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- LIST OF EXPERIMENTS
A) DESIGN AND SIMULATION IN SIMULATION LAB USING
MULTISIM:
1.TWO STAGE RC COUPLED AMPLIFIER.2. COMMON SOURCE AMPLIFIER.
3. TWO STAGE RC COUPLED AMPLIFIER.
4. RC PHASE SHIFT OSCILLATOR USING TRANSISTORS.
5. CLASS A POWER AMPLIFIER.
6. CLASS B COMPLEMENTARY SYMMETRY AMPLIFIER.
7. CURRENT SHUNT FEEDBACK AMPLIFIER.
B) TESTING IN THE HARDWARE LABORATORY:
8. SINGLE TUNED VOLTAGE AMPLIFIER.9. HARTLEY & COLPITT’S OSCILLATORS.
10. CLASS A POWER AMPLIFIER.
11. COMMON EMITTER AMPLIFIER.
12. COMMON SOURCE AMPLIFIER.
13. TWO STAGE RC COUPLED AMPLIFIER.
HARDWARE LAB
EXPERIMENT NO- 1
TWO STAGE RC COUPLED AMPLIFIER
AIM:
1.
To study & plot the
frequency response of a RC coupled amplifier with a pair of shunted emitter
capacitors of 10μF and 100μF.
2.
To calculate maximum gain.
3.
To calculate bandwidth.
4.
To verify AV< AV1.
AV2
COMPONENTS & EQUIPMENT REQUIRED:
S.No
Device
Range/
Rating
Quantity
(in No.s)
1.
Trainer
Board containing
(a)
DC Supply voltage.
(b)
NPN Transistor.
(c)
Resistors.
d) Capacitors.
12 V
BC 107
47 KΩ
2.2 KΩ
1 KΩ
10 KΩ
10mF
100mF.
1
2
2
2
5
2
6
2
2.
Cathode Ray Oscilloscope.
(0-20)MHz
1
3.
Function Generator.
0.1 Hz-10 MHz
1
4
BNC Connector
2
5
Connecting Wires
5A
10
THEORY:
When
the voltage gain provided by a single stage stage is not sufficient, we have to
go for more than one stage in the amplifier. In the circuit diagram, two stages
are shown connected through a link. When the link is open, the voltage gain of
first stage is high. However, when the link is closed the gain is reduced. This
is because of the input impedance of the second stage will be in parallel with
the load resistance of the first stage. Therefore the input impedance is
reduced. Hence overall gain is decreased.
The fall in amplifier gain at low frequencies is due to the effect of
coupling and bypass capacitor.
At medium and high frequencies, the factor
‘f’ makes Xc is very small, so that all by coupling and bypass
capacitors behaves as short circuit. There are also stray capacitance ‘Cs’,
which are capacitances between connecting wires and ground. All these
capacitances values are very small so that at low and medium frequencies, there
impedances are very high. As the frequency increases the reactance of the stray
capacitance fall. When these reactance becomes small enough they begin to shunt
away some of the input and output currents are thus reduces the current gain. Even
if no external stray capacitance is present, the device internal capacitances
through the semiconductor material limit the circuit frequency response.
A cut-off frequency is the frequency at
which the transistor gain falls to 0.707 of its maximum gain.
The range of frequencies over which the
gain of an amplifier is equal to or greater than 70.7% of its maximum gain is
known as maximum gain.
PROCEDURE:
1. Connect the circuit as shown in figure for 10 μF.
2. Adjust input signal amplitude in the function generator and observe
an amplified voltage at the output without distortion.
3. By keeping input signal voltage, say at 50 mV, vary the input signal
frequency from 0-1 MHz as shown in tabular column and note the corresponding
output voltage.
4. Repeat the same procedure for C=100μF.
PRECAUTIONS:
-
1. No loose contacts at the
junctions.
2. Check the connections before
giving the power supply
3. Observations should be taken
carefully.
RESULT:
1.
Frequency Response of RC
coupled (2 stage) amplifier for 10μF and 100 μF is plotted.
2.
For C=10 μF, Gain=
Bandwidth =fH – fL =
3.
For C=100μF, Gain=
Bandwidth =fH – fL
=
4. AV< AV1. AV2
is verified.
VIVA QUESTIONS:
1. What are the advantages and
disadvantages of multi-stage amplifiers?
2. Why gain falls at HF and LF?
3. Why the gain remains
constant at MF?
4. Explain the function of
emitter bypass capacitor, Ce?
5. How the band width will
effect as more number of stages are cascaded?
6. Define frequency response?
7. Give the formula for
effective lower cut-off frequency, when N-number of stages are cascaded.
8. Explain the effect of
coupling capacitors and inter-electrode capacitances on overall gain.
9. By how many times effective
upper cut-off frequency will be reduced, if three identical stages
are cascaded?
10.
Mention the applications of two-stage RC-coupled amplifiers.
TABULAR FORM:
C=10μF Vin = 50 mV
S.No
Frequency
(in Hz)
Vo
(in volts)
Gain A =
Vo/ Vi
Gain(dB) Av =
20 log(Vo/ Vi )
1
100
2
200
3
400
4
800
5
1K
6
2K
7
4K
8
8K
9
10K
10
20K
11
40K
12
80K
13
100K
14
200K
15
300K
16
500K
17
700K
18
900K
19
1M
TABULAR FORM:
C=100μF Vin = 50 mV
S.No
Frequency
(in Hz)
Vo
(in volts)
Gain A =
Vo/ Vi
Gain(dB) Av =
20 log(Vo/ Vi )
1
100
2
200
3
400
4
800
5
1K
6
2K
7
4K
8
8K
9
10K
10
20K
11
40K
12
80K
13
100K
14
200K
15
300K
16
17
700K
18
900K
19
1M
C=100 μF:
Frequency
(in Hz)
V01
V02
Av1
Av2
Av= Av1x
Av2
Theoretical
value
Av
(Practical
value)
10K
20K
@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO 2
VOLTAGE SERIES AND CURRENT SHUNT FEEDBACK
AMPLIFIER
AIM:
1. Current shunt feedback
amplifier
a) Without a shunt Capacitor.
b) With a shunt Capacitor.
2. Voltage series feedback amplifier.
a) With a Resistor.
COMPONENTS & EQUIPMENT
REQUIRED:
S.No
Device
Range/
Rating
Quantity
(in No.s)
1.
Trainer
Board containing
(a)
DC Supply voltage.
(b)
NPN Transistor.
(c)
Resistors.
d) Capacitors.
12 V
BC 107
47 KΩ
2.2 KΩ
1 KΩ
10 KΩ
100 KΩ
0.1mF
22 mF.
1
2
2
2
1
1
1
1
1
1
2.
Cathode Ray
Oscilloscope.
(0-20)MHz
1
3.
Function Generator.
0.1 Hz-10 MHz
1
4
BNC Connector
2
5
Connecting Wires
5A
5
PROCEDURE:
1.
Current series (Without a shunt
Capacitor).
a) Connect the circuit diagram as shown
fig.
b) Adjust input signal amplitude in the
function generator and observe an
amplified voltage at the output
without distortion.
c) By keeping input signal voltage, say at 50
mV, vary the I/P signal
frequency from 50Hz to 1 MHz in step
as shown in tabular column and
note the corresponding O/P voltage.
2.
For Voltage series feedback
amplifier (with & without resistance, Rf ), repeat the above procedure.
PRECAUTIONS:
1. No loose contacts at the
junctions.
2. Check the connections before giving the power
supply
3. Observations should be taken
carefully.
RESULT:
Hence the frequency response for voltage
series and current shunt amplifiers are studied and plotted
a).
Voltage series.
Bandwidth =fH – fL =
b).
Current series (with & without Capacitor)
Bandwidth =fH –
fL =
VIVA QUESTIONS:
1.
What is feedback and what are feedback amplifiers?
2.
What is meant by positive and negative feedback?
3.
What are the advantages and disadvantages of negative feedback?
4.
Differentiate between voltage and current feedback in amplifiers?
5.
Define sensitivity & define De-sensitivity?
6.
Give the topology of current amplifier with current shunt feedback?
TABULAR COLUMN 1: Vin = 50
mV
Current
Shunt:
With
Feedback
Without
Capacitor
Without
Feedback
With
Capacitor
Frequency (in Hz)
Output
Voltage (Vo)
Gain
(in dB) =
20log10(Vo/Vi)
Output
Voltage
(Vo)
Gain
(in dB) =
20log10(Vo/Vi)
20
40
80
100
500
1000
2000
5000
10K
50K
100K
200K
400K
600K
800K
1000K
TABULAR COLUMN 2: Vin = 50 mV
Voltage Series:
Frequency (in Hz)
Output
Voltage (Vo)
Gain
(in dB) =
20log10(Vo/Vi)
20
40
80
100
500
1000
2000
5000
10K
50K
100K
200K
400K
600K
800K
1000K
CIRCUIT DIAGRAM:
EXPECTED GRAPH:
EXPERIMENT NO-3
RC PHASE SHIFT OSCILLATOR
AIM:To find practical frequency of RC phase shift
oscillator and to compare it with theoretical frequency for R=10KW and C = 0.01mF, 0.0022mF & 0.0033mF
respectively
COMPONENTS AND EQUIPMENTS REQUIRED:
S.No
Device
Range/
Rating
Quantity
(in No.s)
1
RC phase shift oscillator trainer board
containing
a) DC supply
voltage
b) Capacitor
c) Resistor
d) NPN
Transistor
12V
1000mF
0.047mF
0.01mF
0.0022mF
0.0033mF
1KW
10KW
47KW
100KW
BC 107
1
1
1
3
3
3
1
2
1
1
1
2
CRO
(0-20) MHz
1
3.
BNC Connector
1
3
Connecting wires
5A
6
PROCEDURE:1.. Connect the circuit as shown in figure.
2. Connect the 0.0022 mF capacitors in the circuit and observe the waveform.
3. Time period of the waveform is to be noted and
frequency should be calculated by the formula f = 1/T.
4. Now fix the capacitance to 0.033 mF and 0.01mF and
calculate the frequency and tabulate as shown.
5. Find theoretical frequency from the formula f
= 1/2PRCÖ6 and compare theoretical and practical frequencies.
PRECAUTIONS: -1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT: -
1.
For C = 0.0022mF & R=10KW
Theoretical frequency=
Practical frequency=
2.
For C = 0.0033mF & R=10KW
Theoretical frequency=
Practical frequency=
3.
For C = 0.01mF & R=10KW
Theoretical frequency=
Practical frequency=
VIVA QUESTIONS:
1. What are the conditions of
oscillations?
2. Give the formula for
frequency of oscillations?
3. What is the total phase
shift produced by RC ladder network?
4. What are the types of
oscillators?
5. What
is the gain of RC phase shift oscillator?
