On National Teacher Day, meet the 2024-25 Kenan Fellows
Ec ii lab manual
1. DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
IV SEMESTER - ECE ‘B’
EC2257 – ELECTRONIC CIRCUITS – II AND SIMULATION LAB
Cycle – I
1. Current Series feedback amplifier: Frequency response, Input and output impedance
calculation
2. Voltage Shunt feedback amplifier: Frequency response, Input and output impedance
calculation
3. RC Phase shift oscillator
4. Hartley Oscillator
5. Colpitts Oscillator
6. CMOS Inverter, NAND and NOR using PSPICE
7. Active filter: Butterworth II order LPF using PSPICE
8. Differential amplifier using PSPICE
Cycle – II
9. Frequency Response of Single Tuned Amplifier
10. Astable multivibrator
11. Monostable multivibrator
12. Bistable multivibrator
13. Wave Shaping Circuits (Integrator, Differentiator, Clippers and Clampers)
14. D/A converter using PSPICE
15. Astable multivibrator using PSPICE
16. Monostable multivibrator using PSPICE
Powered to www.technoscriptz.com
2. Expt No: 1 CURRENT-SERIES FEEDBACK AMPLIFIER
Aim:
To design and test the current-series feedback amplifier and to calculate the
following parameters with and without feedback.
1. Mid band gain.
2. Bandwidth and cut-off frequencies.
3. Input and output impedance.
Components & Equipment required:
Range /
Sl. No. Components / Equipment Quantity
Specifications
1 Power supply (0-30)V 1
2 Function generator (0-20M)Hz 1
3 CRO 1
4 Transistor BC107 1
5 Resistors
6 Capacitors
7 Connecting wires Accordingly
Circuit diagram:
(i) Without feedback:
Vcc = +12V
R1 RC
Cout
C
Cin
RL
B BC107
CRO
E
Vin=50mV
f=(1-3M)Hz R2 RE
CE
0
2
3. (ii) With feedback:
Vcc = +12V
R1 RC
Cout
C
Cin
RL
B BC107
CRO
E
Vin=50mV
f=(1-3M)Hz R2 RE
0
Theory:
The current series feedback amplifier is characterized by having shunt sampling
and series mixing. In amplifiers, there is a sampling network, which samples the output
and gives to the feedback network. The feedback signal is mixed with input signal by
either shunt or series mixing technique. Due to shunt sampling the output resistance
increases by a factor of ‘D’ and the input resistance is also increased by the same factor
due to series mixing. This is basically transconductance amplifier. Its input is voltage
which is amplified as current.
Design:
(i) Without feedback:
VCC = 12V; IC = 1mA; fL = 50Hz; S = 2; RL = 4.7K; hfe =
re = 26mV / IC = 26;
hie = hfe re =
VCE= Vcc/2 (transistor Active) =
VE = IERE = Vcc/10
Applying KVL to output loop, we get
VCC = ICRC + VCE + IERE
3
4. RC =
Since IB is very small when compare with IC,
IC IE
RE = VE / IE =
S = 1+ RB / RE = 2
RB =
VB = VCC R2 / (R1 + R2)
RB = R1 || R2
R1 = R2 =
XCi = Zi / 10 = (hie || RB) / 10 =
Ci = 1 / (2f XCi) =
Xco = (RC || RL)/10 =
Co = 1 / (2f XCo) =
XCE = RE/10 =
CE = 1 / (2f XCE) =
(ii) With feedback (Remove the Emitter Capacitor, CE):
Feedback factor, = -RE =
Gm = -hfe / (hie + RE) =
Desensitivity factor, D = 1 + Gm =
Transconductance with feedback, Gmf = Gm / D =
Input impedance with feedback, Zif = Zi D
Output impedance with feedback, Z0f = Z0 D
Procedure:
1. Connect the circuit as per the circuit diagram.
2. Keeping the input voltage constant, vary the frequency from 50Hz to 3MHz in
regular steps and note down the corresponding output voltage.
3. Plot the graph: Gain (dB) Vs Frequency
4. Calculate the bandwidth from the graph.
5. Calculate the input and output impedance.
6. Remove Emitter Capacitance, and follow the same procedures (1 to 5).
4
5. Tabular column:
(i) Without feedback:
Vi =
Frequency Output Voltage Gain = 20 log(V0/Vi)
Sl. No Gain = V0/Vi
(Hz) (V0) (volts) (dB)
(ii) With feedback:
Vi =
Frequency Output Voltage Gain = 20 log(V0/Vi)
Sl. No Gain = V0/Vi
(Hz) (V0) (volts) (dB)
Model graph: (frequency response)
Gain in dB
Without feedback
With feedback
Frequency in Hz
Result:
Thus the current series feedback amplifier is designed and constructed and the
following parameters are calculated.
Theoretical Practical
With feedback Without feedback With feedback Without feedback
Input
impedance
Output
impedance
Gain
(midband)
Bandwidth
5
6. Expt. No. 2 VOLTAGE SHUNT FEEDBACK AMPLIFIER
Aim:
To design and test the voltage-shunt feedback amplifier and to calculate the
following parameters with and without feedback.
