COLLEGE OF ENGINEERING ROORKEE
LAB MANUAL
ANALOG INTEGRATED CIRCUIT LAB
PEC 451
LIST OF EXPERIMENT
ANALOG INTEGRATED CIRCUIT LAB(PEC 451)
1. Study of operational amplifier as an Adder and Subtractor.
2. Study of operational amplifier as an Integrator and Diffrentiator.
3. Study of Low pass filter and high pass filter using op-amp.
4. Study of Band pass filter and notch pass filter using op-amp.
5. To study the performance of op-amp V-I converter.
6. To study the working of 741IC OP-AMP and to conduct following experiment.
(a) To measure input offset current
(b) To measure input bias current
7. To study a 4- bit R-2-R Ladder digital to analog converter.
8. To study the process of 4-bit analog to digital conversion by counter method .
EXPERIMENT NO.1
AIM-:Study of operational amplifier as an Adder and Subtractor.
APPRATUS REQUIRED-:Experimental kit, multimeter, connecting wires.
THEORY-: Op-Amp as a SummerIn a given circuit, the feedback current if is a function of sums of input current i1+i2+ib. since ibis
very small for close loop operation thus the relation becomes i1+i2= ib .the input currents are
fraction of input voltages V1 and V2 and resistance R1+R2, since node (- input) accros voltage
zero.
Thus the above relation should be written as –v0/ Rf = V1/R1+V2/R2, the sign ahed of V0 inverting
action if the resistance Rf= R1= R2=R, then the o/p written as ,-V0= Va+Vb( as summing relation).
Op-Amp as a subtractor:The operational amplifier has two input which are designated as (-) inverting and (+) noninverting input. If the voltage of same polarity is fed through these inputs simultaneously then
there will be no output assessing that common node rejection is infinite the property of op-amp is
infinite for realization of subtracting action.an op-amp is fed with two voltage source in to both
inputs.the output voltage is a difference b/w input V2 and V1.since all resistance network is R a
relation may be written for the operation as
V0=V2-V1 ( asubtractor relation)
CIRCUIT DIAGRAM:-
PROCEDURE AS A SUMMER -:1. Connect the feedback resistance RF with the output as
show in fig.1 note the value of input resistance connected with inverting input of op-amp .as
show in fig.1 that are resistors are equal in value R1=R2=RF Neglect the non – inverting input
resistance and it may be show as short circuit with ground.
2. Connect given voltmeter with the output as show in fig 1, with respective polarity. Note any
deflection occurs (0).
3. Connect given another variable DC power supply (+2volt) as V2 with the another input
resistance .adjust supply to obtain – 4volt at output measure the input voltage at V2 variable= +2
volt. Verify the relation 1, as 0-V0=V2+V1.
4. Change the polarity of V2 to other direction by mean of pot (-2volt) and note output voltage
(0volt). Write again the relation and put values for verification. – V0=2+(0-2)=0
PROCEDURE AS A SUBSTRACTOR:-1.Remain the circuit as shown fig.2 note the values of
all resistance in network. Note that only one of the input resistance should be accounted for
inverting input and given R S=R.
2. Connect voltmeter with the polarity show in fig 2. Not any output voltage appears at output(0)
3. Connect supply V1 (+ 2volt) with the non-inverting input through R. note the output voltage
(+ 2 volt).
4. Connect the supply V2 (adjust for +1 volt) with the inverting input note the output voltage as
V0= V1-V2(=1 Volt).
5. Change the polarity of the non- inverting input supply. Note the output voltage .write the
output relation as Vo= V1- (-V2).
OBSERVATION TABLE:As a summer -
S.NO
1.
2.
3.
4.
5.
VA
VB
VO= VA-VB
As a subtractorS.NO
VA
VB
VO= VA+VB
1.
2.
3.
4.
5.
Result:-Studied op-amp as a adder and subtractor is verified.
PRECAUTIONS:- (1) All the connections should be right and tight.
(2) Take the readings carefully.
(3) Power supply should be switched off when not in use.
VIVA QUESTIONS:1)
2)
3)
4)
5)
6)
What is an op amp?
What are ideal characteristics of op amp?
What is the function of adder?
