Download 1. Study of OP AMPs - IC 741, IC 555, IC 565, IC 566, IC 1496

DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
1. Study of OP AMPs - IC 741, IC 555, IC 565, IC 566,
IC 1496-functioning, parameters and specifications
IC 741
General Description
The IC 741 is a high performance monolithic operational amplifier constructed
using the planer epitaxial process. High common mode voltage range and absence
of latch-up tendencies make the IC 741 ideal for use as voltage follower. The high
gain and wide range of operating voltage provide superior performance in integrator,
summing amplifier and general feed back applications.
Internal Block Diagram of Op-Amp
Fig.1: Block diagram of op-amp
Pin Configuration
Fig.2: Pin diagram of IC741
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Features
1. No frequency compensation required.
2. Short circuit protection
3. Offset voltage null capability
4. Large common mode and differential voltage ranges
5. Low power consumption
6. No latch-up
Specifications
1. Voltage gain A = α typically 2,00,000
2. I/P resistance RL = α Ω, practically 2MΩ
3. O/P resistance R =0, practically 75Ω
4. Bandwidth = α Hz. It can be operated at any frequency
5. Common mode rejection ratio = α
(Ability of op amp to reject noise voltage)
6. Slew rate + α V/µsec
(Rate of change of O/P voltage)
7. When V1 = V2, VD=0
8. Input offset voltage (Rs ≤ 10KΩ) max 6 mv
9. Input offset current = max 200nA
10. Input bias current : 500nA
11. Input capacitance : typical value 1.4pF
12. Offset voltage adjustment range : ± 15mV
13. Input voltage range : ± 13V
14. Supply voltage rejection ratio : 150 µV/V
15. Output voltage swing: + 13V and – 13V for RL > 2KΩ
16. Output short-circuit current: 25mA
17. supply current: 28mA
18. Power consumption: 85mW
19. Transient response: rise time= 0.3 µs
Overshoot= 5%
Applications
1. AC and DC amplifiers
2. Active filters
3. Oscillators
4. Comparators
5. Regulators
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IC 555
General Description
The operation of SE/NE 555 timer directly depends on its internal function.
The three equal resistors R1, R2, R3 serve as internal voltage divider for the source
voltage. Thus one-third of the source voltage VCC appears across each resistor.
Comparator is basically an Op amp which changes state when one of its
inputs exceeds the reference voltage.
comparator is +1/3 VCC.
The reference voltage for the lower
If a trigger pulse applied at the negative input of this
comparator drops below +1/3 VCC, it causes a change in state. The upper comparator
is referenced at voltage +2/3 VCC. The output of each comparator is fed to the input
terminals of a flip flop.
The flip-flop used in the SE/NE 555 timer IC is a bistable multivibrator. This
flip flop changes states according to the voltage value of its input. Thus if the voltage
at the threshold terminal rises above +2/3 VCC, it causes upper comparator to cause
flip-flop to change its states. On the other hand, if the trigger voltage falls below +1/3
VCC, it causes lower comparator to change its states. Thus the output of the flip flop
is controlled by the voltages of the two comparators. A change in state occurs when
the threshold voltage rises above +2/3 VCC or when the trigger voltage drops below
+1/3 Vcc.
The output of the flip-flop is used to drive the discharge transistor and the
output stage. A high or positive flip-flop output turns on both the discharge transistor
and the output stage. The discharge transistor becomes conductive and behaves as
a low resistance short circuit to ground. The output stage behaves similarly. When
the flip-flop output assumes the low or zero states reverse action takes place i.e., the
discharge transistor behaves as an open circuit or positive VCC state. Thus the
operational state of the discharge transistor and the output stage depends on the
voltage applied to the threshold and the trigger input terminals.
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Block Diagram of IC 555
Fig.3: Block diagram of IC 555
Pin Configuration
Fig.4: Pin diagram of IC555
Function of Various Pins
Pin (1) of 555 is the ground terminal; all the voltages are measured with respect to
this pin.
Pin (2) of 555 is the trigger terminal, if the voltage at this terminal is held greater than
one-third of VCC, the output remains low. A negative going pulse from Vcc to less than
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Vec/3 triggers the output goes to High. The amplitude of the pulse should be able to
make the comparator (inside the IC) change its state. However the width of the
negative going pulse must not be greater than the width of the expected output pulse.
Pin (3) is the output terminal of IC 555. There are 2 possible output states. In the
low output state, the output resistance appearing at pin (3) is very low (approximately
10 Ω). As a result the output current will goes to zero , if the load is connected from
Pin (3) to ground , sink a current I Sink (depending upon load) if the load is connected
from Pin (3) to ground, and sinks zero current if the load is connected between +VCC
and Pin (3).
Pin (4) is the Reset terminal. When unused it is connected to +Vcc. Whenever the
potential of Pin (4) is drives below 0.4V, the output is immediately forced to low state.
The reset terminal enables the timer over-ride command signals at Pin (2) of the IC.
Pin (5) is the Control Voltage terminal. This can be used to alter the reference levels
at which the time comparators change state. A resistor connected from Pin (5) to
ground can do the job. Normally 0.01µF capacitor is connected from Pin (5) to
ground.
This capacitor bypasses supply noise and does not allow it affect the
threshold voltages.
Pin (6) is the threshold terminal. In both astable as well as monostable modes, a
capacitor is connected from Pin (6) to ground. Pin (6) monitors the voltage across
the capacitor when it charges from the supply and forces the already high O/p to Low
when the capacitor reaches +2/3 VCC.
Pin (7) is the discharge terminal. It presents an almost open circuit when the output
is high and allows the capacitor charge from the supply through an external resistor
and presents an almost short circuit when the output is low.
Pin (8) is the +Vcc terminal. 555 can operate at any supply voltage from +3 to +18V.
Features
1. The load can be connected to o/p in two ways i.e. between pin 3 & ground 1 or
between pin 3 & VCC (supply)
2. 555 can be reset by applying negative pulse, otherwise reset can be connected
to +Vcc to avoid false triggering.
3. An external voltage effects threshold and trigger voltages.
4. Timing from micro seconds through hours.
5. Monostable and bistable operation
6. Adjustable duty cycle
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7. Output compatible with CMOS, DTL, TTL
8. High current output sink or source 200mA
9. High temperature stability
10. Trigger and reset inputs are logic compatible.
Specifications
1. Operating temperature
:
SE 555--
-55oC to 125oC
NE 555--
0o to 70oC
2. Supply voltage
:
+5V to +18V
3. Timing
:
µSec to Hours
4. Sink current
:
200mA
5. Temperature stability
:
50 PPM/oC change in temp or 0-005% /oC.
Applications
1. Monostable and Astable Multivibrators
2. dc-ac converters
3. Digital logic probes
4. Waveform generators
5. Analog frequency meters
6. Tachometers
7. Temperature measurement and control
8. Infrared transmitters
9. Regulator & Taxi gas alarms etc.
IC 565
General Description
The Signetics SE/NE 560 series is monolithic phase locked loops. The SE/NE 560,
561, 562, 564, 565, & 567 differ mainly in operating frequency range, power supply
requirements and frequency and bandwidth adjustment ranges.
The device is
available as 14 Pin DIP package and as 10-pin metal can package.
Phase
comparator or phase detector compare the frequency of input signal fs with frequency
of VCO output fo and it generates a signal which is function of difference between the
phase of input signal and phase of feedback signal which is basically a d.c voltage
mixed with high frequency noise. LPF remove high frequency noise voltage. Output
is error voltage. If control voltage of VCO is 0, then frequency is center frequency (fo)
and mode is free running mode. Application of control voltage shifts the output
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frequency of VCO from fo to f. On application of error voltage, difference between fs
& f tends to decrease and VCO is said to be locked. While in locked condition, the
PLL tracks the changes of frequency of input signal.
Block Diagram of IC 565
Fig.5: Block Diagram of IC 565
Pin Configuration
Fig.6: Pin diagram of IC 565
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Specifications
1. Operating frequency range
:
0.001 Hz to 500 KHz
2. Operating voltage range
:
±6 to ±12V
3. Inputs level required for tracking
:
10mV rms minimum to 3v (p-p)
4. Input impedance
:
10 KΩ typically
5. Output sink current
:
1mA typically
6. Drift in VCO center frequency
:
300 PPM/oC typically
:
1.5%/V maximum
8. Triangle wave amplitude
:
typically 2.4 VPP at ± 6V
9. Square wave amplitude
:
typically 5.4 VPP at ± 6V
10. Output source current
:
10mA typically
11. Bandwidth adjustment range
:
<±1 to >± 60%
max.
