Analog Electronics Circuits Lab Dept.

of ECE

N2 Multisim Simulation software

Multisim, a circuit simulation and design tool developed by National Instruments, is widely recognized
for its powerful features that cater to electrical and electronics engineers. Its versatility makes it ideal
for teaching, designing, and analyzing electrical circuits. Within this ecosystem, N2 Multisim emerges
as a specialized iteration that retains Multisim's core functionalities while focusing on simplified user
experiences for specific applications like education or prototyping.

Key Features of N2 Multisim

1. User-Friendly Interface:
N2 Multisim's intuitive graphical interface enables users to design and simulate circuits without
requiring extensive technical expertise. The drag-and-drop functionality simplifies the process
of placing and connecting components.
2. Comprehensive Component Library:
It offers a robust library of pre-built components, including resistors, capacitors, diodes,
transistors, operational amplifiers, and integrated circuits. This variety ensures versatility in
creating circuits ranging from basic to complex.
3. Interactive Simulation:
N2 Multisim integrates real-time simulation tools, allowing users to observe circuit behavior
dynamically. Features like voltage, current, and power analysis enhance learning and debugging
capabilities.
4. Virtual Instruments:
The software includes virtual versions of standard laboratory instruments like oscilloscopes,
function generators, digital multimeters, and power supplies. These tools allow users to test and
analyze circuits as if they were using physical hardware.
5. SPICE-Based Simulation:
N2 Multisim utilizes SPICE (Simulation Program with Integrated Circuit Emphasis) as its
simulation engine. SPICE ensures accurate analysis of analog, digital, and mixed-signal circuits.
6. Schematic Capture:
Users can create detailed schematics, which serve as blueprints for circuit design. These
schematics can be exported for PCB layout creation in associated software like Ultiboard.

Applications of N2 Multisim

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 Analog Electronics Circuits Lab Dept. of ECE
 Education: N2 Multisim is widely used in academic institutions to teach basic and advanced
electrical engineering concepts. Its interactive simulations make it easier for students to
understand circuit behavior without requiring physical components.
 Prototyping: Engineers use it to design and test circuits virtually before committing to physical
prototyping, saving time and resources.
 Research and Development: The software's extensive analysis tools help in studying circuit
performance under various conditions, aiding innovation and optimization.

Tools in N2 Multisim

1. Analysis Tools:
N2 Multisim provides AC, DC, transient, and Fourier analysis, enabling detailed insights into
circuit performance.
2. Probing Tool:
Users can place virtual probes at various points in the circuit to measure parameters like voltage,
current, and resistance in real time.
3. Fault Simulation:
This tool allows users to introduce faults like open or short circuits deliberately. It is particularly
beneficial in troubleshooting and educational settings.
4. Measurement and Graphing:
The integrated measurement tools generate plots and waveforms for analyzing signals, such as
frequency response, signal distortion, and noise.
5. Interactive Breadboard:
The virtual breadboard feature replicates a physical breadboard, helping users to visualize and
simulate circuit assembly.

Below is an illustration of the N2 Multisim interface

1. Toolbar: Contains essential commands like open, save, undo, and redo.
2. Component Library: A searchable library where users can find components to drag and drop
onto the schematic workspace.
3. Workspace: The central area where circuits are designed and connections are made.
4. Instruments Panel: Hosts virtual instruments like oscilloscopes and function generators.
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 Analog Electronics Circuits Lab Dept. of ECE
5. Simulation Controls: Start, pause, or stop simulations using this control panel.

Benefits of Using N2 Multisim

 Cost-Effective Learning: The software eliminates the need for physical components in the
initial stages of learning and design.
 Efficient Prototyping: Virtual simulations reduce errors and iterations in the design process.
 Real-World Problem Solving: With accurate SPICE models and real-time analysis, the
software mirrors real-world circuit behavior.

Conclusion

N2 Multisim is a vital tool for anyone involved in electronics design and education. Its comprehensive
suite of features ensures accurate simulation, detailed analysis, and a smooth design experience.
Whether for a student learning Ohm's law or an engineer designing cutting-edge electronics, N2
Multisim proves to be an invaluable resource.

