unit ii: applications of op-amp part -a (2 marks) 1....

EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 1

UNIT – II: APPLICATIONS OF OP-AMP

PART -A (2 Marks)

1. An input of 3v is fed to the non-inverting terminal of an op-amp. The amplifier has

a Ri of 10KΩ and Rf of 10KΩ .Find the output voltage[AUC April 2004]

V0 =[

2. Draw an integrator circuit using op-amp.[ AUC April 2004]

3. Mention two characteristics of instrumentation amplifier.[ AUC May 2005]

high gain accuracy

high CMRR

high gain stability with low temperature coefficient

low output impedance

4. Mention two applications of Schmitt trigger.[ AUC May 2005]

To convert sine wave to square wave.

Over voltage and over current protection circuit.

On /off temperature controllers.

5. Draw an adder circuit using op amp to get the output expression as vo=-

(0.1+v2+5v3)[ AUC May 2006]


6. Draw the circuit of a voltage follower using op-amp and prove that its gain is

exactly equal to unity.?[ AUC May 2006]

Va = Vb = vin

Node a is directly connected to output

V0= Va

V0=Vin

For the circuit voltage gain is unity.

7. Draw the block diagram of a multiplier using log and antilog amplifiers.[ AUC May

2006]


8. State the requirements of an instrumentation amplifier. [AUC Nov 2006]

high gain accuracy

high CMRR

high gain stability with low temperature coefficient

low output impedance

9. Name four applications of operational amplifier based comparator.[ AUC Nov

2006,April 2008]

ero crossing detector

Window detector

Marker generator

Phase meter

10. What is an antilog amplifier? Draw the circuit diagram of an antilog amplifier.[ AUC

Nov 2007]

Amplifier that converts logarithmic numbers back to decimal numbers is called antilog

amplifiers.

11. What is the principle of regenerative comparator.[ AUC Nov2007]

If positive feedback is added to the comparator circuit, gain can be increased. Hence ,the

transfer curve of comparator becomes more closer to ideal curve. If the loop gain is adjusted

to unity then the gain becomes infinity. This results in an abrupt transition between the

extreme values of the output voltage.

0 0

U3

AD741

3

2

74

6

1

5+

-

V+

V-

OUT

OS1

OS2

B

D1

D1N4001

RF

1k

A

IVO

V4

FREQ = VAMPL = VOFF =

I


12. Draw the circuit diagram of a non-inverting amplifier ?[ AUC May 2008]

13. Draw the circuit of an integrator[AUC Nov 2008]

14. What is a V to I converter?[ AUC Nov 2008]

Converter that produces output current which is directly proportional to the input voltage is

called as V to I converter.


15. Draw the Schmitt trigger and give its application?[ AUC May2009]

Applications

o To convert sine wave to square wave.

o Over voltage and over current protection circuit.

o On /off temperature controllers.

16. Draw the input and output waveforms for an integrator for square wave input.[

AUC May 2010]

0R1

1k

0

R2

1k

R3

1k

V1


ROM

1k

U1

AD741

3

2

74

6

1

5+

-

V+

V-

OUT

OS1

OS2

vo

0


17. Mention the applications of an instrumentation amplifier.[ AUC May 2009]

Data acquisition system.

Temperature indicator

Temperature controller.

Light intensity meter.

18. In what way, a precision rectifier using op-amp is superior to a conventional

rectifier. [AUC Nov09 ,MAY 2011]

Conventional rectifier cannot rectify voltages below 0.6V.

Precision rectifiers rectify voltages having amplitude less than 0.7V

19. Draw an op-amp subtractor circuit. . [AUC Nov09]

20. Design an inverter using op-amp [AUC MAY 2010 ,MAY 2011]


21. Design a peak detector using op-amp [AUC MAY 2010]

22. Distinguish between active and passive filter. [AUC MAY 2011]

ACTIVE FILTER PASSIVE FILTER

Tuning is easy Difficult to tune

Does not cause loading effect Causes loading effect to source or load

More economical Cost is high

23. Why active guard drive is necessary for an instrumentation amplifier?[AUC MAY

2012]

The common ground is shared by variety of circuits.

Due to ground loop interference , additional voltage drop develops and lead to

error in low voltage measurement.