(fout) with temperature
7. Drif in VCO centre frequency with
supply voltage
Center frequency fout = 1.2/4R1C1 Hz
= free running frequency
FL = ± 8 fout/V Hz
V = (+V) – (-V)

fL
3
 2Π (3.6) x10 xC 2
fc = ± 
]1 / 2
Applications
1. Frequency multiplier
2. Frequency shift keying (FSK) demodulator
3. FM detector
IC 566
General Description
The NE/SE 566 Function Generator is a voltage controlled oscillator of
exceptional linearity with buffered square wave and triangle wave outputs.
The
frequency of oscillation is determined by an external resistor and capacitor and the
voltage applied to the control terminal. The oscillator can be programmed over a ten
to one frequency range by proper selection of an external resistance and modulated
over a ten to one range by the control voltage with exceptional linearity.
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Block Diagram of IC566
Fig.7: Block diagram of IC566
Pin diagram
Fig.8 : Pin diagram of IC566
Specifications
Maximum operating Voltage ---
26V
Input voltage
---
3V (P-P)
Storage Temperature
---
-65oC to + 150oC
Operating temperature
---
0oC to +70oC for NE 566
-55oC to +125oC for SE 566
Power dissipation
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---
300mv
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Applications
1. Tone generators.
2. Frequency shift keying
3. FM Modulators
4. clock generators
5. signal generators
6.
Function generator
IC 1496
General Description
IC balanced mixers are widely used in receiver IC’s. The IC versions are
usually described as balanced modulators. Typical example of balanced IC
modulator is MC1496. The circuit consists of a standard differential amplifier (formed
by Q5 _ Q6 combination) driving a quad differential amplifier composed of transistor
Q1 – Q4.
The modulating signal is applied to the standard differential amplifier
(between terminals 1 and 4). The standard differential amplifier acts as a voltage to
current converter. It produces a current proportional to the modulating signal. Q7 and
Q8 are constant current sources for the differential amplifier Q5 – Q6. The lower
differential amplifier has its emitters connected to the package pins ( 2 & 3) so that an
external emitter resistance may be used. Also external load resistors are employed
at the device output (6 and 12 pins).The output collectors are cross-coupled so that
full wave balanced multiplication takes place. As a result, the output voltage is a
constant times the product of the two input signals.
Schematic of IC1496
Fig.9: Pin diagram of IC1496
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Pin Configuration
Fig.10 : Pin diagram of IC1496
Applications of MC 1496
•
Balanced modulator
•
AM Modulator
•
Product Modulator
•
AM Detector
•
Mixer
•
Frequency Doublers.
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2. OP AMP Applications – Adder, Subtractor,
Comparator Circuits
Aim:
To design Adder, Subtractor and Comparator circuits by using operational
amplifier.
Apparatus required
S.No
Equipment/Component name
Specifications/Value
1
IC 741
Refer page no 2
1
2
Resistor
1kΩ
4
3
Diode
0A79
2
4
Regulated Power supply
(0 – 30V),1A
2
5
Function Generator
(.1 – 1MHz), 20V p-p
1
6
Cathode Ray Oscilloscope
(0 – 20MHz)
1
7
Multimeter
3
½
digit display
Quantity
1
Theory
Adder:
A two input summing amplifier may be constructed using non-inverting
mode or inverting mode. The gain of this summing amplifier is 1 as all the resistors
are equal in value; any scaling factor can be obtained by selecting proper external
resistors.
Subtractor: A basic differential amplifier can be used as a subtractor as shown in
fig.3. In this circuit all external resistors are equal in value. Hence the gain of
amplifier is equal to one. The output voltage Vo is equal to the difference voltage
between the non-inverting terminal and the inverting terminal; hence the circuit is
called a subtractor.
Comparator:
The circuit diagram shows an op-amp used as a comparator. A
fixed reference voltage Vref is applied to the one terminal and the varying signal
voltage Vin is applied to the other terminal. Depending upon the application of Vin to
the terminal this can be non inverting or inverting, if input Vin is connected to the non
inverting terminal of the op-amp then it is called as non- inverting comparator as vice
versa. Depending upon the levels of Vin and Vref, the circuit produces output +Vsat or
-Vsat
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Circuit Diagrams
Adder:
Fig. 1: Non- inverting Adder
Fig. 2: Inverting Adder
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Subtractor
Fig. 3: Subtractor
Comparator
Fig. 4: Non-inverting Comparator
Fig. 5: Inverting Comparator
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Designing & Model Calculations
Adder
Non-inverting
Vo = (V1 + V2)
If V1 = 2.5V and V2 = 2.5V, then
Vo = (2.5+2.5) = 5V.
Inverting
Vo = - (V1 + V2)
If V1 = 2.5V and V2 = 2.5V, then
Vo = - (2.5+2.5) = -5V.
Subtractor
Vo = V2 – V1
If V1=2.5 and V2 = 3.3, then
Vo = 3.3 – 2.5 = 0.8V
Comparator
Non-inverting
If Vin < Vref, Vo = -Vsat ≅ - VEE
Vin > Vref, Vo = +Vsat = +VCC
Inverting
If Vin < Vref, Vo = +Vsat ≅ +VCC
Vin > Vref, Vo = -Vsat = - VEE
Procedures
Adder
•
Connect the circuit as per the diagram shown in Fig.1 and Fig.2.
•
Apply the supply voltages of +15V to pin7 and pin4 of IC741 respectively.
•
Apply the inputs V1 and V2 as shown in Fig.1 and Fig.2.
•
Apply two different signals (DC/AC ) to the inputs
•
Vary the input voltages and note down the corresponding output at pin 6 of the IC
741 adder circuit.
•
Notice that the output is equal to the sum of the two inputs.
Subtractor
1. Connect the circuit as per the diagram shown in Fig.3.
2. Apply the supply voltages of +15V to pin7 and pin4 of IC741 respectively.
3
Apply the inputs V1 and V2 as shown in Fig.3.
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4. Apply two different signals (DC/AC) to the inputs
5. Vary the input voltages and note down the corresponding output at pin 6 of the IC
741 subtractor circuit.
7. Notice that the output is equal to the difference of the two inputs.
Comparator
1. A fixed reference voltage Vref and varying voltage Vin is applied as shown in Fig.3
and Fig.4.
2. Vary the input voltage above and below the Vref and note down the output at pin
6 of 741 IC.
3. Observe that,
when Vin is less than Vref,
Output voltage is
Non-inverting
Inverting
-Vsat ( ≅ - VEE)
+Vsat (≅+VCC)
Non-inverting
Inverting
+Vsat (≅+VCC)
-Vsat ( ≅ - VEE)
when Vin is greater than Vref,
Output voltage is
Observations
Adder
V1(V)
V2(V)
Inverting adder
Non-Inverting
Vo(V)
adder Vo(V)
Subtractor
V1(V)
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V2(V)
Vo(V)
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Comparator
Vin(V)
Vref(V)
Non-Inverting
Inverting
comparator
comparator
Vo(V)
Vo(V)
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Results
Inferences
Questions
1. What is the saturation voltage of 741 in terms of VCC?
2. What is the maximum voltage that can be given at the inputs?
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3. Integrator and Differentiator Circuits using IC 741
Aim:
To design and verify the operation of an integrator and differentiator for a
given input.
Apparatus required
S.No
Equipment/Component name
Specifications/Value
Quantity
1
741 IC
Refer page no. 2
2
Capacitors
0.1µf, 0.01µf
Each one
3
Resistors
159Ω, 1.5kΩ
Each one
4
Regulated Power supply
(0 – 30)V,1A
1
5
Function generator
(1Hz – 1MHz)
1
6
Cathode Ray Oscilloscope
(0 – 20MHz)
1
1
Theory
Integrator
In an integrator circuit, the output voltage is integral of the input signal. The
t
output voltage of an integrator is given by Vo = -1/R1Cf
∫ Vidt
o
At low frequencies the gain becomes infinite, so the capacitor is fully charged and
behaves like an open circuit. The gain of an integrator at low frequency can be
limited by connecting a resistor in shunt with capacitor.
Differentiator
In the differentiator circuit the output voltage is the differentiation of the input
voltage. The output voltage of a differentiator is given by Vo = -RfC1
dVi
.The input
dt
impedance of this circuit decreases with increase in frequency, thereby making the
circuit sensitive to high frequency noise and the circuit may become unstable. Hence
an input resistor is connected in series with the capacitor and a capacitor is
connected in parallel with the feedback resistor.