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 Analog Electronics Circuits Lab Dept. of ECE

Analog Electronics Circuits Laboratory

PART-B (Simulation)
Experiment No. 7

Passive RC Low-Pass Filter and Passive RL High-Pass Filter

AIM:

To study the characteristics of passive filters by obtaining the frequency response of low pass RC filter and high
pass RL filter.

Theory:

The impedance of an inductor is proportional to frequency and the impedance of a capacitor is inversely
proportional to frequency. These characteristics can be used to select or reject certain frequencies of an input
signal. This selection and rejection of frequencies is called filtering, and a circuit which does this is called a filter.

If a filter passes high frequencies and rejects low frequencies, then it is a high-pass filter. Conversely, if it passes
low frequencies and rejects high ones, it is a low-pass filter. Filters, like most things, aren't perfect. They don't
absolutely pass some frequencies and absolutely reject others. A frequency is considered passed if its magnitude
(voltage amplitude) is within 70% or 1/sqrt(2) of the maximum amplitude passed and rejected otherwise. The
70% frequency is called corner frequency, roll-off frequency or half-power frequency.

Design:
The corner frequencies for RC filter and RL filter are as follows:

For RC filters (Passive Low-Pass Filter):

Let R = 1 KΩ and C = 1 μF

Therefore, fc = 159 Hz.

For RL filters (Passive High-Pass Filter):

Therefore, fc = 7.95 KHz.

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 Analog Electronics Circuits Lab Dept. of ECE
Circuit Diagram

Figure 1.1: Passive Low Pass RC filter.

Tabular column:
Vin=1V (P) Constant

Frequency (Hz) Vo (p) Av=Vo(p)/Vin(p) [Av]dB=20Log (Av)

20
40
60
80
100
120
140
160
180
200
400
600
800
1k
1.2k
1.4k
1.6k
1.8k
2k

PROCEDURE:

1. Make the connections as shown in the figure 1.

2. Apply the sinusoidal input signal of amplitude 1Vp or 2Vp-p and a frequency of 20Hz
and measure the output signal voltage Vo(p-p)
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 Analog Electronics Circuits Lab Dept. of ECE
3. keeping the input signal amplitude constant throughout the frequency response,
increase the input signal frequency in steps of 20Hz till 200Hz and note down the
corresponding output Vo(p)
4. Increase the input signal frequency in steps of 100Hz till 2KHz and note down the
corresponding output signal voltage Vo(p)
4. Plot the frequency response on a semi log graph sheet with a voltage gain (Av) in dB on Y axis and
the frequency (f) on X axis as shown in the figure.
5. In a graph, locate -3dB point from (Av) max in dB and then identify the practical cut off frequency
(fc)
6. Identify the practical values of pass-band gain and bandwidth from the graph (frequency response)
from the same graph, measure the roll-of-factor (ROF) by using following relationship:
Roll-of-factor(ROF)=(Av in dB at 10fc)-(Av in dB at fc)
7. Observe the frequency response of Low-Pass filter in multisim tool (Simulate-------->
Analyses and Simulation ----> AC Sweep -----> Frquency Parameters (Change the parameters as
required here) ----> Output---> select the output variable in the list ---->Add -----> Save ----> Run).
Expected Output at corner or cut-off frequency of 159 Hz.

Fig 1.2 Response of Low pass RC filter at 159Hz

Expected frequency response of Low-Pass Filter

Fig 1.3 Frequency response of low pass RC filter

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 Analog Electronics Circuits Lab Dept. of ECE

Circuit Diagram

Figure 1.4: High Pass RL filter.

Tabular column:
Vin=1V (P) Constant

Frequency (Hz) Vo (p) Av=Vo(p)/Vin(p) [Av]Db=20Log (Av)

500
1k
2k
3k
4k
5k
6k
7k
8k
9k
10k
20k
30k
40k
50k
60k
70k
80k
90k
100k

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 Analog Electronics Circuits Lab Dept. of ECE
PROCEDURE:

