G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
Department: - Computer Science & Engineering
Branch: -3rd Semester
Subject: - Analog Circuits
List of Experiments
Cycle I :
1.
To study full wave rectifier with and without filter.
2.
To study zener diode as a shunt voltage regulator.
3.
To study diode as a clipper.
4.
To study diode as a clamper.
5.
To study common base configuration of transistor.
Cycle II :
6.
To study inverting and non-inverting amplifier.
7.
To study adder-subtractor circuit.
8.
To study differentiator and integrator.
9.
To study IC-741 as a triangular wave generator.
10.
To study instrumentation amplifier.
Experiment No. 1
AIM: To study center tap full wave rectifier.
Objective:
1) Study the full wave rectifier circuit without filter.
2) Study the full wave rectifier circuit with filter.
Circuit Diagram:
Theory:
Rectifier is an electronic circuit. Rectifier is a circuit which is used to convert AC into
DC. For positive half cycle P is connected to +ve terminal of battery and N is connected
to –ve terminal of battery hence D1 is forward biased. Thus D1 is shorted while D2 is
reversed biased because P is connected to –ve and N is connected to +ve terminal of
battery. Hence D2 is open circuit.
For negative half cycle P is connected to –ve terminal of battery and N is connected
to +ve terminal of battery hence now D1 is reversed biased that mean it is open circuit.
Negative is given to N type & positive is given to P . Hence D2 is forward biased. Thus it
is shorted.i.e. in case of Full wave rectifier we get output for +ve half cycle as well as –ve
half cycle. This output is called as DC pulsating output.
This rectified output contains ac,and dc. Filters are used to filter out pure dc signal from
rectified output.
Any one of the following filter can be used for filtering.
1)Capacitor Filter
2)Inductor Filter
3)LC Filter.
4) π Filter.
Capacitor Filter: Capacitor is connected in parallel to load resistor R.Rectified signal
contains ac+dc components. Capacitor is short circuit for ac & open circuit for dc. Hence
ac current passes through Capacitor & dc current cannot pass. So dc current passes
through resistor R and it produces nearly dc output voltage. Thus capacitor filter produces
dc output from the rectified signal
Procedure:
1) Connect the circuit as shown in the circuit diagram.
2) Apply ac signal to the circuit.
3) Note down the input as well as output readings
4) Connect capacitor in parallel with the resistor and note down the input & output
readings.
5) Draw the input waveform , rectified waveform and filtered output for a center tap
full wave rectifier.
Observations:
Vin: ______
Vout : (Vdc)(without filter)
Vout : (Vdc) (with filter)
Ac components present in rectified output(Vac)
Result: Thus the full wave rectifier circuit is studied. Ripple factor is_____.
Viva Questions:
1) What is pulsating DC?
2) What is transformer utilization factor for full wave rectifier?
Experiment No. 2
AIM: To study zener diode as a Regulator.
Objectives:
1.To study zener diode as voltage regulator
2. To calculate % line regulation
Equipments:
1.Variable regulated DC power supply: (O-10V),
2. Milliammeter :( O-10OmA),
3. Voltmeter :( O-1OV), (0-1V)
4. Zener diode kit.
Circuit Diagram :
Theory:
Manufactures provide a specification sheet for each type of zener diode.
Ratings include,
1. Zener voltage.
2. Tolerance range of zener voltage.
3. Zener current limits.
4. Maximum power dissipation.
5. Maximum operating temperature.
6. Maximum zener impedance in ohms.
7. Reverse leakage current.
Voltage Regulator: A voltage regulator circuit is required to maintain a constant dc
output voltage across the load terminals in spite of the variation:
1) Variation in input mains voltage
2)Change in the load current
3)Change in the temperature
The voltage regulator circuit can be designed using zener diode.
For that purpose, zener diode is operated always in reverse biased condition. Here,
zener is operated in breakdown region and is used to regulate the voltage across a
load when there are variations in the supply voltage or load current.