CIRCUIT DIAGRAM:
EXPECTED WAVEFORM:
TABULAR COLUMN:
S.No
C
(mF)
R
(W)
Theoretical Frequency
(KHz)
Practical Frequency
(KHz)
Vo (p-p)
(Volts)
1
0.0022
10K
2
0.0033
10K
3
0.01
10K
@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO-4
(A)
HARTLEY OSCILLATOR
AIM:
To find practical frequency of a Hartley oscillator and to compare it
with theoretical frequency for L = 10mH and C = 0.01mF, 0.033mF and
0.047mF.
COMPONENTS AND EQUIPMENTS REQUIRED:
S.No
Device
Range/
Rating
Quantity
(in No.s)
1
Hartley Oscillator trainer board
containing
a) DC supply
voltage
b) Inductors
c) Capacitor
d) Resistor
e) NPN
Transistor
12V
5mH
0.22mF
0.01mF
0.033mF
0.047mF
1KW
10KW
47KW
BC 107
1
2
2
1
1
1
1
1
1
1
2
Cathode Ray Oscilloscope
(0-20) MHz
1
3.
BNC Connector
1
4
Connecting wires
5A
4
PROCEDURE:
1.
Connect the circuit as shown in
figure.
2.
Connect 0.01mF capacitor in the circuit and observe the waveform.
3.
Time period of the waveform is
to be noted and frequency is to be calculated by the formula f = 1/T .
4.
Now fix the capacitance to
0.033 mF and 0.047mF and calculate the frequency and tabulate the readings as shown.
5.
Find the theoretical frequency
from the formula
Where LT = L1
+ L2 = 5 mH + 5mH = 10 mH and
compare theoretical
and practical values.
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
- For C = 0.01mF, & LT = 10 mH;
Theoretical
frequency =
Practical
frequency =
2. For C = 0.033mF, & LT = 10 mH;
Theoretical frequency =
Practical frequency
=
3. For C = 0.047mF, & LTs = 10 mH;
Theoretical
frequency =
Practical
frequency =
CIRCUIT DIAGRAM:
EXPECTED WAVEFORM:
TABULATIONS:
S.No
LT(mH)
C (mF)
Theoretical frequency (KHz)
Practical frequency (KHz)
Vo (peak to peak)
1
10
0.01
2
10
0.033
3
10
0.047
@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO-4
(B)
COLPITTS OSCILLATOR
AIM:
To find practical frequency of Colpitt’s oscillator and to compare it
with theoretical
Frequency for L= 5mH and C= 0.001mF, 0.0022mF, 0.0033mF respectively.
COMPONENTS & EQIUPMENT REQUIRED:
-
S.No
Device
Range/
Rating
Quantity
(in No.s)
1
Colpitts Oscillator trainer board
containing
a) DC supply
voltage
b) Inductors
c) Capacitor
d) Resistor
e) NPN Transistor
12V
5mH
0.01mF
0.1mF
100 mF
1KW
1.5KW
10KW
47KW
BC 107
1
1
1
1
1
1
1
1
1
1
2
Cathode Ray Oscilloscope
(0-20) MHz
1
3.
BNC Connector
1
4
Connecting wires
5A
4
PROCEDURE:-1. Connect the
circuit as shown in the figure
2. Connect C2=
0.001mFin the circuit and observe the
waveform.
3. Time period
of the waveform is to be noted and frequency should be calculated
by the formula f=1/T
4. Now, fix the
capacitance to 0.002 mF and
then to 0.003 mF and calculate the
frequency and tabulate the reading as
shown.
6.
Find theoretical frequency from
the formula
and compare theoretical and practical values.
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
Frequency
of oscillations of Colpitts oscillator is measured practically and compared
with theoretical values .
1. For C=0.0022mF & L= 5mH
Theoretical frequency =
Practical frequency =
2. For C=0.0033mF & L= 5mH
Theoretical frequency =
Practical frequency =
3. For C=0.001mF & L= 5mH
Theoretical frequency =
Practical frequency =
VIVA QUESTIONS:
1.
What are the applications of LC oscillations?
2. What type of feedback is
used in oscillators?
3. What is the expression for the frequency of oscillations of
Colpitt’s and Hartley oscillator?
4.
Whether an oscillator is dc to ac converter. Explain?
5.
What is the loop gain of an oscillator?
6.
What is the difference between amplifier and oscillator?
7.
What is the condition for sustained oscillations?
8.
How many inductors and capacitors are used in Hartley Oscillator?
9.
How the oscillations are produced in Hartley oscillator?
10.
What is the difference between damped oscillations undamped oscillations?
11. How does Colpitt’s differ from Hartley?
CIRCUIT DIAGRAM: -
COLPITTS
OSCILLATOR
EXPECTED WAVEFORM:
TABULAR COLUMN:
S.NO
L(mH)
C1 (mF)
C2 (mF)
CT (mF)
Theoretical
Frequency
(KHz)
Practical
Frequency
(KHz)
Vo(V)
Peak to peak
1
5
0.01
0.001
2
5
0.01
0.0022
3
5
0.01
0.0033
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO- 5
CLASS A POWER AMPLIFIER
AIM:
1.
To study and plot the frequency
response of a Class A Power Amplifier.
2.
To calculate efficiency of
Class A Power Amplifier.
COMPONENTS & EQUIPMENT REQUIRED:
S.No
Apparatus
Range/
Rating
Quantity
(in No.s)
1.
Trainer
Board containing
(a)
DC Supply voltage.
(b)
NPN Transistor.
(c)
Resistors.
d) Capacitor.
e) Inductor.
12 V
BC 107
560Ω
100KΩ
470Ω
22 mF.
50mH
1
1
1
1
1
1
1
2.
D.C
Milliammeter
0-100mA
1
3.
Cathode Ray
Oscilloscope.
(0-20)MHz
1
4.
Function Generator.
0.1 Hz-10 MHz
1
5.
BNC Connector
2
6.
Connecting Wires
5A
5
THEORY:
In CLASS A amplifier, the
transistor is biased such that the output current flows, ie. Transistor is ON
for the full cycle (3600) of the input a.c. signal. In CLASS A Power
amplifier there is no distortion when compared with other amplifiers. The
maximum value of theoretical efficiency is 25% for a series fed and 50% for a
transformer coupled power amplifier.
Conversion
Efficiency:
It is the measure of
the ability of an active device in converting the d.c. Power of the supply into
the a.c. power delivered to the load. Conversion efficiency is also referred to
as theoretical efficiency and collector circuit efficiency (for transistor
amplifier) and is denoted by η. By definition, the percentage efficiency is
η.
= Signal power delivered to the load
x 100 %
d.c.
power supplied to output circuit
PROCEDURE:
1.
Connect the circuit as shown in
figure.
2.
Adjust input signal amplitude
in the function generator and observe an amplified voltage at the output
without distortion.
3.
By keeping input signal
voltage, say at 50 mV, vary the input signal frequency from 0-1 MHz as shown in
tabular column and note the corresponding output voltage.
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
- Frequency Response of CLASS-A Power amplifier is plotted.
- Efficiency of CLASS A Power amplifier is found to be___________
- Bandwidth fH – fL = ____________
VIVA QUESTIONS:
1. Differentiate between voltage amplifier and power amplifier
2. Why power amplifiers are considered as large signal amplifier?
3. When does maximum power dissipation happen in this circuit ?.
4. What is the maximum theoretical efficiency?
5. Sketch wave form of output current with respective input
signal.
6. What are the different types of class-A power amplifiers
available?
7. What is the theoretical efficiency of the transformer coupled
class-A power amplifier?
8. What is difference in AC, DC load line?.
9. How do you locate the Q-point ?
10. What are the applications of class-A power amplifier?
CIRCUIT DIAGRAM:
EXPECTED GRAPH:
TABULAR FORM:
Vin = 50 mV
S.No
Frequency
(in Hz)
Vo
(in volts)
Gain A =
Vo/
Vi
Gain(dB) Av =
20 log(Vo/ Vi )
1
100
2
200
3
400
4
800
5
1K
6
2K
7
4K
8
8K
9
10K
10
20K
11
40K
12
80K
13
100K
14
200K
CALCULATIONS:
When signal is
removed, Vi=0
Zero signal
current, Ic =
Input Power, Pin=Vcc
x Ic
Output Power,
=
Efficiency, η =
=
=
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO- 6
SINGLE TUNED VOLTAGE AMPLIFIER
AIM:
1. To study & plot the frequency response
of a Single Tuned voltage amplifier
1.
To find the resonant frequency.
2.
To calculate gain and
bandwidth.
COMPONENTS
& EQUIPMENT REQUIRED:
S.No
Apparatus
Range/
Rating
Quantity
(in No.s)
1.
Trainer
Board containing
(a)
DC Supply voltage.
(b)
NPN Transistor.
(c)
Resistors.
d) Capacitor.
e) Inductor.
12 V
BC 107
47 KΩ
150Ω
1 KΩ
10 KΩ
10mF
22 mF.
0.022 mF.
0.033mF.
1mH
1
1
1
1
1
2
2
1
1
1
1
2.
Cathode Ray
Oscilloscope.
(0-20)MHz
1
3.
Function Generator.
0.1 Hz-10 MHz
1
4.
BNC Connector
2
5.
Connecting Wires
5A
5
THEORY:
A Tuned Amplifier uses a parallel
tuned circuit a parallel tuned circuit has high input impedance at its
frequency of resonance, and the impedance falls off sharping as the frequency
departs from the frequency of resonance. Hence the gain Vs frequency curve of a
tuned amplifier is very similar to the impedance Vs frequency curve of a tuned
amplifier are therefore, used for amplification of a narrow band of
frequencies.
A resonant circuit generally uses
either a variable inductor or variable capacitor for adjusting the resonant
frequency at the center of the band of frequencies to be amplified. Over this
narrow band of frequency, the gain of the tuned amplifier remains more or less
constant.
An important example is the radio
frequency (R.F) amplifier, which amplifies either a single radio frequency
signal or a narrow band or frequencies center about the resonant frequency. The
tuned circuit formed by L & C resonates at the frequency of operation.
PROCEDURE:
1.
Connect the circuit as shown in
figure.
2.
Connect the 0.022μF capacitor
3.
Adjust input signal amplitude
in the function generator and observe an amplified voltage at the output
without distortion.
4.
By keeping input signal
voltage, say at 50 mV, vary the input signal frequency from 0-1 MHz as shown in
tabular column and note the corresponding output voltage.
5.
Repeat the same procedure
for 0.033μF capacitor.
6.
Plot the graph: gain (Vs)
frequency.
7.
Calculate the Ft and
Fp
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
1.
Frequency response of RF Tuned
voltage amplifier is plotted.
2.
For 0.022μF, gain = ________dB
Bandwidth= _________
3. For 0.033μF, gain = ________dB
Bandwidth= _________
VIVA QUESTIONS:
1.
What is the purpose of tuned amplifier?
2.
What is Quality factor?
3.
Why should
we prefer parallel resonant circuit in tuned amplifier.
4.
What type of tuning we need to increase gain and bandwidth.?
5.
What are the limitations of single tuned amplifier?