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Circuit Diagrams
Fig. 1: Integrator
Fig. 2: Differentiator
Design equations
Integrator
Choose T = 2πR1Cf
Where T= Time period of the input signal
Assume Cf and find R1
Select Rf = 10R1
Vo (p-p) =
−1
R1C f
T /2
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∫
Vi ( p − p ) dt
o
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Differentiator
Select given frequency fa = 1/(2πRfC1), Assume C1 and find Rf
Select fb = 10 fa = 1/2πR1C1 and find R1
From R1C1 = RfCf, find Cf
Model Calculations
Integrator
For T= 1 msec
f a= 1/T = 1 KHz
fa = 1 KHz = 1/(2πR1Cf)
Assuming Cf= 0.1µf, Rf is found from R1=1/(2πfaCf)
R1=1.59 KΩ
Rf = 10 R1
Rf= 15.9kΩ
Differentiator
For T = 1 msec
f= 1/T = 1 KHz
fa = 1 KHz = 1/(2πRfC1)
Assuming C1= 0.1µf, Rf is found from Rf=1/(2πfaC1)
Rf=1.59 KΩ
fb = 10 fa = 1/2πR1C1
for C1= 0.1µf; R1 =159Ω
Procedures
Integrator
1. Connect the circuit as per the diagram shown in Fig.1
2. Apply a square wave/sine input of 4V(p-p) at 1KHz
3. Observe the output at pin 6.
4. Draw input and output waveforms as shown in Fig.3.
Differentiator
1. Connect the circuit as per the diagram shown in Fig. 2
2. Apply a square wave/sine input of 4V(p-p) at 1KHz
3. Observe the output at pin 6
4. Draw the input and output waveforms as shown in Fig.4
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Wave Forms
Integrator
Fig. 3: Input and output waves forms of integrator
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Differentiator
Fig. 4 : Input and output waveforms of Differentiator
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Observations
Integrator
Input –Square wave
Output - Triangular
Amplitude(VP-P)
Time period
Amplitude (VP-P)
Time period
(V)
(ms)
(V)
(ms)
Input –sine wave
Output - cosine
Amplitude(VP-P)
Time period
Amplitude (VP-P)
Time period
(V)
(ms)
(V)
(ms)
Differentiator
Input –square wave
Output - Spikes
Amplitude (VP-P)
Time period
Amplitude (VP-P)
Time period
(V)
(ms)
(V)
(ms)
Input –sine wave
Output - cosine
Amplitude (VP-P)
Time period
Amplitude (VP-P)
Time period
(V)
(ms)
(V)
(ms)
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Results
Inferences
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Questions
1. What are the problems of ideal differentiator?
2. What are the problems of ideal integrator?
3. What are the applications of differentiator and integrator?
4. What is the need for Rf in the circuit of integrator?
5. What is the effect of C1 on the output of a differentiator?
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4. Active Filter Applications – LPF, HPF (first order)
Aim:
To design and obtain the frequency response of
i) First order Low Pass Filter (LPF)
ii) First order High Pass Filter (HPF)
Apparatus required
S.No
Equipment/Component name
Specifications/Value Quantity
1
IC 741
Refer page no. 2
1
2
Resistors
10k ohm
3
Variable Resistor
20kΩ pot
1
3
capacitors
0.01µf
1
4
Cathode Ray Oscilloscope
(0 – 20MHz)
1
5
Regulated Power supply
(0 – 30V),1A
1
6
Function Generator
(1Hz – 1MHz)
1
Theory
LPF
A LPF allows frequencies from 0 to higher cut of frequency, fH. At fH the gain
is 3 db from maximum gain and after fH gain decreases at a constant rate with an
increase in frequency.
The gain decreases 20dB each time the frequency is
increased by 10. Hence the rate at which the gain rolls off after fH is 20dB/decade or
6 dB/ octave, where octave signifies a two fold increase in frequency. The frequency
f=fH is called the cut off frequency because the gain of the filter at this frequency is
down by 3 dB from 0 Hz. Other equivalent terms for cut-off frequency are -3dB
frequency, break frequency, or corner frequency.
HPF
The frequency at which the magnitude of the gain is 0.707 times the maximum
value of gain is called low cut off frequency. Obviously, all frequencies higher than fL
are pass band frequencies with the highest frequency determined by the closed –
loop band width all of the op-amp. A HPF allows frequiences grater than lower cut of
frequency fL. At fL gain is 3 db down from maximum gain. The frequency f= fL is
called the lower cut off frequency.
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Circuit diagrams
Fig. 1: Low pass filter
Fig. 2: High pass filter
Design equations
•
Chose a value of high cut off frequeinces f H =1/( 2πRC )
•
Assume c <01 µF and calculate R using R= 1/(2πfHC) Ω
•
Select value of R1 and RF depending on the desired pass band gain AF = 1+
(RF/R1)
First Order LPF: To design a Low Pass Filter for higher cut off frequency fH = 4 KHz
and pass band gain of 2
fH = 1/( 2πRC )
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Assuming C=0.01 µF, the value of R is found from
R= 1/(2πfHC) Ω =3.97KΩ
The pass band gain of LPF is given by
AF = 1+ (RF/R1)= 2
Assuming R1=10 KΩ, the value of RF is found from
RF=( AF-1) R1=10KΩ
First Order HPF: To design a High Pass Filter for lower cut off frequency
fL = 4 KHz and pass band gain of 2
fL = 1/( 2πRC )
Assuming C=0.01 µF,the value of R is found from
R= 1/(2πfLC) Ω =3.97KΩ
The pass band gain of HPF is given by
AF = 1+ (RF/R1)= 2
Assuming R1=10 KΩ, the value of RF is found from
RF=( AF-1) R1=10KΩ
Procedure
First Order LPF
1. Connections are made as per the circuit diagram shown in Fig.1.
2. Apply sinusoidal wave of constant amplitude as the input such that op-amp does
not go into saturation.
3. Vary the input frequency and note down the output amplitude at each step as
shown in Table (a).
4. Plot the frequency response as shown in Fig.3.
First Order HPF
1.
Connections are made as per the circuit diagrams shown in Fig. 2.
1.
Apply sinusoidal wave of constant amplitude as the input such that op-amp does
not go into saturation.
2.
Vary the input frequency and note down the output amplitude at each step as
shown in Table (b).
4.
Plot the frequency response as shown in Fig 4.
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Tabular Form Observations
Input voltage Vin = 0.5V
Table (a) First order LPF
Table (b) First order HPF
Frequency
O/P
Voltage
Gain
Frequency
O/P
Voltage
Gain
(Hz)
Voltage(V)
Gain
(dB)
(Hz)
Voltage(V)
Gain
(dB)
Vo/Vi
Vo/Vi
Model graphs
Fig. 3: Frequency response of LPF
IC APPLICATIONS LABORATORY
Fig. 4:Frequency response of HPF
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Results
Inferences
Questions
1. What is meant by frequency scaling?
2. How do you convert an original frequency (cut off) fH to a new cut off frequency
fH?
3. What is the effect of order of the filter on frequency response characteristics?
4. What modifications in circuit diagrams require to change the order of the filter?
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5. Active Filter Applications – BPF & Band Reject
(Wideband) and Notch Filters
Aim: To design and obtain the frequency response of
i)
Wide Band pass filter
ii)
Wide Band reject filter
iii)
Notch filter
Apparatus required
S.No
Equipment/Component name
Specifications/Value
Quantity
1
741 IC
Refer page no 2
3
2
Resistors
5.6kΩ
9
Resistors
39kΩ
2
3
Resistors
(20kΩ pot)
2
4
Capacitors
0.01µf
2
Capacitors
0.1µf
2
Capacitors
0.2µf
5
Regulated Power supply
(0 – 30)V,1A
1
1
6
Function Generator
(1Hz – 1MHZ)
1
7
Cathode Ray Oscilloscope
(0 – 20MHz)
1
Theory
Filter is a circuit that accepts certain band of frequencies and rejects other band of
frequencies
Band pass filter:
A band pass filter has a pass band between two cutoff
frequencies fH and fL such that fH > f L. Any input frequency outside this pass band is
attenuated. There are two types of band-pass filters. Wide band pass and Narrow
band pass filters. We can define a filter as wide band pass if its quality factor Q <10.
If Q>10, then we call the filter a narrow band pass filter. A wide band pass filter can
be formed by simply cascading high-pass and low-pass filter the order of band pass
filter depends on the order of high pass and low pass filter.
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Band Rejection Filter:
The band-reject filter is also called a band-stop or
band-elimination filter. In this filter, frequencies are attenuated in the stop band while
they are passed outside this band. Band reject filters are classified as wide bandreject narrow band-reject. Wide band-reject filter is formed using a low pass filter, a
high-pass filter and summing amplifier. To realize a band-reject response, the low
cut off frequency fL of high pass filter must be larger than high cut off frequency f H of
low pass filter. The pass band gain of both the high pass and low pass sections must
be equal.