The figure shows the zener voltage regulator, it consists of a current limiting resistor
RS connected in series with the input voltage Vs and zener diode is connected in
parallel with the load RL in reverse biased condition. The output voltage is always
selected with a breakdown voltage Vz of the diode.
The input source current, IS = IZ + IL………….. (1)
The drop across the series resistance, Rs = Vin – Vz …….. (2)
And current flowing through it, Is = (Vin – VZ) / RS ………….. (3)
From equation (1) and (2) we get, (Vin - Vz )/Rs = Iz +IL ………… (4)
Regulation with a varying input voltage (line regulation): It is defined as the
change in regulated voltage with respect to variation in line voltage. It is denoted by
‘LR’.
Zener Diode Regulator : In this, input voltage varies but load resistance remains
constant hence, the load current remains constant. As the input voltage increases,
form equation (3) Is also varies accordingly. Therefore, zener current Iz will
increase. The extra voltage is dropped across the Rs. Since, increased Iz will still
have a constant Vz and Vz is equal to Vout. The output voltage will remain constant.
If there is decrease in Vin, Iz decreases as load current remains constant and voltage
drop across Rs is reduced. But even though Iz may change, Vz remains constant
hence, output voltage remains constant.
Observation Table :
Sr.No.
1
Vi(volts)
Vo(volts)
2
3
4
5
6
7
8
9
10
Result: As the input voltage increases the output voltage remains constant i.e the
voltage is regulated.
Viva Questions:
1) Why zener diodes are used in reversed biased condition?
2) What is the difference between avalanche breakdown & zener breakdown?
Experiment No. 3
AIM: To study diode as a clipper.
Objective:
1. To study diode as Clipper.
Equipments:
1.
2.
3.
4.
5.
Breadboard kit .
AC signal source.
Diode.
Resistor.
DC power supply(0-20v)
Circuit Diagram:
Figure 1: Schematic of a clipper circuit.
Theory:
The diode is the simplest of semiconductor devices but plays a very vital role in
electronic systems, having characteristics that closely match those of a simple switch. It
will appear in a range of applications, extending from the simple to the very complex.
The diode has many applications. There are variety of diode networks called clippers
that have the ability to “clip” off a portion of the input signale without distorting the
remaining part of the alternating waveform. The half-wave rectifier is an example of the
simplest form of diode clipper . Depending on the orientation of the diode, the positive or
negative region of the input signal is “clipped” off.
Clipper circuit limits an input voltage to certain minimum and maximum values. In the
circuit in Figure 1, one can see that as long as Vin is less than VB1, then the diode will be
reverse biased(an open circuit). In this case, the output voltagewill track the input
voltage. If Vi exceeds VB1 then the diode turns on and then Vo will beVB1 thus this circuit
limits the output voltage to less than VB1. By rearranging the components, variations on
this circuit can be achieved.
PROCEDURE:
a) Connect the circuit in Figure 1 .R = 100 kohms and a 1N5404
diode.
b) For the input signal, use 5Vp-p, 1kHz sine wave and use power
supply to provide the battery voltage.
c) Measure and sketch the input and output waveforms.
Observations:
Vin: _______
Vout: _______
Result : Thus diode as clipper circuit is studied. Clipping voltage______volts.
Viva Questions :
1) How the anode & cathode terminals of diode can be identified?
2) What are the applications of the clipper?
Experiment No. 4
AIM: To study diode as a Clamper.
Objective:
1. To study diode as Clamper.
Equipments:
1.
2.
3.
4.
5.
Breadboard kit .
AC signal source.
Capacitor.
Resistor.
DC power supply(0-20v)
Circuit Diagram:
Theory:
The diode is the simplest of semiconductor devices but plays a very vital role in
electronic systems, having characteristics that closely match those of a simple switch. It
will appear in a range of applications, extending from the simple to the very complex.
The clamping network is one that will “clamp” a signal to a different DC level. The
network must have a capacitor , a diode, and a resistive element, but it can also employ
an independent DC supply to introduce an additional shift. The magnitude of R and C
must be chosen such that the time constant τ = RC is large enough to ensure that the
voltage across the capacitor does not discharge significantly during the interval the diode
is non conducting.
