the ammeter will measure AC instead of
DC current. Attach a BNC-to-banana adapter
to the Ch 1 input of the oscilloscope, and
connect its positive input to the positive output
of the function generator. The TTL output of
the function generator should remain connected
to the A input of the frequency counter.
L
C
A
R
~
Oscilloscope
Ch. 1
Fig. 7
2. Calculate the theoretical values of the resonance frequency f o, the bandwidth and
the quality from the values of resistance, inductance and capacitance.
3. Measure the root-mean-square current I on the multimeter over a range of
frequencies, starting with f = 2000 Hz and increasing by 500-Hz intervals.
Maintain Vtotal at 0.40 V peak-to-peak by monitoring the amplitude of the sine
wave on the oscilloscope and adjusting the amplitude of the function generator. A
setting of 0.2 Volts/div gives a convenient size to the sine wave, and should peak
2 squares above and 2 squares below the center-line. The wave can be centered
by rotating the POSITION dial on the oscilloscope. Place the measurements in
Data Table 3.
4. Plot current I vs. frequency f on a piece of graph paper. Obtain extra readings as
necessary near the resonance frequency to clearly define the resonance peak. Use
a French curve to connect the data points as smoothly as possible.
5. Find f1 and f2 from this graph as the frequencies for which I = 0.707 Imax , as
shown in Figure 4. Calculate fo as the average of f1 and f2, and calculate the
bandwidth and quality from the values of f 1 and f2. Find the percent difference of
these three quantities from their theoretical values.
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LABORATORY REPORT: AC CIRCUITS AND RESONANCE
Part A:
Data Table 1: Capacitive Reactance
Frequency
(Hz)
Voltage across
function generator
(V)
Voltage across
resistor
(V)
Voltage across
capacitor
(V)
Current
(A)
Measured
reactance
()
Calculated
reactance
()
2000
4000
6000
8000
10,000
4000
6000
8000
10,000
Percent difference
Data Table 2: Inductive Reactance
Frequency
(Hz)
Voltage across
function generator
(V)
Voltage across
resistor
(V)
Voltage across
inductor
(V)
Current
(A)
Measured
reactance
()
Calculated
reactance
()
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Percent difference
Data for Part B:
R = ___________
__________
L =
C = ___________
fo =
1
2 LC
=
__________
Bandwidth =
__________
R
= ___________
2L
Q=
2f o L
=
R
Data Table 3
Source
Frequency f
(Hz)
I
(A)
Source
Frequency f
(Hz)
I
(A)
2000
2500
3000
3500
4000
4500
5000
5500
6000
6500
7000
7500
8000
Experimental values from graph:
f1 = ________________ Hz f2 = ________________ Hz
Resonance frequency = fo = ________________ Hz
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______________
Bandwidth = f2  f1 = ________________ Hz
______________
Quality =
______________
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fo
= ________________
f 2  f1
% difference =
% difference =
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Experiment 11
B-H Curve
Aim: To study Hysteresis property of the given magnetic material and hence to
determine
a) Energy loss /cycle/ unit volume b) Remnant Flux Density and c) Coercive Field
Strength.
Apparatus: Specimen, B-H Curve tracer unit, Cathode Ray Oscilloscope (CRO).
Formula:
Energy loss is determined using the formula S S A
PL
ENvH
0.5 J/ per cycle/unit volume,
Where N is the Number of turns in the coil (300), P is the resistance in series with the coil
(65), L = Length of the coil (0.033m) and SH and SV are the horizontal, vertical
sensitivities
of the CRO and A= Area of the loop.
The Coercive field is determined from the formula
A turns m-1
PL
H N OC SH
C
The Remnant flux density = 2
0 0.5 B OB S WbmV
Procedure: Initially the following settings are made for CRO. The CRO is switched on
and is set to X-Y mode. The bright spot is adjusted to the centre of the display with the
help of Horizontal and Vertical shift knobs. Both the channels (X-Channel
(Horizontal,CH1) and Y-Channel (Vertical, CH2 ) are set to AC mode.
One terminal of the magnetizing coil is connected to point C of the main unit and the
other Terminal to any of the point between V1 to V3 (V3 is recommended). Outputs X &
Y of the main unit are connected respectively to CH1 & CH2 of the CRO. IC probe and
the Supply (P.S) are connected to the main kit. The main kit is switched on. The
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resistance (P) is set for maximum
value with the help of the given knob. With no specimen, the horizontal gain of the CRO
is adjusted until a convenient X deflection is obtained on the CRO display. Specimen is
inserted
through the coil such that it touches only the probe at the centre not the conducting
tracks.