Notch Filter or Narrow Band Reject Filter:
The narrow band reject
filter, often called the notch fitter is commonly used for the rejection of a single
frequency. The most commonly used notch filter is the twin-T network .This is a
passive filter composed of two T-shaped networks. One T network is made up of two
resistors and a capacitor, while the other uses two capacitors and a resistor. The
passive twin-t network has a relatively low figure of merit can be increased
significantly by using active notch filter as shown in fig.3.The notch-out frequency is
the frequency at which maximum attenuation occurs and is given by
fN = 1/( 2πRC )
Circuit diagrams
Fig. 1: Wideband pass filter
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Fig. 2: Wideband reject filter
Fig. 3: Notch filter
Design
Band pass filter: To design a band pass filter having fH =400Hz and fL = 4KHz
and pass band gain of 2. As shown in Fig.1,the first section consisting of OpAmp,RF,R1,R and C is the high pass filter and second consisting of low pass filter.
The design of low pass and high pass filters.
Low Pass Filter Design
Assuming C’=0.01µf, the value of R’ is found from
R’ = 1/(2πfH C’) Ω =3.97KΩ
The pass band gain of LPF is given by ALPF = 1+ (R’ F / R’1) =2
Assuming R’ 1=5.6 KΩ, the value of R’F is found from R’F = ( AF-1) R’1=5.6KΩ
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High Pass Filter Design
Assuming C=0.01µf, the value of R is found from
R = 1/(2πfLC) Ω =39.7KΩ
The pass band gain of HPF is given by AHPF = 1+ (RF / R1 )=2
Assuming R1=5.6 KΩ, the value of RF is found from
RF = ( AF-1) R1=5.6KΩ
Band reject filter:
To design a band reject filter with fH = 400Hz, fL = 4000Hz
and pass band gain of 2
Low Pass Filter Design
Assuming C’=0.01µf, the value of R’ is found from
R’ = 1/(2πfH C’) Ω =39KΩ
The pass band gain of LPF is given by ALPF = 1+ (R’ F / R’ 1 )=2
Assuming R’ 1=5.6 KΩ, the value of R’F is found from
R’F =( AF-1) R’ 1=5.6KΩ
High Pass Filter Design
Assuming C=0.01µf, the value of R is found from
R = 1/ (2πfLC) Ω =3.9KΩ
The pass band gain of HPF is given by AHPF = 1+ (RF / R1) =2
Assuming R1=5.6 KΩ, the value of RF is found from
RF = (AF-1) R1=5.6KΩ
Adder circuit design: Select all resistors equal value such that gain is unity.
Assume R2=R3=R4=5.6 KΩ
Notch Filter Design:
fN = 400Hz
Assuming C=0.1µf,the value of R is found from
R = 1/ (2πfNC)=39 KΩ
Procedure
Wide Band Pass Filter
1. Connect the circuit as per the circuit diagram shown in Fig.1
2. Apply sinusoidal wave of 0.5V amplitude as input such that op-amp does not go
into saturation (depending on gain).
3. Vary the input frequency from 100 Hz to 100 KHz and note down the output
amplitude at each step as shown in Table (a).
4. Plot the frequency response as shown in Fig.4.
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Wide Band Reject Filter
1. Connect the circuit as per the circuit diagram shown in Fig.2
2. Apply sinusoidal wave of 0.5V amplitude as input such that opamp does not go
into saturation (depending on gain).
3. Vary the input frequency from 100 Hz to 100 KHz and note down the output
amplitude at each step as shown in Table( b).
4. Plot the frequency response as shown in Fig.5.
Notch Filter
1. Connect the circuit as per the circuit diagram shown in Fig 3
2. Apply sinusoidal wave of 2Vp-p amplitude as input such that opamp does not go
into saturation (depending on gain).
3. Vary the input frequency from 100 Hz to 4 KHz and note down the output
amplitude at each step as shown in Table( c).
4. Plot the frequency response as shown in Fig 6.
Observations
Input voltage (Vi) = 0.5V
Table(a) Band pass filter
Frequency
Table(b) Band Reject Filter
O/P
Gain
Gain in
Voltage
Vo/Vi
dB
Frequency
O/P
Gain
Gain in
Voltage(V)
Vo/Vi
dB
Vo(V)
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Table(c) Notch filter
Input voltage=2Vp-p
Frequency
O/P Voltage(V)
Vo/Vi
Gain in
dB
Model graphs
Fig. 4 : Frequency response of
wide band pass filter
IC APPLICATIONS LABORATORY
Fig. 5 : Frequency response
of wide band reject filter
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Fig. 6: Frequency response of notch filter
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Results
Inferences
Questions
1. What is the relation between fC & f H, fL?
2. How do you increase the gain of the wideband pass filter?
3. What is the application of Notch filter?
4. What is the order of the filter (each type)?.What modifications you suggest for
the circuit diagram to increase the order of the filter?
5. What is the gain roll off outside the pass band?
6. What is the difference between active and passive filters?
7. What are the advantages of active filters over passive filters?
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
6. IC 741 Oscillator Circuits
Phase Shift and Wien Bridge Oscillators
Aim:
To design (i) phase shift and (ii) Wien Bridge oscillators for the given
frequency of oscillation and verify it practically.
Apparatus required
S.No
Equipment/Component name
Specifications/Value
1
IC 741
Refer page no. 2
1
2
Resistors
1.3 KΩ,3.18 KΩ
Each Three
13KΩ, ,31.8 KΩ
Each one
3
Quantity
Variable Resistor
500 KΩ pot
Capacitors
0.1 µF
1
3
0.01 µF
2
4
Regulated Power supply
(0 – 30V),1A
1
5
Cathode Ray Oscilloscope
(0 -20MHz)
1
Theory
The function of an oscillator is to generate alternating current or voltage
waveforms.i,e. an oscillator circuits generates a repetitive wave form of fixed
amplitude and frequency without any external input signal. In oscillators positive
feedback is used. There are two conditions for oscillators
•
magnitude of closed loop gain Aβ≥1
•
Total phase shift of the loop gain Aβ=00 or 3600.
In the phase shift oscillator, out of 360o phase shift, 180o phase shift is
provided by the op-amp and another 180o is by 3 RC networks. In the Weinbridge
oscillator, the balancing condition of the bridge provides the total 360o phase shift.
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Circuit Diagrams
Fig. 1 : RC Phase shift oscillator
Fig. 2: Wien Bridge oscillator
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Design
1. Phase shift oscillator
To design a phase shift oscillator with fo =500 Hz
fo = 1/(2πRC 6 )
and gain= RF/R1= 29
Assuming C = 0.1 µF,the value of R is found from
R = 1/ (2π foC 6 ) = 1.3 KΩ
Take R1 = 10R =13 KΩ
RF = 29R1 (use 500K pot)
2. Wien Bridge Oscillator
To design a Wien bridge oscillator with fo =5 KHz
fo = 1/2πRC and RF = 2R1
Assuming C = 0.01 µF, the value of R is found from
R= 1/2πfc= 3.18 KΩ
Take R1 = 10 R=31.8 KΩ
RF = 2R1 (use 100K pot)
Procedures
1. Phase shift oscillator
1. Connect the circuit as per the circuit diagram shown in Fig.1
2. Observe the output waveform on the CRO.
3. Vary the potentiometer to get the undistorted waveform as shown Fig.a.
4. Measure the time period and amplitude of the output waveform.
5. Plot the waveforms on a graph sheet.
2. Wien Bridge Oscillator
1. Connect the circuit as per the circuit diagram shown in Fig.2
2. Observe the output waveform on the CRO.
3. Vary the potentiometer to get the undistorted waveform as shown in Fig.b
4. Measure the time period and amplitude of the output waveform.
5. Plot the waveforms on a graph sheet
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Waveforms
Fig.a: RC Phase Shift Oscillator
Fig.b: Wien Bridge Oscillator
Tabular form and Observations
1. Phase shift oscillator
S.No
Amplitude(VP-P)
Time period
Practical
Theoretical
(ms)
frequency
frequency (Hz)
(Hz)
2. Wien Bridge Oscillator
S.No
Amplitude(VP-P)
Time period
Practical
Theoretical frequency
(ms)
frequency
(Hz)
(Hz)
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Results
Inferences
[
Questions
1. What is an oscillator?
2. How do you change the frequency of oscillation in RC phase shift and Wien
bridge oscillators?