1. Make the connections as shown in the figure 2.

2. Apply the sinusoidal input signal of amplitude 1V(p) and a frequency of 500Hz and
measure the output signal voltage Vo(p) keeping the input signal amplitude constant throughout the
frequency response, increase the input signal frequency in steps of 500Hz till 1kHz and note down the
corresponding output Vo (p).
3. Increase the signal frequency in steps of 1KHz till 10KHz and note down the corresponding output
signal voltage(p). Again increase the signal frequency in steps of 10KHz till 100KHz and note down the
corresponding output signal voltage(p).
4. Plot the frequency response on a semi log graph sheet with a voltage gain (Av) in dB on
Y axis and the frequency (f) on X axis as shown in the figure.
5. In a graph, locate -3dB point from (Av) max in dB and then identify the practical cut off frequency (fc)
6. Identify the practical values of pass-band gain and bandwidth from the graph frequency response).
From the same graph, measure the roll of factor (rof) by using following relationship
Roll of factor=(Av in dB at fc)-(Av in dB at fc/10)
7 Observe the frequency response of Low-Pass filter in multisim tool (Simulate-------->
Analyses and Simulation ----> AC Sweep -----> Frquency Parameters (Change the parameters as
required here) ----> Output---> select the output variable in the list ---->Add -----> Save ----> Run).

Expected Output at corner or cut-off frequency of 7.95KHz.

Fig1.5 Response of high pass RL filter at 7.95KHz

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 Analog Electronics Circuits Lab Dept. of ECE
Expected frequency response of High-Pass Filter

Fig 1.6 Frequency response of high pass RL filter

Result

Measured cutoff frequency of the low-pass filter____

Measured cutoff frequency of the high-pass filter____

Viva voce Questions

1. What is a low-pass filter, and what is its purpose?
2. How does an RC low-pass filter work?
3. What is the formula for the cutoff frequency of an RC low-pass filter?
4. Define the terms "cutoff frequency" and "time constant" in an RC circuit.
5. Why is the resistor and capacitor configuration important in a low-pass filter?

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 Analog Electronics Circuits Lab Dept. of ECE

Experiment No. 8
Half wave rectifier and full wave rectifier using bridge rectifier
AIM: Design and simulate the half wave rectifier circuit to verify the ripple factor in both with and
without filter

Theory: A half-wave rectifier is a circuit that converts alternating current (AC) into direct current (DC)
by allowing only one half-cycle of the AC voltage waveform to pass through. The other half is blocked.
This process results in a pulsating DC output with a frequency equal to the input AC frequency.
A full-wave rectifier converts the entire AC waveform (both positive and negative half-cycles) into a
pulsating DC output. The bridge rectifier is a popular configuration for a full-wave rectifier. It uses four
diodes arranged in a bridge configuration to rectify both halves of the input AC signal.
Circuit diagram

Fig 2.1 Circuit diagram of Half wave rectifier without filter

Design:
Assumptions,
Let Id=10mA,Vd=0.7V,Vs=12V,Vp=120V
Transformer ratio design
WKT Vs/Vp=N2/N1
12/120=N2/N1, therefore N2/N1=1/10
Where Vs=secondary voltage,Vp=primary voltage,N1=Number of turns at primary end of
transformer,N2=Number of turns at secondary end of transformer.℮
Design of RL
Using the KVL for the circuit shown in the figure 1
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 Analog Electronics Circuits Lab Dept. of ECE
Vs-Vd-IdRL=0
RL=(Vs-Vd)/RD=(12-0.7)/10*10-3=1.3kΩ
Expected waveform

Fig 2.2 Waveform of Half wave rectifier without filter

Procedure
a) Without filter
i) Connect the circuit as shown in the fig 1 in multisim software application.
ii) Run and measure the output ‘Vm’ on CRO.
iii) Calculate the value and fill the tabular column.
iv) Compare practical value of ripple factor with the theoretical value.

ϒ= 2
-1
Vm in volts Vdc=Vm/π Vrms=Vm/2

T P

Table 1

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 Analog Electronics Circuits Lab Dept. of ECE
ii) with filter

Circuit diagram

Fig 2.3 Circuit diagram of a Half wave rectifier with filter

Design
Design of capacitor C

WKT, ϒ=1/2 fCRL

Let f=50Hz, RL=1.3KΩ, ϒ=0.06
Therefore, C=74µF (By substituting)
Expected waveform

Fig 2.4 Waveform of Half wave rectifier with filter

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 Analog Electronics Circuits Lab Dept. of ECE
Tabular column

Vm Vrpp Vdc=Vm-Vrpp/2 Vrms=Vrpp/2 ϒ=Vrms/Vdc

T P

Table 2
Procedure
b) With filter
i) Connect the circuit as shown in the fig 2 in Multisim.
ii) Run and measure Vm and Vrpp values from CRO, notedown the same in the tabular
column.
iii) Compare practical value of ripple factor with the theoretical value.