Clamper circuit works by allowing the capacitor to charge up and act like a battery. This
is the voltage across the capacitor depends on the input waveform, the output maximum(
or the minimum depending on the orientation of the diode) will be clamped to a fixed
reference point (in this case, ground potential). The only design constraint is that 2πRC
be five times larger than the period of the input waveform.
The circuit for a positive clamper is shown in the figure. During the negative half
cycle of the input signal, the diode conducts and acts like a short circuit.The output
voltage Vo = 0V. The capacitor is charged to the peak value of input voltage Vm. and it
behaves like a battery. During the positive half of the input signal, the diode does not
conduct and acts as an open circuit. Hence the output voltageVo=Vm+Vm. This gives a
positively clamped voltage.
PROCEDURE:
a) Connect the circuit in Figure 1 .R = 100 kohms and a 1N5404
diode.
b) For the input signal, use 5Vp-p, 1kHz sine wave and use power
supply to provide the battery voltage.
c) Measure and sketch the input and output waveforms.
Observations:
Vin : _________
Vout :_________
Result : Thus diode as clamper circuit is studied. Clamping voltage______volts.
Viva Question:
1) How the required clamping can be achieved?
2) What are the applications of the clamper?
Experiment No. 5
Aim :- To Plot I/P & O/P Characteristics of Common Base Transistor Configuration in
Active Region .Find I/P & O/P resistance ,Current Gain .
Apparatus:Transistor- SL 100
Resistor – 2.2 KOhm, 2.2KOhm
Ammeter - 0-15mA , 0-1mA
Voltmeter- 0-1.5V , 0-15V
Power Supply-0-30V
Circuit Diagram :-
Theory :In this configuration, emitter current Ie is the i/p current & collector current Ic is the o/p
current. The i/p signal is applied between the emitter & base whereas, o/p is taken out
from the collector & base as shown in circuit diagram. The o/p characteristics can be
plotted between Ic(Y axis) & Vcb(X axis) for constant Ie. In active region, emitter to
base junction Je is forward biased while collector to base junction Jc is reversed biased.
It may be noted that i/p characteristics is linear in the upper region. But
nonlinear in the lower region. Therefore, the AC i/p resistance (dynamic resistance)
depends upon the location of the operating point selected along the curve. It’s value in the
linear region of curve is about 50Ω.
Procedure :I/P Char: 1) Bias transistor in active region.
2) Fix Vcb at 1,2,3 V
3) Increase Vbe in steps of 0.1V & note down corresponding Ie.(10 reading)
4) Plot Graph of Vbe vs.Ie.
O/P Char: 1) Keep Ie constant at 1,2mA .
2) Increase Vcb in steps of 1V & note down the corresponding of Ic. .(10
reading)
3) Plot Graph of Vcb vs.Ic.
Formulae:
Ri = Vbe/Ie at constant Vcb
Ro = Vcb/Ic at constant Ie.
Current Gain = Ic/Ie.
Observation Table:I/p Char: At constant Vcb
O/p Char: At constant Ie
Result :
Ri=
Ro=
Current Gain=
Precaution :1) Do not heat-up the transistor more than the specified time duration.
2) Measure the value of currents and voltages accurately because the change in some
parameter values may be minute.
Viva Questions:1) What is the range in volts for VBE between cut in & saturation for a Si transistor?
2) Define Base spreading resistance for a transistor.
Experiment No. 6A
Aim:- Design and verify gain and frequency response of Inverting and
Noninverting amplifier using IC741.
IC 741 As Inverting Amplifier.
Objective:(1) To study IC 741 as Inverting Amplifier.
(2) To see the effect on O/P by changing R1 & RF.
(3) To Study,” Why this Amplifier is called an Inverting Amplifier”.
Apparatus:- Signal generator, CRO- dual channel, and Patch chord.
Components :R11 = 10K, R12 = 1K, R13 = 100Ω, RF1 = 10K , RF2 = 100K
RF3 = 33K, R2 → 10K, R2 → 10K, R3 → 1K, R4 → 100Ω ,
RL = 10K , IC741.