3. What are the applications of oscillators?
4. What is the advantage of using opamp in the oscillator circuit?
5. How do you achieve fine variations in fo ?
6. How do you achieve coarse variations in fo ?
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
7. Function Generator using OPAMPs
Aim: To generate square wave and triangular wave forms by using OPAMPs.
Apparatus required
S.No Equipment/Component name
Specifications/Value
Quantity
1
741 IC
Refer page no.2
2
2
Capacitors
0.01µf,0.001µf
Each one
3
Resistors
86kΩ ,68kΩ ,680kΩ
Each one
Resistors
100kΩ
2
4
Regulated Power supply
(0 – 30V),1A
1
5
Cathode Ray Oscilloscope
(0 -20MHz)
1
Theory: Function generator generates waveforms such as sine, triangular, square
waves and so on of different frequencies and amplitudes. The circuit shown in Fig.1
is a simple circuit which generates square waves and triangular waves
simultaneously.
Here the first section is a square wave generator and second
section is an integrator. When square wave is given as input to integrator it produces
triangular wave.
Circuit Diagram
Fig.1: Function generator
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Design
Square wave Generator
T= 2RfC ln (2R2 +R1/ R1)
Assume R1 = 1.16 R2
Then T= 2RfC
Assume C and find Rf
Assume R1 and find R2
Integrator
Take f= 1/(2π R3 Cf)
Assume Cf find R3
R4 =10 R3
Model Calculations
Square wave Generator
For T= 2 m sec
T = 2 Rf C
Assuming C= 0.01µf
Rf = 2.10-3/ 2.01.10-6
= 10 KΩ
Assuming R1 = 100 KΩ
R2 = 86 KΩ
Integrator
Assume Cf = 0.01µf
R3= 1/(2π f Cf) = 3.18KΩ take R3= 5KΩ
R4 = 50 KΩ
Procedure
1. Connect the circuit as per the circuit diagram shown above.
2. Obtain square wave at A and Triangular wave at Vo2 as shown in Fig.1.
3. Draw the output waveforms as shown in Fig.2(a) and (b).
Observations
Square Wave:
Vp-p = 26 V(p-p)
T = 1.8 msec
Triangular Wave:
Vp-p = 1.3 V
T= 1.8 msec
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Wave Forms
Fig. 2 (a): Output at ‘A’
Fig.2(b): Output at V02
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
.
Results
Inferences
Questions
1. How do you change the frequency of square wave?
2. What are the applications of function generator?
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
8. IC 555 Timer-Monostable Operation Circuit
Aim: To generate a pulse using Monostable Multivibrator by using IC555
Apparatus required
S.No
Equipment/Component
Specifications/Value
Quantity
name
1
555 IC
Refer page no 6
2
Capacitors
0.1µf,0.01µf
3
Resistor
10kΩ
1
4
Regulated Power supply
(0 – 30V),1A
1
5
Function Generator
(1HZ – 1MHz)
1
6
Cathode ray oscilloscope
(0 – 20MHz)
1
Theory:
1
Each one
A Monostable Multivibrator, often called a one-shot Multivibrator, is a
pulse-generating circuit in which the duration of the pulse is determined by the RC
network connected externally to the 555 timer. In a stable or stand by mode the
output of the circuit is approximately Zero or at logic-low level. When an external
trigger pulse is obtained, the output is forced to go high ( ≅ VCC). The time for which
the output remains high is determined by the external RC network connected to the
timer. At the end of the timing interval, the output automatically reverts back to its
logic-low stable state. The output stays low until the trigger pulse is again applied.
Then the cycle repeats. The Monostable circuit has only one stable state (output
low), hence the name monostable.
Normally the output of the Monostable
Multivibrator is low.
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Circuit Diagram
Fig.1: Monostable Circuit using IC555
Design
Consider VCC = 5V, for given tp
Output pulse width tp = 1.1 RA C
Assume C in the order of microfarads & Find RA
Typical values
If C=0.1 µF, RA = 10k then tp = 1.1 mSec
Trigger Voltage must be greater than 1/3 Vcc let it be = 4 V
Procedure
1. Connect the circuit as shown in the circuit diagram.
2. Apply Negative triggering pulses at pin 2 of frequency 1 KHz.
3. Observe the output waveform and measure the pulse duration T high.
4. Compare it with given tp value.
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Waveforms
Fig. 2(a): Trigger signal
(b) Output Voltage
(c) Capacitor Voltage
Readings
Trigger
Output wave
Capacitor output
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Results
Inferences
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Questions
1. Is the triggering given is edge type or level type? If it is edge type, trailing or
raising edge?
2. What is the effect of amplitude and frequency of trigger on the output?
3. How to achieve variation of output pulse width over fine and course ranges?
4. What is the effect of Vcc on output?
5. What are the ideal charging and discharging time constants (in terms of R and C)
of capacitor voltage?
6. What is the other name of monostable Multivibrator? Why?
7. What are the applications of monostable Multivibrator?
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
9. IC 555 Timer - Astable Operation Circuit
Aim:
To generate pulse and square waveforms using IC555.
Apparatus required
S.No
Equipment/Component
Specifications/Value
Quantity
name
1
IC 555
Refer page no. 6
1
2
Resistors
3.6kΩ,7.2kΩ
Each one
3
Capacitors
0.1µf,0.01µf
Each one
4
Diode
OA79
1
5
Regulated Power supply
(0 – 30V),1A
1
6
Cathode Ray Oscilloscope
(0 – 20MHz)
1
Theory
When the power supply VCC is connected, the external timing capacitor ‘C”
charges towards VCC with a time constant (RA+RB) C. During this time, pin 3 is high
(≈VCC) as Reset R=0, Set S=1 and this combination makes Q =0 which has
unclamped the timing capacitor ‘C’.
When the capacitor voltage equals 2/3 VCC, the upper comparator triggers the
control flip flop on that Q =1. It makes Q1 ON and capacitor ‘C’ starts discharging
towards ground through RB and transistor Q1 with a time constant RBC. Current also
flows into Q1 through RA. Resistors RA and RB must be large enough to limit this
current and prevent damage to the discharge transistor Q1. The minimum value of
RA is approximately equal to VCC/0.2 where 0.2A is the maximum current through the
ON transistor Q1.
During the discharge of the timing capacitor C, as it reaches VCC/3, the lower
comparator is triggered and at this stage S=1, R=0 which turns Q =0. Now Q =0
unclamps the external timing capacitor C.
The capacitor C is thus periodically
charged and discharged between 2/3 VCC and 1/3 VCC respectively. The length of
time that the output remains HIGH is the time for the capacitor to charge from 1/3 VCC
to 2/3 VCC.
The capacitor voltage for a low pass RC circuit subjected to a step input of
VCC volts is given by VC = VCC [1- exp (-t/RC)]
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Circuit Diagram
Fig.1: 555 Astable Circuit
Design
Charging Time t c = 0.69 (RA + 2RB) C
Discharging Time t d = 0.69 (RB) C
Total time period T = 0.69 (RA + 2 RB) C
f= 1/T = 1.44/ (RA + 2RB) C
The maker of the 555 timer Signetics defined
Duty cycle (D) = td/T = RB/(RA+2RB)
For a given value of frequency assume C and Duty cycle then find RA, RB
Model calculations
Given f=1 KHz. Assuming c=0.1µF and D=0.25
∴1 KHz = 1.44/ (RA+2RB) x 0.1x10-6 and 0.25 =RB/ (RA+2RB)
Solving both the above equations, we obtain RA & RB as
RA = 7.2K Ω
RB = 3.6K Ω
Procedure
i) Pulse Generator
1. Connect the circuit as per the circuit diagram shown without connecting the
diode OA79.
2. Observe and note down the waveform at pin 6 and across timing capacitor.
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
3.
Measure the frequency of oscillations and duty cycle and then compare with
the theoretical values.
4. Sketch both the waveforms to the same time scale.
ii) Square wave generator
1. Connect the diode OA79 as shown in Figure to get D=0.5 or 50%.
2. Choose Ra=Rb = 10KΩ and C=0.1µF
3. Observe the output waveform, measure frequency of oscillations and the duty
cycle and then sketch the o/p waveform.