What is scaling amplifier?
What are the applications of subtractor?
What is the use of offset null compensating network in adder?
EXPERIMENT NO.2
AIM-:Study of operational amplifier as an Integrator and Differentiator.
APPRATUS REQUIRED-:Experimental kit, CRO, Function generator, connecting wires.
THEORY-:Op-Amp as a integratorThe given circuit produces an output voltage waveform which is integral of the input voltage (
waveform) the charging current of capacitor is ic= c DVC/dt, the voltage at output and input (
across the capacitor is zero). Applying the input current through input resistance
Rfrom a voltage voltage source Vin ,the following equation may be written Vin/R= C.d/dt (-vo) or
output voltage.
V0= - 1/Rc∫Vindt+ C
Where ,C is called integration constant.
Op-Amp as a differentiator:A differentiator circuit as the name implies this circuit perform mathematical operation of
differentiation. Its output voltage (waveform) is the derivative of the input voltage (waveform) in
the circuit the charge current IF. as stated earlier IF is a function of v0 and RF thus a relation may
be written as C.dVin/dt= -VO/RF,or
Vo=-R.C dVin/dt + V
Where v is differentiation constant.
CIRCUIT DIAGRAM:-
PROCEDURE:-AS A INTEGRATOR
1. Connect the circuit as shown in fig 3 connect function generator output with the input resistor
R, note the values of R (10K) and C(0.1µF).
2. Adjust generator output for 6 Vpp output at 250 Hz= 4msec.square wave, connect cro at input
and measure the input as given.
3. Connect the CRO at output and note the waveform (triangular) and its amplitude in VPP.
4. Change the input frequency to 500Hz= 2msec .square wave of similar amplitude 6 Vpp at
input . Note the output voltage amplitude in VPP.
5. Calculate integration constant from the results obtained from the observation.
6. Calculate the time constant R.C given values as: ʎ = R.C = (Vin/c ).f in. V0 in millisecond if
R in ohm and c in µf.
7. Conclude the result. From observation it is shown that the output voltage is a function of
input frequency, its amplitude and time constant RC. The output decrease with increase in
frequency.
PROCEDURE AS A DIFFERENTIATOR:1.Connect the circuit as fig 4. Connect function generator output with the input of capacitor
C. note the values of R (10K) and C (0.1,µf).
2. Feed the triangular pulses of 250 Hz = 4msec of 6 Vpp at the input. Connect CRO at output
and observe the output waveforms. Measure the output amplitude in vpp.
3. Change the input frequency to 500HZ = 2m sec of 6 VPP and observe the differentiated
output waveforms in shape of square wave measure its amplitude in Vpp . Trace it upon the
paper.
4. From experiment it is shown that the output of differentiator decrease with decrease in
frequency. The width of the differentiated wave from depends upon input periodic time t and
time constant of the circuit RC.
RESULT:-Studied op-amp as an integrator and differentiator can be verified by observing the
output waveforms.
PRECAUTIONS:-
(1)Connections should be tight and right.
(2)switch off the power supply when not in use.
VIVA QUESTIONS
1) What is a differentiator and integrator?
2) What is the effect of resistor Rf that is connected across the feedback capacitor in
practical integrator?
3) If input of integrator is DC voltage, then which type of waveform will be obtained at the
output.
4) Give the applications of a differentiator?
5) If input to differentiator is square wave, then what will be the output?
EXPERIMENT NO.3
AIM:-Study of Low pass filter and high pass filter using op-amp.
APPRATUS REQUIRED:-Experimental kit,a general purpose dual trace CRO, Function
generator, connecting leads.