Bridge rectifier
Circuit diagram

Fig 2.5 Circuit diagram of bridge rectifier without filter

i) Without filter
Design of RL

Let Vs secondary voltage be 10V

Choose Vp, primary voltage=230V,f=50 Hz
Therefore N1(No of turns at primary)=230, N2(No of turns at secondary)=1
Vs=(N2/N1)Vp
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 Analog Electronics Circuits Lab Dept. of ECE
Therefore Vs=10V .Choose the transformer ratio as 23:1

WKT , RL=(Vs-Vd1-Vd2)/ID
Let ID=5mA
Therefore RL=1.72kΩ

Expected waveform

Fig 2.6 Waveform of bridge rectifier without filter

Tabular column
ϒ=
Vm Vdc=2Vm/π Vrms=Vm/ 2-1 η=Pdc/Pac=(Vdc2/Vrms2)*100

T P
T P

Table 1

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 Analog Electronics Circuits Lab Dept. of ECE

ii) With filter

Circuit diagram

Fig 2.7 Circuit diagram of Bridge wave rectifier with filter

Design
Design of RL
WKT , RL=(Vs-Vd1-Vd2)/ID
Let ID=5mA
Therefore RL=1.72kΩ

Design of C
Let ϒ=3%=0.03(ripple factor)

WKT, ϒ=1/4 fCRL

Therefore ϒ=54.94µF 56µF

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 Analog Electronics Circuits Lab Dept. of ECE
Expected waveform

Fig 2.8 Waveform of Bridge wave rectifier with filter

Tabular column
ϒ=Vrms/Vdc
Vm Vrpp Vdc=Vm- Vrms=Vrpp/2
Vrpp/2
T P

Table 2

Result:
Half-Wave Rectifier:The half-wave rectifier allowed only one half-cycle of the AC input to pass
through, blocking the other half.
Full-Wave Rectifier (Bridge Rectifier):The full-wave rectifier using the bridge configuration allowed
both half-cycles of the AC input to pass through, converting both into a pulsating DC output.
In conclusion, the full-wave rectifier provides higher efficiency, smoother output, and better
performance, making it more suitable for practical applications than the half-wave rectifier.

Viva voce Questions

1. How does a diode conduct in a half-wave rectifier circuit?
2. What happens to the output during the negative half-cycle of the input AC signal in a half-wave
rectifier?
3. What is the role of the diode in a half-wave rectifier?
4. Why is a filter capacitor used in conjunction with a half-wave rectifier?
5. How does the shape of the output waveform differ from the input signal in a half-wave rectifier?

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 Analog Electronics Circuits Lab Dept. of ECE

Experiment No. 9
A. Precision Half Wave Rectifier
Aim: To design and simulate the Precision half wave rectifier for the specified output voltage.

Components required:

Sl.No. Name of the components Description

1 Resistors 10kΩ
2 1kΩ
3 OP-AMP 741 IC
4 AC Input source 0.2Vpk, f=1KHz
5 DC Power Supply Vcc=15V , -Vee=-
15V
6 Diode IN4007G
Circuit Digram:
R2

D1 10kΩ XSC1

1N4007G Ext Trig

VEE +
_
-15.0V A B
4

+ _ + _
R1 U1
2 D2
1kΩ 6
3 1N4007G
7
5
1

741
R3
V1 1kΩ
0.2Vpk VCC
1kHz
15.0V
0°

Figure 9.1 Circuit diagram of Precision half wave rectifier

Theory

A precision half-wave rectifier, often called an active rectifier, is an electronic circuit designed to
rectify small amplitude AC signals with high accuracy. Traditional rectifiers using diodes face
limitations due to the forward voltage drop (approximately 0.7V for silicon diodes and 0.3V for
Schottky diodes). These voltage drops become significant when working with low-level signals, leading
to reduced efficiency and accuracy. Precision rectifiers address this limitation by using an operational
amplifier (op-amp) in conjunction with diodes to minimize voltage losses and enable rectification of
signals as small as a few millivolts.