6.
What is meant by Stagger tuning?
7.
What is the conduction angle of an tuned amplifier if it is operated in class B mode?
8.
What are the applications of tuned amplifier
9. What
are the different types of tuned
circuits ?
10. State
relation between resonant frequency and bandwidth of a Tuned amplifier.
11. Differentiate
between Narrow band and Wideband tuned amplifiers ?
12. Calculate
bandwidth of a Tuned amplifier whose resonant frequency is 15KHz and Q-factor
is 100.
13. Specify
the applications of Tuned amplifiers.
CIRCUIT DIAGRAM:
EXPECTED GRAPH:
TABULAR COLUMN -1:
Vin = 50 mV
C= 0.022μF
S.No
Frequency
(in Hz)
Vo
(in volts)
Gain A =
Vo/
Vi
Gain(dB) Av =
20 log(Vo/ Vi )
1
100
2
200
3
400
4
800
5
1K
6
2K
7
4K
8
8K
9
10K
10
20K
11
40K
12
80K
13
100K
14
200K
TABULAR COLUMN-2:
Vin = 50 mV
C= 0.033μF
S.No
Frequency
(in Hz)
Vo
(in volts)
Gain A =
Vo/
Vi
Gain(dB) Av =
20 log(Vo/ Vi )
1
100
2
200
3
400
4
800
5
1K
6
2K
7
4K
8
8K
9
10K
10
20K
11
40K
12
80K
13
100K
14
200K
@@@@@@@@@@@@@@@@@@@@@@@@@@@@
SIMULATION LAB
EXPERIMENT NO: 1
CE AMPLIFIER
AIM:
To plot
the frequency response of CE amplifier and calculate gain bandwidth.
SOFTWARE REQUIRED: MultiSim Software
COMPONENTS
& EQUIPMENTS REQUIRED: -
S.No
Apparatus
Range/
Rating
Quantity
(in No.s)
1.
CE
Amplifier trainer Board with
DC
power supply
DC
power supply
NPN
transistor
Carbon
film resistor
(e)Carbon film resistor
(f) Capacitor.
BC 107
100KW, 1/2W
2.2KW, 1/2W
0.1µF
1
1
1
1
1
2
2.
Cathode Ray Oscilloscope.
(0-20)MHz
1
3.
Function
Generator.
0.1 Hz-10 MHz
1
4.
BNC
Connector
2
5.
Connecting
Wires
5A
5
PROCEDURE: -
1. Connect the
circuit diagram as shown in figure.
2. Adjust input signal amplitude in the function generator and observe an amplified voltage at the
output without distortion.
3. By keeping input signal voltages at 50mV,
vary the input signal frequency from 0 to 1MHz in steps as shown in tabular column and note the corresponding output voltages.
PRECAUTIONS:
-
1. Oscilloscope probes negative terminal should
be at equipotential points (i.e.ground voltage= 0), because both terminals are internally shorted in dual trace
oscilloscope.
2. Ensure that output voltage is exactly an
amplified version of input voltage
without any distortion (adjust input voltage amplitude to that extent)
3. No loose connections at the junctions.
RESULT:
-
Frequency
response of CE amplifier is plotted.
Gain, AV
= ________dB.
Bandwidth=
fH--fL =________Hz.
VIVA QUESTIONS
1. What are the advantages and
disadvantages of single-stage amplifiers?
2. Why gain falls at HF and LF?
3. Why the gain remains
constant at MF?
4. Explain the function of
emitter bypass capacitor, Ce?
5. How the band width will
effect as more number of stages are cascaded?
6. Define frequency response?
7. What is the phase difference
between input and output waveforms of a CE amplifier?
8. What
is early effect?
TABULAR
COLUMN:
Input voltage: Vi = 50mV
Frequency
(in Hz)
Output (Vo)
(Peak to Peak)
Gain
AV=V0/Vi
Gain (in dB) =
20 log 10 VO/ Vi
20
600
1K
2K
4K
8K
10K
20K
30K
40K
50K
60K
80K
100K
250K
500K
750K
1000K
CIRCUIT DIAGRAM:
Bandwidth
= fH-fL
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO: 2
COMMON SOURCE AMPLIFIER
AIM:
-
a) To Plot the frequency response of a common
source amplifier.
b) Calculate gain.
c) Calculate bandwidth.
SOFTWARE REQUIRED: Multi Sim Software
COMPONENTS
& EQUIPMENTS REQUIRED: -
S.No
Device
Range/Rating
Quantity
(in No.s)
1.
FET amplifier Trainer Board with
(a) DC supply voltage
(b) FET
(c) Capacitors
(d) Resistors
12V
BFW 11
0.1mF
47mF
1.5KW
4.7 KW
1MW
1
1
2
1
1
1
1
2.
Signal generator
0.1Hz-1MHz
1
3.
CRO
0Hz-20MHz
1
4.
Connecting wires
5A
4
PROCEDURE:
-
1.Connect the circuit diagram as shown in
figure.
2.Adjust input signal amplitude in the
function generator and observe an amplified voltage at the output without 3.distortion.
4.By keeping input signal voltage, say at
50mV, vary the input signal frequency from 0 to 1MHz in steps as shown in
tabular column and note the corresponding output voltages.
PRECAUTIONS:
Oscilloscopes
probes negative terminal should be at equipotential points(i.e. ground voltage
is zero) because both terminals are internally shorted in dual trace
oscilloscope.
RESULT:
-
Hence,
the frequency response of FET (CS) amplifier is plotted.
Gain = _______dB (maximum).
3.
Bandwidth= fH--fL =
_________Hz.
VIVA QUESTIONS:
1. What is the difference between FET and BJT?
2. FET is unipolar or bipolar?
3. Draw the symbol of FET?
4. What are the applications of FET?
5. FET is voltage controlled or current controlled?
TABULAR
COLUMN:Input = 50mV
Frequency (in Hz)
Output
Voltage (Vo)
Gain
Av=Vo/Vi
Gain
(in dB) =
20log10(Vo/Vi)
20
40
80
100
500
1000
5000
10K
50K
100K
200K
400K
600K
800K
CIRCUIT DIAGRAM:
COMMON SOURCE AMPLIFIER
EXPECTED GRAPH:
EXPERIMENT NO- 3
TWO STAGE RC COUPLED
AMPLIFIER
AIM:
1.To plot the frequency response of a RC
coupled amplifier with a pair of shunted emitter capacitors of 10 μF and 100μF.
2. To calculate gain.
3. To calculate bandwidth.
SOFTWARE REQUIRED: MultiSim Software
COMPONENTS & EQUIPMENT REQUIRED:
S.No
Device
Range/
Rating
Quantity
(in No.s)
1.
Trainer Board containing
a) DC Supply voltage.
b) NPN Transistor.
c) Resistors.
d) Capacitors.
12 V
BC 107
47 KΩ
2.2 KΩ
1 KΩ
10 KΩ
100mF
10mF.
1
2
2
2
5
2
6
2.
Cathode
Ray Oscilloscope.
(0-20)MHz
1
3.
Function
Generator.
0.1 Hz-10 MHz
1
4
BNC
Connector
2
5
Connecting
Wires
5A
10
THEORY:
When the voltage gain provided by a single stage stage is not
sufficient, we have to go for more than one stage in the amplifier. In the
circuit diagram, two stages are shown connected through a link. When the link
is open, the voltage gain of first stage is high. However, when the link is
closed the gain is reduced. This is because of the input impedance of the
second stage will be in parallel with the load resistance of the first stage.
Therefore the input impedance is reduced. Hence overall gain is decreased. The fall in amplifier gain at low frequencies
is due to the effect of coupling and bypass capacitor.
At medium and high frequencies, the factor
‘f’ makes Xc is very small, so that all by coupling and bypass
capacitors behaves as short circuit. There are also stray capacitance ‘Cs’,
which are capacitances between connecting wires and ground. All these
capacitances values are very small so that at low and medium frequencies, there
impedances are very high. As the frequency increases the reactance of the stray
capacitance fall. When these reactance becomes small enough they begin to shunt
away some of the input and output currents are thus reduces the current gain. Even
if no external stray capacitance is present, the device internal capacitances
through the semiconductor material limit the circuit frequency response.
A cut-off frequency is the frequency at
which the transistor gain falls to 0.707 of its maximum gain.
The range of frequencies over which the
gain of an amplifier is equal to or greater than 70.7% of its maximum gain is
known as maximum gain
PROCEDURE:
1. Connect the circuit as shown in figure
for 10 μF.
2. Adjust input signal amplitude in the
function generator and observe an amplified voltage at the output without
distortion.
3. By keeping input signal voltage, say at
50 mV, vary the input signal frequency from 0-1 MHz as shown in tabular column
and note the corresponding output voltage.
4. Repeat the same procedure for C=100μF.
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
-
Hence, the
frequency Response of RC coupled (2 stage) amplifier for 10μF and 100 μF
is plotted.
1.
For C=10 μF,
Gain=
Bandwidth =fH – fL =
2.
For C=100μF
Gain=
Bandwidth =fH – fL = CIRCUIT DIAGRAM:
TWO STAGE RC COUPLED
AMPLIFIER
EXPECTED GRAPH:
TABULAR FORM:
C=10μF Vin
= 50 mV
S.No
Frequency
(in Hz)
Vo
(in volts)
Gain A =
Vo/ Vi
Gain(dB) Av =
20 log(Vo/ Vi )
1
100
2
200
3
400
4
800
5
1K
6
2K
7
4K
8
8K
9
10K
10
20K
11
40K
12
80K
13
100K
14
200K
15
300K
16
500K
17
700K
18
900K
19
1M
TABULAR FORM:
C=100μF Vin
= 50 mV
S.No
Frequency
(in Hz)
Vo
(in volts)
Gain A =
Vo/ Vi
Gain(dB) Av =
20 log(Vo/ Vi )
1
100
2
200
3
400
4
800
5
1K
6
2K
7
4K
8
8K
9
10K
10
20K
11
40K
12
80K
13
100K
14
200K
15
300K
16
500K
17
700K
18
900K
19
1M
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT - 4
CLASS B POWER AMPLIFIER
(COMPLEMENTARY SYMMETRY)
AIM:
To study the frequency response of CLASS B
Complementary Symmetry amplifier and to calculate its
efficiency.
SOFTWARE REQUIRED: MultiSim Software
APPARATUS REQUIRED:
S.No
Device
Range/
Rating
Qty
(in No.)
1
CLASS B Power amplifier Trainer Kit
1
2
Signal Generator
(0Hz-1 MHz)
1
3
Cathode Ray Oscilloscope
(0 Hz –20 MHz)
1
4
D.C Milliammeter
(0-50) mA
1
5
Decade Resistance Box
(1-100) KΩ
1
6
BNC Connector
2
7
Connecting Wires
5A
6
THEORY:
In a CLASS B amplifier the transistor is
biased almost at cut off, so that it remains forward biased only for one half
cycle of the input signal. Hence its conduction angle is only 1800. It
has the following two advantages over CLASS A power amplifier. Possible to
obtain greater power output Efficiency is higher. Negligible power loss (as no
output current flows) at no input signal. And also it eliminates the
disadvantages of Push-Pull configuration like bulky and expensive output
transformer. The maximum theoretical efficiency of a CLASS B power amplifier is
78.5 %.