Waveforms
Fig. 2 (a) Pulse output (b) Capacitor voltage of Unsymmetrical square wave output
(c)Square wave output
Observations
Parameter
Unsymmetrical
Symmetrical
Voltage VPP
Time period T
Duty cycle
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
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Results
Inferences
Questions
1. What is the effect of C on the output?
2. How do you vary the duty cycle?
3. What are the applications of 555 in astable mode?
4. What is the function of diode in the circuit?
5. On what parameters Tc and Td designed?
6. What are charging and discharging times
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
10. Schmitt Trigger Circuits- using IC 741 & IC 555
Aim: To design the Schmitt trigger circuit using IC 741 and IC 555
Apparatus required
S.No
Equipment/Component name
Specifications/Value
Quantity
1
IC 741
Refer page no. 2
1
2
555IC
Refer page no 6
1
3
Cathode Ray Oscilloscope
(0 – 20MHz)
1
4
Multimeter
5
Resistors
1
100 Ω
2
56 KΩ
1
6
Capacitors
0.1 µf, 0.01 µf
7
Regulated power supply
(0 -30V),1A
Each one
1
Theory
The circuit shows an inverting comparator with positive feed back. This circuit
converts orbitrary wave forms to a square wave or pulse. The circuit is known as the
Schmitt trigger (or) squaring circuit. The input voltage Vin changes the state of the
output Vo every time it exceeds certain voltage levels called the upper threshold
voltage Vut and lower threshold voltage Vlt.
When Vo= - Vsat , the voltage across R1 is referred to as lower threshold
voltage, Vlt . When Vo=+Vsat , the voltage across R1 is referred to as upper threshold
voltage Vut.
The comparator with positive feed back is said to exhibit hysterisis, a dead
band condition.
Hysteresis width VH = VUT-VLT
Circuit Diagrams
Fig. 1: Schmitt trigger circuit using IC 741
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Fig. 2: Schmitt trigger circuit using IC 555
Design
Vutp = [R1/(R1+R2 )](+Vsat )
Vltp = [R1/(R1+R2 )](-Vsat)
Vhy = Vutp – Vltp
=[R1/(R1+R2)] [+Vsat – (-Vsat)]
Procedure
1. Connect the circuit as shown in Fig.1 and Fig.2.
2. Apply an arbitrary waveform (sine/triangular) of peak voltage greater than UTP to
the input of a Schmitt trigger.
3. Observe the output at pin6 of the IC 741 and at pin3 of IC 555 Schmitt trigger
circuit by varying the input and note down the readings as shown in Table 1 and
Table 2
4. Find the upper and lower threshold voltages (Vutp, VLtp) from the output wave
form.
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Wave forms
Fig. 3(a) Schmitt trigger input wave form (b) Schmitt trigger output wave form
Observations
Table 1
Parameter
Input
741
Output
555
741
555
Voltage( Vp-p)
Time period(ms)
Table 2
Parameter
741
555
Vutp
Vltp
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Results
Inferences
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Questions
1. What is the other name for Schmitt trigger circuit?
2. In Schmitt trigger which type of feed back is used?
3. What is meant by hysteresis?
4. What are effects of input signal amplitude and frequency on output?
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
11. IC 565- PLL Applications
Aim:
To design a frequency multiplier using IC 565
Apparatus required
S.No
Equipment/Component name
Specifications/Value
1
IC 565
Refer page no. 8
1
2
IC 555
Refer page no. 6
1
3
Resistors
12KΩ,54.5 KΩ
4
Capacitors
0.01µF
2
0.1 µF
1
10µF
Quantity
Each one
5
Regulated power supply
(0 -30V),1A
1
1
6
Cathode Ray Oscilloscope
(0 – 20MHz)
1
Theory
The frequency divider is inserted between the VCO and the phase
comparator of PLL. Since the output of the divider is locked to the input frequency fIN,
the VCO is actually running at a multiple of the input frequency .The desired amount
of multiplication can be obtained by selecting a proper divide– by – N network ,where
N is an integer. To obtain the output frequency f OUT=2fIN, N = 2 is chosen. One must
determine the input frequency range and then adjust the free running frequency fOUT
of the VCO by means of R1 and C1 so that the output frequency of the divider is
midway within the predetermined input frequency range. The output of the VCO now
should be 2fIN . The output of the VCO should be adjusted by varying potentiometer
R1. A small capacitor is connected between pin7 and pin8 to eliminate possible
oscillations. Also, capacitor C2 should be large enough to stabilize the VCO
frequency.
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Circuit diagram
Fig.1: PLL as Frequency Multiplier
Design
If C= 0.01µF and the frequency of input trigger signal is 2KHz, output pulse
width of 555 in monostable mode is given by
1.1RAC = 1.2T =1.2/f
RA= 1.2/(1.1Cf)=54.5KΩ
fIN=fOUT/N
Under locked conditions,
fOUT = NfIN = 2fIN = 4KHz
Procedure
1. The circuit is connected as per the circuit diagram.
2. Apply a square wave input to the pin2 of the 565
3. Observe the output at pin4 of 565 under locked condition.
4. Give the output of 565 to the pin2 of 555 IC.
5. Observe the output of 555 at pin3.
6. Now give the output of 555 as feedback to the pin5 of the 565.
7. Observe the frequency of output signal fo at pin4 of 565 IC.
8. Draw the wave forms.
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Wave forms
Fig. 2(a): Input
(b): PLL output under locked conditions without 555
(c): Output at pin4 of 565 with 555 connected in the feedback
Observations
Parameter
Input
Output
Amplitude (Vp-p)
Frequency (KHz)
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Result
Inferences
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Questions
1. Out of capture and lock ranges, which is smaller?
2. What is the function of VCO in a PLL?
3. What does happen if frequency divider network -by- 4 is placed in the
feedback?
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12. IC 566 – VCO Applications
Aim: i) To observe the applications of VCO-IC 566
ii) To generate the frequency modulated wave by using IC 566
Apparatus required
S.No
Equipment/Component Name
Specifications/Value
1
IC 566
Refer page no.10
1
2
Resistors
10KΩ
2
1.5KΩ
1
0.1 µF
1
100 pF
1
3
Capacitors
Quantity
4
Regulated power supply
0-30 V, 1 A
1
5
Cathode Ray Oscilloscope
0-20 MHz
1
6
Function Generator
0.1-1 MHz
1
Theory
The VCO is a free running Multivibrator and operates at a set frequency fo
called free running frequency. This frequency is determined by an external timing
capacitor and an external resistor. It can also be shifted to either side by applying a
d.c control voltage vc to an appropriate terminal of the IC. The frequency deviation is
directly proportional to the dc control voltage and hence it is called a “voltage
controlled oscillator” or, in short, VCO.
The output frequency of the VCO can be changed either by R1, C1 or the
voltage VC at the modulating input terminal (pin 5). The voltage VC can be varied by
connecting a R1R2 circuit. The components R1 and C1 are first selected so that VCO
output frequency lies in the centre of the operating frequency range.
Now the
modulating input voltage is usually varied from 0.75 VCC which can produce a
frequency variation of about 10 to 1.
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Circuit Diagram
Fig.1: Voltage Controlled Oscillator
Design
1. Maximum deviation time period =T.
2. fmin = 1/T.
where fmin can be obtained from the FM wave
3. Maximum deviation, ∆f= fo - fmin
4. Modulation index β = ∆f/fm
5. Band width BW = 2(β+1) fm = 2 (∆f+fm)
6.
Free running frequency,fo = 2(VCC -Vc) / R1C1VCC
Procedure
1. The circuit is connected as per the circuit diagram shown in Fig1.
2. Observe the modulating signal on CRO and measure the amplitude and
frequency of the signal.
3. Without giving modulating signal, take output at pin 4, we get the carrier
wave.
4. Measure the maximum frequency deviation of each step and evaluate the
modulating Index.
mf = β = ∆f/fm
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Waveforms:
Fig. 2 (a): Input wave of VCO (b) Output of VCO at pin3 (c) Output of VCO at pin4
Observations
VCC=+12V; R1=R3=10KΩ; R2=1.5KΩ; fm=1KHz
Free running frequency, fo = 26.1KHz
fmin = 8.33KHz
∆f= 17.77 KHz
β = ∆f/fm = 17.77
Band width BW ≈ 36 KHz
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Result
Inferences
Questions
1. What are the applications of VCO?
2. What is the effect of C1 on the output?
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13. Voltage Regulator using IC723
Aim: To design a low voltage variable regulator of 2 to 7V using IC 723.
Apparatus required
S.No
Equipment/Component name
Specifications/Value
1
IC 723
Refer appendix A
2
Resistors
3.3KΩ,4.7KΩ,
Quantity
1
Each one
100 Ω
3
Variable Resistors
1KΩ, 5.6KΩ
4
Regulated Power supply
0 -30 V,1A
5
Multimeter
3
½
digit display
Each one
1
1
Theory
A voltage regulator is a circuit that supplies a constant voltage regardless of
changes in load current and input voltage variations. Using IC 723, we can design
both low voltage and high voltage regulators with adjustable voltages.