Due to distributed cable capacitances degradation of CMRR occurs.

The active guard drive eliminates all these problems.

24. What is comparator? [AUC MAY 2012]

Comparator is a circuit that compares the voltage applied at one of its input to that

applied at its other input and to produce an output voltage which is either Vh or vl


-vee

vin

I0

0

VD0

I0 ib=0I0

vee

R4

1k

vf

U1

AD741

3

2

74

6

1

5+

-

V+

V-

OUT

OS1

OS2

RL

1k

V2

25. Draw a non –inverting amplifier with a voltage gain of 3 [AUC NOV 2013]

26. Give an application for each of the following circuits : voltage follower ,peak

detector ,Schmitt trigger , clamper

Voltage follower : used in isolator circuit

Peak detector : used for amplitude modulation in communication applications.

Schmitt Trigger : square wave generators , ON –OFF controllers.

Clamper : DC restorer in television receivers.

PART-B (16 marks)

1. a)Explain briefly about the working of voltage to current converter[AUC May 2004]

Voltage to current converter

This circuit converts voltage to current

Circuit diagram

Vin +

-

U2

AD741

3

26

7 14 5

Ri

1k

Rf

2k


Operation

Input voltage is applied to the non inverting terminal .

Load resistance is connected in place of feedback resistor rf.

This circuit is also called as negative feedback amplifier(current series) R1 is directly

proportional to Io

Applying KVL to the input loop

Vin=vd+vf

But open loop gain Av of op-amp is very large

Vd ~ 0

Vin=Vf

But Vin=R1*I0

I0=Vin /r1

Thus vin is converted into proportional output current (I0 = vin / R1)

Applications

Low voltage dc and ac voltmeters

LCD and zener diode testers

b) )Explain briefly about the working of triangular wave generator .

Triangular wave generators

This circuit consist of a square wave generator and an integrator.

The square wave generator generate square wave at its output. This square wave is integrated

by the integrator to generate a triangular waveform.


Circuit diagram

Waveform

Triangular wave is generated by alternatively charging and discharging a

capacitor with a constant current source.

Assume V0’ is high at +Vsat .

This forces a constant current (+Vsat / R4) through C.

When V0’ is low at –Vsat .

This forces a constant current (-Vsat / R4) through C.

The frequency of the triangular wave is same as the square wave frequency.

The amplitude of the triangular wave will decrease as the frequency increases.

0

0

R9

1k

V02

U2

AD741

3

2

74

6

1

5+

-

V+

V-

OUT

OS1

OS2

C2

1n

C1

1n

R7

1k

0

R4

1k

R5

1kR6

1k

U3

AD741

3

2

74

6

1

5+

-

V+

V-

OUT

OS1

OS2

R8

1k


2. a)Draw the circuit diagram of an op-amp differentiator and derive an expression for

the output in terms of the input. [AUC Nov,04]

Differentiator

A differentiator circuit produces differentiated version of the input voltage applied to it

Circuit diagram

Waveforms

V2

15Vdc

0

V

0

V

15.00V

R3

1k

-15.00V19.30uV

V10

TD = 0.5ms

TF = 1msPW = 0.5msPER = 0.5ms

V1 = 5v

TR = 1ms

V2 = 0v

-60.47uV5.000V

0

U1

AD741

3

2

74

6

1

5+

-

V+

V-

OUT

OS1

OS2

C1

0.01u

R2

1k

0

-79.72uV

V3

15Vdc


Analysis

Ic = IB +IF

IB ~ 0

IC = IF

w.k.t IC =C DV/DT

In this circuit C is replaced by C1

IC=C1 DVC/DT

Voltage across c is given by

Vc =C1 D/DT(VIN-V2)

But IF=V2- VO / RF

But IC and IF are equal because IB =0

V2-VO /RF =C1 D/DT(VIN-V2)

V1=V2=0

-VO/RF=C1D/DT(VIN)

VO=-RFC1 D/DT(VIN)

Thus the output is –rfc1 times the time derivative of the input voltage

Applications :

1. P-I-D Controllers

2. High pass filter

3. Wave shaping circuit to generate narrow pulses


b)Design a schmitt trigger for UTP=0.5v and LTP=-0.5v

Vut =+0.5V , Vlt =-0.5V , Supply +/- 15V

=

Op-amp 741 with Ib (max) =500nA

Let I2 be much higher than Ib(max)

I2 = 100 Ib(max) = 50

R2 = Vut /I2 = 0.5 / 50 =10 KΩ

R1=V0-Vut / I2 =13.5 – 0.5 /50 = 260 KΩ =270 KΩ (practically)

3. a)State the advantages and limitations of active filters.