PROCEDURE:
1. Switch ON the CLASS B amplifier trainer.
2. Connect Milliammeter to (A) terminals
and DRB to the RL terminals and fix RL=50Ω.
3. Apply the input voltage from the signal
generator to the Vs terminals.
4. Connect channel 1 of CRO to the Vs
terminals and channel 2 across the load.
5. By varying the input voltage, observe
the maximum distortion less output waveform.
And note down the voltage reading.
6. Calculate the efficiency.
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
Thus efficiency of CLASS B (Complementary
symmetry) amplifier is calculated and
frequency response is studied.
VIVA QUESTIONS:
1. Differentiate between
voltage amplifier and power amplifier?
2. Explain impedance matching
provided by transformer?
3. Under what condition power
dissipation is maximum for transistor in this circuit?
4. What is the maximum
theoretical efficiency?
5. Sketch current waveform in
each transistor with respective input signal?
6. How do you test matched
transistors required for this circuit with DMM?.
7. What is the theoretical
efficiency of the complementary stage amplifier.
8. How do you measure DC and AC
out put of this amplifier?
9. Is this amplifier working in
class A or B. ?
10. How
can you reduce cross over distortion?
CLASS B POWER AMPLIFIER (COMPLEMENTARY SYMMETRY)
OBSERVATIONS:
Vs=2v
FREQUENCY
Vo (volts)
Idc (mA)
Efficiency
10 KHz
CALCULATIONS:
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO-5
RC PHASE SHIFT OSCILLATOR
AIM:
To find practical frequency of RC phase shift oscillator and to compare
it with theoretical frequency for R=10KW and C = 0.01mF, 0.0022mF & 0.0033mF
Respectively
SOFTWARE
REQUIRED: MultiSim Software
COMPONENTS AND EQUIPMENTS REQUIRED:
S.No
Device
Range/
Rating
Quantity
(in No.s)
1
RC phase shift oscillator trainer board
containing
a) DC supply voltage
b) Capacitor
c) Resistor
d) NPN Transistor
12V-----------
1000mF-------
0.047mF------
0.01mF--------
0.0022mF------
0.0033mF----
1KW-----------
10KW---------
47KW----------
100KW---------
BC 107--------
1
1
1
3
3
3
1
4
1
1
1
2
CRO
(0-20) MHz
1
3.
BNC Connector
1
3
Connecting wires
5A
6
PROCEDURE:
1. Connect the circuit as shown in figure.
2. Connect the 0.0022 mF capacitors in the circuit and observe the waveform.
3. Time period of the waveform is to be
noted and frequency should be calculated by the formula f = 1/T.
4. Now fix the capacitance to 0.033 mF and 0.01mF and
calculate the frequency and tabulate as shown.
5. Find theoretical frequency from the
formula f = 1/2PRCÖ6 and compare theoretical and practical frequencies.
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
-
For C = 0.0022mF & R=10KW
Theoretical frequency=
Practical frequency=
For C = 0.0033mF & R=10KW
Theoretical frequency=
Practical frequency=
For C = 0.01mF & R=10KW
Theoretical frequency=
Practical frequency=
VIVA QUESTIONS:
1. What are the conditions of
oscillations?
2. Give the formula for
frequency of oscillations?
3. What is the total phase
shift produced by RC ladder network?
4. What are the types of oscillators?
5. What
is the gain of RC phase shift oscillator?
CIRCUIT DIAGRAM:
RC PHASE SHIFT OSCILLATOR
EXPECTED WAVEFORM:
TABULAR
COLUMN:
S.No
C
(mF)
R
(W)
Theoretical Frequency
(KHz)
Practical Frequency
(KHz)
Vo (p-p)
(Volts)
1
0.0022
10K
2
0.0033
10K
3
0.01
10K
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO- 6
SINGLE TUNED VOLTAGE AMPLIFIER
AIM:
1. To study & plot the frequency
response of a Single Tuned voltage amplifier
2. To find the resonant frequency.
3. To calculate gain and bandwidth.
SOFTWARE REQUIRED: MultiSim Software
COMPONENTS & EQUIPMENT REQUIRED:
S.No
Apparatus
Range/
Rating
Quantity
(in No.s)
1.
Trainer Board containing
DC Supply voltage.
NPN Transistor.
Resistors.
d) Capacitor.
e) Inductor.
12 V
BC 107
47 KΩ
150Ω
1 KΩ
10 KΩ
10mF
22 mF.
0.022 mF.
0.033mF.
1mH
1
1
1
1
1
2
2
1
1
1
1
2.
Cathode
Ray Oscilloscope.
(0-20)MHz
1
3.
Function
Generator.
0.1 Hz-10 MHz
1
4.
BNC
Connector
2
5.
Connecting
Wires
5A
5
THEORY:
A Tuned Amplifier uses a parallel tuned circuit a parallel tuned circuit
has high input impedance at its frequency of resonance, and the impedance falls
off sharping as the frequency departs from the frequency of resonance. Hence
the gain Vs frequency curve of a tuned amplifier is very similar to the
impedance Vs frequency curve of a tuned amplifier are therefore, used for
amplification of a narrow band of frequencies.
A resonant circuit generally uses either a variable inductor or variable
capacitor for adjusting the resonant frequency at the center of the band of
frequencies to be amplified. Over this narrow band of frequency, the gain of
the tuned amplifier remains more or less constant.
An important example is the
radio frequency (R.F) amplifier, which amplifies either a single radio
frequency signal or a narrow band or frequencies center about the resonant
frequency. The tuned circuit formed by L & C resonates at the frequency of
operation.
PROCEDURE:
1. Connect the circuit as shown in figure.
2. Connect the 0.022μF capacitor
3. Adjust input signal amplitude in the
function generator and observe an amplified voltage at the output without
distortion.
4. By keeping input signal voltage, say at
50 mV, vary the input signal frequency from 0-1 MHz as shown in tabular column
and note the corresponding output voltage.
5. Repeat the same procedure for 0.033μF capacitor.
6 Plot the graph: gain (Vs) frequency.
1.
Calculate the Ft and
Fp
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
Frequency response of RF Tuned voltage
amplifier is plotted.
For 0.022μF, gain = ________dB
Bandwidth= _________
3.
For 0.033μF, gain = ________dB
Bandwidth= _________
CIRCUIT DIAGRAM:
SINGLE TUNED VOLTAGE AMPLIFIER
EXPECTED GRAPH:
TABULAR FORM-1:
Vin =
50 mV
C= 0.022μF
S.No
Frequency
(in Hz)
Vo
(in volts)
Gain A =
Vo/ Vi
Gain(dB) Av =
20 log(Vo/ Vi )
1
100
2
200
3
400
4
800
5
1K
6
2K
7
4K
8
8K
9
10K
10
20K
11
40K
12
80K
13
100K
14
200K
TABULAR FORM-2:
Vin =
50 mV
C= 0.033μF
S.No
Frequency
(in Hz)
Vo
(in volts)
Gain A =
Vo/ Vi
Gain(dB) Av =
20 log(Vo/ Vi )
1
100
2
200
3
400
4
800
5
1K
6
2K
7
4K
8
8K
9
10K
10
20K
11
40K
12
80K
13
100K
14
200K
BELOW ARE VIVA QUESTIONS
1. What is the type of capacitor used in RC coupled amplifier for a) coupling two
phases b) by pass emitter
Ans. Generally electrolytic capacitors are used
2. What is signal source used for experiment of an RC coupled amplifier and how
much maximum voltage it could give
Ans. A function generator with 1 MHz highest frequency 0-30 V P – P output
is used in sine wave mode
3. What is the pin configuration on bread board used in the lab
Ans. In the bread boards there are two rows of horizontally shorted pin bank
on top and bottom, there are vertically shorted rows of pins in two halfs up
and down from the center of the bread board.
4. For Class – A amplifier How do you bring operating point of amplifier at center
of supply voltage
Ans. By adjusting the value of resistor used from base to supply.
If Vc < Vcc/2 base resistor is increased and if Vc > Vcc/2 it is decreased
5. What are the transistors used in complementary push pull experiment give type
number
Ans. A matched pair of NPN (CL100 or SL100) and PNP (CK100 or SK100)
used. Matching can be done by testing above in diode position of DMM for
same drop and HFE sockets of NPN, PNP for β.
6. How do you determine AC power output in class A amplifier i.e., do you measure
current or voltage and how?
Ans. P-P voltage is measured using CRO since AC current cannot be
measured.
7. How much current do you pass through reference zener in series regulated power
supply experiment
Ans. Iz min approximately equal to 5 mA is used.
8. In shunt regulator how is the value of resistor between base and emitter of shunt
transistor determined
Ans. VBE = 0.7 V, Resistor between BE = 0.7 / Iz min. practically 100 is
used
9. How do you determine Q of oil used in tuned amplifier experiment
Ans. Q = ωL /Rs where ω is Resonant frequency; R is series resistance of coil
10. What is the transistor used in shunt regulator and what is its Case style
Ans. NPN transistor SL100 is used the case style is TO2
11. Name one NPN and PNP transistors we have used in ECA lab
Ans. BC 107 NPN, CK 100 PNP
12. What ammeter do you need to measure DC power to class – C amplifier (DMM,
Analog coil meter or Analog moving iron meter)
Ans. Analog coil meter
13. What is the input resistance of oscilloscope you have used
Ans. 1 M , ± 1 %
14. What is the lamp used on all instruments to show presence of mains
Ans. In all modern equipment LED(Light emitting diode) is used in various
colors
EXPERIMENT NO- 1
TWO STAGE RC COUPLED AMPLIFIER
AIM:
1.
To study & plot the
frequency response of a RC coupled amplifier with a pair of shunted emitter
capacitors of 10μF and 100μF.
2.
To calculate maximum gain.
3.
To calculate bandwidth.
4.
To verify AV< AV1.
AV2
COMPONENTS & EQUIPMENT REQUIRED:
S.No
|
Device
|
Range/
Rating
|
Quantity
(in No.s)
|
1.
|
Trainer
Board containing
(a)
DC Supply voltage.
(b)
NPN Transistor.
(c)
Resistors.
d) Capacitors.
|
12 V
BC 107
47 KΩ
2.2 KΩ
1 KΩ
10 KΩ
10mF
100mF.
|
1
2
2
2
5
2
6
2
|
2.
|
Cathode Ray Oscilloscope.
|
(0-20)MHz
|
1
|
3.
|
Function Generator.
|
0.1 Hz-10 MHz
|
1
|
4
|
BNC Connector
|
2
|
|
5
|
Connecting Wires
|
5A
|
10
|
THEORY:
When
the voltage gain provided by a single stage stage is not sufficient, we have to
go for more than one stage in the amplifier. In the circuit diagram, two stages
are shown connected through a link. When the link is open, the voltage gain of
first stage is high. However, when the link is closed the gain is reduced. This
is because of the input impedance of the second stage will be in parallel with
the load resistance of the first stage. Therefore the input impedance is
reduced. Hence overall gain is decreased.