The Y gain of the CRO is adjusted to get appropriate Loop. Trace the loop on the graph
paper
by reading coordinates of the points A, B, C, D, E, F on the loop in CRO and area of the
loop is
Measured.
a) Energy loss is determined using the formula S S A
PL
ENvH
0.5 J/ per cycle/unit
Volume,
b) OC is measured from the graph. The Coercive field is determined from the
formulaA turns m1
PL
H N OC SH
C
c) OB is measured from the graph. The Remnant flux density is determined using the
Formula 20 B 0.5 OB S WbmV
Result: The Energy loss in the specimen= ……………..J/cycle/cubic meter
The Remnant Induction=……………..Wb m-2
The Coercive field = ……………...A turns m-2
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Experiment no.12
Transistor characteristics BJT
Aim: To study the input and output characteristics of a PNP/ NPN transistors in common
base configuration
Equipment: Power Supply ( 0-15V), DMMs and potentiometer and other components.
Theory
Circuit Diagrams:
VCC
Transistor used CK100
Fig: Common Base Configuration of PNP transistor
Procedure:
A. Common Base configuration
1. For the input characteristics of common base configuration, set-up the ckt of fig
4.1. Set VEE around 5-6 V and do not change it during the experiment (why?).
Now vary VEB with the potentiometer and record IE keeping VCB at zero volt.
Observations may be recorded up to a maximum of 30mA emitter current. Repeat
the observations (for four sets) for various values of VCB from 0 to 10 V.
2. For output characteristics follow the ckt of fig 4.1. Set V EE around 5-6 V and do
not change it during the experiment. Set IE =0 with the pot and record the IC as a
function of VCB (from 0-10V). Record at least four sets of O/p curve by varying IE
from 0-20mA.
3. Plot all the curves. Plot the load line. Find input and output resistances and current
gain dc.
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Experiment
Title: to verify the working of LVDT
Objectives:
Measuring voltage with displacement variation using Linear Variable Differential
Transformer (LVDT).
Apparatus:
LVDT
Micrometer for LVDT
Voltmeter
Theory:
The Linear Variable Differential Transformer is a position sensing device that provides
an AC output voltage proportional to the displacement of its core passing through its
windings. LVDTs provide linear output for small displacements where the core remains
within the primary coils. The exact distance is a function of the geometry of the LVDT.
An LVDT is much like any other transformer in that it consists of a primary coil,
secondary coils, and a magnetic core. An alternating current, known as the carrier signal,
is produced in the primary coil. The changing current in the primary coil produces a
varying magnetic field around the core. This magnetic field induces an alternating (AC)
voltage in the secondary coils that are in proximity to the core. As with any transformer,
the voltage of the induced signal in the secondary coil is linearly related to the number of
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coils. The basic transformer relation is:
where:
Vout is the voltage at the output,
Vin is the voltage at the input,
Nout is the number of windings of the output coil, and
Nin is the number of windings of the input coil.
As the core is displaced, the number of coils in the secondary coil exposed to the coil
changes linearly. Therefore the amplitude of the induced signal varies linearly with
displacement.
The LVDT indicates direction of displacement by having the two secondary coils whose
outputs are balanced against one another. The secondary coils in an LVDT are connected
in the opposite sense (one clockwise, the other counter clockwise). Thus when the same
varying magnetic field is applied to both secondary coils, their output voltages have the
same amplitude but differ in sign. The outputs from the two secondary coils are summed
together, usually by simply connecting the secondary coils together at a common center
point. At an equilibrium position (generally zero displacement) a zero output signal is
produced.
The induced AC signal is then demodulated so that a DC voltage that is sensitive to the
amplitude and phase of the AC signal is produced.
Procedure:
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1. Connect the LVDT with the micrometer.
2. Connect the Voltmeter with the LVDT signal conditioner.
3. Connect the LVDT signal conditioner with the power supply of 110 Volts.
4. Set the position of LVDT such that a range of voltage from +10 to -10 volts can
be achieved.