Advantages :

Gain and Frequency adjustment flexibility:

Since the op-amp is capable of providing a gain, the i/p signal is not attenuated as it

is in a passive filter. [Active filter is easier to tune or adjust].

No loading problem:

Because of the high input resistance and low o/p resistance of the op-amp, the active

filter does not cause loading of the source or load.

Cost:

Active filters are more economical than passive filter. This is because of the variety of

cheaper op-amps and the absence of inductors.


Limitations :

Finite bandwidth limits high frequency operation.

Sensitivity : Active filters are sensitive to the temperature and environmental

changes.

Requirement of dc power supply.

b)A second order low pass filter at cut off frequency 1KHz [AUC Dec 2006]

fc =

Assume :R1=R2=R , C1=C2=C

f=

Given f=1 Khz ; Af= ; Af =3-1.1414 ; Af=1.586

Choose C=0.1µF

R= = =1.59 KΩ

Af=1+

0.586=

Rf =0.586 *10 =5.86 KΩ

Let Ri=10KΩ


4. a)Explain the working of an instrumentation amplifier. [NOV 2013]

Instrumentation amplifiers

An instrumentation amplifier is basically a differential amplifier which meets certain

requirements.

In industries the measurement and control of quantities such as temperature , pressure is

required

A transducer is used to convert these quantities into a proportional electrical signal.

Output of transducer is then applied to an amplifier called instrumentation amplifier

This amplifier amplifies the low level output signal of the transducer to such a level it can drive

the indicator or display.

Requirements

Precise

Low level signal amplification

High CMRR

Low noise

High input resistance

Low power dissipation

High slew rate

Circuit diagram


Analysis

Voltages at nodes A and B are given as

Va=V2 and Vb=V1

Expression for current is

I = Va-vb/ r2 = v2-v1 / r2

Output voltage of op-amp 2 is given by

V02 = Va + ir1

=V2+V2-V1/r2 * r1

=v2r2 +(v2-v1)r1 / r2 = v2(r2+r1) – v1 r1 / r2

Similarly output of op-amp1 is given by

V01=(r1+r2) v1 – r1v2 / v2

Output of first stage is given by

V02 – V01 = v2(r2+r1) – v1 r1 / r2 - (r1+r2) v1 – r1v2 / v2

=(2r1+r2)(v2-v1) /r2

Vo2-v01 = [1+ 2r1/r2][v2-v1]

Av1=vo2-v01/ v2-v1 = 1+ 2r1/r2

Similarly Av2= r4/r3

Overall gain of instrumentation amplifier is given by

Av=Av1 * Av2 =(1+2r1/r2) * r4/r3

Output voltage is then given by

Vo=av(v2-v1)

Applications


Temperature controller

Light intensity meter

Medical equipments


5. a)Draw the circuit of a second order butterworth low pass filter and derive its

transfer function.

The above diagram is transferred to S domain

A second order LPF having a gain 40dB/decade in stop band. A First order LPF can be

converted into a II order type simply by using an additional RC network.

The gain of the II order filter is set by R1 and RF, while the high cut off frequency fH is

determined by R2,C2,R3 and C3.


Writing Kirchoff’s current law at node VA (S) .