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 Analog Electronics Circuits Lab Dept. of ECE

Design:
Consider Vi=0.2Vpk and V0=2V
During positive half cycle of the input diode D1 Will conduct whereas the diode D2 will not conduct.
Therefore, Vo=0V
During Negative half cycle of the input diode D1 Will not conduct whereas the diode D2 conduct.
Therefore, Vo is given by

Vo= )(-Vi)
To Obtain, Vo=2V, choose R2=10KΩ and R1=1KΩ

Therefore Vo=
Vo=2V
R3=10KΩ||1KΩ=909Ω
Choose R3=1KΩ
Procedure:
1. Simulate the circuit as shown in figure 1.
2. Observe the half wave rectified output in the CRO.
3. Measure and note down the Vo of the half wave rectified output.
4. Compare the practical Vo with the theoretical Vo.
Expected waveform:

Figure 9.2 Input and Output waveforms of precision half wave rectifier

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 Analog Electronics Circuits Lab Dept. of ECE
Result:

1. Theoretical Vo of half wave rectified output = _____ V

2. Practical Vo of half wave rectified output = _____ V

The experiment is simulated and verified the half wave rectified output successfully as per the design.

B. Precision Full Wave Rectifier

Aim: To design and simulate the precision full wave rectifier for the specified output voltage:

Components required:

Sl.No. Name of the components Description Quantity

1 Resistors 1k Ω 5

2 10k Ω 1

3 0.5k Ω 1

4 OP-AMP 741 IC 2

5 Diode 1N4007G 2

6 AC Input Source 0.2V pk , f=1KHz 1

7 DC Power supply Vcc=+15V , Vee=-15V 2 each

8 CRO ---- 1

Circuit Diagram:
R5 R6

1kΩ R2 10kΩ

D1 1kΩ XSC1

VEE
1N4007G VEE Ext Trig
+
-15.0V
-15.0V _
A B
4

_ _
4

+ +
R1
2 D2 R4
1kΩ A1 6
2

0.5kΩ A2 6
3 1N4007G
3
V1 R3
7
5
1

741
7
5
1

0.2Vpk 1kΩ 741

1kHz VCC R7 VCC
0°
1kΩ
15.0V 15.0V

Figure 9.3 Ciruit daigram of precision full wave rectifier

Design:

Let Vi=0.2Vpk and V0=2V

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 Analog Electronics Circuits Lab Dept. of ECE
During positive half cycle of the input diode D1 will not conduct whereas the diode D2 conducts.
Therefore the voltage at positive terminal of diode D2 (Let, VB), is given by

VB=
Where R2=1KΩ=R1

( )is called as the gain (closed loop of op-amp A1

VB=( ) (O.2)
VB=-0.2V
At the same time the voltage at the input end of resistor R5 (Let, VA) is given by
VA=0.2V
The voltages VA and VB are the two inputs for the inverting summing circuit. Therefore, the output of
summing circuit is given by

Vo=( + --------------(2)

Choose,

Where. Closed loop gain of an op-amp A2 when the input ‘VB’ is considered

Where. Closed loop gain of an op-amp A2 when the input ‘VA’ is considered

Let us assume, 10. To achieve this gain, assume R6=10KΩ, and R5=1KΩ.

Therefore, 20. To achieve this gain, assume R4=0.5KΩ.

Substituting these values in equation ------ (2)

Vo= [(-20)(-0.2)]+[(-10)(0.2)]
Vo =4-2
Vo =2. This is the output obtained during positive half cycle of the input.
Similarly, during -ve half cycle of the input diode D1 will conduct whereas the diode D2will not
conduct. Therefore the voltage at the positive terminal of diode D2 (Let VB) is zero
i,e VB=0V
At the same time, the voltage the input end of resistor R5 (Let, VA) is -0.2V
i,e VA= - 0.2V
Therefore the output of inverting summing circuit is given by

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 Analog Electronics Circuits Lab Dept. of ECE

Vo=( VB+ VA

Vo = (20)(0) + (-10)(-0.2)
Vo=0+2
Vo=2V------ This is the output voltage obtained during the –ve half cycle of the input.