For a low voltage regulator, the output VO can be varied in the range of
voltages Vo < Vref, where as for high voltage regulator, it is VO > Vref. The voltage Vref
is generally about 7.5V. Although voltage regulators can be designed using Opamps, it is quicker and easier to use IC voltage Regulators.
IC 723 is a general purpose regulator and is a 14-pin IC with internal short
circuit current limiting, thermal shutdown, current/voltage boosting etc. Furthermore
it is an adjustable voltage regulator which can be varied over both positive and
negative voltage ranges. By simply varying the connections made externally, we can
operate the IC in the required mode of operation. Typical performance parameters
are line and load regulations which determine the precise characteristics of a
regulator. The pin configuration and specifications are shown in the Appendix-A.
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Circuit Diagram
Fig.1: Voltage Regulator
Design of Low voltage Regulator
Assume Io= 1mA,VR=7.5V
RB = 3.3 KΩ
For given Vo
R1 = ( VR – VO ) / Io
R2 = VO / Io
Procedure
Line Regulation:
1. Connect the circuit as shown in Fig.1.
2. Make Vin=10V and by varying the 5.6 KΩ Potentiometer. Set the output
Voltage to 5V by keeping RL =1KΩ
3. By varying Vn from 2 to 20V, measure the output voltage Vo.
4. Draw the graph between Vn and Vo as shown in model graph (a)
5. Repeat the above steps for Vo=3V
Load Regulation: For Vo=5V
1. Set Vi such that VO= 5 V
2. By varying RL, measure IL and Vo
3. Plot the graph between IL and Vo as shown in model graph (b)
4. Repeat above steps 1 to 3 for VO=3V.
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Observations
Line Regulation:
Vo set to 5V
Vi(V)
Vo(V)
Vo set to 3V
Vi(V)
Vo(V)
Load Regulation:
Vo set to 5V
IL (mA)
Vo(V)
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Vo set to 3V
IL (mA)
Vo(V)
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Model graphs
Line Regulation
Load Regulation
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Results
Inferences
Questions
1. What is the effect of R1 on the output voltage?
2. What are the applications of voltage regulators?
3. What is the effect of Vi on output?
.
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14. Three Terminal Voltage Regulators- 7805, 7809, 7912
Aim:
To obtain the regulation characteristics of three terminal voltage regulators.
Apparatus required
S.No
Equipment/Component Name
Specifications/Values
Quantity
1
Bread board
1
2
IC7805
Refer appendix A
1
3
IC7809
Refer appendix A
1
4
IC7912
Refer appendix A
1
5
Multimeter
3
6
Milli ammeter
0-150 mA
1
7
Regulated power supply
0-30 V
1
8
Connecting wires
9
Resistors pot
½
digit display
100Ω ,1k Ω
1
Each one
Theory
A voltage regulator is a circuit that supplies a constant voltage regardless of
changes in load current and input voltage. IC voltage regulators are versatile,
relatively inexpensive and are available with features such as programmable output,
current/voltage boosting, internal short circuit current limiting, thermal shunt down
and floating operation for high voltage applications.
The 78XX series consists of three-terminal positive voltage regulators with
seven voltage options. These IC’s are designed as fixed voltage regulators and with
adequate heat sinking can deliver output currents in excess of 1A.
The 79XX series of fixed output voltage regulators are complements to the
78XX series devices. These negative regulators are available in same seven voltage
options.
Typical performance parameters for voltage regulators are line regulation,
load regulation, temperature stability and ripple rejection. The pin configurations and
typical parameters at 250C are shown in the Appendix-B.
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Circuit Diagrams
Fig. 1: Positive Voltage Regulator
Fig. 2: Negative Voltage Regulator
Procedure
Line Regulation
1. Connect the circuit as shown in Fig.1 by keeping S open for 7805.
2. Vary the dc input voltage from 0 to 10V in suitable stages and note down the
output voltage in each case as shown in Table1 and plot the graph between
input voltage and output voltage.
3. Repeat the above steps for negative voltage regulator as shown in Fig.2 for
7912 for an input of 0 to -15V.
4. Note down the dropout voltage whose typical value = 2V and line regulation
typical value = 4mv for Vin =7V to 25V.
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Load regulation
1. Connect the circuit as shown in the Fig. 1 by keeping S closed for load
regulation.
2. Now vary R1 and measure current IL and note down the output voltage Vo in
each case as shown in Table 2 and plot the graph between current IL and Vo.
3. Repeat the above steps as shown in Fig 2 by keeping switch S closed for
negative voltage regulator 7912.
Output Resistance
Ro= (VNL – VFL) Ω
IFL
VNL -
load voltage with no load current
VFL -
load voltage with full load current
IFL -
full load current.
Observations
Line regulation
Load Regulation
1) IC 7805
Input Voltage
Output Voltage
Load Current
Output Voltage
Vi,(V)
Vo(V)
IL(mA)
Vo(V)
2) IC 7809
Input Voltage
Output
Load Current
Output Voltage
Vi,(V)
Voltage Vo(V)
IL(mA)
Vo(V)
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3) IC 7912
Input Voltage
Output
Load Current
Output Voltage
Vi,(V)
Voltage Vo(V)
IL(mA)
Vo(V)
Model Graphs
Fig.3: IC 7805
Fig.4: IC 7809
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Fig.5: IC 7912
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Result
Inferences
Questions
1. Mention the IC number for a negative fixed three terminal voltage regulator of
12V.
2. Explain the significance of IC regulators in power supply
3. What is drop-out voltage?
4. What is the role of C1 and C2?
5. What are C1 and C2 called?
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15. 4 bit DAC using OP AMP
Aim:
To design 1) weighted resistor DAC
2) R-2R ladder Network DAC
Apparatus required
S.No.
Equipment/Component name
Specifications/Value
1
741 IC
Refer page no. 2
2
Resistors
1KΩ,2KΩ,4KΩ, 8KΩ
3
Regulated Power supply
0-30 V , 1A
4
Multimeter (DMM)
5
Connecting wires
6
Digital trainer Board
Theory:
3
½
digit display
Quantity
1
Each one
1
1
1
Digital systems are used in ever more applications, because of their
increasingly efficient, reliable, and economical operation with the development of the
microprocessor, data processing has become an integral part of various systems
Data processing involves transfer of data to and from the micro computer via
input/output devices. Since digital systems such as micro computers use a binary
system of ones and zeros, the data to be put into the micro computer must be
converted from analog to digital form.
On the other hand, a digital-to-analog
converter is used when a binary output from a digital system must be converted to
some equivalent analog voltage or current. The function of DAC is exactly opposite
to that of an ADC.
A DAC in its simplest form uses an op-amp and either binary weighted
resistors or R-2R ladder resistors. In binary-weighted resistor op-amp is connected
in the inverting mode, it can also be connected in the non inverting mode. Since the
number of inputs used is four, the converter is called a 4-bit binary digital converter.
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Circuit Diagrams
Fig. 1: Binary weighted resistor DAC
Fig. 2: R – 2R Ladder DAC
Design and Model Calculations
1. Weighted Resistor DAC
b
b
b
b
Vo = -Rf  A + B + c + D
 8R 4 R 2 R R
]
For input 1111, Rf = R = 4.7KΩ
Vo = -
Rf
1 1 1
+
+
+
1
]
x5
8 4 2
R

Vo = - 9.375 V
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2.R-2R Ladder Network:
b
b
b
b
Vo = -Rf  A + B + c + D
16R 8R 4 R 2R
] X5
For input 1111, Rf = R= 1KΩ
Procedure
1. Connect the circuit as shown in Fig.1.
2. Vary the inputs A, B, C, D from the digital trainer board and note down the output
at pin 6. For logic ‘1’, 5 V is applied and for logic ‘0’, 0 V is applied.
3. Repeat the above two steps for R – 2R ladder DAC shown in Fig.2.
Observations
Weighted resistor DAC
S.No.
D
C
(MSB)
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B
A
Theoretical Voltage(V)
Practical Voltage(V)
(LSB)
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R-2R Ladder DAC
S.No.
D
C
(MSB)
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B
A
Theoretical Voltage(V)
Practical Voltage(V)
(LSB)
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DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Model Graph
Decimal Equivalent of Binary inputs
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Results
Inferences
Questions
1. How do you obtain a positive staircase waveform?
2. What are the drawbacks of binary weighted resistor DAC?
3. What is the effect of number of bits on output ?
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Additional Experiments
1. Adder - Subtractor
Aim:
To design adder-subtractor circuit by using operational amplifier.