I1 = I2 + I3 ------------(1)

V in - V A / R2 = V A-V 0 / 1/SC2 + V A-V1/R3 ------------------2

using voltage divider rule,

V 1 = [1/SC3] / [R3 + 1/SC2] VA

V A =V1/[R3C3 s + 1]

V1 =V A [ R3 C3 S + 1]

Substituting the value of VA in eqn 2 and solving for V1 , we get,

V in-V A [1+R3 C3 S] / R2=V A[1+s R3 C3]-V0 / [1/sc2] + V A[1+s R3 C3]-VA/ R3

]

]

]

For an op-amp in non inverting configuration

Vo=Af VA

Af=

VA= The voltage at the non inverting terminal

Vo=Af ]

= V0[1- ]

= V0

The denominator quadratic in the gain (V0/Vin) eqn must have two real and equal roots. This means that w 2 H = 1/R2 R3 C2 C3 wH = 1/R2 R3 C2 C3 q (2 πf H)2 = 1/ R2 R3 C2 C3 Fh =1/ 2π√R2 R3 C2 C3 For a second-order LP Butterworth response, the volt gain magnitude eqn is, V 0 /Vin =AF / √1+(f/fh)4


b)Draw the circuit of an astable multivibrator using op-amp and derive an

expression for its frequency of oscillation[AUC June 2006]

Expression for frequency of oscillation

Similarly

Total time period is given by T= t1+t2=

= [a]

=

= [b]

Subtracting 1 on both sides in equation[a]

= -1

= [c]

Dividing equation [b] and [c] we get

V0U5

AD741

3

2

74

6

1

5+

-

V+

V-

OUT

OS1

OS2

0

V18

-15Vdc

R3

1k

R2

1k

V19

15Vdc

V

C1

0.01u

vd

0

0

R

1k

0


0 0

VIRTUAL GROUND

U3

AD741

3

2

74

6

1

5+

-

V+

V-

OUT

OS1

OS2

D1

D1N4001

R1

1k

I1

VO

V4


V01

T=T1+T2 =

f0 =

6. Explain briefly about log amplifiers[AUC Dec 2007]

Log amplifier :

These circuits are used in analog computers.

Circuit diagram

PN junction diode is connected in the feedback path. Output voltage is nothing but the

voltage across the diode.

V0= - Vf-------1

Due to infinite input impedance, the current going into the inverting terminal is zero.

i1 = if= Vin/R-------2

Generally,

Vf = ή vt[log (if) –log(i0)]-----3

Substituting 3 in 1 we get

V0=- ή vt[log (if) –log(i0)]

= ή vt[log (Vin/r) –log(i0)]

Vo= -ή vt log [ vin / r *i0]

Vo= -ή vt log [ vin / vref]


Where vref = r.Io

Vo is directly proportional to the logarithm of the input voltage.

7. With diagram explain the working principle of ICL 8038 function generator. [AUC

Dec 2007]

Functional Diagram

Output Waveform :


It consists of two current sources, two comparators, two buffers, one FF and a sine wave

converter.

Pin description:

Pin 1 & Pin 12: Sine wave adjusts:

The distortion in the sine wave output can be reduced by adjusting the 100KΩ pots connected

between pin12 & pin11 and between pin 1 & 6.

Pin 2 Sine Wave Output:

Sine wave output is available at this pin. The amplitude of this sine wave is 0.22 Vcc. Where

5V ≤ Vcc ≤ 15 V.

Pin 3 Triangular Wave output:

Triangular wave is available at this pin. The amplitude of the triangular wave is 0.33Vcc. Where

5V ≤ Vcc ≤ 15 V.

Pin 4 & Pin 5 Duty cycle / Frequency adjust:

The symmetry of all the output wave forms & 50% duty cycle for the square wave output is

adjusted by the external resistors connected from Vcc to pin 4. These external resistors &

capacitors at pin 10 will decide the frequency of the output wave forms.

Pin 6 + Vcc:

Positive supply voltage the value of which is between 10 & 30V is applied to this pin.

Pin 7 : FM Bias:

This pin along with pin no8 is used to TEST the IC 8038.

Pin9 : Square Wave Output:

A square wave output is available at this pin. It is an open collector output so that this pin can be

connected through the load to different power supply voltages. This arrangement is very useful

in making the square wave output.

Pin 10 : Timing Capacitors:

The external capacitor C connected to this pin will decide the output frequency along with the

resistors connected to pin 4 & 5.

Pin 11 : -VEE or Ground:

If a single polarity supply is to be used then this pin is connected to supply ground & if ( )

supply voltages are to be used then (-) supply is connected to this pin.