1. Mid band gain.
2. Bandwidth and cut-off frequencies.
3. Input and output impedance.
Components & Equipment required:
Range /
Sl. No. Components / Equipment Quantity
Specifications
1 Power supply (0-30)V 1
2 Function generator (0-20M)Hz 1
3 CRO 1
4 Transistor BC107 1
5 Resistors
6 Capacitors
7 Connecting wires
Circuit Diagram:
(i) Without Feedback:
Vcc = +12V
R1 RC
Cout
C
Cin
B BC107
E CRO
Vin=50mV
f=(1-3M)Hz R2 RE
CE
0
6
7. (ii) With Feedback:
Vcc = +12V
R1 RC
Rf Cf Cout
C
Cin
BC107
B CRO
E
Vin=50mV
f=(1-3M)Hz R2 RE
CE
0
Theory:
In voltage shunt feedback amplifier, the feedback signal voltage is given to
the base of the transistor in shunt through the base resistor RB. This shunt connection
tends to decrease the input resistance and the voltage feedback tends to decrease the
output resistance. In the circuit RB appears directly across the input base terminal and
output collector terminal. A part of output is feedback to input through RB and increase in
IC decreases IB. Thus negative feedback exists in the circuit. So this circuit is also called
voltage feedback bias circuit. This feedback amplifier is known an transresistance
amplifier. It amplifies the input current to required voltage levels. The feedback path
consists of a resistor and a capacitor.
Design
(i) Without Feedback:
VCC = 12V; IC = 1mA; AV = 30; Rf = 2.5K; S = 2; hfe = ; =1/ Rf = 0.0004
re = 26mV / IC = 26;
hie = hfe re =
VCE= Vcc/2 (transistor Active) =
VE = IERE = Vcc/10 =
7
8. Applying KVL to output loop, we get VCC = ICRC + VCE + IERE
RC =
Since IB is very small when compare with IC, IC IE
RE = VE / IE =
S = 1+ RB / RE
RB =
VB = VCC R2 / (R1 + R2)
RB = R1 || R2
R1 = R2 =
(ii) With feedback:
RO = RC || Rf =
Ri = (RB || hie ) Rf =
Rm = -(hfe (RB || Rf) (RC || Rf)) / ((RB || Rf) + hie) =
Desensitivity factor, D = 1 + Rm
Rif = Ri / D =
Rof = Ro / D =
Rmf = Rm / D =
XCi = Rif /10 =
Ci = 1 / (2f XCi) =
Xco = Rof /10 =
Co = 1 / (2f XCo) =
RE’ = RE || ((RB + hie) / (1+hfe))
XCE = RE’/10 =
CE = 1 / (2f XCE) =
XCf = Rf/10
Cf = 1 / (2f XCf) =
Procedure:
1. Connect the circuit as per the circuit diagram.
2. Keeping the input voltage constant, vary the frequency from 50Hz to 3MHz in
regular steps and note down the corresponding output voltage.
3. Plot the graph: Gain (dB) Vs Frequency
8
9. 4. Calculate the bandwidth from the graph.
5. Calculate the input and output impedance.
6. Remove Emitter Capacitance, and follow the same procedures (1 to 5).
Tabular Column:
(i) Without Feedback:
Vi = 10 mV
Frequency V0 Gain = V0/Vi Gain (dB) = 20 log(V0/Vi)
(volts)
(ii) With Feedback:
Vi = 10 mV
Frequency V0 Gain = V0/Vi Gain (dB) = 20 log(V0/Vi)
(volts)
Model graph: (frequency response)
Gain in dB
Without feedback
With feedback
Frequency in Hz
Result:
Thus the current series feedback amplifier is designed and constructed and the
following parameters are calculated.
Theoretical Practical
With feedback Without feedback With feedback Without feedback
Input
impedance
Output
impedance
Gain
(midband)
Bandwidth
9
10. Expt. No. 3 RC PHASE SHIFT OSCILLATOR
Aim:
To design and construct a RC phase shift oscillator for the given frequency (f0).
Components & Equipment required:
Range /
Sl. No. Components / Equipment Quantity
Specifications
1 Power supply (0-30)V 1
2 Function generator (0-20M)Hz 1
3 CRO 1
4 Transistor BC107 1
5 Resistors
6 Capacitors
7 Connecting wires Accordingly
Circuit Diagram:
Vcc = +12V
R1 RC
Cout
C
Cin
B BC107
E RL
CRO
R2 RE
CE
0
C C C
R R R
0
10
11. Theory:
In the RC phase shift oscillator, the required phase shift of 180˚ in the feedback
loop from the output to input is obtained by using R and C components, instead of tank
circuit. Here a common emitter amplifier is used in forward path followed by three
sections of RC phase network in the reverse path with the output of the last section being
returned to the input of the amplifier. The phase shift Ф is given by each RC section
Ф=tanˉ1 (1/ωrc). In practice R-value is adjusted such that Ф becomes 60˚. If the value
of R and C are chosen such that the given frequency for the phase shift of each RC
section is 60˚. Therefore at a specific frequency the total phase shift from base to
transistor’s around circuit and back to base is exactly 360˚ or 0˚. Thus the Barkhausen
criterion for oscillation is satisfied
Design:
VCC = 12V; IC = 1mA; C = 0.01F; fo = ; S = 2; hfe =
re = 26mV / IC = 26;
hie = hfe re =
VCE= Vcc/2 (transistor Active) =
VE = IERE = Vcc/10
Applying KVL to output loop, we get
VCC = ICRC + VCE + IERE
RC =
Since IB is very small when compare with IC,
IC IE
RE = VE / IE =
S = 1+ RB / RE = 2
RB =
VB = VBE + VE =
VB = VCC R2 / (R1 + R2)
RB = R1 || R2
R1 = R2 =
Gain formula is given by,
h fe R Leff
AV = (Av = -29, design given)
h ie
Effective load resistance is given by, Rleff = Rc || RL
RL =
11
12. XCi = {[hie+(1+hfe)RE] || RB}/10 =
Ci = 1 / (2f XCi) =
Xco = Rleff /10 =
Co = 1 / (2f XCo) =
XCE = RE/10 =
CE = 1 / (2f XCE) =
Feedback Network:
f0 = ; C = 0.01f;
1
fo =
2 6RC
R=
Procedure:
1. Connections are made as per the circuit diagram.
2. Switch on the power supply and observe the output on the CRO (sine wave).
3. Note down the practical frequency and compare with its theoretical frequency.
Model Graph:
Vout (Voltage)
Time(ms)
Result:
Thus RC phase shift oscillator is designed and constructed and the output sine
wave frequency is calculated as
Theoretical Practical
Frequency
12
13. Expt. No.4 HARTELY OSCILLATOR
Aim:
To design and construct the given oscillator for the given frequency (fO).
Components & Equipment required:
Range /
Sl. No. Components / Equipment Quantity
Specifications
1 Power supply (0-30)V 1
2 Function generator (0-20M)Hz 1
3 CRO 1
4 Transistor BC107 1
5 Resistors
6 Capacitors
7 DIB
8 DCB
9 Connecting wires Accordingly
Circuit Diagram:
Vcc = +12V
R1 RC
Cout
C
Cin
B
BC107
CRO
E
RL
R2 RE
CE
0
C
L1 L2
0
13
14. Theory:
Hartley oscillator is a type of sine wave generator. The oscillator derives its
initial output from the noise signals present in the circuit. After considerable time, it
gains strength and thereby producing sustained oscillations. Hartley Oscillator have two
major parts namely – amplifier part and feedback part. The amplifier part has a typically
CE amplifier with voltage divider bias. In the feedback path, there is a LCL network.
The feedback network generally provides a fraction of output as feedback.
Design:
VCC = 12V; IC = 1mA; fo = ; S = 2; hfe =
re = 26mV / IC = 26;
hie = hfe re =
VCE= Vcc/2 (transistor Active) =
VE = IERE = Vcc/10
Applying KVL to output loop, we get
VCC = ICRC + VCE + IERE
RC =
Since IB is very small when compare with IC, IC IE
RE = VE / IE =
S = 1+ RB / RE = 2
RB =
VB = VBE + VE =
VB = VCC R2 / (R1 + R2)
RB = R1 || R2
R1 = R2 =
Gain formula is given by,
h fe R Leff
AV = (Av = -29, design given)
h ie
Effective load resistance is given by, Rleff = Rc || RL
RL =
XCi = {[hie+(1+hfe)RE] || RB}/10 =
Ci = 1 / (2f XCi) =
Xco = Rleff /10 =
14
15. Co = 1 / (2f XCo) =
XCE = RE/10 =
CE = 1 / (2f XCE) =
Feedback Network:
f0 = ; L1 = 1mH; L2 = 10mH
1 L
A= = 1
L2
1
f=
2 L1 L 2 C
C=
Procedure:
1. Connections are made as per the circuit diagram.
2. Switch on the power supply and observe the output on the CRO (sine wave).
3. Note down the practical frequency and compare with its theoretical frequency.
Model Graph:
Vout (Voltage)
Time(ms)
Result:
Thus Hartley oscillator is designed and constructed and the output sine wave
frequency is calculated as
Theoretical Practical
Frequency
15
16. Expt. No.5 COLPITTS OSCILLATOR
Aim: To design and construct the given oscillator at the given operating frequency.
Equipments required:
Range /
Sl. No. Components / Equipment Quantity
Specifications
1 Power supply (0-30)V 1
2 Function generator (0-20M)Hz 1
3 CRO 1
4 Transistor BC107 1
5 Resistors
6 Capacitors
7 DIB
8 DCB
9 Connecting wires
Circuit Diagram:
Vcc =
+12V
R1 RC
Cout
C
Cin
B
BC107
CRO
E
RL
R2 RE
CE
0
L
C1 C2
0
16
17. Theory:
A Colpitts oscillator is the electrical dual of a Hartley oscillator. In the Colpitts
circuit, two capacitors and one inductor determine the frequency of oscillation. The
oscillator derives its initial output from the noise signals present in the circuit. After
considerable time, it gains strength and thereby producing sustained oscillations. It has
two major parts namely – amplifier part and feedback part. The amplifier part has a
typically CE amplifier with voltage divider bias. In the feedback path, there is a CLC
network. The feedback network generally provides a fraction of output as feedback.
Design:
VCC = 12V; IC = 1mA; fo = ; S = 2; hfe =
re = 26mV / IC = 26;
hie = hfe re =
VCE= Vcc/2 (transistor Active) =
VE = IERE = Vcc/10
Applying KVL to output loop, we get
VCC = ICRC + VCE + IERE
RC =
Since IB is very small when compare with IC, IC IE
RE = VE / IE =
S = 1+ RB / RE = 2
RB =
VB = VBE + VE =
VB = VCC R2 / (R1 + R2)
RB = R1 || R2
R1 = R2 =
Gain formula is given by,
h fe R Leff
AV = (Av = -29, design given)
h ie
Effective load resistance is given by, Rleff = Rc || RL
RL =
XCi = {[hie+(1+hfe)RE] || RB}/10 =
Ci = 1 / (2f XCi) =
Xco = Rleff /10 =
17
18. Co = 1 / (2f XCo) =
XCE = RE/10 =
CE = 1 / (2f XCE) =
Feedback Network:
f0 = ; C1 = ; C2 =
1 C
A= = 2
C1
1 C1 C 2
f=
2 LC1C 2
L=
Procedure:
1. Rig up the circuit as per the circuit diagrams (both oscillators).
2. Switches on the power supply and observe the output on the CRO (sine wave).
3. Note down the practical frequency and compare with its theoretical frequency.
Model Graph:
Vout (Voltage)
Time(ms)
Result:
Thus Colpitts oscillator is designed and constructed and the output sine wave
frequency is calculated as
Theoretical Practical
Frequency
18
19. Expt. No. 6 CMOS Inverter, NAND and NOR using PSPICE
Aim:
To plot the transient characteristics of output voltage for the given CMOS inverter,
NAND and NOR from 0 to 80s in steps of 1s. To calculate the voltage gain, input
impedance and output impedance for the input voltage of 5V.
Parameter Table:
Parameters PMOS NMOS
L 1 1
W 20 5
VTO -2 2
KP 4.5E-4 2
CBD 5p 5p
CBS 2p 2p
RD 5 5
RS 2 2
RB 0 0
RG 0 0
RDS 1Meg 1Meg
CGSO 1p 1p
CGDO 1p 1p
CGBO 1p 1p
Circuit Diagram:
(i) Inverter:
VDD = +5V
3
S
G Q1
D
1
Vin 2 Vout
D
Q2 RL = 100K
G S
0
19
20. (ii) NAND
VDD = +5V
2
S S
G G
Q2
Q1
D D
4 Vout
D
Q3
G
Vin1 S
1
RL = 100K
D
Q4
G
Vin2 3 S
(iii) NOR
VDD = +5V
3
Vin1 1 G S
Q1
D
2 G S
Vin2 Q2
D
4 Vout
D D
Q3 Q4
G G RL = 100K
S S
20
21. Theory:
(i) Inverter
CMOS is widely used in digital IC’s because of their high speed, low power
dissipation and it can be operated at high voltages resulting in improved noise immunity.
The inverter consists of two MOSFETs. The source of p-channel device is connected to
+VDD and that of n-channel device is connected to ground. The gates of two devices are
connected as common input.
(ii) NAND
It consists of two p-channel MOSFETs connected in parallel and two n-channel
MOSFETs connected in series. P-channel MOSFET is ON when gate is negative and N-
channel MOSFET is ON when gate is positive. Thus when both input is low and when
either of input is low, the output is high.
(iii) NOR
It consists of two p-channel MOSFETs connected in series and two n-channel
MOSFETs connected in parallel. P-channel MOSFET is ON when gate is negative and
N-channel MOSFET is ON when gate is positive. Thus when both inputs are high and
when either of input is high, the output is low. When both the inputs are low, the output
is high.
Truth Table:
(i) Inverter
Input Output
0 1
1 0
(ii) NAND
V1 V2 Output
0 0 1
0 1 1
1 0 1
1 1 0
(iii) NOR
V1 V2 Output
0 0 1
0 1 0
1 0 0
1 1 0
21
22. Model Graph:
(i) Inverter
Voltage
Input Waveform
5V
time (s)
0 10 20 30 40 50 60 70 80
Output Waveform
5V
time (s)
0 10 20 30 40 50 60 70 80
(ii) NAND
Voltage
Input Waveform
time (s)
0 10 20 30 40 50 60 70 80
Output Waveform
time (s)
0 10 20 30 40 50 60 70 80
time (s)
0 10 20 30 40 50 60 70 80
22
23. (iii) NOR
Voltage
Input Waveform
time (s)
0 10 20 30 40 50 60 70 80
Output Waveform
0 10 20 30 40 50 60 70 80 time (s)
time (s)
0 10 20 30 40 50 60 70 80
Output:
(i) Inverter
Gain = V(2)/Vin =
Input Resistance at Vin =
Output Resistance at V(2) =
(ii) NAND
Gain = V(4)/Vin1 = V(4)/Vin2 =
Input Resistance at Vin1 =
Input Resistance at Vin2 =
Output Resistance at V(4) =
(iii) NOR
Gain = V(4)/Vin1 = V(4)/Vin2 =
Input Resistance at Vin1 =
Input Resistance at Vin2 =
Output Resistance at V(4) =
Result:
Thus the transient characteristics of output voltage for the given CMOS inverter,
NAND and NOR is plotted and the voltage gain, input impedance and output impedance
are calculated.
23
24. Expt. No.7 SECOND ORDER BUTTERWORTH - LOW PASS FILTER
Aim:
To design and implement the second order butterworth Low pass filter using
PSPICE.
Circuit Diagram:
RIN 1K RF 586 ohm
2
0 V+
7
2 -
R1 R2 6 6 VOUT
5 3 3 +
1 LM741
1.59K 1.59K 4
RL
V-
VIN C2 0.1u 10K
C1 0.1u
1V
(100 - 10K)Hz
0
Theory:
A Low pass filter has a constant gain from 0 to fH. Hence the bandwidth of the
filter is fH. The range of frequency from 0 to fH is called pass band. The range of
frequencies beyond fH is completely attenuated and it is called as stop band.
Design:
fH = 1000HZ C1= C2 =0.1F RIN=1000
fH = 1 / 2RC
R = 1 / 2CfH
R = R1 = R2 = 1592
24
25. Gain = 1.586
1.586 = 1 + (RF / RIN)
RF = 586
Model Graph:
Gain (dB)
3dB
Frequency (HZ)
fH
Result:
Thus Low pass filter is designed and implemented using PSPICE.