The fall in amplifier gain at low frequencies is due to the effect of
coupling and bypass capacitor.
At medium and high frequencies, the factor
‘f’ makes Xc is very small, so that all by coupling and bypass
capacitors behaves as short circuit. There are also stray capacitance ‘Cs’,
which are capacitances between connecting wires and ground. All these
capacitances values are very small so that at low and medium frequencies, there
impedances are very high. As the frequency increases the reactance of the stray
capacitance fall. When these reactance becomes small enough they begin to shunt
away some of the input and output currents are thus reduces the current gain. Even
if no external stray capacitance is present, the device internal capacitances
through the semiconductor material limit the circuit frequency response.
A cut-off frequency is the frequency at
which the transistor gain falls to 0.707 of its maximum gain.
The range of frequencies over which the
gain of an amplifier is equal to or greater than 70.7% of its maximum gain is
known as maximum gain.
PROCEDURE:
1. Connect the circuit as shown in figure for 10 μF.
2. Adjust input signal amplitude in the function generator and observe
an amplified voltage at the output without distortion.
3. By keeping input signal voltage, say at 50 mV, vary the input signal
frequency from 0-1 MHz as shown in tabular column and note the corresponding
output voltage.
4. Repeat the same procedure for C=100μF.
PRECAUTIONS:
-
1. No loose contacts at the
junctions.
2. Check the connections before
giving the power supply
3. Observations should be taken
carefully.
RESULT:
1. Frequency Response of RC coupled (2 stage) amplifier for 10μF and 100 μF is plotted.
2. For C=10 μF, Gain=
Bandwidth =fH – fL =
3. For C=100μF, Gain=
Bandwidth =fH – fL
=
4. AV< AV1. AV2
is verified.
VIVA QUESTIONS:
1. What are the advantages and
disadvantages of multi-stage amplifiers?
2. Why gain falls at HF and LF?
3. Why the gain remains
constant at MF?
4. Explain the function of
emitter bypass capacitor, Ce?
5. How the band width will
effect as more number of stages are cascaded?
6. Define frequency response?
7. Give the formula for
effective lower cut-off frequency, when N-number of stages are cascaded.
8. Explain the effect of
coupling capacitors and inter-electrode capacitances on overall gain.
9. By how many times effective
upper cut-off frequency will be reduced, if three identical stages
are cascaded?
10.
Mention the applications of two-stage RC-coupled amplifiers.
TABULAR FORM:
C=10μF Vin = 50 mV
S.No
|
Frequency
(in Hz)
|
Vo
(in volts)
|
Gain A =
Vo/ Vi
|
Gain(dB) Av =
20 log(Vo/ Vi )
|
1
|
100
|
|||
2
|
200
|
|||
3
|
400
|
|||
4
|
800
|
|||
5
|
1K
|
|||
6
|
2K
|
|||
7
|
4K
|
|||
8
|
8K
|
|||
9
|
10K
|
|||
10
|
20K
|
|||
11
|
40K
|
|||
12
|
80K
|
|||
13
|
100K
|
|||
14
|
200K
|
|||
15
|
300K
|
|||
16
|
500K
|
|||
17
|
700K
|
|||
18
|
900K
|
|||
19
|
1M
|
TABULAR FORM:
C=100μF Vin = 50 mV
S.No
|
Frequency
(in Hz)
|
Vo
(in volts)
|
Gain A =
Vo/ Vi
|
Gain(dB) Av =
20 log(Vo/ Vi )
|
1
|
100
|
|||
2
|
200
|
|||
3
|
400
|
|||
4
|
800
|
|||
5
|
1K
|
|||
6
|
2K
|
|||
7
|
4K
|
|||
8
|
8K
|
|||
9
|
10K
|
|||
10
|
20K
|
|||
11
|
40K
|
|||
12
|
80K
|
|||
13
|
100K
|
|||
14
|
200K
|
|||
15
|
300K
|
|||
16
|
||||
17
|
700K
|
|||
18
|
900K
|
|||
19
|
1M
|
C=100 μF:
Frequency
(in Hz)
|
V01
|
V02
|
Av1
|
Av2
|
Av= Av1x
Av2
Theoretical
value
|
Av
(Practical
value)
|
10K
|
||||||
20K
|
@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO 2
VOLTAGE SERIES AND CURRENT SHUNT FEEDBACK
AMPLIFIER
AIM:
1. Current shunt feedback
amplifier
a) Without a shunt Capacitor.
b) With a shunt Capacitor.
2. Voltage series feedback amplifier.
a) With a Resistor.
COMPONENTS & EQUIPMENT
REQUIRED:
S.No
|
Device
|
Range/
Rating
|
Quantity
(in No.s)
|
1.
|
Trainer
Board containing
(a)
DC Supply voltage.
(b)
NPN Transistor.
(c)
Resistors.
d) Capacitors.
|
12 V
BC 107
47 KΩ
2.2 KΩ
1 KΩ
10 KΩ
100 KΩ
0.1mF
22 mF.
|
1
2
2
2
1
1
1
1
1
1
|
2.
|
Cathode Ray
Oscilloscope.
|
(0-20)MHz
|
1
|
3.
|
Function Generator.
|
0.1 Hz-10 MHz
|
1
|
4
|
BNC Connector
|
2
|
|
5
|
Connecting Wires
|
5A
|
5
|
PROCEDURE:
1.
Current series (Without a shunt
Capacitor).
a) Connect the circuit diagram as shown
fig.
b) Adjust input signal amplitude in the
function generator and observe an
amplified voltage at the output
without distortion.
c) By keeping input signal voltage, say at 50
mV, vary the I/P signal
frequency from 50Hz to 1 MHz in step
as shown in tabular column and
note the corresponding O/P voltage.
2.
For Voltage series feedback
amplifier (with & without resistance, Rf ), repeat the above procedure.
PRECAUTIONS:
1. No loose contacts at the
junctions.
2. Check the connections before giving the power
supply
3. Observations should be taken
carefully.
RESULT:
Hence the frequency response for voltage
series and current shunt amplifiers are studied and plotted
a).
Voltage series.
Bandwidth =fH – fL =
b).
Current series (with & without Capacitor)
Bandwidth =fH –
fL =
VIVA QUESTIONS:
1.
What is feedback and what are feedback amplifiers?
2.
What is meant by positive and negative feedback?
3.
What are the advantages and disadvantages of negative feedback?
4.
Differentiate between voltage and current feedback in amplifiers?
5.
Define sensitivity & define De-sensitivity?
6.
Give the topology of current amplifier with current shunt feedback?
TABULAR COLUMN 1: Vin = 50
mV
Current
Shunt:
With
Feedback
Without
Capacitor
|
Without
Feedback
With
Capacitor
|
|||
Frequency (in Hz)
|
Output
Voltage (Vo)
|
Gain
(in dB) =
20log10(Vo/Vi)
|
Output
Voltage
(Vo)
|
Gain
(in dB) =
20log10(Vo/Vi)
|
20
|
||||
40
|
||||
80
|
||||
100
|
||||
500
|
||||
1000
|
||||
2000
|
||||
5000
|
||||
10K
|
||||
50K
|
||||
100K
|
||||
200K
|
||||
400K
|
||||
600K
|
||||
800K
|
||||
1000K
|
TABULAR COLUMN 2: Vin = 50 mV
Voltage Series:
Frequency (in Hz)
|
Output
Voltage (Vo)
|
Gain
(in dB) =
20log10(Vo/Vi)
|
20
|
||
40
|
||
80
|
||
100
|
||
500
|
||
1000
|
||
2000
|
||
5000
|
||
10K
|
||
50K
|
||
100K
|
||
200K
|
||
400K
|
||
600K
|
||
800K
|
||
1000K
|
CIRCUIT DIAGRAM:
EXPECTED GRAPH:
EXPERIMENT NO-3
RC PHASE SHIFT OSCILLATOR
AIM:To find practical frequency of RC phase shift
oscillator and to compare it with theoretical frequency for R=10KW and C = 0.01mF, 0.0022mF & 0.0033mF
respectively
COMPONENTS AND EQUIPMENTS REQUIRED:
S.No
|
Device
|
Range/
Rating
|
Quantity
(in No.s)
|
1
|
RC phase shift oscillator trainer board
containing
a) DC supply
voltage
b) Capacitor
c) Resistor
d) NPN
Transistor
|
12V
1000mF
0.047mF
0.01mF
0.0022mF
0.0033mF
1KW
10KW
47KW
100KW
BC 107
|
1
1
1
3
3
3
1
2
1
1
1
|
2
|
CRO
|
(0-20) MHz
|
1
|
3.
|
BNC Connector
|
1
|
|
3
|
Connecting wires
|
5A
|
6
|
PROCEDURE:1.. Connect the circuit as shown in figure.
2. Connect the 0.0022 mF capacitors in the circuit and observe the waveform.
3. Time period of the waveform is to be noted and
frequency should be calculated by the formula f = 1/T.
4. Now fix the capacitance to 0.033 mF and 0.01mF and
calculate the frequency and tabulate as shown.
5. Find theoretical frequency from the formula f
= 1/2PRCÖ6 and compare theoretical and practical frequencies.
PRECAUTIONS: -1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT: -
1.
For C = 0.0022mF & R=10KW
Theoretical frequency=
Practical frequency=
2.
For C = 0.0033mF & R=10KW
Theoretical frequency=
Practical frequency=
3.
For C = 0.01mF & R=10KW
Theoretical frequency=
Practical frequency=
VIVA QUESTIONS:
1. What are the conditions of
oscillations?
2. Give the formula for
frequency of oscillations?
3. What is the total phase
shift produced by RC ladder network?
4. What are the types of
oscillators?
5. What
is the gain of RC phase shift oscillator?
CIRCUIT DIAGRAM:
EXPECTED WAVEFORM:
TABULAR COLUMN:
S.No
|
C
(mF)
|
R
(W)
|
Theoretical Frequency
(KHz)
|
Practical Frequency
(KHz)
|
Vo (p-p)
(Volts)
|
1
|
0.0022
|
10K
|
|||
2
|
0.0033
|
10K
|
|||
3
|
0.01
|
10K
|
EXPERIMENT NO-4
(A)
HARTLEY OSCILLATOR
AIM:
To find practical frequency of a Hartley oscillator and to compare it
with theoretical frequency for L = 10mH and C = 0.01mF, 0.033mF and
0.047mF.