THEORY:-LOW PASS FILTER
A low pass filter has property to pass out low frequency band which should be below the cut-off
frequency fc. These are many type of electrical circuit to construct these filter but buffer worth
filter is common. In fig 1.(a) base low pass filter is shown in which R-C has finite is show cut-off
characteristics which can be realized asV0= Vin /(1+2𝜋frc)
The above expression is true if the output Vo is terminated at very high impedance. The filter cut
of rate will be -3db octave since. It is a low pass filter .if two successive stage are combined then
it will first order filter. The cut-off rate will be (3) + (-3)= -6db octave. The cut off frequency is
solely depends upon the R-C. an unusual way is made while connecting C1 with the output of opamp rather than ground in practice the output signal is fed back to the node a with a phase
difference produced by the network R2-C2 the phase cancellation effect the cut of rate which
produce shaper slope of 12db/octave the gain of the amplifier is made 3db more higher then it is
required at cut off frequency for buffer worth response. The cut off frequency of the filter is
fc=1/2π√R1R2C1C2
the equation reduced to below if R1=R2 and C1=C2
fc= 1/2πRC
AV = 1+R4/R3= 1.55
HIGH PASS FILTER:In active 2nd order high pass filter is shown. The cut off frequency solely depend upon the C-R
AN unusual way is made while connecting R1with the output of op-amp rather than ground in
practice the output signal is feedback to the node a, with a phase difference produced by the
network C2-R2. The phase cancellation effect the cut-off rate which produce sharper slope of
12dboctave.the gain of the amplifier is made 3 db more higher then it is required at cut-off
frequency. The cut-off frequency of the filter is
fc=1/2π√R1R2C1C2
If R in kΩ and C in µf then f in KHZ Equation reduced to below if R1=R2 and C1=C2
fc = 1/2πRC
the value of R3 and R4 should be to give +3 db gain at fc equal to
Av= 1+R4/R3
= 1.55
CIRCUIT DIAGRAM:-
PROCEDURE AS A LOW PASS FILTER_:-1. Switch on the power connect CRO channel A
at the output connect channel B with the input of low pass filter .trigger CRO with channel B.
2. Adjust function generator for 100 HZ sine wave. Set amplitude control to display 6 vpp sine
wave signal at channel B.
3. Increase frequency of the function generator in 100Hz steps and note the output voltage each
time up to frequency of 10khz more.
4. Tabulate the observations. Plot the band pass band characteristics between input frequency
and output voltage upon a semilog paper and calculate the gain. And find out cut-off frequency.
5. Compute the theoretical frequency and compare it with the obtained f0.
PROCEDURE AS A HIGH PASS FILTER.1. Switch on the power. Connect CRO channel A
the output. Connect channel B with the input of high pass filter .trigger CRO with channel B.
2. Adjust function generator for 10 khz Sine wave. Set amplitude control to display 6 V pp sine
wave signal at channel B.
3. Decrease frequency of the function generator in 500 Hz steps and note the output voltage each
time up to frequency of 1000 Hz take steps of 100 HZ after it up to 50HZ
4. Tabulate the observation. Plot the pass band characteristics between input frequency and
output voltage upon a semilogpaper .calculate the gain. And find out cut- out frequency fc from
the plot.
5. Compute the theoretical frequency and compare it with obtained fc.
RESULT-Studied respective low pass and high pass filter.
PRECAUTIONS:(1) All the connections should be right and tight.
(2) Take the readings carefully.
VIVA QUESTIONS
1. What is the function of the filter?
2. What are the different types of filters?
3. Define pass band and stop band of filters?
4. Define cut off frequency?
5. What is the difference between HPF&LPF?
EXPERIMENT NO.4
AIM:- Study of Band pass filter and notch pass filter using op-amp.
APPRATUS REQUIRED:- Experimental kit,a general purpose dual trace CRO, Function
generator, connecting leads.
THEORY:-
Band PASS FILTER
The pass band frequency is depend upon the C-R combination and Q of the filter. An unusual
way is made while connecting C1and R3 with the output of op-amp. Rather than ground in
practice the output signal is fed back to the node a and b, with a gain/ phase difference produced
by the network C2- R2. The circuit is referred as multiple feedbacks. The result of this feedback
is very sharp pass band within few hertz the fc1 and fc2 are depend upon another factor called Q
of the filter which depends upon C2 and R3. The q of the filter will be
Q= πfc C R3
fc is in KHZ, C in µf and R in kΩ
the centre frequency fc is
fc= Q /2π √R1R2( C1+ C2)/2
fc= Q /2π√ R1R2C
If C1=C2 in µf,R in kΩ,then fc is in KHZ.