5. Change the LVDT displacement and record the voltmeter reading in the table.
6. Plot the graph voltage versus displacement.
Table 1
S. No.
Displacement
(inch)
1
a=
Resultant Displacement
(x-a)
(inch)
Voltage
(Volts)
2
3
4
5
6
7
8
9
10
11
12
13
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EXPERIMENT NO 14
Title: To study the working of thermocouple
Objectives:
To examine the thermocouple voltage and find corresponding temperature under the
following conditions:
1. To measure voltage of thermocouple without considering the intermediate
thermocouple effect of measurement setup.
2. To measure voltage of thermocouple with ice-point reference junction and
fine corresponding voltage using the thermocouple table.
3. To measure voltage of thermocouple using ambient reference block and
calculate the corrected voltage and then find the corresponding temperature.
Apparatus:
J type thermocouples
4-1/2 digit DVM.
Temperature Indicator.
Ice point water
Boiling water
Theory
Thermocouple:
Thermocouples plays very important role in industry. They are used as transducers to
produce electromotive force to actuate equipment. They are used directly in such devices
as furnace valves, recorders, and temperature-recording instruments.
The simplest electrical temperature-sensitive device is the thermocouple. It consists of a
pair of wires of dissimilar metals joined together at one end. The other ends are
connected to an appropriate meter or circuit. The joined ends are known as the hot
junction and the other ends are the cold ones. When the hot junction is heated, a
measurable voltage is generated across the cold ends.
With proper selection of the wires, the voltage varies in relationship to the temperature
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being measured. Because of this, the thermocouple can be considered a thermoelectric
transducer because of its characteristic of converting thermal energy into electrical
energy. Figure 1 shows a typical circuit using a thermocouple to record temperature
changes in a heat chamber.
When the thermocouple is heated at the hot junction, while the cold junction is at a
relatively constant temperature, the difference in temperature of the two junctions causes
the meter to indicate a current. The indication of the meter is calibrated to be proportional
to temperature.
Procedure:
1. Setup the experiment as Figure 4 in theory sheets and measure voltage V.
V = _____________________
T = _____________________
2. Setup the experiment as Figure 6 of theory sheets and measure voltage V and
calculate V1. Find temperature (T1) corresponding to V1 from table.
V  V1  V2
V   (T1  T2 )
where,
V1  t1
V2  t 2
t ( 0 C )  T ( 0 K )  273.15
V
V1
T
T1
3. Setup the experiment according to figure 12.
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a. Note reference temperature, which will be ambient temperature from
temperature indicator.
b. Measure V and find V1.
V  V1  TREF
c. Find the temperature from table corresponding to V.
TREF
V
V1
T
T1
4. Compare voltages from setup 1, 2 and 3 and write your conclusions below.
Conclusions:
Explain:
Which set up gives the correct temperature?
Which set up gives maximum error?
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EXPERIMENT NO 15
Title: To study the working of thermocouple
Objectives:
1. To find the effect of loading on strain gauge resistance using Wheatstone bridge.
2. To find the effect of loading on strain gauge and find voltage difference using
bridge circuit.
3. To find unknown load by using results and graphs obtained in part 1 and 2.
Apparatus:
Strain gauge
Different Weights 1 kg, 2k, 5 kg.
4-1/2 digit DVM.
Wheatstone bridge
Three resistances of 120 ohms.
Power supply
Theory:
The strain gauge is a transducer employing electrical resistance variation to sense the
strain produced by a force or weight. It is a very versatile detector for measuring weight,
pressure, mechanical force, or displacement.
Strain, being a fundamental engineering phenomenon, exists in all matters at all times,
due either to external loads or the weight of the matter itself. These strains vary in
magnitude, depending upon the materials and loads involved. Engineers have worked for
centuries in an attempt to measure strain accurately, but only in the last decade we have
achieved much advancement in the art of strain measurement. The terms linear deformation and strain are synonymous and refer to the change in any linear dimension of a
body, usually due to the application of external forces. The strain of a piece of rubber,
when loaded, is ordinarily apparent to the eye. However, the strain of a bridge strut as a
locomotive passes may not be apparent to the eye. Strain as defined above is often spoken
of as "total strain." Average unit strain is the amount of strain per unit length and has
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somewhat greater significance than does total strain. Strain gauges are used to determine
unit strain, and consequently when one refers to strain, he is usually referring to unit
strain. As defined, strain has units of inches per inch.
