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EXPERIMENT 1
ACTIVE SECOND ORDER FILTERS:
LOW PASS FILTER AND HIGH PASS FILTER
Aim: a)Todesign a Second order Butterworth Low pass filter for a given cut off frequency,
fH= 1KHz .Draw the frequency response.
b) Todesign a Second order Butterworth High pass filter for a given cut off frequency,
fL= 1KHz .Draw the frequency response.
Components: IC 741 op amp, resistor, multi output, power supply, signal generator, CRO.
Design and circuit:
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Low pass filter
Let Vcc=12v , fH =1 kHz , gain Af= 2, to simplify the design calculations set
Set R1 = Rf= 10k, C1 = C2 = 0.01F = C
FH =
, Let R2 = R3 = R
R =
= 15k
C= 0.01f now Af= 1+ Rf/ R [Af= 2]
Rf= 10k let Rf= R1= 10k
Waveform:
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High pass filter:
Let Vcc = 12v, fL = 1kHz, gain Af= 2
To simplify the design calculation
Set R1 = Rf= 10k C1 = C2 = 0.01F = C
FL =
Let R2 = R3 = R
R =
= 15k
C= 0.01f now Af= 1+ Rf/ R -- [Af= 2]
21 = Rf / R1 hence Rf = R1 let Rf = 10k = R1= 10k
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Waveform:
Procedure:
Connections are made as per the circuit diagram. Set Vin= 2 to 5V P-P in the function generation. By varying the frequency in the function generator, note down the P-P voltage of the output waveform
the oscilloscope.
Plot the frequency response in the given semi log sheet. Find the cut off frequency.
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Tabular Column
Low pass filter :
Vi=2V
Frequency (Hz) Vo Gain=Vo/Vi Gain (dB)
20log10(vo/vin)
Result:
Cut off frequency (theoretical) = 1KHz.
Cut off frequency (practical) =-----------
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High Pass Filter:
Tabular column
Vi=2V
Frequency (Hz) Vo Gain=Vo/Vi Gain (dB)
20log10(Vo/Vin)
Result:
Cut off frequency (theoretical) = 1KHz.
Cut off frequency (practical) = ----------
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EXPERIMENT 2-a
ACTIVE WIDE BANDPASS FILTER
Aim: To design an active wide band pass filter and to plot the frequency response characteristics
given frequency v/s voltage gain.
Components: IC 741 op amp, resistor, capacitor, power supply, signal generator, CRO.
Design and circuit:
Given FL = 5kHz FH = 9kHz
BW = 9kHz - 5kHz = 4kHz
Fc = = = 5.91kHz
For second order band pass filter
Af= 1 +
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Procedure:
Rig up the circuit as shown. Given a sinusoidal input of 2Vp-p Vary the frequency of sinusoidal input from 100 Hz to 100 kHz without changing the input voltage level
At each frequency note down the output peak to peak voltage from CRO. Find the gain in dB at each frequency using the formula.
20 log (Vo/Vin).
Frequency response of the filter in obtained by plotting gain in dB v/s frequency.
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Waveform:
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Tabular Column
Vi=5V
Frequency (Hz) Vo in volt Gain=Vo/Vi Gain (dB)
20log10(vo/vin)
Result:
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EXPERIMENT 2-b
ACTIVE BANDSTOP FILTER
Aim: To design an active wide band stop filter and to plot the frequency response characteristics
given frequency v/s voltage gain.
Components: IC 741 op amp, resistor, capacitor, power supply, signal generator, CRO.
Design and circuit:
FL = 5kHz fH = 9kHz
Pass band gain of second order filter is given byAF = 1.586
Af= 1 +
Let Rf= R1= 10k
(AF - 1) RP = RF = ?
RF = (1.586 -1) 10 = 5.86
RF= 5.86k
For HPF fC =
C= 0.01F
R =
FC = = = 6.7kHz
R=
X 6.7 X 10
3X 0.01 X 10
-6
= 2.3k
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Procedure:
Rig up the circuit as shown. Given a sinusoidal input of 2Vp-p Vary the frequency of sinusoidal input from 5 kHz to 9 kHz without changing the input voltage level. At each frequency note down the output peak to peak voltage from CRO. Find the gain in dB at each frequency using the formula.
20 log (Vo/Vin).
Frequency response of the filter in obtained by plotting gain in dB v/s frequency.
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Tabular Column
Vi=5V
Frequency (Hz) Vo in volt Gain=Vo/Vi Gain (dB)
20log10(vo/vin)
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Waveform:
Result:
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EXPERIMENT 3
SCHMIT TRIGGER CIRCUIT
Aim: Design and testing of the Schmitt trigger circuit using op-amp
Components: Op-amps (IC-741), Resistors, Capacitors, CRO, Signal generators, probes, connecting wi
and Power chords.