Pin 13 & Pin 14: NC (No Connection)

Important features of IC 8038:

1. All the outputs are simultaneously available.

2. Frequency range : 0.001Hz to 500kHz

3. Low distortion in the output wave forms.

4. Low frequency drift due to change in temperature.

5. Easy to use.


Parameters:

(i) Frequency of the output wave form:

The output frequency dependent on the values of resistors R1 & R2 along with the external

capacitor C connected at pin 10.

If RA= RB = R & if RC is adjusted for 50% duty cycle then fo = RC 0.3 ; RA = R1, RB = R3, RC

= R2

(ii) Duty cycle / Frequency Adjust : (Pin 4 & 5):

Duty cycle as well as the frequency of the output wave form can be adjusted by controlling the

values of external resistors at pin 4 & 5.

The values of resistors RA & RB connected between Vcc * pin 4 & 5 respectively along with the

capacitor connected at pin 10 decide the frequency of the wave form.The values of RA & RB

should be in the range of 1kΩ to 1MΩ.

(iii) FM Bias:

· The FM Bias input (pin7) corresponds to the junction of resistors R1 & R2.

· The voltage Vin is the voltage between Vcc & pin8 and it decides the output frequency.

· The output frequency is proportional to Vin as given by the following expression

For RA = RB (50% duty cycle).

fo = CRAVcc

1.5Vin ; where C is the timing capacitor

· With pin 7 & 8 connected to each other the output frequency is given by fo = RC0.3

where R = RA = RB for 50% duty cycle.

(iv) FM Sweep input (pin 8):

· This input should be connected to pin 7, if we want a constant output frequency.

· But if the output frequency is supposed to vary, then a variable dc voltage should be applied to

this pin.

· The voltage between Vcc & pin 8 is called Vin and it decides the output frequency as,

1.5 Vin

fo = C RA Vcc

A potentiometer can be connected to this pin to obtain the required variable voltage

required to change the output frequency.


8. Design a differentiator to differentiate an input signal that varies in frequency from

10hz to about 1khz.[ AUC April 2008]

Steps to be followed to design a differentiator

Select fa equal to the highest frequency of the input signal.

Assume a value C1 < 1 µf and calculate Rf.

Select fb =20fa and calculate the values of R1 and Cf so that R1C1 =RfCf

Step 1 : select f and calculate Rf

fa= 1 KHz as the highest input frequency is 1 KHz.

Let C1 =0.1 µf

But fa =

Rf =

Step 2 : Calculate R1 and Cf

fb = 20 fa =20 KHz

fb =

R1 =

As R1 C1 = Rf cf

Cf =

Rcomp =Rf =

Designed circuit


9. Explain with neat sketch about the working of an instrumentation amplifier and

show with derivation that gain of this amplifier can be varied by using a variable

resistance R.[ AUC June 2009]

Instrumentation amplifiers


requirements.


required





Requirements

Precise


High CMRR

Low noise



High slew rate

Circuit diagram


Analysis


Va=V2 and Vb=V1


I = Va-vb/ r2 = v2-v1 / r2


V02 = Va + ir1

=V2+V2-V1/r2 * r1

=v2r2 +(v2-v1)r1 / r2 = v2(r2+r1) – v1 r1 / r2


V01=(r1+r2) v1 – r1v2 / v2


V02 – V01 = v2(r2+r1) – v1 r1 / r2 - (r1+r2) v1 – r1v2 / v2

=(2r1+r2)(v2-v1) /r2

Vo2-v01 = [1+ 2r1/r2][v2-v1]

Av1=vo2-v01/ v2-v1 = 1+ 2r1/r2



Av=Av1 * Av2 =(1+2r1/r2) * r4/r3


Vo=av(v2-v1)

Applications




Medical equipments


10. Show with help of circuit diagram an op-amp used as i) Scale changer ii) Phase

Shifter iii)inverting adder iv)Non-inverting adder.

Scale changer

Applying KCL

Phase shifter

The circuit that introduces phase shift as signal transmits from output to input is called

phase shifter.


Applying KCL at Node B

Applying KCL at Node A

Since VA =VB


Inverting adder

The summing amplifier is same as the inverting amplifier.

It has several inputs.

The feedback applied through Rf from the output to the input terminal is negative.