Procedure:
1. Simulate the circuit as shown in figure 1.
2. Observe the full wave rectified output in the CRO.
3. Measure and note down the Vo of the full wave rectified output.
4. Compare the practical Vo with the theoretical Vo.

Expected Waveform:

Figure 9.4 Input and Output waveforms of precision full wave rectifier
Result:

1. Theoretical Vo of full wave rectified output = _____ V

2. Practical Vo of full wave rectified output = _____ V

The experiment is simulated and verified the full wave rectified output successfully as per the design.

Viva voce Questions

1. What is the principle behind a precision half-wave rectifier?

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 Analog Electronics Circuits Lab Dept. of ECE
2. How does a precision half-wave rectifier differ from a standard diode half-wave rectifier?

3. Why is an op-amp used in the precision rectifier circuit?

4. What is the function of the diode in the precision rectifier circuit?

5. Why do we need to choose a specific op-amp for a precision half-wave rectifier

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 Analog Electronics Circuits Lab Dept. of ECE

Experiment No. 10
Schmitt trigger
Aim: To convert an analog signal into a digital signal by introducing hysteresis using scmitt trigger,
thereby ensuring a stable and noise-immune output.
Components required:

Sl.No. Name of the components Description

1 Resistors 5kΩ
2 1kΩ
3 OP-AMP 741 IC
4 AC Input source 5Vpk, f=100Hz
5 DC Power Supply Vcc=13V , Vee=-13V

Theory
A Schmitt trigger is an electronic circuit that converts an analog input signal into a digital output
signal. It is a type of comparator circuit with hysteresis—a property that introduces two distinct
threshold voltages (upper threshold VUTV_{UT}VUT and lower threshold VLTV_{LT}VLT). The
primary function of a Schmitt trigger is to provide noise immunity and produce a clean square wave
output from a noisy or slowly varying input signal.
Design:
Case 1: When UTP is +ve and LTP is -ve
Let UTP= VUT=2V and VLT=-2V=LTP

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 Analog Electronics Circuits Lab Dept. of ECE
Circuit Diagram:

Figure 10.1: Circuit diagram of Schmitt trigger

Let Vcc=13V, Vsat=Vcc-1=12V,
-Vsat=Vcc+1=-12V
When Vin is -ve, Vo=Vsat
From the Fig 1,UTP=VUT=(R1/(R1+R2))Vsat
1/6=R1/ (R1+R2),R2=5R1,If R1=1KΩ
R2=5KΩ
When Vin =+ve and Vo=-Vsat
LTP=VLT= (R1/(R1+R2))(-Vsat)
1/6=R1/ (R1+R2)
Therefore R2=5R1, If R1=1kΩ
R2=5kΩ

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 Analog Electronics Circuits Lab Dept. of ECE
Waveforms

Fig 10.2 Waveform for UTP is +ve and LTP is -ve

Case 2: When UTP=+4V and LTP=-2V

Circuit diagram

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 Analog Electronics Circuits Lab Dept. of ECE

Fig 10.3 Circuit of Schmitt trigger

Design
WKT, UTP=VUT=(R1/(R1+R2))Vsat
4=(R1/R1+R2)*12 , R1=2R2
Therefore if R2=1KΩ
R1=2KΩ
LTP=VLT=(R3/R2+R3)(-Vsat)
-2=(R3/R2+R3)(-12) , R3=5R2
Therefore IF R2=1KΩ
R3=5KΩ

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 Analog Electronics Circuits Lab Dept. of ECE

Waveform

Fig 10.4 Waveform for UTP=+4V and LTP=-2V

Case3:When UTP=+ve and LTP =+ve
Let UTP=4V and LTP=2V
Circuit diagram

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 Analog Electronics Circuits Lab Dept. of ECE

Fig 10.5 Circuit diagram of schmitt trigger

Design
From fig2, WKT
UTP= (R1/(R1+R2))Vref + (R2/(R1+R2))Vsat -------1
LTP= (R1/(R1+R2))Vref - (R2/(R1+R2))Vsat -------2
Adding Equations 1 and 2
UTP+LTP=2R1/(R1+R2)Vref -------3
Subtracting equations 1 and 2
UTP-LTP= (2R2/(R1+R2))Vsat ---------4
4-2= (2R2/(R1+R2))*11
1/11=R2/R1+R2
R1+R2=11R2
Let R2=1KΩ, Therefore R1=1KΩ
From eq 3,
4+2=(2(10K)/(10K+1K))Vref
Vref=3.3V