Theory: An op-Amp can be used for number of application like Amplifier,
Adder, Subtractor , Rectifier ,Multivibrators and Analog computer etc. Here we are
using 741 as a inverting amplifier. It is called as inverting amplifier because here input is
connected at inverting input i.e. pin no.2 So we get inverted signal of the input at the
output The basic ckt of inverting amplifier is shown below. In this mode of operation the
positive input terminal of the amplifier is grounded and the input signal vi is applied to
the negative input terminal via resistor Rr1. The feedback applied through Rf from the
input terminal, is negative. This helps to in maintaining gain stable. The inverting
operation performed by circuit is determined by RF & R1.
Circuit Diagram:-
Procedure:(1) Connect the ckt. as shown in fig.
(2) Select proper R1 & Rf.
(3) Connect 1 – channel of CRO at the o/p & other at I/p.
(4) Connect signal generator at I/p. Adjust I/p at 200m.vp-p.
(5) Observe the o/p with respect .to I/p.
(6) Observe the change in o/p by change in resistor between pin 3 & gnd.
(7) Calculate theoretical & practical gain.
(8) Draw the waveform on graph.
Observation Table: -
Simulation Results:-
Experiment No. 6B
IC 741 As Non-Inverting Amplifier.
OBJECTIVES;(1) To study IC 741 as Non-inverting amplifier
(2) To see the effect on o/p by changing Rj & Rf.
Apparatus:Signal generator, CRO, patch chord = 3 no.
Component value:R11 = 1K R12 = 10K R13 = 107 pot RL = 10K
RF1 = 10K RF2 = 100K RF3 = 33K IC = 741.
Theory: An OP-AMP can be used for number of application like Amplifier, Adder, Substractor,
Rectifier,Multivibrators, and Analog computer etc. Here we are going to study 741 as a
non-inverting amplifier. Itis called as non-inverting amplifier because input is applied at
pin no.3 i.e. non-inverting input. So we geto/p signal in phase with input signal. In this
case the i/p signal is applied directly to the non-inverting (+ve) i/p terminal of the
amplifier & the feed back resistor ‘RF’ is connected between the o/p terminal &
negative I/p terminal. The ‘R’ is connected between the inverting terminal & ground.
Note that Vi is not equal to zero in this case, meaning that non-inverting ckt has to virtual
ground at one of it’s i/p terminals.Thus the closed loop gain of a non-inverting amplifier
is always greater than or equal to unity & it is determined by R1 & Rf.
Circuit Diagram :
Procedure:(1) Connect the ckt. as shown in fig.
(2) Select proper R1 & Rf.
(3) Connect 1- channel of CRO at o/p & other at i/p.
(4) Connect signal generator at i/p.
(5) Observe the change in o/p by changing R1, Rf and frequency of i/p.
(6) Draw the waveform on graph paper.
Observation Table:-
Input-output waveforms:-
Result:The I/p signal is amplified at the o/p & is in phase with I/p signal.
The input signal is amplified & inverted at the output of inverting amplifier.
The input signal is amplified & non-inverted at the output of inverting amplifier.
Viva Questions :1) What is virtual ground concept?
2) What is feedback? Which type of feedback is used in linear application?
Experiment No.7
Aim:- Design and verify op-amp application as adder and substractor.
Apparatus:-IC 741,Bread Board, Power Supply, Connecting Wires, resistors.
Components Values: For (adder)
Ra = 1K, Rb = 1K, RF = 1K, R = 250Ω, IC 741.
For (substractor)
Ra = 1K, Rb = 1K, RF = 1K, R = 1K, IC 741
Circuit Diagram:-
ADDER
SUBTRACTOR
Theory: Figure shows inverting configuration with two inputs V1, V2. The circuit can
be verified by examining the expression for the voltage V0, which is obtained KCL at
node V2. From figure shown,
Ia + Ib = IB + I F ---------(1)
Since Ri and A of the op-amp are ideally infinity, IB = 0A and V1 = V2 = 0V.Therefore,
If in the above circuit Ra = Rb = Rf The equation can be written as,
Vo = -(Rf / R) (V1+V2)
This means that the output voltage is equal to negative sum of all the inputs times the
gain of the circuit -hence, is called as a summing amplifier. When the gain of the circuit
is unity that is Ra = Rb = Rf, the output voltage is equal to negative sum of all input
voltage. Negative gain in this equation indicates that there is a phase shift of 180 between
the input and output.