Applying KCL

Due to virtual ground node B and node A is at ground potential

Output voltage is equal to

Non inverting Summer

Circuit diagram

Applying KCL


In non inverting amplifier output voltage must be equal to

The output voltage is equal to

Draw an op-amp circuit whose output is v1+v2-v3-v4.[ AUC June 2009,nov 09]

11. Draw the circuit of a fourth order butterworth low pass filter having an upper cutoff

frequency of 1 Khz.[ AUC May2009]

Solution :

The general equation of butterworth filter denomination contains

i

ii


In the fourth order filter n=4

Consider equation i

The second order filter general equation is

Let Ri1 =Ri2 =10KΩ


12. Explain voltage follower, operation of op-amp in detail[AUC Nov 2009]

Voltage Follower: [Non-Inverting Buffer]

A circuit in which the output voltage follows the input voltage is called voltage follower.

Va=Vb =Vin

Due to virtual ground node b is at potential vin and node a is also at the potential Vin.

Vo=Va

V0=Vin

The voltage gain is unity.

13. Draw the circuit diagram and explain the working of precision rectifier. [AUC Nov

2009,2013]

Precision Rectifier:

The signal processing applications with very low voltage, current and power levels require

rectifier circuits. The ordinary diodes cannot rectify voltages below the cut-in-voltage of the

diode.

A circuit which can act as an ideal diode or precision signal – processing rectifier circuit for

rectifying voltages which are below the level of cut-in voltage of the diode can be designed

by placing the diode in the feedback loop of an op-amp.

Precision diodes:


Half – wave Rectifier:

A non-saturating half wave precision rectifier circuit is shown in figure. When Vi > 0V ,

the voltage at the inverting input becomes positive, forcing the output VOA to go

negative.

This results in forward biasing the diode D1 and the op-amp output drops only by ≈ 0.7V

below the inverting input voltage.

Diode D2 becomes reverse biased. The output voltage V0 is zero when the input is

positive. When Vi > 0, the op-amp output VOA becomes positive, forward biasing the

diode D2 and reverse biasing the diode D1 .

The circuit then acts like an inverting amplifier circuit with a nonlinear diode in the

forward path. The gain of the circuit is unity when Rf = Ri .

Figure shows the arrangement of a precision diode. It is a single diode arrangement

and functions as a non-inverting precision half – wave rectifier circuit.

If V1 in the circuit of figure is positive,the op-amp output VOA also becomes positive.

Then the closed loop condition is achieved for the op-amp and the output voltage V0

= Vi .

when Vi < 0, the voltage V0A becomes negative and the diode is reverse biased. The

loop is then broken and the output V0

= 0.


Full wave Rectifier: The Full wave Rectifier circuit commonly used an absolute value circuit is shown in figure. The first part of the total circuit is a half wave rectifier circuit considered earlier in figure. The second part of the circuit is an inverting.


For positive input voltage Vi > 0V and assuming that RF =Ri = R, the output voltage VOA = Vi .

The voltage V0 appears as (-) input to the summing op-amp circuit formed by A2 ,

The gain for the input V’0 is R/(R/2), as shown in figure.

The input Vi also appears as an input to the summing amplifier. Then, the net output is V0 = -Vi -

2V’0 = -Vi -2(-Vi ) = Vi

Since Vi > 0V, V’0 will be positive, with its input output characteristics in first quadrant.

For negative input Vi < 0V, the output V’0 of the first part of rectifier circuit is zero. Thus, one input

of the summing circuit has a value of zero. However, Vi is also applied as an input to the summer

circuit formed by the op-amp A2 .

The gain for this input id (-R/R) = -1, and hence the output is V0 = -Vi . Since Vi is negative, v0

will be inverted and will thus be positive.

This corresponds to the second quadrant of the circuit.

To summarize the operation of the circuit,

V0 = Vi when Vi < 0V and V0 = Vi for Vi > 0V, and hence V0 = |Vi |

It can be observed that this circuit is of non-saturating form.