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 Analog Electronics Circuits Lab Dept. of ECE
Waveforms

Fig 10.6 Waveform of UTP=+ve and LTP =+ve

Case 4:When UTP=-ve and LTP=-ve

Let UTP=-2V and LTP=-4V
Circuit diagram

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 Analog Electronics Circuits Lab Dept. of ECE

Fig 10.7 Circuit diagram of Schmitt trigger

Design
From the fig3
UTP= (R1/(R1+R2))Vref + (R2/(R1+R2))Vsat --------1
LTP= (R1/(R1+R2))Vref - (R2/(R1+R2))Vsat --------2
Adding 1 and 2
UTP+LTP=(2R1/(R1+R2))Vref ------3
Subtracting 1 and 2
UTP-LTP=(2R2/(R1+R2))Vsat -------4
-2-(-4)=(2R2/(R1+R2))*11
1/11=R2/(R1+R2)
R1=11R2-R2
R1=10R2
If R2=1KΩ, Then R1=1KΩ
From the equation 3,
-2-4= (2R1/(R1+R2))Vref
-6= (2*10K/(10K+1K))Vref
Vref=-3.3V
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Waveform

Fig 10.8 Waveform of UTP=-ve and LTP=-ve

Procedure
1. Connect the power supply to the op-amp .
2. Connect the feedback resistor between the output of the op-amp and the non-inverting input.
3. Connect the resistor (R2) between the non-inverting input and ground. This establishes the
threshold voltages for switching.
4. Apply a voltage of 5V to the non-inverting input of the op-amp.
5. Connect the oscilloscope to the output and observe the waveform
Result
The output switches between high and low levels crossing the threshold levels.
Viva voce Questions
1. What is a Schmitt trigger, and why is it used in circuits?
2. What is meant by "hysteresis" in a Schmitt trigger?
3. How does a Schmitt trigger work in terms of threshold voltages?
4. How is the hysteresis effect created in a Schmitt trigger circuit?
5. What components are typically used to build a Schmitt trigger?

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Comparators
Aim: To design and study the working of a comparator circuit using an operational amplifier (op-
amp) or a dedicated comparator IC, and to observe how the circuit compares two input voltages and
produces a corresponding digital output.

Theory:
A comparator is an electronic circuit that compares two input voltages and produces a digital output
indicating which input is higher. Comparators are widely used in applications that require decision-
making based on voltage levels, such as signal processing, analog-to-digital conversion, and threshold
detection.

i) Non inverting comparator with zero reference voltage ,Vref=0V

Circuit diagram

Fig 10.9 Circuit diagram of a non inverting comparator with Vref=0V

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 Analog Electronics Circuits Lab Dept. of ECE
Waveform

Fig 10.10 Waveform of non inverting comparator with Vref=0V

Procedure
1. The op-amp is configured as a comparator.
2. The non-inverting terminal (+) is connected to the input signal .
3. The inverting terminal (−) is connected to ground (0V reference voltage).
4. The op-amp output is connected to an oscilloscope to observe the result.

Result
1. When the input voltage at the non-inverting terminal is greater than 0V (positive), the output of
the comparator will swing to the positive supply voltage (+Vcc), indicating a high state.
2. When the input voltage is less than 0V (negative), the output will swing to the negative supply
voltage (−Vcc), indicating a low state.

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 Analog Electronics Circuits Lab Dept. of ECE

ii) Non inverting comparater with reference voltage not equal to zero
Circuit diagram

Fig 10.11 Circuit diagram of a Non inverting comparater with Vref≠0

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 Analog Electronics Circuits Lab Dept. of ECE
Waveform

Fig 10.12 Waveform of Non inverting comparater with Vref≠0

Procedure
1. The non-inverting terminal (+) of the op-amp is connected to the input signal.
2. The inverting terminal (−) is connected to a reference voltage.
3. The output of the op-amp is connected to an oscilloscope.

Result
1. When the input voltage at the non-inverting terminal is greater than the reference voltage (Vref),
the output of the comparator will switch to the positive supply voltage (+Vcc), indicating a high
state.
2. When the input voltage is less than the reference voltage (Vref), the output will switch to the
negative supply voltage (−Vcc), indicating a low state.