25
26. Expt. No.8 DIFFERENTIAL AMPLIFIER
Aim:
To implement the differential amplifier using PSPICE.
Circuit Diagram:
RF 10K
V+
RIN 10K 7
VIN 1 2 2 - 6
1 LM741
6
3 + Vout
VIN
2 5 R2 10K 3 4
RCOMP V-
10K
Theory:
A differential amplifier amplifies the difference between two voltages V1 and V2.
The output of the differential amplifier is dependent on the difference between two
signals and the common mode signal since it finds the difference between two inputs it
can be used as a subtractor. The output of differential amplifier is
RF
VO = (V2 – V1)
R1
26
27. Model Graph:
Voltage
V1
time
V2
time
V3
time
Calculation:
V1 = 5V V2 = 10V
RF 10K
VO = (V2 – V1) = (10 – 5)
R1 10K
VO = 5V
Output:
VO = 5V
Result:
Thus a differential amplifier is implemented using operational amplifier.
27
28. Expt. No.9 FREQUENCY RESPONSE OF SINGLE TUNED AMPLIFIER
Aim:
To design and construct a single tuned amplifier and to plot the frequency
response.
Equipment Required:
Range /
Sl. No. Components / Equipment Quantity
Specifications
1 Power supply (0-30)V 1
2 Function generator (0-20M)Hz 1
3 CRO 1
4 Transistor BC107 1
5 Resistors
6 Capacitors
7 DIB
8 DCB
9 Connecting wires
Circuit Diagram:
VCC = +10V
R1 L
C
Cout
C
Cin
B
BC107
E RL
CRO
V=50mV R2
RE
f=(1-3M)Hz CE
28
29. Design:
VCC = 12V; IC = 1mA; fo = ; S = 2; hfe =
Q = 5; L = 1mH
re = 26mV / IC = 26;
hie = hfe re =
VCE= Vcc/2 (transistor Active) =
VE = IERE = Vcc/10
Applying KVL to output loop, we get
VCC = ICRC + VCE + IERE
RC =
Since IB is very small when compare with IC, IC IE
RE = VE / IE =
S = 1+ RB / RE = 2
RB =
VB = VBE + VE =
VB = VCC R2 / (R1 + R2)
RB = R1 || R2
R1 = R2 =
RL =
XCi = {[hie+(1+hfe)RE] || RB}/10 =
Ci = 1 / (2f XCi) =
Xco = (RC||RL) /10 =
Co = 1 / (2f XCo) =
XCE = RE/10 =
CE = 1 / (2f XCE) =
Q = RL / L
RL =
1
f0 =
2 LC
C=
29
30. Procedure:
1. Connect the circuit as per the circuit diagram.
2. Set Vi = 50 mV (say), using the signal generator.
3. Keeping the input voltage constant, vary the frequency from 0Hz to3MHz in
regular steps and note down the corresponding output voltage.
4. Plot the graph: Gain (dB) Vs Frequency
Tabular Column:
Vi = 50 mV
Frequency V0 (volts) Gain (dB) = 20 log(V0/Vi)
Model Graph: (Frequency Response)
GAIN
(dB)
|A|
(dB)
fc FREQUENCY
Result:
Thus single tuned amplifier is designed and constructed for the given
operating frequency and the frequency response is plotted.
30
31. Expt. No.10 ASTABLE MULTIVIBRATOR
Aim:
To design and construct an astable multivibrator using transistor and to plot the
output waveform.
Components / Equipments Required:
Range /
Sl. No. Item name Quantity
Specification
1 Transistor BC107 2
2 Resistors 4.9K, 1.6M 2 each
3 Capacitors 0.45nF 2
4 CRO (0-20M)Hz 1
5 Power Supply (0-30)V 1
6 Connecting Wires Accordingly
Circuit Diagram:
Vcc = +12V
RC R R RC
5.9K 1.5M 1.5M 5.9K
C
Vo1 C
0.48nF Vo2
0.48nF
C
B C
BC107
BC107
B
E
E
0
31
32. Theory:
Astable multivibrator is also known as free running multivibrator. It is
rectangular wave shaping circuit having non-stable states. This circuit does not need an
external trigger to change state. It consists of two similar NPN transistors. They are
capacitor coupled. It has 2 quasi-stable states. It switches between the two states without
any applications of input trigger pulses. Thus it produces a square wave output without
any input trigger. The time period of the output square wave is given by, T = 1.38RC.
Design Procedure:
VCC = 10V; IC = 2mA; VCE (sat) = 0.2V; f = 1KHz; hfe =
VCC - VCE (sat) 12 – 0.2
RC = = = 5.9K
IC 0.002
R hfe RC = 315 * 5.9 * 103 = 1.85M
R = 1.5M
T = 1.38RC
C = T / (1.38R) = (1 * 10-3) / (1.38 * 1.5 * 106)= 0.48nF
Procedure:
1. Connections are made as per the circuit diagram.
2. Switch on the power supply.
3. Note down the output TON, TOFF and output voltage from CRO.
4. Plot the output waveform in the graph.
Tabular Column:
Amplitude TON TOFF Frequency
(in volts) (ms) (ms) (in Hz)
Vo1
Vo2
32
33. Model Graph:
Vo1 (Volts)
Time (ms)
Vo2 (Volts)
Time (ms)
RESULT:
Thus the astable multivibrator is designed and constructed using transistor and its
output waveform is plotted.