14. Draw and explain the commonly used three op-amp instrumentation amplifiers and

derive expression for its gain [AUC MAY 2010]


requirements.


required





Requirements

Precise


High CMRR

Low noise



High slew rate

Circuit diagram


Analysis


Va=V2 and Vb=V1


I = Va-vb/ r2 = v2-v1 / r2


V02 = Va + ir1

=V2+V2-V1/r2 * r1

=v2r2 +(v2-v1)r1 / r2 = v2(r2+r1) – v1 r1 / r2


V01=(r1+r2) v1 – r1v2 / v2


V02 – V01 = v2(r2+r1) – v1 r1 / r2 - (r1+r2) v1 – r1v2 / v2

=(2r1+r2)(v2-v1) /r2

Vo2-v01 = [1+ 2r1/r2][v2-v1]

Av1=vo2-v01/ v2-v1 = 1+ 2r1/r2




Av=Av1 * Av2 =(1+2r1/r2) * r4/r3


Vo=av(v2-v1)

Applications




Medical equipments

15. Explain the working principles of regenerative comparator with necessary

diagrams[AUC MAY 2010,2011]

Schmitt trigger

Comparator which use positive feedback is known as Schmitt trigger.

Circuit diagram

This circuit converts an irregular shaped waveform to a square wave or pulse. The

circuit is known as Schmitt Trigger or squaring circuit.

The input voltage Vin triggers (changes the state of) the o/p V0 every time it exceeds

certain voltage levels called the upper threshold Vut and lower threshold voltage. These

threshold voltages are obtained by using the voltage divider R1 –R2, where the voltage

across R1 is feedback to the (+) input.

The voltage across R1 is variable reference threshold voltage that depends on the value

of the output voltage.

When V0 = +Vsat, the voltage across R1 is called “upper threshold” voltage Vut.

0R1

1k

0

R2

1k

R3

1k

V1


ROM

1k

U1

AD741

3

2

74

6

1

5+

-

V+

V-

OUT

OS1

OS2

vo

0


The input voltage Vin must be more positive than Vut in order to cause the output V0 to

switch from +Vsat to –Vsat. As long as Vin < Vut , V0 is at +Vsat, using voltage divider

rule,

V ut =(R1 / R1 + R2 ) +V sat

Similarly, when V0 = -Vsat, the voltage across R1 is called lower threshold voltage Vlt .

the vin must be more negative than Vlt in order to cause V0 to switch from –Vsat to

+Vsat.

In other words, for Vin > Vlt , V0 is at –Vsat. Vlt is given by the following eqn. V lt =

(R1/R1 + R2) -V sat

Thus, if the threshold voltages Vut and Vlt are made larger than the input noise voltages,

the positive feedback will eliminate the false o/p transitions. Also the positive feedback,

because of its regenerative action, will make V0 switch faster between +Vsat and –Vsat.

Resistance Rcomp tR1 || R2 is used to minimize the offset problems.

The comparator with positive feedback is said to exhibit hysteresis, a dead band

condition. (i.e) when the input of the comparator exceeds Vut its output switches from

+Vsat to –Vsat and reverts to its original state, +Vsat

when the input goes below Vlt. The hysteresis voltage is equal to the difference between

Vut and Vlt.

Therefore

Vref = Vut – Vlt

Vref = R1 /(R1 + R2) [+Vsat -(-Vsat)]

Operation :

Ref voltage v1 is developed across r2

V1=r2/r1+r2 * v0

Two different triggering voltages are defined for Schmitt trigger

Upper threshold voltage (Vut)

Lower threshold voltage (Vlt)

Vut forces transition from + Vsat to – Vsat

Vlt forces transition from –vsat to + Vsat

Vut = r2/r1+r2 * Vsat

Vlt = r2/r1+r2 * -vsat


Waveforms

Hysteresis

The center of the hysteresis loop may be shifted by choosing a centre voltage which is the

average of Vut and Vlt

Vcenter = [Vut+Vlt] /2


Applications

To convert sine wave to square wave.

Over voltage and over current protection circuit.

On /off temperature controllers.

16. What are the limitations of an ideal op-amp differentiator ?How are these

limitations overcome in practical differentiator ? Explain with necessary diagrams.