Apparatus required
S.No
Equipment/Component name
Specifications/Value
1
IC 741
Refer page no. 2
1
2
Resistor
1kΩ
5
4
Regulated Power supply
(0 – 30V),1A
3
6
Cathode Ray Oscilloscope
(0 – 20MHz)
1
7
Multimeter
3
Theory: The circuit diagram shows
½
digit display
Quantity
1
an op-amp used as an adder-subtractor. Two
inputs are given for inverting and two for non-inverting terminal of opamp and these
inputs are applied through resistors. The output of the circuit is difference between
sum of non-inverting inputs and sum of inverting inputs.
Circuit Diagram
Fig.1 : Adder-subtractor
Procedure
•
Connect the circuit as per the diagram shown in Fig 1.
•
Apply the supply voltages of +15V to pin7 and pin4 of IC741 respectively.
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•
Apply the inputs V1, V2 ,V3 and V4 as shown in Fig 1.
•
Vary the input voltages and note down the corresponding output at pin 6 of the IC
741 adder circuit.
Notice that the output is equal to the difference between sum of non-inverting inputs
and sum of inverting inputs.
Observations
V1(V)
V2(V)
V3(V)
V4(V)
V0(V)
Model Calculations
Vo = (V1 + V2) - (V3 + V4)
If V1 = 2.5V , V2 = 2.5V, V3 = 2.5V and V4=3V then
Vo = -0.5V
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Result
Inferences
Questions
1. What is the saturation voltage of 741 in terms of VCC?
2. What is the maximum voltage that can be given at the inputs?
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2. Voltage- to- Current Converter
Aim:
To design voltage to current converter with floating load and grounded load
using op amp
Apparatus required
S.No
Equipment/Component name
Specifications/Value
1
741 IC
Refer page no. 2
1
2
Resistors
10 KΩ
5
1KΩ
1
1
3
Regulated Power supply
(0-30V),1A
4
Multimeter
3
5
Ammeter
(0 – 30) µA
6
Digital trainer Board
½
digit display
Quantity
1
1
1
Theory
In many applications we must convert the given voltage into current. The two
types of voltage to current converters are
1. V to I converters with floating load
2. V to I converters with grounded load.
Floating load V – I converters are used as low voltage ac and dc voltmeters, diode
match finders, light emitting diodes and zener diode testers. V to I converters
Grounded load are used in testing such devices as zeners and LEDs forming a
ground load.
Circuit Diagrams
Fig. 1: V – I converter with grounded load
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Fig. 2: V – I converter with floating load
Design
V – I converter with grounded load
I1+I2=IL
(Vin-V1)/R+(Vo-V1)/R=IL
Vin+Vo-2Vi=ILR
Since op-amp is non inverting
Gain=1+(R/R)=2
Vo=2Vi
Vin=Vo-Vo+ILR
IL=Vin/R
V – I converter with floating load
I1=IL
IL = Vin/R1
Procedure
V – I converter with grounded load
1. Connect the circuit as per the circuit diagram shown in Fig. 1.
2. Set ac input to any desired value.
3. Switch on the dual trace supply and note down the readings of ammeter
4. Repeat the above procedure for varies values input voltages.
V – I converter with floating load
1. Connect the circuit as per the circuit diagram in Fig. 2.
2. Apply input voltage to the non-inverting terminal of 741.
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3. Observe the output from CRO and note down the ammeter reading for various
values of input voltage.
Observations
V – I converter with floating load
Vin(V)
Current (mA)
RL=1KΩ
RL=10KΩ
V – I converter with grounded load
Vin
Current(mA)
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Results
Inferences
Questions
1. What is the effect of RL on the output current in V-to-I converter with
floating load?
2. What is the effect of R on the output current in V-to-I converter with
grounded load?
3. For what ranges of currents the circuits are useful?
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3. Precision Half wave Rectifier
Aim: To obtain a precision rectifier (half wave rectifier using IC 741).
Apparatus required
S.No.
Equipment/Component name
Specifications/Value
Quantity
1
741 IC
Refer page no. 2
1
2
Resistors
10 KΩ
5
1KΩ
1
3
Regulated Power supply
(0-30V),1A
1
5
Cathode Ray Oscilloscope
(0-20MHz)
1
6
Digital trainer Board
1
Theory
There are two types of half wave rectifiers. One is inverting half wave rectifier
and second one is non-inverting half wave rectifier. The below circuit show the noninverting half wave rectifier with diode (0A79) in the feed back loop of an op-amp.
Circuit diagram
Fig. 1: Precision Half wave Rectifier
Procedure
1. Connect the circuit as per the circuit diagram.
2. Give the sinusoidal input of 100mVp-p, 1 KHz from function generator.
3. Switch on the dual power supply of + 15V.
4. Note down the output from CRO.
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Model Graphs
Fig. a) Input waveform to the half wave rectifier
b ) Output waveform
Sample readings
Parameter
Input
Output
Amplitude (V),Vp-p
Time period (ms)
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Results
Inferences
Questions
1. What is the output if the diode is reversed?
2. What is a super diode?
3. What is precision rectifier?
4. What modifications you suggest to get negative half cycles at output?
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4. Clipper Circuits using IC 741
Aim:
To obtain the clipped waveforms of the input using IC741.
Apparatus required
S.No
Equipment/Component name
Specifications/Value Quantity
1
741 IC
Refer page no 2
1
2
Resistors
10 KΩ
1
3
Regulated Power supply
(0-30V),1A
1
4
Function generator
(0-1MHz)
1
5
Diode
0A79
1
6
Cathode Ray Oscilloscope
(0-20MHz)
1
Theory
A clipper is a circuit that removes positive or negative parts of the input
signal. In this circuit the op-amp is basically used as a voltage follower with a diode in
the feed back path. The clipping level is determined by the reference voltage Vref
which should be less than input voltage range of op-amp. Additionally +Vref is
derived from the positive supply voltage and -Vref is derived from the negative supply
voltage.
Circuit diagrams
Fig. 1: Positive Clipper
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Fig. 2: Negative Clipper
Procedure
Positive clipper
1. Connect the circuit as per the circuit diagram shown in Fig 1.
2. Apply the reference voltage of 1V.
3. Apply a 6Vp-p of sine wave as input.
4. Note down the output waveform as shown in Fig 3(a) and 3(b).
Negative clipper
1. Connect the circuit as per the circuit diagram shown in Fig 2.
2. Apply the reference voltage of 1V.
3. Apply a 6Vp-p of sine wave as input.
4. Note down the output waveform as shown in Fig 3(c) and 3(d).
Waveforms
Positive clipper
Fig. 3 (a) : Input wave form
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(b) : Output wave form
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Negative clipper
Fig. 3 (c): Input waveform
(d): Output waveform
Observations
Positive clipper
Parameter
Input Voltage
Output Voltage
Input Voltage
Output Voltage
Amplitude (V),Vp-p
Time period (ms)
Negative clipper
Parameter
Amplitude (V), Vp-p
Time period (ms)
Precautions
Check the connections before giving the power supply.
Readings should be taken carefully.
Result
Inferences
Questions
1. What is the effect of Vref on the output?
2. How do you change a positive clipper into negative clipper?
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APPENDIX-A
IC723
Pin Configuration
Specifications of 723
Power dissipation
:
1W
Input Voltage
:
9.5 to 40V
Output Voltage
:
2 to 37V
Output Current
:
150mA for Vin-Vo = 3V
10mA for Vin-Vo = 38V
Load regulation
:
0.6% Vo
Line regulation
:
0.5% Vo
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APPENDIX-B
Pin Configurations
78XX
79XX
Plastic package
Typical parameters at 25oC
Parameter
LM 7805
LM 7809
LM 7912
Vout,V
5
9
-12
Imax,A
1.5
1.5
1.5
Load Reg,mV
10
12
12
Line Reg,mV
3
6
4
Ripple Rej,dB
80
72
72
Dropout
2
2
2
Rout,mΩ
8
16
18
ISL, A
2.1
0.45
1.5
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REFERENCES
1. D.Roy Choudhury and Shail B.Jain, “Linear Integrated Circuits”, 2nd edition,
New Age International.
2. James M. Fiore, “Operational Amplifiers and Linear Integrated Circuits”:
Theory and Application, WEST.
3. “Malvino, Electronic Principles”, 6th edition, TMH
4. Ramakant A. Gayakwad, “Operational and Linear Integrated Circuits”,4th
edition, PHI.
5. Roy Mancini, “OPAMPs for Everyone”, 2nd edition, Newnes.
6. S. Franco, “Design with Operational Amplifiers and Analog Integrated
Circuits”, 3rd edition, TMH.
7. William D. Stanley, “Operational Amplifiers with Linear Integrated Circuits”, 4th
edition, Pearson.
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