COMPONENTS AND EQUIPMENTS REQUIRED:
S.No
|
Device
|
Range/
Rating
|
Quantity
(in No.s)
|
1
|
Hartley Oscillator trainer board
containing
a) DC supply
voltage
b) Inductors
c) Capacitor
d) Resistor
e) NPN
Transistor
|
12V
5mH
0.22mF
0.01mF
0.033mF
0.047mF
1KW
10KW
47KW
BC 107
|
1
2
2
1
1
1
1
1
1
1
|
2
|
Cathode Ray Oscilloscope
|
(0-20) MHz
|
1
|
3.
|
BNC Connector
|
1
|
|
4
|
Connecting wires
|
5A
|
4
|
PROCEDURE:
1.
Connect the circuit as shown in
figure.
2.
Connect 0.01mF capacitor in the circuit and observe the waveform.
3.
Time period of the waveform is
to be noted and frequency is to be calculated by the formula f = 1/T .
4.
Now fix the capacitance to
0.033 mF and 0.047mF and calculate the frequency and tabulate the readings as shown.
5.
Find the theoretical frequency
from the formula
Where LT = L1
+ L2 = 5 mH + 5mH = 10 mH and
compare theoretical
and practical values.
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
- For C = 0.01mF, & LT = 10 mH;
Theoretical
frequency =
Practical
frequency =
2. For C = 0.033mF, & LT = 10 mH;
Theoretical frequency =
Practical frequency
=
3. For C = 0.047mF, & LTs = 10 mH;
Theoretical
frequency =
Practical
frequency =
CIRCUIT DIAGRAM:
EXPECTED WAVEFORM:
TABULATIONS:
S.No
|
LT(mH)
|
C (mF)
|
Theoretical frequency (KHz)
|
Practical frequency (KHz)
|
Vo (peak to peak)
|
1
|
10
|
0.01
|
|||
2
|
10
|
0.033
|
|||
3
|
10
|
0.047
|
EXPERIMENT NO-4
(B)
COLPITTS OSCILLATOR
AIM:
To find practical frequency of Colpitt’s oscillator and to compare it
with theoretical
Frequency for L= 5mH and C= 0.001mF, 0.0022mF, 0.0033mF respectively.
COMPONENTS & EQIUPMENT REQUIRED:
-
S.No
|
Device
|
Range/
Rating
|
Quantity
(in No.s)
|
1
|
Colpitts Oscillator trainer board
containing
a) DC supply
voltage
b) Inductors
c) Capacitor
d) Resistor
e) NPN Transistor
|
12V
5mH
0.01mF
0.1mF
100 mF
1KW
1.5KW
10KW
47KW
BC 107
|
1
1
1
1
1
1
1
1
1
1
|
2
|
Cathode Ray Oscilloscope
|
(0-20) MHz
|
1
|
3.
|
BNC Connector
|
1
|
|
4
|
Connecting wires
|
5A
|
4
|
PROCEDURE:-1. Connect the circuit as shown in the figure
2. Connect C2=
0.001mFin the circuit and observe the
waveform.
3. Time period
of the waveform is to be noted and frequency should be calculated
by the formula f=1/T
4. Now, fix the
capacitance to 0.002 mF and
then to 0.003 mF and calculate the
frequency and tabulate the reading as
shown.
6.
Find theoretical frequency from
the formula
and compare theoretical and practical values.
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
Frequency
of oscillations of Colpitts oscillator is measured practically and compared
with theoretical values .
1. For C=0.0022mF & L= 5mH
Theoretical frequency =
Practical frequency =
2. For C=0.0033mF & L= 5mH
Theoretical frequency =
Practical frequency =
3. For C=0.001mF & L= 5mH
Theoretical frequency =
Practical frequency =
VIVA QUESTIONS:
1.
What are the applications of LC oscillations?
2. What type of feedback is
used in oscillators?
3. What is the expression for the frequency of oscillations of
Colpitt’s and Hartley oscillator?
4.
Whether an oscillator is dc to ac converter. Explain?
5.
What is the loop gain of an oscillator?
6.
What is the difference between amplifier and oscillator?
7.
What is the condition for sustained oscillations?
8.
How many inductors and capacitors are used in Hartley Oscillator?
9.
How the oscillations are produced in Hartley oscillator?
10.
What is the difference between damped oscillations undamped oscillations?
11. How does Colpitt’s differ from Hartley?
CIRCUIT DIAGRAM: -
COLPITTS
OSCILLATOR
EXPECTED WAVEFORM:
TABULAR COLUMN:
S.NO
|
L(mH)
|
C1 (mF)
|
C2 (mF)
|
CT (mF)
|
Theoretical
Frequency
(KHz)
|
Practical
Frequency
(KHz)
|
Vo(V)
Peak to peak
|
1
|
5
|
0.01
|
0.001
|
||||
2
|
5
|
0.01
|
0.0022
|
||||
3
|
5
|
0.01
|
0.0033
|
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO- 5
CLASS A POWER AMPLIFIER
AIM:
1.
To study and plot the frequency
response of a Class A Power Amplifier.
2.
To calculate efficiency of
Class A Power Amplifier.
COMPONENTS & EQUIPMENT REQUIRED:
S.No
|
Apparatus
|
Range/
Rating
|
Quantity
(in No.s)
|
1.
|
Trainer
Board containing
(a)
DC Supply voltage.
(b)
NPN Transistor.
(c)
Resistors.
d) Capacitor.
e) Inductor.
|
12 V
BC 107
560Ω
100KΩ
470Ω
22 mF.
50mH
|
1
1
1
1
1
1
1
|
2.
|
D.C
Milliammeter
|
0-100mA
|
1
|
3.
|
Cathode Ray
Oscilloscope.
|
(0-20)MHz
|
1
|
4.
|
Function Generator.
|
0.1 Hz-10 MHz
|
1
|
5.
|
BNC Connector
|
2
|
|
6.
|
Connecting Wires
|
5A
|
5
|
THEORY:
In CLASS A amplifier, the
transistor is biased such that the output current flows, ie. Transistor is ON
for the full cycle (3600) of the input a.c. signal. In CLASS A Power
amplifier there is no distortion when compared with other amplifiers. The
maximum value of theoretical efficiency is 25% for a series fed and 50% for a
transformer coupled power amplifier.
Conversion
Efficiency:
It is the measure of
the ability of an active device in converting the d.c. Power of the supply into
the a.c. power delivered to the load. Conversion efficiency is also referred to
as theoretical efficiency and collector circuit efficiency (for transistor
amplifier) and is denoted by η. By definition, the percentage efficiency is
η.
= Signal power delivered to the load
x 100 %
d.c.
power supplied to output circuit
PROCEDURE:
1.
Connect the circuit as shown in
figure.
2.
Adjust input signal amplitude
in the function generator and observe an amplified voltage at the output
without distortion.
3.
By keeping input signal
voltage, say at 50 mV, vary the input signal frequency from 0-1 MHz as shown in
tabular column and note the corresponding output voltage.
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
- Frequency Response of CLASS-A Power amplifier is plotted.
- Efficiency of CLASS A Power amplifier is found to be___________
- Bandwidth fH – fL = ____________
VIVA QUESTIONS:
1. Differentiate between voltage amplifier and power amplifier
2. Why power amplifiers are considered as large signal amplifier?
3. When does maximum power dissipation happen in this circuit ?.
4. What is the maximum theoretical efficiency?
5. Sketch wave form of output current with respective input
signal.
6. What are the different types of class-A power amplifiers
available?
7. What is the theoretical efficiency of the transformer coupled
class-A power amplifier?
8. What is difference in AC, DC load line?.
9. How do you locate the Q-point ?
10. What are the applications of class-A power amplifier?
CIRCUIT DIAGRAM:
EXPECTED GRAPH:
TABULAR FORM:
Vin = 50 mV
S.No
|
Frequency
(in Hz)
|
Vo
(in volts)
|
Gain A =
Vo/
Vi
|
Gain(dB) Av =
20 log(Vo/ Vi )
|
1
|
100
|
|||
2
|
200
|
|||
3
|
400
|
|||
4
|
800
|
|||
5
|
1K
|
|||
6
|
2K
|
|||
7
|
4K
|
|||
8
|
8K
|
|||
9
|
10K
|
|||
10
|
20K
|
|||
11
|
40K
|
|||
12
|
80K
|
|||
13
|
100K
|
|||
14
|
200K
|
CALCULATIONS:
When signal is
removed, Vi=0
Zero signal
current, Ic =
Input Power, Pin=Vcc
x Ic
Output Power,
=
Efficiency, η =
=
=
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO- 6
SINGLE TUNED VOLTAGE AMPLIFIER
AIM:
1. To study & plot the frequency response
of a Single Tuned voltage amplifier
1.
To find the resonant frequency.
2.
To calculate gain and
bandwidth.
COMPONENTS
& EQUIPMENT REQUIRED:
S.No
|
Apparatus
|
Range/
Rating
|
Quantity
(in No.s)
|
1.
|
Trainer
Board containing
(a)
DC Supply voltage.
(b)
NPN Transistor.
(c)
Resistors.
d) Capacitor.
e) Inductor.
|
12 V
BC 107
47 KΩ
150Ω
1 KΩ
10 KΩ
10mF
22 mF.
0.022 mF.
0.033mF.
1mH
|
1
1
1
1
1
2
2
1
1
1
1
|
2.
|
Cathode Ray
Oscilloscope.
|
(0-20)MHz
|
1
|
3.
|
Function Generator.
|
0.1 Hz-10 MHz
|
1
|
4.
|
BNC Connector
|
2
|
|
5.
|
Connecting Wires
|
5A
|
5
|
THEORY:
A Tuned Amplifier uses a parallel
tuned circuit a parallel tuned circuit has high input impedance at its
frequency of resonance, and the impedance falls off sharping as the frequency
departs from the frequency of resonance. Hence the gain Vs frequency curve of a
tuned amplifier is very similar to the impedance Vs frequency curve of a tuned
amplifier are therefore, used for amplification of a narrow band of
frequencies.
A resonant circuit generally uses
either a variable inductor or variable capacitor for adjusting the resonant
frequency at the center of the band of frequencies to be amplified. Over this
narrow band of frequency, the gain of the tuned amplifier remains more or less
constant.
An important example is the radio
frequency (R.F) amplifier, which amplifies either a single radio frequency
signal or a narrow band or frequencies center about the resonant frequency. The
tuned circuit formed by L & C resonates at the frequency of operation.
PROCEDURE:
1.
Connect the circuit as shown in
figure.
2.
Connect the 0.022μF capacitor
3.
Adjust input signal amplitude
in the function generator and observe an amplified voltage at the output
without distortion.
4.
By keeping input signal
voltage, say at 50 mV, vary the input signal frequency from 0-1 MHz as shown in
tabular column and note the corresponding output voltage.
5.
Repeat the same procedure
for 0.033μF capacitor.
6.
Plot the graph: gain (Vs)
frequency.
7.
Calculate the Ft and
Fp
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
1.
Frequency response of RF Tuned
voltage amplifier is plotted.
2.