33
34. Expt. No.11 MONOSTABLE MULTIVIBRATOR
Aim:
To design and construct monostable multivibrator using transistor and to plot the
output waveform.
Components / Equipments Required:
Sl. No. Item name Range / Specification Quantity
1 Transistor BC107 2
2 Resistors 4.9K, 1.6M 2 each
3 Capacitors 0.45nF 2
4 CRO (0-20M)Hz 1
5 Power Supply (0-30)V 1
6 Connecting Wires Accordingly
Circuit Diagram:
VCC = +12V
RC
5.9K
RC R
5.9K 1.13M 10K R1
Vo1 C
D1
C1
1N4007 1.28nF Vo2
25nF
C
C
B
B BC107
BC107
E
E R2
100K
VBB = -2V
0
34
35. Theory:
Monostable multivibrator has two states which are (i) quasi-stable state and (ii)
stable state. When a trigger input is given to the monostable multivibrator, it switches
between two states. It has resistor coupling with one transistor. The other transistor has
capacitive coupling. The capacitor is used to increase the speed of switching. The
resistor R2 is used to provide negative voltage to the base so that Q1 is OFF and Q2 is
ON. Thus an output square wave is obtained from monostable multivibrator.
Design Procedure:
VCC = 12V; VBB = -2V; IC = 2mA; VCE (sat) = 0.2V; f = 1KHz; hfe =
VCC - VCE (sat) 12 – 0.2
RC = = = 5.9K
IC 0.002
IB2(min) = IC2 / hfe =
Select IB2 > IB2(min)
IB2 =
VCC – VBE (sat)
R= =
I B2
T = 0.69RC
C = T / 0.69R =
-VBBR1 VCE (sat) R2
VB1 = +
R1 + R2 R1 + R2
VBBR1 VCE (sat) R2
= (since, V B1 is very less)
R1 + R2 R 1 + R2
VBBR1 = VCE (sat) R2
R2 =10R1 (since, VBB = 2V and VCE (sat) = 0.2V)
Let R1 = 10K, then R2 = 100K
Choose C1 = 25pF.
Procedure:
1. Connections are made as per the circuit diagram.
2. Switch on the power supply.
35
36. 3. Observe the output at collector terminals.
4. Trigger Monostable with pulse and note down the output TON, TOFF and voltage
from CRO.
5. Plot the waveform in the graph.
Tabular Column:
Width Input Output
(ms) TON TOFF Voltage TON TOFF Voltage
(ms) (ms) (Volts) (ms) (ms) (Volts)
Model Graph:
Vo1 (Volts)
Time (ms)
Vo2 (Volts)
Time (ms)
Result:
Thus the monostable multivibrator is designed and constructed using transistor
and its output waveform is plotted.
36
37. Expt. No.12 BISTABLE MULTIVIBRATOR
Aim:
To design a bistable multivibrator and to plot its output waveform.
Components / Equipments Required:
Sl. No. Item name Range / Specification Quantity
1 Transistor BC107 2
2 Resistors 4.9K, 1.6M 2 each
3 Capacitors 0.45nF 2
4 CRO (0-20M)Hz 1
5 Power Supply (0-30)V 1
6 Connecting Wires Accordingly
Circuit Diagram:
VCC = +10V
RC RC
Pulse Trigger
5.9K (VT < VCC) 5.9K
D1 1N4007 1N4007 D2
C 50pF C 50pF
R1 10K R1 10K
Vo1 Vo2
C
C B
B BC107
BC107
E
E
R2 R2
1.8M 1.8M
VBB=-2V
0
37
38. Theory:
The bistable multivibrator has two stable states. The multivibrator can exist
indefinitely in either of the twostable states. It requires an external trigger pulse to
change from one stable state to another. The circuit remains in one stable state until an
external trigger pulse is applied. The bistable multivibrator is used for the performance
of many digital operations such as counting and storing of binary information. The
multivibrator also finds an applications in generation and pulse type waveform.
Design:
VCC =12V; VBB = -12V; IC = 2mA; VCE (sat) = 0.2V; VBE (sat) = 0.7V
VCC - VCE (sat) 12 – 0.2
RC = = = 5.9K
IC 0.002
R2 hfe RC = 315 * 5.9 * 103 = 1.85M
R2 = 1.8M
Let R1 = 10K, C1 = C2 = 50pF
Procedure:
1. Connections are made as per the circuit diagram.
2. Set the input trigger using trigger pulse generator.
3. Note the output waveform from CRO and plot the graph.
Tabular Column:
Input Width Input Output
Voltage (ms) TON TOFF Voltage TON TOFF Voltage
(Volts) (ms) (ms) (Volts) (ms) (ms) (Volts)
38
39. Model Graph:
Vo1 (Volts)
Time (ms)
Vo2 (Volts)
Time (ms)
Result:
Thus bistable multivibrator has been constructed and its output waveforms are
studied.
39
40. Expt. No. 13 WAVE SHAPING CIRCUITS
(Differentiator, Integrator, Clipper and Clamper)
Aim:
To design and implement different wave shaping circuits (Differentiator,
Integrator, Clipper and Clamper).