Procedure:For Adder Circuit:
1) Connect the circuit as shown in figure.
2) Give the supply voltage to op-amp.
3) Apply the input signals Va, Vb, to the inverting input terminal of op-amp.
4) Note the output for the corresponding voltages.
5) Take the reading for several input voltages Va, Vb, Vc.
6) Calculate the theoretical and practical output voltage.
Observation Table:-
For subtractor Circuit:1) Connect the circuit as shown in figure.
2) Give the supply voltage to op-amp.
3) Apply the input signals Va, Vb, to the inverting input terminal of op-amp.
4) Note the output for the corresponding voltages.
5) Take the reading for various of input voltages Va, Vb,
6) Calculate the theoretical and practical output voltage.
Observation Table: -
Result:- Thus theoretical and practical values are verified.
Viva Questions :1) State two practical application of adder/substractor ckt.
2)Which type of Op-Amp mode is used for adder and why?
Experiment No. 8
Aim:- Design and verify gain and frequency response of Integrator and
Differentiator ckt. using IC741.
Apparatus:Op- Amp 741, Resistances , Capacitor, Function Generator.
Theory:Integrator and Differentiator: A circuit in which the output voltage is the integration of the input voltage is called as the
integrator or the integration amplifier. Such a circuit is obtained by using a basic
inverting amplifier configuration if the feedback resistor RF is replaced by a capacitor CF
The expression for the output voltage Vo can be obtained by writing Kirchhoff′s current
equation at node V2 I =IB + IF (Since IB is negligibly small, I ≅ IF) Recall that the
relationship between current through and voltage across the capacitor is Ic = C dv/dt .
Therefore (Vin - V2)/R1 = CF (d/dt)(V2-Vo)
However, V2 ≅0 (virtual ground Concept) because A is very large.
Therefore, Vin/R1 = CF d/dt(-Vo)
The output voltage can be obtained by integrating both sides with respect to time:
INTEGRATOR CIRCUIT:-
Simulation Result :a) For Sine wave as input
b) Square wave as input
DIFFERENTIATOR CIRCUIT:
Simulation Results :-
Procedure:- (Integrator)
1) Switch on the power supply.
2) Apply input from function generator i.e. sine or square the input should be
such that T> RF. (CF)
3) Connect C1 in parallel with RF by patch cords.
4) Connect output of integrator to CRO by probe and observe the waveforms.
5) Vary input frequency and observe the change in output waveforms.
Procedure:- (Differentiator)
1) Switch on the power supply.
2) Apply input from function generator i.e. sine or square (the input should be such that
T> RF.(CF).
3) Connect C1 in parallel with RF by patch cords
4) Connect output of differentiator to CRO by probe and observe the waveforms
5) Vary input frequency and observe the change in output waveforms
Result:- Thus output waveforms of integrator and differentiator are studied.
Viva Questions :1)What do you mean by cut off frequency.
2) Explain integrator and diffentiator
Experiment No. 9
Aim : To study IC-741 as a triangular wave generator.
Objectives:1) To study IC 741 as triangular wave generator.
2) To measure it’s time period.
Apparatus:
Instruments required:- CRO , Multimeter etc.
Components used :A1=A2=IC 741 , R1=100K(Pot)+47K , R2=10K,
R3=47K, C1=0.01uf
Theory:A waveform generator is a device that generates Sine wave ,square wave &
triangular wave. The square and triangular wave can be generated using the principle,
Which involved in charging & discharging of a capacitor. One of the application of
741 is in waveform generation.