Viva Questions

1. What is a comparator, and how does it differ from an operational amplifier (op-amp)?
2. What are the typical uses of a comparator circuit?
3. What is the basic functionality of a comparator circuit?
4. What are the main components used in a comparator circuit?
5. Why do comparators typically not have feedback resistors like op-amps?

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 Analog Electronics Circuits Lab Dept. of ECE

EXPERIMENT 11
Design and simulation of a common source amplifier using JFET

Aim: To design and analyze a high-performance common-source amplifier using an EPC

(EPC stands for Efficient Power Conversion) transistor to achieve high gain, high-speed
operation, and efficiency for RF, analog, or mixed-signal applications.

Theory

An Eficcient Power conversion Common Source (EPC CS) amplifier is a configuration that
combines two well-known amplifier stages—emitter follower (common collector) and
common source amplifier—to leverage the advantages of each while mitigating their
limitations. Here's a detailed theory of the EPC CS amplifier.

Circuit diagram

Fig 11.1 Circuit of common source amplifier using JFET

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 Analog Electronics Circuits Lab Dept. of ECE

Waveform

Fig 11.2 Waveform response of common source amplifier.

Design

Assume VDD=12V ,VGS=-1V,ID=2Ma

Let VDS=VDD/2=12/2=6V

Let Rs=10% of VDD=(10/100)12=1.2V

Let VRD=40% OF VDD=(40/100)12=4.8V

Therefore RD=VRD/ID=4.8/2mA=2.4kΩ

Rs=VRS/ID=1.2/2mA=600Ω

WKT VG-VGS-VRS=0

VG=VGS+VRS=-1+1.2=0.2V

Using voltage divider rule

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 Analog Electronics Circuits Lab Dept. of ECE
VR2=(VDD*R2)/(R1+R2)

Assume R1=1MΩ foe minimum gate current

Therefore VR2=(12R2)/(1MΩ+R2)

0.2=(12R2)/(1MΩ+R2)

R2=0.2MΩ/11.8=16.94kΩ

Design of C1,C2 and Cs

Let Xc1=R1||R2/10=(1MΩ*16.94kΩ)/10=1.66kΩ

Therfore Xc1=1/2πfc1,c1=1/2πfXc1=1/2π*2.*1.66kΩ=4.79µf

Let Xc2=RL/10=4.7k/10=470Ω

Also Xc2=1/2πfc2c2=1/2πfXc2=1/2π*20*470=16.93µf

Let Xcs=Rs/10=600/10=60Ω at 20Hz

XCs=1/2πfCsCs=1/2πfXCs, Cs=1/2π*20*60=132.62µf

Fig11.3 Frequency response output at mid band frequency

Therefore,Midband voltage gain at 20kHz,Av=Vo/Vi=0.26/50MaAv=5

In decibels,[Av]dbmax=20logAv=20log5=13.97⁓=14db

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Procedure

1. Place the JFET model in the circuit workspace.

2. Connect the components as per the typical CS amplifier schematic:
3. Drain to RD, which connects to the DC power supply
4. Source to Rs, optionally bypassed by CSC_SCS, and then to ground.
5. Gate to ground through Rg.
6. Connect an input signal source (AC source) to the gate through Cin.
7. Connect a load resistor RL at the output, with Cout coupling the amplifier output to it.

Use the following procedure to get the frequency response.

1. Click on simulate .

2.Select Frequency parameters

Start freq:1Hz

Stop freq:300GHz

Sweep type:decade

No of points per decade:10

Vertical scale:decade

3.Go to the circuit and click on output line and get the net name.

Therefore gain-bandwidth product of an amplifier or figure of merit[FOM]=Midband

gain*Bandwidth=14dB*100GHz=1.4*1012dBHz

Result

1.Drain voltage=____

2. Gate voltage=____

3.Source voltage=____

Viva Questions

1. What is the role of a JFET in a common source amplifier?

2. Why is the configuration called a "common source" amplifier?
3. Explain the function of RSRD, and RG in the circuit.
4. What is the purpose of coupling capacitors Cin and Cout?
5. What is the role of the bypass capacitor CS across RS?

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