Components / Equipments Required:
Range /
Sl. No. Components / Equipment Quantity
Specifications
1 Function / Pulse generator (0 – 3M)Hz 1
2 CRO (0-20M)Hz 1
3 Resistor 1K / 100K 1
4 Capacitor 0.1F 1
5 Connecting wires Accordingly
Circuit Diagram:
(i) Differentiator:
C
0.1uF
Vin=5V R
f= 1KHz 1K /
CRO
100K
0
(ii) Integrator:
R 1K / 100K
Vin=5V
f= 1KHz C
0.1uF CRO
0
40
41. (iii) Clipper:
(a) Series Positive Clipper:
D
1N4007 CRO
Vin=5V R
f=1KHz 10K
0
(b) Shunt Positive Clipper:
R 10K
D
Vin=5V
f=1KHz
1N4007 CRO
0
(c) Series Negative Clipper:
D
1N4007 CRO
Vin=5V R
f=1KHz 10K
0
(d) Shunt Negative Clipper:
R 10K
CRO
D
Vin=5V
f=1KHz
1N4007
0
41
42. (e) Positive Biased Series Positive Clipper:
D
1N4007
R
10K
CRO
Vin=5V
f=1KHz VB
2V
0
(f) Positive Biased Shunt Positive Clipper:
R 10K
1N4007 D
Vin=5V CRO
f=1KHz VB
2V
0
(g) Positive Biased Series Negative Clipper:
D
1N4007
R
10K
CRO
Vin=5V
f=1KHz VB
2V
0
42
43. (h) Positive Biased Shunt Negative Clipper:
R 10K
1N4007 D
Vin=5V CRO
f=1KHz VB
2V
0
(i) Negative Biased Series Positive Clipper:
D
1N4007
R
10K CRO
Vin=5V
f=1KHz VB
2V
0
(j) Negative Biased Shunt Positive Clipper:
R 10K
D
1N4007
Vin=5V CRO
f=1KHz VB
2V
0
43
44. (k) Negative Biased Series Negative Clipper:
D
1N4007
R
10K CRO
Vin=5V
f=1KHz
VB
2V
0
(l) Negative Biased Shunt Negative Clipper:
R 10K
D
1N4007
Vin=5V CRO
f=1KHz VB
2V
0
(m) Combinational Clipper
R 10K
D D
1N4007 1N4007
Vin=5V CRO
f=1KHz VB
VB
2V
2V
0
44
45. (iv) Clamper:
(a) Positive Clamper:
C
0.1uF
R CRO
Vin=5V D
1N4007 10K
f=1KHz
0
(b) Negative Clamper:
C
0.1uF
D R CRO
Vin=5V
1N4007 10K
f=1KHz
0
Theory:
(i) Differentiator:
The high pass RC network acts as a differentiator whose output voltage depends
upon the differential of input voltage. Its output voltage of the differentiator can be
expressed as,
d
Vout = Vin
dt
45
46. (ii) Integrator:
The low pass RC network acts as an integrator whose output voltage depends
upon the integration of input voltage. Its output voltage of the integrator can be
expressed as,
Vout = Vin dt
(iii) Clipper:
This circuit is basically a rectifier circuit, which clips the input waveform
according to the required specification. The diode acts as a clipper. There are several
clippers like positive clipper, negative clipper, etc. Depending upon the connection of
diode it can be classified as series and shunt.
(iv) Clamper:
The clamper circuit is a type of wave shaping circuit in which the DC level of the
input signal is altered. The DC voltage is varied accordingly and it is classified as
positive clamper or negative clamper accordingly.
Design:
(i) Differentiator:
f = 1KHz
= RC = 1ms
If C = 0.1F
Then R = 10K
For T << , Choose R = 1K and
For T >> , Choose R = 100K
(ii) Integrator:
f = 1KHz
= RC = 1ms
If C = 0.1F
Then R = 10K
For T << , Choose R = 1K and
For T >> , Choose R = 100K
Procedure:
1. Connect the circuit as per the circuit diagram.
2. Set Vin = 5V and f = 1KHz.
3. Observe the Output waveform and plot the graph.
46
47. Model Graph:
(i) Differentiator
Vin (Volts)
Time (ms)
Vout (Volts)
Time (ms)
Time (ms)
(ii) Integrator
Vin (Volts)
Time (ms)
Vout (Volts)
Time (ms)
Time (ms)
47
48. (iii) Clipper:
Vin (Volts)
Time (ms)
Vout (Volts)
Series Positive Clipper
Time (ms)
Shunt Positive Clipper
Time (ms)
Series Negative Clipper
Time (ms)
Shunt Negative Clipper
Time (ms)
Positive Biased Series Positive Clipper
2V
Time (ms)
48
49. Positive Biased Shunt Positive Clipper
2V
Time (ms)
Positive Biased Series Negative Clipper
2V
Time (ms)
Positive Biased Shunt Negative Clipper
2V
Time (ms)
Negative Biased Series Positive Clipper
Time (ms)
-2V
Negative Biased Shunt Positive Clipper
Time (ms)
-2V
Negative Biased Series Negative Clipper
Time (ms)
-2V
49
50. Negative Biased Shunt Negative Clipper
Time (ms)
-2V
Combinational Clipper
2V
Time (ms)
-2V
(iv) Clamper:
Positive Clamper:
Time (ms)
Negative Clamper:
Time (ms)
Result:
Thus different wave shaping circuits are studied and their output waveforms are
plotted.
50
51. EXPT NO.14 DIGITAL TO ANALOG CONVERTER
(R – 2R LADDER TYPE)
Aim:
To construct a 8 – bit digital to analog converter using R – 2R ladder type.
Circuit Diagram:
20K
V+
10K 10K 10K 7
4 -
1 2 3 6 VOUT
LM741
5 +
20K
20K 20K 20K 8
V-
0
0
0 0 0
9 -10V
Theory:
A DAC accepts an n – bit input word b1, b2, ……, bn in binary and produces an
analog signal that is proportional to the input. In this type of DAC, reference voltage is
applied to one switch and the other switches are grounded. It is easier to build and
number of bits can be expanded by adding more R – 2R sections. The circuit slow down
due to stray capacitance.