In integrated circuit the output is triangular if it’s input is square wave. From this
we can construct triangular wave generator, simply by giving the output of square wave
generator to an integrator. Figure shows the circuit diagram of triangular wave generator.
The OP-AMP A1 compares the voltage V1 continuously with inverting input voltage,
which is 0V. When the voltage V1 goes slightly below or above volt , the output of A1
will goes negative or positive saturation level respectively, means output having only
two stages,0V and +Vcc , which the square wave output .
Circuit Diagram:
When input to an integrator A2, is ‘+Vcc’,the output of A2 decreases linerly until it
reaches the lower trip point of the comparator which is -VTR .
The comparator output then switches rapidly to its negative saturation level –V and the
integrator output then increases linearly. When the integrator output reaches the upper
trip point , ‘+VTR’, comparator again switches and the operation repeated again & again.
-VTR = - R2/R3(+Vsat) --------------------(1)
+VTR = - R2/R3(-Vsat) --------------------(2)
From the above equations the peak to peak output amplitude of the triangular wave is ,
VO(p-p) = +VTR - (-VTR)
VO(p-p) = (R2/R3)(Vsat)
Where Vsat = +VTR - ( -VTR)
Fig(2) shows that output waveforms ,
The time period of the triangular wave is ,
T = (4R1C1R2)/R3
----------------------(3)
Frequency of oscillator given F = 1/T
By varying the value of R1 output frequency can be varied.
Procedure :1. To study the circuit given on the panel of the kit.
2. Connect the one channel of CRO to the output of opamp(1) to observe the square
wave and second channel of CRO to the output of the circuit , to observe the
triangular wave.
3. Measure the amplitude and frequency of the square wave and triangular wave.
4. Vary the pot R1 and observe the change in both the outputs.
5. Draw both the waveforms on graph paper and compare the theoretical and
observed frequency and amplitude.
Observation:Sr.
R1
No.
1.
Min.
2.
Med.
3.
Max.
Calculated
Freq.
Observed
Frequency
Calculated
Amplitude(P-P)
Observed
Amplitude(P-P)
Result:The frequency of the triangular waveform can be changed by changing R1.
Viva Vice :
1) What is the value of resistor R1 used in this experiment?
2) Which type of feedback is used in triangular wave generator ?
Experiment No. 10
Aim: - To Study instrumentational Amplifier .
Apparatus:DC power supply
Op-amplifiers
CRO
Function generator
Circuit diagram:-
Theory:- It is beneficial to be able to adjust the gain of the amplifier circuit without
having to change more than one resistor value. This intimidating circuit is constructed
from a buffered differential amplifier stage with three new resistors linking the two buffer
circuits together. Consider all resistors to be of equal value except for Rgain. The negative
feedback of the upper-left op-amp causes the voltage at point 1 (top of Rgain) to be equal
to V1. Likewise, the voltage at point 2 (bottom of Rgain) is held to a value equal to V2.
This establishes a voltage drop across Rgain equal to the voltage difference between V1
and V2. That voltage drop causes a current through Rgain, and since the feedback loops of
the two input op-amps draw no current, that same amount of current through Rgain must
be going through the two "R" resistors above and below it. This produces a voltage drop
between points 3 and 4 .
The regular differential amplifier on the right-hand side of the circuit then takes this
voltage drop between points 3 and 4, and amplifies it by a gain of 1 (assuming again that
all "R" resistors are of equal value). Though this looks like a cumbersome way to build a
differential amplifier, it has the distinct advantages of possessing extremely high input
impedances on the V1 and V2 inputs (because they connect straight into the noninverting
inputs of their respective op-amps), and adjustable gain that can be set by a single
resistor. Manipulating the above formula a bit, we have a general expression for overall
voltage gain in the instrumentation amplifier.
Procedure:1. Assemble the circuit.
2. Apply input signal .
3. At different input signals observe output .
Observations:
Vout1 =_________
Vout2 =_________
Vout3 =_________
Result:The ouput voltage is found to be _________
Viva Questions:1. what are the applications of instrumentation amplifier?
2. what are the ideal characteristics of op-amp?