CIRCUIT DIAGRAM:-
THE ACTIVE NOTCH FILTER:In fig another R-C is show which has two paths for input signal. Actually it is passive filter,
where one T has made up with two capacitors and resister and other T has two resistors and one
capacitor. The R2=2R1 and C1=2C2 .when it is perfectly balanced it acts as a frequency depend
attenuator that produce nearly zero output at its centre frequency f, a finite output at all other
frequencies. The characteristics reject frequency of this filter is,
f=1/2πRC
Unfortunately the twin T network has a relatively low figure of merit, because of very tight
component tolerance and difficult to adjust it is also used in fixed frequency applications such as
notch filter or narrow band reject filter.
PROCEDURE AS A BAND PASS FILTER-: 1.Switch on the power. Connect CRO channel A
at the output connect channel B with the input of high pass filter. Trigger CRO with channel B .
2. Adjust function generator for 1 khz sine wave.
3. Decrease frequency of function generator in 50Hz steps and note the output voltage each time
up to frequency of 10 Hz.
4. Increse input frequency beyond 12KHZ to khz in step note the output voltage in each step.
5. Tabulate the observation .plot the pass band chracterstics between input frequency and output
voltage upon a semilog paper. Calculate the gain and find the frequency fc from the plot.
PROCEDURE AS A NOTCH FILTER:- 1.Switch on the power. Connect CRO channel B.at
the output connect channel B with the input of notch filter. Trigger CRO with channel A.
2. Adjust function generator for 1 khz sine wave
3. Increase frequency of function generator in 100Hz steps and note the output voltage each time
up to frequency of 10 Hz.
4. Tabulate the observation .plot the rejection chracterstics between input frequency and output
voltage upon a semilog paper. Calculate the gain and find the frequency fn from the plot.
RESULT- The respective Band Pass filter and notch filter waveforms are observed and drawn.
PRECAUTIONS:(1) All the connections should be right and tight.
(2) Switch off the power supply when not in use.
VIVA QUESTIONS
1. What are the differences between active filters & passive filters?
2. Why are the active filters not preferred at higher frequencies?
3. What is the difference between filter, attenuator and equaliser?
4. What is meant by all pass filter?
5. Define Slew rate, SVRR, input bias current, input offset current, input offset voltage.
EXPERIMENT NO.05
AIM:-To study the performance of op-amp V-I converter.
APPRATUS REQUIRED :-Experimental kit ,connecting wires, multimeter.
THEORY:-In circuit, the load is in form of milimeter is connected with ground at non-inverting
terminal. The op-amp has two loops are around the inverting terminal as usual and another
around the non inverting terminal. Assume input current I0 = 0, Then as KCL equation can be
written as.
At node B,
Iin+I2 = IL
(1)
[( Vin- V2)/R]2 + [(Vb-V2)/R]=IL
SO Vin+Vb-2V2= I1R
But the op-amp is in non-inverting mode. Hence the gain is RF/R1=2
Then the o/p voltage will be V0= 2V2= Vin+V0-ILR
Vin= IL R or IL= Vin/R→
(2)
Eq.2 tells that the value of current through the load depends on input voltage Vin& R. in this
case all RB should have equal values these type of circuit employed in photo process and
exposure meters
We know
V= IR
I= V/R
I=V/2
V= 2I
CIRCUIT DIAGRAM:-
PROCEDURE:-1.Switch on the power. Keep care of polarity of meters and input DC source.
2. Keep D.C supply control at minimum switch ON the power .
3. IncreseDCsupply slowly in steps and note the corresponding current from the millimeter.
Tabulate the readings.
4. Connect a patch cords across R1 AND RD points. ( R =1k)
OBSERVATION TABLE-:
S.NO
VOLTAGE (V)
CURRENT(mA)
RESULT-Result is verified using op-amp V-I converter.
PRECAUTION-(1).Connection should be right and tight.
(2) Power supply should be switched of when not in use.
(3) Readings should be taken carefully.
VIVA QUESTIONS
1.
2.
3.
4.
What are the applications of V-I converter?
What do you mean by input biased current?
What do you mean by voltage follower?
What do you mean by an open loop configuration of an op-amp?
EXPERIMENT NO.06
AIM- :-To study the working of 741IC OP-AMP and to conduct following experiment.