Design and circuit diagram:
UTP = 4v and LTP = 2v
UTP > LTP
Wrt UTP =
Vref +
(Vsat)
Vsat = 12v
LTP =
Vref -
Vsat
UTP + LTP = 2.
Vref
0.6 =
----------------------1
UTPLTP =
Vsat
2.0 =
Vsat ---------------------- 2
From equation 2
= 0.0625
R1 + R2 = 16R2
From equation 1
6.0 = 2[
] Vref
Vref = 1.87V ==1.9V
Chose R2 = 10k and R1 = 90k
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Transfer characteristics:
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Procedure:
Make the connections as shown in the circuit. Apply input check square wave output. Use the X-Y mode of CRO to obtain transfer characteristics note UTP and LTP values. Compare with expected value.
Result: output is observed and analyzed.
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EXPERIMENT 4
R2R LADDER
Aim: design and testing of R-2R ladder using UA-741 OP-amp.
Components: IC 741 op amp, resistor, capacitor, multi meters, spring board, patch cords, power supp
signal generator, CRO.
Design and circuit:
D0 , D1, D2, D3 are digital input and may be high (1) or low (0)
VR(0) = 0
VR(1) = VR= reference voltage can be selected depending on max analog output required
R = (Full scale analog output voltage) / (2N
+ 1)
N- No of digital input
R = 2N
or % R =
Let Vr= 500mV
For a 4 bit DAC
Vo = (23
D3 + 22
D2 + 21
D1 + 20
D0)
VR= 0.5 X 24 = +12v
VR(1) = 12V and VR(0) = 0V
Vo min = 0 Vo max = 0.5 (8 + 4 +2 + 1 ) = 7.5v
AV =
= 3.12 when Vr = 5v
Vo max =
= 3.125
= 3.12 - 1 = 2.12
Let R1 = 1k Rf = 2.12k
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Procedure:
Connect the circuit R-2R ladder network as shown in figure.To measure Vminset all digital input to logic 0
i.e. D0 = D1 = D2 = D3 = 0
then V0 =0 theoretically and verify it practically suppose if the inp
from the digital trainer has min of 0.2V which is logic 0 its V0 min
0.12V.
To measure the resolution is defined as the smallest incremental changLet. D3 = D2 = D1 = D0 = 0 & let LSB D0 = 1
Vo = (VR /24) [0 + 0 + 0 + 1] = 0.2083V.
R = 0.2083V [theory].
Verify it practically using digital voltmeter.
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To measure full scale output voltage full scale output voltage obtaineby setting all the input to logic high.
i.e. D0 = D1 = D2 = D3 = +5vVo max = (VR /24) [8 + 4 + 2 + 1] = 3.125V.
Theoretically calculated value is verified by measuring its practically
Tabular Column
Decimal Digital Theoretical Experimental
Result:
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EXPERIMENT 5-a
ASTABLE MULTIVIBRATORS
ASYMMETRICAL AND SYMMETRICAL
Aim: Design an Astable Multivibrator using IC 555 timer to generate a clock signal of
i) Frequency 1 KHz with 0.75 duty cycle. (Asymmetrical)
ii) Frequency 1 KHz with 0.50 duty cycle. (Symmetrical)
Components: Resistors, Diode, Connecting wires, Capacitors, IC 555 (timer), Bread Board / Linear
kit, Power Supply, CRO, Probes and connecting wires.
Design and Circuit diagram:
a) AsymmetricVcc = 12V , f =1 kHz T = 1ms duty cycle 75% or .75
T =
=
= 1ms = TH + TL
d =
= 0.75 TH = 0.75ms T = 1ms
therefore TL = 0.25ms
W. K. T TL = 0.693 RB C
Let c = 0.01F so RB = 3.6k choose RB = 3.3k
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Similarly TH =0.693(RA + RB ) C = 0.75ms
(RA + RB) =0.75ms/0.693 X 0.1F = 10.82k , RA =7.2k
Choose RA = 6.8k
Check Vcc = 12V Voc = 2/3 Vcc = 8V VLT = 1/3 Vcc 4V
b) Symmetrical
f = 1kHz T=1ms duty cycle d = 50% or 0.5
T = =
= 1ms = TH + TL
w. k. T. TH + TL = 0.693RC , RA = RB = R
R =
Let C = 0.01F therefore R = 7.2k
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Choose RA = RB = R = 6.8k
Procedure:
Asymmetrical: Frequency 1 KHz with 0.75 duty cycle.
Verify the components and patch chords whether they are in good condition. Connect the Astable multivibrator circuit using IC 555 timer as shown in the ckt as per the design.
Switch on the DC power supply unit Vcc=12V. Observer the output waveform at pin no 6 on CRO.(capacitor output) Also observe the output waveform at pin no 3 on CRO.(Multivibrator output) For the capacitor output at pin no 6 , measure the maximum and minimum voltage levels. Verify that V
=2/3Vcc and VLT= 1/3 Vcc.