Strain gauges work on the principle that as a piece of wire is stretched, its resistance
changes. A strain gauge of either the bonded or the unbonded type is made of fine wire
wound back and forth in such a way that with a load applied to the material it is
fastened to, the strain gauge wire will stretch, increasing its length and decreasing its
cross-sectional area. The result will be an increase in its resistance, because the
resistance, R, of a metallic conductor varies directly with length, L, and inversely with
cross-sectional area, A. Mathematically the relationship is
R
L
A
where  is a constant depending upon the type of wire, L is the length of the wire in the
same units as  , and A is the cross-sectional area measured in units compatible with  .
Four properties of a strain gauge are important to consider when it is used to measure the
strain in a material. They are:
1. Gauge configuration.
2. Gauge sensitivity.
3. Gauge backing material.
4. Method of gauge attachment.
The sensitivity of a strain gauge is a function of the conductive material, size,
configuration, nominal resistance, and the way the gauge is energized.
Strain-gauge conductor materials may be either metal alloys or semiconductor material.
Nickel-chrome-iron-alloys tend to yield high gauge sensitivities as well as have long
gauge life. These alloys are quite good when used for dynamic strain measurements, but
because of a high temperature coefficient, they are not as satisfactory for static strain
measurements.
Copper-nickel alloys are generally use when temperatures are below 500 to 600°F. They
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are less sensitive to temperature changes and provide a less sensitive gauge factor than
the nickel-chrome-iron alloys. Nickel-chrome alloys are useful in the construction of
strain gauges for high temperature measurements.
In using electric strain gauges, two physical qualities are of particular interest, the change
in gauge resistance and the change in length (strain). The relationship between these two
variables is dimensionless and is called the "gauge factor" of the strain gauge and can be
expressed mathematically as:
GF 
GF 
R
L
R
L
R
R
Strain
In this relationship R and L represent, respectively, the initial resistance and the initial
length of the strain gauge wire, while  R and  L represent the small changes in
resistance and length which occur as the gauge is strained along with the surface to which
it is bonded. The gauge factor of a strain gauge is a measure of the amount of resistance
change for a given strain and is thus an index of the strain sensitivity of the gauge. With
all other variables remaining the same, the higher the gauge factor, the more sensitive the
gauge and the greater the electrical output.
The most common type of strain gauge used today for stress analysis is the bonded
resistance strain gauge shown below.
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These gauges use a grid of fine wire or a constantan metal foil grid encapsulated in a thin
resin backing. The gauge is glued to the carefully prepared test specimen by a thin layer
of epoxy. The epoxy acts as the carrier matrix to transfer the strain in the specimen to the
strain gauge. As the gauge changes in length, the tiny wires either contract or elongate
depending upon a tensile or compressive state of stress in the specimen. The crosssectional area will increase for compression and decrease in tension. Because the wire has
an electrical resistance that is proportional to the inverse of the cross-sectional area,
R
1
, a measure of the change in resistance will produce the strain in the material.
A
Procedure:
(A)
Using Wheatstone bridge:
1. Connect strain gauge with Wheatstone bridge and find the resistance of strain
gauge with no load and record the value in the table.
2. Find the resistance of strain gauge with loads, 1 kg, 2 kg, 3 kg, 4 kg and 5 kg,
through Wheatstone bridge and record the values in the table.
Load (kg)
Resistance (ohms)
0
1
2
3
4
5
Unknown
3. Plot Resistance versus Load in the graph paper and write your conclusions.
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(B)
Department- AS
Using Bridge Circuit:
1. Connect strain gauge with the bridge circuit as shown the following figure. Set the
power supply to 10 Volts and all three resistances are 120 ohms.
R1
R2
V
V s =10V
a
b
DVM
R3
Strain
Gauge
2. Find the voltage difference ( V ) across nodes “a” and “b” using digital voltmeter (DVM) with no load and record the value in the table.
3. Find the voltage difference ( V ) using digital volt-meter (DVM) with loads, 1
kg, 2 kg, 3 kg, 4 kg and 5 kg, and record the values in the following table.
Load (kg)
V Voltage Difference (mV)
0
1
2
3
4
5
Unknown
4. Plot the voltage difference V versus Load in the graph paper and write your
conclusions.
(C)
Find Unknown Load using Graphs:
1. Find the unknown load using resistance versus load graph (obtained in part A).
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Unknown Load : _____________________ kg.
2. Find the voltage difference using voltage difference versus load graph (obtained in part
B).
Unknown Load : _____________________ kg.
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EXPERIMENT
Title: Full-Wave Rectification
Objectives:
To calculate and draw the DC output voltages of half-wave and full-wave rectifiers.