(a)Input biasing current measurement.
(b)Input offset current measurement.
APPRATUS REQUIRED-:OP-AMP model-741,dc power supply(0-10v),connecting wire,
THEORY-:An-op-amp is a high gain direct coupled amplifier having low o/p impedance and
high input impedance. The first stage of an op-amp consists of a differential amplifier resulting
in two inputs. They are characterized by (-ve)& (+ve) sign respectively. An op-amp operates
from two power supply (+Vcc)&(-Vcc).
Bias and offset current-:The two transistor of the first stage of an op-amp must draw source
base current for a purpose of operating condition. if the non inverting terminal draw a base
current for a proper operation of non inverting terminals draw a base current of Ib1 & (-ve)
terminals draw a base current of Ib2to the i/p .
I/p bias current = (Ib1+ Ib2)/2
Op-amp are designed in such a way that the input bias current is very small so that the input
signal does not get loaded the theoretically because of the differential amplifier stage the two
base current Ib1&Ib2 should be same.
In practice, however there are never exactly identical this inspires that offset current is defined as
input offset current I0 =Ib1-Ib2
This should always flow and would cause a non zero output voltage from amplifier without an
input this applying a potential at any one of input or preferably by using internal offset balanced
terminal.
CIRCUIT DIAGRAM:-
PROCEDURE INPUT BAIS CURRENT MEASUREMENT-:1.Switch off the offset null
potentiometer.
2.Set R1=1 M ohm, connect 2 to ground and connect 1to 8 and sketch the resulting circuit
diagram.
3. Switch on the power supply wait for 2 minutes and then read the output voltage eout.
4. Calculate the non inverting input bias current as Ib1= eout/31.3µA.
5. Repeat R2= 1 m ohm, connect 2 to ground and 3 to 8 and 9 to meter I/P. Ib2= eout/31.3µA.
6. Calculate the input bais current. Ib= Ib1+Ib2 /2 µA.
PROCEDURE INPUT OFFSETCURRENT MEASUREMENT-:1.Switch off the offset null
potentiometer.
2. Set R1=1 M ohm, R2= 1m ohm connect 2 to ground and connect 1to 8 and sketch the resulting
circuit diagram.
3. Switch on the power supply wait for 2 minutes and then read the output voltage eout.
4. Calculate the input offset current as I0= eout/31.3µA.
5. Compare this value of I0 with the following. I0= (Ib1-Ib2)µA
OBSERVATIONS :FOR Ib1:- EOUT = .862
Ib1 :- eout/31.3 =
FOR Ib2:- EOUT = .281
Ib2= Eout/31.3
Offset current= Ib1 – Ib2
Bias current = Ib1+ Ib2 /2
RESULT-The value of input of bais current is…………
The input offset current is…………….
PRECAUTION- (1).Connection should be right and tight.
(2) power supply should be switched of when not in use.
(3)readings should be taken carefully.
VIVA QUESTIONS
1.
2.
3.
4.
5.
6.
What is OPAMP?
Draw the pin configuration of IC741.
List out the ideal characteristics of OPAMP?
Mention some of the linear applications of op – amps?
Mention some of the non – linear applications of op-amps?
Define input offset current.
EXPERIMENT NO.07
AIM:-To study a 4- bit R-2-R Ladder digital to analog converter.
APPRATUS REQUIRED:-Experimental board on 8-bit R- 2R Digital to analog converter,
digital voltmeter range 0-20 volt.
THEORY-:A circuit which can convert a binary number in to a corresponding voltage/current is
called D to A converter( D/A or DAC).the input to a DAC will be a N-bit digital word and its
output is an analog voltage or current. The various o/p in a DAC are in terms of a full- scale o/p
voltage V0 (Fs) .the highest of 4-bit word is 1111and the number next to this is 10000.The o/p
voltage corresponding to the LSB input(…………….001) is V0 (LSB).similarly for MSB input (
10000……..) it is Vo ( MSB) and for max input ( 1111…….) it is vo (max)
V0( LSB) = VO(Fs)/1
V0( MSB) = VO(Fs)/2
V (max)= V0(Fs)-V0(LSB)
If V0 (Fs)= 10V
V0(LSB)= 10/16= 0.625 Volt
Vo(MSB)= 10/2= 5 volt
V(max)= 10-0.625
=9.375
The methods of D to A conversion are based on generating current proportional to the positional
value of a bit in the binary word. If the bit is high the current is allowed to flow and if the bit is
low –then the current is dissolved.