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Compare the capacitor voltage Vc with output waveform Vo and note that capacitor charges and Vc riexponentially when output is high. The capacitor C discharges through RB and the diode and Vc fa
exponentially when output is low.
Verify the designed value of frequency matches with practical value.
Symmetrical: Frequency 1 KHz with 0.50 duty cycle
Verify the components and patch chords whether they are in good condition. Connect the Astable multivibrator circuit using IC 555 timer as shown in the ckt as per the design. Switch on the DC power supply unit Vcc=12V. Observer the output waveform at pin no 6 on CRO.(capacitor output) Also observe the output waveform at pin no 3 on CRO.(Multivibrator output) For the capacitor output at pin no 6 , measure the maximum and minimum voltage levels. Verify that V
=2/3Vcc and VLT= 1/3 Vcc.
Compare the capacitor voltage Vc with output waveform Vo and note that capacitor charges and Vc riexponentially when output is high. The capacitor C discharges through RB and the diode and Vc fa
exponentially when output is low.
Verify the designed value of frequency matches with practical value.
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Waveform:
Result: The Astable Multivibrators (Asymmetrical and Symmetrical) are constructed for the giv
design , the theoretical and practical values are verified for the obtained waveforms.
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EXPERIMENT 5-b
MONOSTABLE MULTIVIBRATOR
Aim: To design a Monostable Multivibrator using IC 555 (timer).
Components: Resistor, Capacitors, IC-555, Bread Board, Power supply, CRO, connecting wires, pa
chords, signal generator and probes.
Design and circuit diagram:
Time delay T= 1ms
T = 1.1 RC assume C = 0.1F
1ms = 1.1 X R X 0.1F
R =
R = 0.09k
Choose R = 10k
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Procedure:
Verify all the components and patch chords. Connect the Monostable Multivibrator circuit using IC 555-timer as like shown in ckt. Switch on the DC power supply unit Vcc-12V and apply periodic input trigger pulse at pin no 2 usi
signal generator as the source.
Adjust the input frequency of signal generator to 80 HZ and adjust the input pulse amplitude to 12V soto obtain proper waveform across the capacitor C..
Observe the timer output waveform at pin no 3 and measure its higher and lower voltage levels and outpacross C on CRO.
Measure the output frequency f using CRO and verify that it is equal to the designed frequency. Verify whether the theoretical values are matching with practical values and observe the outputs.
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Waveform:
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Result: The Monostable Multivibrator is constructed for the given design and the theoretical and pract
values are verified for the obtained waveforms.
EXPERIMENT 6-a
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AMPLITUDE MODULATION: EMITTER MODULATION
Aim: To construct an Emitter Modulator to generate Amplitude Modulation (AM) waves at a carr
frequency of 50KHZ and to determine the % modulation index at modulating frequency of 1KHZ a
plot the variation of modulation index versus peak amplitude of modulating signal.
Components Required: Transistor (SL100), Resistors, Capacitors, CRO, Signal Generators,
Transformer, Inductor, Power Supply, Probes, Wires and Power chords.
Design and Circuit Diagram:
Emitter modulation
Let Vcc= 10V , =75 , Ic = 2mA VBE = 0.7V VCE = 5V RE= 100
IC= IB therefore IB =
= 26.6A
RB =
= 26.3k
Choose 22k = RB
Fcarrier= 50kHz =
let L = 1mH then C = 0.01F
choose C1 = C2 = 0.01F and C3 = 47F
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Envelop detector :
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Waveform:
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Procedure:
Rig up the circuit as shown in the figure. Verify the design by measuring the VCE and IC , without switching on the RF and AF generators. Switch on the RF signal generator and AF signal generator and adjust fm=1 KHz and fc=50 K
respectively.
Observe the output on the CRO, which is the AM waveform. Note down Emax and Emin from the output waveform. Calculate the % modulation index by using the formula,
% m = Emax-Emin/Emax+Emin * 100
Vary the amplitude of AF signal that is VAF , correspondingly tabulate the Emax and Emin readingscalculate the % modulation index (%m) .
Plot the graph of VAF versus % modulation index (%m) for different VAF.Tabular column:
VAF (volts) Emax(volts) Emin(volts) %m
Result: An AM waveform is generated and observed for the given frequencies and %m is verified from
tabular column.
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EXPERIMENT 7
PULSE MODULATIONS: PULSE AMPLITUTED MODULATION
Aim: To conduct the experiment to generate PAM signal and also design a circuit to demodulate
PAM signal plot the relative waveform
Components : Transistor SL100, Diode 0A79, Resistors, Capacitors, CRO, Signal generators, DC pow
supplies, Linear IC trainer kit, probes, connecting wires and Power chords.