For 0.022μF, gain = ________dB
Bandwidth= _________
3. For 0.033μF, gain = ________dB
Bandwidth= _________
VIVA QUESTIONS:
1.
What is the purpose of tuned amplifier?
2.
What is Quality factor?
3.
Why should
we prefer parallel resonant circuit in tuned amplifier.
4.
What type of tuning we need to increase gain and bandwidth.?
5.
What are the limitations of single tuned amplifier?
6.
What is meant by Stagger tuning?
7.
What is the conduction angle of an tuned amplifier if it is operated in class B mode?
8.
What are the applications of tuned amplifier
9. What
are the different types of tuned
circuits ?
10. State
relation between resonant frequency and bandwidth of a Tuned amplifier.
11. Differentiate
between Narrow band and Wideband tuned amplifiers ?
12. Calculate
bandwidth of a Tuned amplifier whose resonant frequency is 15KHz and Q-factor
is 100.
13. Specify
the applications of Tuned amplifiers.
CIRCUIT DIAGRAM:
EXPECTED GRAPH:
TABULAR COLUMN -1:
Vin = 50 mV
C= 0.022μF
S.No
|
Frequency
(in Hz)
|
Vo
(in volts)
|
Gain A =
Vo/
Vi
|
Gain(dB) Av =
20 log(Vo/ Vi )
|
1
|
100
|
|||
2
|
200
|
|||
3
|
400
|
|||
4
|
800
|
|||
5
|
1K
|
|||
6
|
2K
|
|||
7
|
4K
|
|||
8
|
8K
|
|||
9
|
10K
|
|||
10
|
20K
|
|||
11
|
40K
|
|||
12
|
80K
|
|||
13
|
100K
|
|||
14
|
200K
|
TABULAR COLUMN-2:
Vin = 50 mV
C= 0.033μF
S.No
|
Frequency
(in Hz)
|
Vo
(in volts)
|
Gain A =
Vo/
Vi
|
Gain(dB) Av =
20 log(Vo/ Vi )
|
1
|
100
|
|||
2
|
200
|
|||
3
|
400
|
|||
4
|
800
|
|||
5
|
1K
|
|||
6
|
2K
|
|||
7
|
4K
|
|||
8
|
8K
|
|||
9
|
10K
|
|||
10
|
20K
|
|||
11
|
40K
|
|||
12
|
80K
|
|||
13
|
100K
|
|||
14
|
200K
|
@@@@@@@@@@@@@@@@@@@@@@@@@@@@
SIMULATION LAB
EXPERIMENT NO: 1
CE AMPLIFIER
AIM:
To plot
the frequency response of CE amplifier and calculate gain bandwidth.
SOFTWARE REQUIRED: MultiSim Software
COMPONENTS
& EQUIPMENTS REQUIRED: -
S.No
|
Apparatus
|
Range/
Rating
|
Quantity
(in No.s)
|
1.
|
CE
Amplifier trainer Board with
DC
power supply
DC
power supply
NPN
transistor
Carbon
film resistor
(e)Carbon film resistor
(f) Capacitor.
|
BC 107
100KW, 1/2W
2.2KW, 1/2W
0.1µF
|
1
1
1
1
1
2
|
2.
|
Cathode Ray Oscilloscope.
|
(0-20)MHz
|
1
|
3.
|
Function
Generator.
|
0.1 Hz-10 MHz
|
1
|
4.
|
BNC
Connector
|
2
|
|
5.
|
Connecting
Wires
|
5A
|
5
|
PROCEDURE: -
1. Connect the
circuit diagram as shown in figure.
2. Adjust input signal amplitude in the function generator and observe an amplified voltage at the
output without distortion.
3. By keeping input signal voltages at 50mV,
vary the input signal frequency from 0 to 1MHz in steps as shown in tabular column and note the corresponding output voltages.
PRECAUTIONS:
-
1. Oscilloscope probes negative terminal should
be at equipotential points (i.e.ground voltage= 0), because both terminals are internally shorted in dual trace
oscilloscope.
2. Ensure that output voltage is exactly an
amplified version of input voltage
without any distortion (adjust input voltage amplitude to that extent)
3. No loose connections at the junctions.
RESULT:
-
Frequency
response of CE amplifier is plotted.
Gain, AV
= ________dB.
Bandwidth=
fH--fL =________Hz.
VIVA QUESTIONS
1. What are the advantages and
disadvantages of single-stage amplifiers?
2. Why gain falls at HF and LF?
3. Why the gain remains
constant at MF?
4. Explain the function of
emitter bypass capacitor, Ce?
5. How the band width will
effect as more number of stages are cascaded?
6. Define frequency response?
7. What is the phase difference
between input and output waveforms of a CE amplifier?
8. What
is early effect?
TABULAR
COLUMN:
Input voltage: Vi = 50mV
Frequency
(in Hz)
|
Output (Vo)
(Peak to Peak)
|
Gain
AV=V0/Vi
|
Gain (in dB) =
20 log 10 VO/ Vi
|
20
|
|||
600
|
|||
1K
|
|||
2K
|
|||
4K
|
|||
8K
|
|||
10K
|
|||
20K
|
|||
30K
|
|||
40K
|
|||
50K
|
|||
60K
|
|||
80K
|
|||
100K
|
|||
250K
|
|||
500K
|
|||
750K
|
|||
1000K
|
CIRCUIT DIAGRAM:
Bandwidth
= fH-fL
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO: 2
COMMON SOURCE AMPLIFIER
AIM:
-
a) To Plot the frequency response of a common
source amplifier.
b) Calculate gain.
c) Calculate bandwidth.
SOFTWARE REQUIRED: Multi Sim Software
COMPONENTS
& EQUIPMENTS REQUIRED: -
S.No
|
Device
|
Range/Rating
|
Quantity
(in No.s)
|
1.
|
FET amplifier Trainer Board with
(a) DC supply voltage
(b) FET
(c) Capacitors
(d) Resistors
|
12V
BFW 11
0.1mF
47mF
1.5KW
4.7 KW
1MW
|
1
1
2
1
1
1
1
|
2.
|
Signal generator
|
0.1Hz-1MHz
|
1
|
3.
|
CRO
|
0Hz-20MHz
|
1
|
4.
|
Connecting wires
|
5A
|
4
|
PROCEDURE:
-
1.Connect the circuit diagram as shown in
figure.
2.Adjust input signal amplitude in the
function generator and observe an amplified voltage at the output without 3.distortion.
4.By keeping input signal voltage, say at
50mV, vary the input signal frequency from 0 to 1MHz in steps as shown in
tabular column and note the corresponding output voltages.
PRECAUTIONS:
Oscilloscopes
probes negative terminal should be at equipotential points(i.e. ground voltage
is zero) because both terminals are internally shorted in dual trace
oscilloscope.
RESULT:
-
Hence,
the frequency response of FET (CS) amplifier is plotted.
Gain = _______dB (maximum).
3.
Bandwidth= fH--fL =
_________Hz.
VIVA QUESTIONS:
1. What is the difference between FET and BJT?
2. FET is unipolar or bipolar?
3. Draw the symbol of FET?
4. What are the applications of FET?
5. FET is voltage controlled or current controlled?
TABULAR
COLUMN:Input = 50mV
Frequency (in Hz)
|
Output
Voltage (Vo)
|
Gain
Av=Vo/Vi
|
Gain
(in dB) =
20log10(Vo/Vi)
|
20
|
|||
40
|
|||
80
|
|||
100
|
|||
500
|
|||
1000
|
|||
5000
|
|||
10K
|
|||
50K
|
|||
100K
|
|||
200K
|
|||
400K
|
|||
600K
|
|||
800K
|
CIRCUIT DIAGRAM:
COMMON SOURCE AMPLIFIER
EXPECTED GRAPH:
EXPERIMENT NO- 3
TWO STAGE RC COUPLED
AMPLIFIER
AIM:
1.To plot the frequency response of a RC
coupled amplifier with a pair of shunted emitter capacitors of 10 μF and 100μF.
2. To calculate gain.
3. To calculate bandwidth.
SOFTWARE REQUIRED: MultiSim Software
COMPONENTS & EQUIPMENT REQUIRED:
S.No
|
Device
|
Range/
Rating
|
Quantity
(in No.s)
|
1.
|
Trainer Board containing
a) DC Supply voltage.
b) NPN Transistor.
c) Resistors.
d) Capacitors.
|
12 V
BC 107
47 KΩ
2.2 KΩ
1 KΩ
10 KΩ
100mF
10mF.
|
1
2
2
2
5
2
6
|
2.
|
Cathode
Ray Oscilloscope.
|
(0-20)MHz
|
1
|
3.
|
Function
Generator.
|
0.1 Hz-10 MHz
|
1
|
4
|
BNC
Connector
|
2
|
|
5
|
Connecting
Wires
|
5A
|
10
|
THEORY:
When the voltage gain provided by a single stage stage is not
sufficient, we have to go for more than one stage in the amplifier. In the
circuit diagram, two stages are shown connected through a link. When the link
is open, the voltage gain of first stage is high. However, when the link is
closed the gain is reduced. This is because of the input impedance of the
second stage will be in parallel with the load resistance of the first stage.
Therefore the input impedance is reduced. Hence overall gain is decreased. The fall in amplifier gain at low frequencies
is due to the effect of coupling and bypass capacitor.
At medium and high frequencies, the factor
‘f’ makes Xc is very small, so that all by coupling and bypass
capacitors behaves as short circuit. There are also stray capacitance ‘Cs’,
which are capacitances between connecting wires and ground. All these
capacitances values are very small so that at low and medium frequencies, there
impedances are very high. As the frequency increases the reactance of the stray
capacitance fall. When these reactance becomes small enough they begin to shunt
away some of the input and output currents are thus reduces the current gain. Even
if no external stray capacitance is present, the device internal capacitances
through the semiconductor material limit the circuit frequency response.
A cut-off frequency is the frequency at
which the transistor gain falls to 0.707 of its maximum gain.
The range of frequencies over which the
gain of an amplifier is equal to or greater than 70.7% of its maximum gain is
known as maximum gain
PROCEDURE:
1. Connect the circuit as shown in figure
for 10 μF.
2. Adjust input signal amplitude in the
function generator and observe an amplified voltage at the output without
distortion.
3. By keeping input signal voltage, say at
50 mV, vary the input signal frequency from 0-1 MHz as shown in tabular column
and note the corresponding output voltage.
4. Repeat the same procedure for C=100μF.
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
-
Hence, the
frequency Response of RC coupled (2 stage) amplifier for 10μF and 100 μF
is plotted.
1.
For C=10 μF,
Gain=
Bandwidth =fH – fL =
2.