CIRCUIT DIAGRAM-:
PROCEDURE 4- BIT R-2R DIGITAL-TO-ANALOG CONVERTOR:- 1. Switch on the
supply and adjust the VREF= 10volt.
2. Connect VREF to T5 and digital voltmeter 0—20, volt between T5 –T6.
3.4-bit R-2R network with VREF=10 volt.
4. Now measure the voltage for each binary from 0000 to 1111 and verify the eqs.(1) to (3). You
will find that the voltage 0.625 volt is added for each next binary digit.
5. Find out V0( LSB), V0(MSB) and V (max) theoretically and compare the results obtained
practically.
6. Plot the observations for each binary no.and compare it with fig. and set any other value of
VREF ( say 10.24 volt) and repeat steps.
OBSERVATION:DIGITAL INPUT
ANALOG
OUTPUT(VOLT)
RESULT-: Studied the 4-bit R-2R Ladder digital to analog converter.
PRECAUTIONS-:
(1).Switch off the power supply when not in use.
(2)All the connections should be right and tight.
(3)Reading Should be taken carefully by pressing the right key
corresponding to specific binary number.
VIVA QUESTIONS
1.
2.
3.
4.
5.
What is meant by resolution of DAC?
What is meant by linearity of DAC?
What is meant by accuracy of DAC?
What are the applications of by DAC?
What are the disadvantages of weighted resistor DAC?
EXPERIMENT NO.8
AIM-:To study the process of 4-bit analog to digital conversion by counter method .
APPRATUS REQUIRED-: A/D Converter kit, connecting wires.
THEORY-:A/D Comparator are the power inter phase b/w digital and analog words ADC in
values transition of analog information in to equivalent digital information(word), there are so
many connection techniques in given band counter method is adopted. The band layout designed
is blocks verify the whole process which is described as follows.
(A)Comparator-: Its basic function is to compare I/p analog voltage in) with the binary ladder
o/p voltage ( reference) o/p signal to reset the gate flip-flop.
(B)Flip-flop_-:It is RS flip-flop which is called gate flip-flop. its instruction is to make count
gate enables and disable, where it is set applying start signal. Its o/p goes logic 1 which allows
count gate to pass I/P clock signal to the up-counter.
(C)Up counter-: It is a 4- bit counter the o/p of which is a digital equivalent of analog I/P
initiated by start pluse.
(D)Binary ladder and switched-: There are the scent of 4- bit digital to analog converter. the
up-counter signal switch guard analog switches b/w applied reference and and ground which
produce the resultant o o/p is the form of reference o/p to the comparator.
(E)Start and set mono- flops-: There are set of two mono – flops used to set the count gate
flip-flop and to reset the counter.
CIRCUIT DIAGRAM
PROCEDURE ANALOG TO DIGITAL CONVERTOR-: 1. Connect the given voltmeter
across the analog voltage sockets and adjust input voltage to 1.00 volt with the given knob
situated just under the voltmeter note the comparator status LED comes on disconnect the
voltmeter from the analog input sockets and connect it with the ref –in socket.
2. Apply a brief push on start key . it will start the conversion which is indicated by another two
LEDs fitted with the gate. Input and output.
3. Note the data output LEDs fitted at the top of the board. Also check the change of voltage in
analog meter.
OBSERVATION TABLE-:
ANALOG INPUT(VOLT)
REFERENCE VOLTAGE DIGITAL OUTPUT
(VOLT)
RESULT-:Studied the 4-bit analog to digital converter is verified.
PRECAUTION-: (1).Switch off the power supply when not in use.
(2).All the connections should be right and tight.
(3)Reading Should be taken carefully .
VIVA QUESTIONS
1.
2.
3.
4.
5.
List the broad classification of ADCs?
List out some integrating type converters.
What are the main advantages of integrating type ADCs?
Define conversion time?
Define resolution of a data converter?