Observation:
Output Voltage
d1 (MSB) d2 d3 (LSB)
VO (Volts)
0 0 0 0
0 0 1 1.25
0 1 0 2.5
0 1 1 3.75
1 0 0 5
1 0 1 6.25
1 1 0 7.5
1 1 1 8.75
51
52. Calculation:
Output Voltage, VO = VR (d12-1 + d22-2 + d32-3 )
For 100, VO = 5V
Output:
VO = 5V
Model Graph:
Voltage
5V
time
Result:
Thus R – 2R ladder type digital to analog converter is implemented.
52
53. EXPT NO.15 ASTABLE MULTIVIBRATOR
Aim:
To plot the transient response of voltages at collector terminals of the two
transistors Q1 and Q2. Initial node voltages at collector and base are zero.
Circuit Diagram:
VCC = +10V
5
R2 R1 R3 R4
4.9K 850K 850K 4.9K
C1 C2
2 4
3 1
Vo1 Vo2
0.9nF 0.9nF
C
C
B Q2
Q1 B BC107
BC107
E
E
0
Theory:
It has two quasi stable states. The transition between the two states occurs
automatically due to charging and discharging of the capacitors and not due to any
external trigger. Thus none of the transistor is allowed to remain in ON or OFF state.
Design:
VCC = 10V; IC = 2mA; VCE = 0.2V; C = 0.9nF
VCC – VCE (sat) 10 – 0.2
RC = = = 4.9K
IC 0.002
R hfe RC = 850K
53
54. T = 1.38 R C
T = 1ms
C = T / (1.38R) = 0.9nF
Model Graph:
Voltage
V01
Time (ms)
0 0.5 1 1.5
V02
Time (ms)
0 0.5 1 1.5
Result:
Thus astable multivibrator is designed and transient response is plotted.
54
55. EXPT NO. 16 MONOSTABLE MULTIVIBRATOR
Aim:
To plot the transient response of voltages at collector terminals of Q1 and Q2.
Initial voltages of base and collector of Q1 transistor is zero.
Circuit Diagram:
VCC = +12V
5
R4
5.9K
R2 R3
5.9K 452K 10K R5
Vo1 C1
D1
7 2 C2
3
1 4
1N4007 3.2nF Vo2
25nF
C
C B Q2
Q1 B BC107
BC107
E
E R1
100K
6
VBB = -2V
0
Theory:
Monostable multivibrator has two states which are (i) quasi-stable state and (ii)
stable state. When a trigger input is given to the monostable multivibrator, it switches
between two states. It has resistor coupling with one transistor. The other transistor has
capacitive coupling. The capacitor is used to increase the speed of switching. The
resistor R2 is used to provide negative voltage to the base so that Q1 is OFF and Q2 is
ON. Thus an output square wave is obtained from monostable multivibrator.
55
56. Model Graph:
Vo1 (Volts)
Time (ms)
Vo2 (Volts)
Time (ms)
Result:
Thus monostable multivibrator is designed and transient response is plotted.
56
57. 13. WEIN BRIDGE OSCILLATOR
AIM: To study and calculate frequency of. Wein Bridge Oscillator.
APPARATUS:
Transistor (BC 107) 2 No
Resistors 10K 4 No
1 K 3 No
2.2 K 2 No
33 K
6.8 K
Capacitors 10 F 2 No
100 F
0.01 F 2 No
RPS (0 – 30 V)
Potentiometer
Bread Boar
CRO
Connecting wires
CIRCUITDIAGRAM:
THEORY:
57
58. The wein bridge oscillator is a standard circuit for generating low
frequencies in the range of 10 Hz to about 1MHz.The method used for getting
+ve feedback in wein bridge oscillator is to use two stages of an RC-coupled
amplifier. Since one stage of the RC-coupled amplifier introduces a phase shift
of 180 deg, two stages will introduces a phase shift of 360 deg. At the
frequency of oscillations f the +ve feedback network shown in fig makes the
input & output in the phase. The frequency of oscillations is given as
f =1/2π√R1C1R2C2
In addition to the positive feedback
PROCEDURE:
1. Connections are made as per the circuit diagram
2. Feed the output of the oscillator to a C.R.O by making adjustments in the
Potentiometer connected in the +ve feedback loop, try to obtain a stable
sine Wave.
3. Measure the time period of the waveform obtained on CRO. & calculate the
Frequency of oscillations.
4. Repeat the procedure for different values of capacitance.
OBSERVATION:
Given R=10kΩ, C=0.01μF
fT = 1/ 2RC
1
fP = =
T
Amplitude,V0 =
MODEL WAVE FORM:
58
59. RESULT:
The frequency of the wein bridge oscillator is calculated and is verified
VIVA QUESTIONS:
1. Give the formula for frequency of oscillations?
2. What is the condition for wien bridge oscillator to generate oscillations?
3. What is the total phase shift provided by the oscillator?
4. What is the function of lead-lag network in Wein bridge oscillator?
5. which type of feedback is used in Wein bridge oscillator
6. What is gain of Wein bridge oscillator?
7. what are the application of Wein bridge oscillator
8. What is the condition for oscillations?
9. What is the difference between damped oscillations undamped
Oscillations?
Wein bridge oscillator is either LC or RC oscillator.
59