Design and circuit diagram:
fc>>
i.e. Rc>>
let fc =15kHz C = 0.1F
R >
R = 666
Select R = 680
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Modulator :
Demodulation:
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Waveform:
Procedure:
Rig up the circuit as shown in figure. Set up carrier amplitude to around 2Vp-p and frequency in range 5 kHz to 15 kHz.
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Set up amplitude to around 1Vp-p and frequency 2 kHz. Connect CRO around emitter of transistor and observe the PAM waveform. To verify sampling thermo keep the modulation signal frequency to 2 kHz and carrier frequency to twice
modulating wave and check the other end the demodulation waveform it should match with m(t).
Result: PAM waveform is observed and analyzed.
EXPERIMENT 8-a
PULSE MODULATIONS: PULSE WIDTH MODULATION
Aim: Todesign a Pulse Width Modulation circuit and transmit an analog signal and also demodulate
generated PWM wave using a suitable Demodulation circuit.
Components: Op-amps (IC-741), Resistors, Capacitors, CRO, Signal generators, DC power suppl
Linear IC trainer kit, probes, connecting wires, patch chords and Power chords.
Design and circuit diagram:
Modulation:
Design
Let RC >> T
Time period T = 0.1ms
R1 C1 = 10T = 1ms
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Let R1= 10k C1 = 0.01F
Choose R1 = R2 =R3= 10k
fc =
=
fc = 1.59kHz
Circuit:
Waveform
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Procedure:
Rig up the Modulation circuit. Set up the message (sine wave) signal of frequency 500HZ and amplitude of 5V P-P.
A square/ramp/saw tooth waveform is used as a carrier signal with a 5V P-P amplitude, and frequency2-3 kHz.
Apply a reference voltage of the range 1-5V. The Op-amp offset is kept at zero value. The square wave signal frequency is adjusted to obtain proper pulse width. PWM should vary on variation of message signal. If the output is distorted then frequency is reduced
get proper output.
The Vref is kept at zero and the wave form is observed varying the amplitude and the change in widthalso observed.
Increase Vref up to 5 V and repeat the same procedure. The output waveform is observed and plotted. The demodulation circuit is rigged
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The PWM output from the modulation circuit is fed as the input to the demodulation circuit. The demodulated output that is the sine wave (message signal) is observed and plotted.
Result: The pulse width modulation is obtained and the signal is analyzed. The demodulated outpu
also obtained and plotted.
EXPERIMENT 8-b
PULSE MODULATIONS: PULSE POSITION MODULATION
Aim: To generate PPM signal of given pulse width for a given modulating signal from a PWM sign
using an IC 555(timer).
Components : Op-amps (IC-741), Resistors, Capacitors, IC 555, Diode IN4007,CRO,Signal generato
DC power supplies, Linear IC trainer kit, probes, connecting wires and Power chords.
Design and circuit diagram:
Let pulse of the PPM signal = 50MHz
T = 1.1RC, let CA = 0.1F
PWM fc(t) = 3kHz CA = 0.1F
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RA =
= 530.5
Choose RA= 470
Waveform
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Procedure:
Rig up PWM modulator ckt and observe the PWM output. Connect the output of the PWM modulator as triggering input of PPM modulator at pin 2 through
capacitor.
The PPM output is observed at pin 3. Draw the PPM waveform with respect to the PWM waveform. Analyze the PPM waveform and observe that for each trailing edge of the PWM wave there is pu
positioned and all the pulses are of equal widths and durations.
Result: PPM waveform is observed and analyzed.
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EXPERIMENT 9
FREQUENCY MODULATION USING IC 8038
Aim: To design and generate FM signal using IC 8038 and demonstrate the generation of frequen
modulated wave.
Components : IC 8038,Resistors,Capacitors,CRO,Signal Generators, Power supply, Probes Wires, Dand Power chords.
Design and circuit diagram:
IC description:
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Test circuit
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FM generation circuit
Design
Let fc = 3kHz
Fc =
Choose Cc = 0.01F
R =
= 10k
R = Ra = Rb = RL= 10k
Cc = C1 = 0.01F
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Procedure:
Rig up the Test circuit and observe the waveforms at pin numbers 2, 3 and 9 generating sine, triangle asquare waves respectively.
Rig up the FM generation circuit. Set the amplitude of the modulating signal to 1V and its frequency to 1KHZ. Observe the FM output at pin number 2 on the CRO.
Wave form:
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Result: The Frequency modulated waveform is observed.
EXPERIMENT 10
HALF WAVE AND FULL WAVE PRECISION RECTIFIERS
Aim: To realize the half wave precision rectifier.
Components: Op-amp UA-741, Diode IN4007, Resistors.
Design and circuit diagram:
Half wave :
Waveform
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Full wave:
Waveform:
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Result: waveform is observed and analyzed.
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