Without smoothing capacitor and with smoothing capacitor
Equipment:
 AC power supply or Function Generator
 2 Voltmeters
 Function Generator
 Oscilloscope
Components
 Diode: Silicon 4×(D1N4002) or Silicon Bridge rectifier
 Resistors: 10 kΩ,
 Capacitor :( 0.47 F)
 Capacitor :( 4.7 F)
 Electrolytic Capacitor :( 2×100 F)
Fig. 5
1. Construct the circuit of Fig.5. Set the supply to 10 V p-p with the frequency of 50
Hz. Put the Channel 1 of the oscilloscope probes at function generator (trainer board)
and sketch the input waveform obtained.
Put Channel 1 of the oscilloscope probes across the resistor and sketch the output
waveform obtained. Measure and record the DC level of the output voltage using
the (Dc bottom in the C.R.O.) oscilloscope.
Keeping the x-y button of the C.R.O. at outside position i.e. do not press it.
Adjust the power supply unit at 10 V, then a sinusoidal or another type from function generator
input voltage Vi will be displayed on the screen of the C.R.O.
Measure the input signal voltage Vinput and the Voutput signal voltage Vo using both the
voltmeter and C.R.O.
V input = 10 volts
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With voltmeter
With oscilloscope
Department- AS
10 Va.c
Vp-p
Vd.c
VP-P
Table 1
Draw the input waveform, Vi : and the output waveform, Vo :
Calculate the effective value of the input signal V rms which is give by:
Vrms = Vpp / (2√2) = Vm / √2
Transfer the graphs into a diagram , and determine the amplitude (peak value) V in
and the frequency f of the Vin(t).
Determine the amplitude and the frequency of the output voltage.
The second of Kirchhoff's Laws is used to calculate the output voltage:
VA (T) = V MAX INPUT – 2× 0.7 V
V(th): threshold voltage across the diode for Si diode=0.7 V
Smoothing and filtering
Objectives
Representing the ripple voltage on the load voltage
Determining the ripple voltage as a function of the charging capacitor and the load
resistor
Measuring and calculating the r.m.s. value of the ripple voltage
Fig. 6
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1. Assemble the circuit as shown in Fig. 6 and apply an a.c. voltage 10 V, f = 50 Hz.
2. Use channel 1 of the oscilloscope to measure the voltage V o across the load resistor.
3. Record the settings of the oscilloscope: Y1 = --- volts/div (dc) t, X = --- ms/div.
trigger switch to line.
4. Insert smoothing (filter) capacitor C = 0.47F, 4.7F,100 F; in the circuit to
terminal parallel to the load one after other record the output voltage as shown in Fig.5,
5. determine the ripple voltage as a function of the charging capacitor and the load
resistor
6. Connect an ammeter between terminals 1 and 3 in Fig.6 to measure the d.c. current.
7. Measure the d.c. component of the load current IL and the peak-to-peak value of the
ripple voltage Vr pp for the combination of charging capacitor and load resistor as
a function of the capacitance value of the smoothing capacitor C=0.47F, 4.7F,
100 F and at the same time measure the ripple voltage Vr pp using CRO and Record
the values.
8. The r.m.s. value of the ripple voltage can be calculated using expression
Vr ( rms ) 
I dc
2.4 I dc 2.4Vdc


.
C
RL C
4 3 fC
Where Idc in mA, C is in F, and
RL is in k
9.
Calculate the ripple frequencies for the values given in table 5 and enter your results
into table 2.
10. Calculate the ripple factor r using the following equation:
V (rms )
1
1
r r

(
)
Vdc
2 3 F R LC
11. Draw the output signal voltage in each case of using Cvalues.
12. Comment on the results you obtained.
C
(F)
0.47
4.7
V2
(voltmeter)
(Volts)
Vp-p
(C.R.O. volts)
I dc
( mA)
T
(msec)
F = l/T
(Hz)
100
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Questions:
Department- AS
Table 2
1. Discuss the mechanism of operation of the Full-wave rectification using
Si-bridge rectifier.
2. What do you notice from the values of the average output voltage in both;
the half- and full wave rectification circuits? Comment.
3. Describe the ripple voltage dependence on the charging capacitor and the
load resistor RL.
4. Is it possible to display both input and output voltages simultaneously with
channel 1 resp. channel 2 give reasons for your answer in Full wave
rectifier circuit (fig. 6) that you used?
5. Determine the peak reverse voltage across diode V 1in full wave rectifier
circuit (fig. 6) that you used.
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