For C=100μF
Gain=
Bandwidth =fH – fL =
TWO STAGE RC COUPLED
AMPLIFIER
EXPECTED GRAPH:
TABULAR FORM:
C=10μF Vin
= 50 mV
S.No
|
Frequency
(in Hz)
|
Vo
(in volts)
|
Gain A =
Vo/ Vi
|
Gain(dB) Av =
20 log(Vo/ Vi )
|
1
|
100
|
|||
2
|
200
|
|||
3
|
400
|
|||
4
|
800
|
|||
5
|
1K
|
|||
6
|
2K
|
|||
7
|
4K
|
|||
8
|
8K
|
|||
9
|
10K
|
|||
10
|
20K
|
|||
11
|
40K
|
|||
12
|
80K
|
|||
13
|
100K
|
|||
14
|
200K
|
|||
15
|
300K
|
|||
16
|
500K
|
|||
17
|
700K
|
|||
18
|
900K
|
|||
19
|
1M
|
TABULAR FORM:
C=100μF Vin
= 50 mV
S.No
|
Frequency
(in Hz)
|
Vo
(in volts)
|
Gain A =
Vo/ Vi
|
Gain(dB) Av =
20 log(Vo/ Vi )
|
1
|
100
|
|||
2
|
200
|
|||
3
|
400
|
|||
4
|
800
|
|||
5
|
1K
|
|||
6
|
2K
|
|||
7
|
4K
|
|||
8
|
8K
|
|||
9
|
10K
|
|||
10
|
20K
|
|||
11
|
40K
|
|||
12
|
80K
|
|||
13
|
100K
|
|||
14
|
200K
|
|||
15
|
300K
|
|||
16
|
500K
|
|||
17
|
700K
|
|||
18
|
900K
|
|||
19
|
1M
|
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT - 4
CLASS B POWER AMPLIFIER
(COMPLEMENTARY SYMMETRY)
AIM:
To study the frequency response of CLASS B
Complementary Symmetry amplifier and to calculate its
efficiency.
SOFTWARE REQUIRED: MultiSim Software
APPARATUS REQUIRED:
S.No
|
Device
|
Range/
Rating
|
Qty
(in No.)
|
1
|
CLASS B Power amplifier Trainer Kit
|
1
|
|
2
|
Signal Generator
|
(0Hz-1 MHz)
|
1
|
3
|
Cathode Ray Oscilloscope
|
(0 Hz –20 MHz)
|
1
|
4
|
D.C Milliammeter
|
(0-50) mA
|
1
|
5
|
Decade Resistance Box
|
(1-100) KΩ
|
1
|
6
|
BNC Connector
|
2
|
|
7
|
Connecting Wires
|
5A
|
6
|
THEORY:
In a CLASS B amplifier the transistor is
biased almost at cut off, so that it remains forward biased only for one half
cycle of the input signal. Hence its conduction angle is only 1800. It
has the following two advantages over CLASS A power amplifier. Possible to
obtain greater power output Efficiency is higher. Negligible power loss (as no
output current flows) at no input signal. And also it eliminates the
disadvantages of Push-Pull configuration like bulky and expensive output
transformer. The maximum theoretical efficiency of a CLASS B power amplifier is
78.5 %.
PROCEDURE:
1. Switch ON the CLASS B amplifier trainer.
2. Connect Milliammeter to (A) terminals
and DRB to the RL terminals and fix RL=50Ω.
3. Apply the input voltage from the signal
generator to the Vs terminals.
4. Connect channel 1 of CRO to the Vs
terminals and channel 2 across the load.
5. By varying the input voltage, observe
the maximum distortion less output waveform.
And note down the voltage reading.
6. Calculate the efficiency.
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
Thus efficiency of CLASS B (Complementary
symmetry) amplifier is calculated and
frequency response is studied.
VIVA QUESTIONS:
1. Differentiate between
voltage amplifier and power amplifier?
2. Explain impedance matching
provided by transformer?
3. Under what condition power
dissipation is maximum for transistor in this circuit?
4. What is the maximum
theoretical efficiency?
5. Sketch current waveform in
each transistor with respective input signal?
6. How do you test matched
transistors required for this circuit with DMM?.
7. What is the theoretical
efficiency of the complementary stage amplifier.
8. How do you measure DC and AC
out put of this amplifier?
9. Is this amplifier working in
class A or B. ?
10. How
can you reduce cross over distortion?
CLASS B POWER AMPLIFIER (COMPLEMENTARY SYMMETRY)
OBSERVATIONS:
Vs=2v
FREQUENCY
|
Vo (volts)
|
Idc (mA)
|
Efficiency
|
10 KHz
|
CALCULATIONS:
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO-5
RC PHASE SHIFT OSCILLATOR
AIM:
To find practical frequency of RC phase shift oscillator and to compare
it with theoretical frequency for R=10KW and C = 0.01mF, 0.0022mF & 0.0033mF
Respectively
SOFTWARE
REQUIRED: MultiSim Software
COMPONENTS AND EQUIPMENTS REQUIRED:
S.No
|
Device
|
Range/
Rating
|
Quantity
(in No.s)
|
1
|
RC phase shift oscillator trainer board
containing
a) DC supply voltage
b) Capacitor
c) Resistor
d) NPN Transistor
|
12V-----------
1000mF-------
0.047mF------
0.01mF--------
0.0022mF------
0.0033mF----
1KW-----------
10KW---------
47KW----------
100KW---------
BC 107--------
|
1
1
1
3
3
3
1
4
1
1
1
|
2
|
CRO
|
(0-20) MHz
|
1
|
3.
|
BNC Connector
|
1
|
|
3
|
Connecting wires
|
5A
|
6
|
PROCEDURE:
1. Connect the circuit as shown in figure.
2. Connect the 0.0022 mF capacitors in the circuit and observe the waveform.
3. Time period of the waveform is to be
noted and frequency should be calculated by the formula f = 1/T.
4. Now fix the capacitance to 0.033 mF and 0.01mF and
calculate the frequency and tabulate as shown.
5. Find theoretical frequency from the
formula f = 1/2PRCÖ6 and compare theoretical and practical frequencies.
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
-
For C = 0.0022mF & R=10KW
Theoretical frequency=
Practical frequency=
For C = 0.0033mF & R=10KW
Theoretical frequency=
Practical frequency=
For C = 0.01mF & R=10KW
Theoretical frequency=
Practical frequency=
VIVA QUESTIONS:
1. What are the conditions of
oscillations?
2. Give the formula for
frequency of oscillations?
3. What is the total phase
shift produced by RC ladder network?
4. What are the types of oscillators?
5. What
is the gain of RC phase shift oscillator?
CIRCUIT DIAGRAM:
RC PHASE SHIFT OSCILLATOR
EXPECTED WAVEFORM:
TABULAR
COLUMN:
S.No
|
C
(mF)
|
R
(W)
|
Theoretical Frequency
(KHz)
|
Practical Frequency
(KHz)
|
Vo (p-p)
(Volts)
|
1
|
0.0022
|
10K
|
|||
2
|
0.0033
|
10K
|
|||
3
|
0.01
|
10K
|
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
EXPERIMENT NO- 6
SINGLE TUNED VOLTAGE AMPLIFIER
AIM:
1. To study & plot the frequency
response of a Single Tuned voltage amplifier
2. To find the resonant frequency.
3. To calculate gain and bandwidth.
SOFTWARE REQUIRED: MultiSim Software
COMPONENTS & EQUIPMENT REQUIRED:
S.No
|
Apparatus
|
Range/
Rating
|
Quantity
(in No.s)
|
1.
|
Trainer Board containing
DC Supply voltage.
NPN Transistor.
Resistors.
d) Capacitor.
e) Inductor.
|
12 V
BC 107
47 KΩ
150Ω
1 KΩ
10 KΩ
10mF
22 mF.
0.022 mF.
0.033mF.
1mH
|
1
1
1
1
1
2
2
1
1
1
1
|
2.
|
Cathode
Ray Oscilloscope.
|
(0-20)MHz
|
1
|
3.
|
Function
Generator.
|
0.1 Hz-10 MHz
|
1
|
4.
|
BNC
Connector
|
2
|
|
5.
|
Connecting
Wires
|
5A
|
5
|
THEORY:
A Tuned Amplifier uses a parallel tuned circuit a parallel tuned circuit
has high input impedance at its frequency of resonance, and the impedance falls
off sharping as the frequency departs from the frequency of resonance. Hence
the gain Vs frequency curve of a tuned amplifier is very similar to the
impedance Vs frequency curve of a tuned amplifier are therefore, used for
amplification of a narrow band of frequencies.
A resonant circuit generally uses either a variable inductor or variable
capacitor for adjusting the resonant frequency at the center of the band of
frequencies to be amplified. Over this narrow band of frequency, the gain of
the tuned amplifier remains more or less constant.
An important example is the
radio frequency (R.F) amplifier, which amplifies either a single radio
frequency signal or a narrow band or frequencies center about the resonant
frequency. The tuned circuit formed by L & C resonates at the frequency of
operation.
PROCEDURE:
1. Connect the circuit as shown in figure.
2. Connect the 0.022μF capacitor
3. Adjust input signal amplitude in the
function generator and observe an amplified voltage at the output without
distortion.
4. By keeping input signal voltage, say at
50 mV, vary the input signal frequency from 0-1 MHz as shown in tabular column
and note the corresponding output voltage.
5. Repeat the same procedure for 0.033μF capacitor.
6 Plot the graph: gain (Vs) frequency.
1.
Calculate the Ft and
Fp
PRECAUTIONS:
-
1. No loose contacts at the junctions.
2. Check the connections before giving
the power supply
3. Observations should be taken
carefully.
RESULT:
Frequency response of RF Tuned voltage
amplifier is plotted.
For 0.022μF, gain = ________dB
Bandwidth= _________
3.
For 0.033μF, gain = ________dB
Bandwidth= _________
CIRCUIT DIAGRAM:
SINGLE TUNED VOLTAGE AMPLIFIER
EXPECTED GRAPH:
TABULAR FORM-1:
Vin =
50 mV
C= 0.022μF
S.No
|
Frequency
(in Hz)
|
Vo
(in volts)
|
Gain A =
Vo/ Vi
|
Gain(dB) Av =
20 log(Vo/ Vi )
|
1
|
100
|
|||
2
|
200
|
|||
3
|
400
|
|||
4
|
800
|
|||
5
|
1K
|
|||
6
|
2K
|
|||
7
|
4K
|
|||
8
|
8K
|
|||
9
|
10K
|
|||
10
|
20K
|
|||
11
|
40K
|
|||
12
|
80K
|
|||
13
|
100K
|
|||
14
|
200K
|
TABULAR FORM-2:
Vin =
50 mV
C= 0.033μF
S.No
|
Frequency
(in Hz)
|
Vo
(in volts)
|
Gain A =
Vo/ Vi
|
Gain(dB) Av =
20 log(Vo/ Vi )
|
1
|
100
|
|||
2
|
200
|
|||
3
|
400
|
|||
4
|
800
|
|||
5
|
1K
|
|||
6
|
2K
|
|||
7
|
4K
|
|||
8
|
8K
|
|||
9
|
10K
|
|||
10
|
20K
|
|||
11
|
40K
|
|||
12
|
80K
|
|||
13
|
100K
|
|||
14
|
200K
|
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