[AUC MAY 2011]

The ideal differentiator suffers from stability and noise problems which are overcome by

means of practical differentiator

Analysis


0 0

VIRTUAL GROUND

U3

AD741

3

2

74

6

1

5+

-

V+

V-

OUT

OS1

OS2

D1

D1N4001

R1

1k

I1

VO

V4


V01

Applications :

Wave shaping circuits

Used as detector in FM demodulators

17. Explain the working of logarithmic and antilogarithmic amplifiers using op-amp

with necessary diagrams . [AUC MAY 2011]

Log amplifier :

These circuits are used in analog computers.

Circuit diagram

PN junction diode is connected in the feedback path. Output voltage is nothing but the

voltage across the diode.

V0= - Vf-------1

Due to infinite input impedance, the current going into the inverting terminal is zero.

i1 = if= Vin/R-------2


Generally,

Vf = ή vt[log (if) –log(i0)]-----3

Substituting 3 in 1 we get

V0=- ή vt[log (if) –log(i0)]

= ή vt[log (Vin/r) –log(i0)]

Vo= -ή vt log [ vin / r *i0]

Vo= -ή vt log [ vin / vref]

Where vref = r.Io

Vo is directly proportional to the logarithm of the input voltage.

Antilog amplifier

Antilog amplifiers are used to produce the antilog or exponential function.

Circuit diagram

Analysis

Node B is connected to ground, A is also at ground potential (virtual ground)

Since ,A is virtual ground , Vin=Vf ,If =I0 . e vin / ήVt

Assume input current of the op-amp will be zero, the current “I” flowing through the feedback

resistor Rf is given by

I=If = -Vo/ rf

- Vo / rf = I0 . e vin / ήVt

Vo=- I0 rf . e vin / ήVt

V0 is directly proportional to the exponential function of vin

0 0

U3

AD741

3

2

74

6

1

5+

-

V+

V-

OUT

OS1

OS2

B

D1

D1N4001

RF

1k

A

IVO

V4


I


(ii) Draw and explain the circuit of a voltage to current converter if the load is (1)

Floating (2) Grounded. [AUC MAY 2012]

Voltage to Current Converter with floating loads (V/I):

Voltage to current converter in which load resistor RL is floating (not connected to ground).

Vin is applied to the non inverting input terminal, and the feedback voltage across R1 devices the inverting input terminal.

This circuit is also called as a current – series negative feedback amplifier.

Because the feedback voltage across R1 (applied Non-inverting terminal) depends on the output current i0 and is in series with the input difference voltage Vid .

Writing KVL for the input loop, Vin =V id +V f V id t 0v , since A is very largeA Vin =V f Vin = R1 i0 or i0 = Vin/R1 From the fig input voltage Vin is converted into output current of Vin/R1 [Vin -> i0 ] . In other words, input volt appears across R1. If R1 is a precision resistor, the output current (i0 = Vin/R1 ) will be precisely fixed. Applications: 1. Low voltage ac and dc voltmeters 2. Diode match finders 3. LED 4. Zener diode testers.


Voltage – to current converter with Grounded load: This is the other type V – I converter, in which one terminal of the load is connected to ground.

Analysis of the circuit: The analysis of the circuit can be done by following 2 steps.

1. To determine the voltage V1 at the non-inverting (+) terminals and

2. To establish relationship between V1 and the load current IL .

Applying KCL at node V1 we can write that,

I L = I1 + I 2 -------------1

But I1 = V in -V 1/R

and I 2 = V 0-V1/R

Sub these values into eqn 1

I L =V in -V1/R +V 0-V 1/R

I L =(Vin R -V1 R +V 0 R-V1 R)R2

=V inR +V 0 R - 2V1 R/R2

=R V in +V 0 -2V1 R /R2

I L =V in +V 0-2V1/R

RI L =V in +V 0 -2V1

V 1 =V in +V 0-I L R/2

the op-amp is connected in the non-inverting node A hence ,gain of the circuit is

0-V2 /R =V2-V0/R , 2V2=V0

The output voltage is given by

V 0 = 2V1

V 0 = 2[ V in +V 0 -I L R/2]

V 0 =V in +V 0 -I L R

V in = I L R

I L =V in /R

The load current I Lis dependent on the input voltageV in and Resistor R


V10

TD = 0.5ms

TF = 1msPW = 0.5msPER = 0.5ms

V1 = 5v

TR = 1ms

V2 = 0v

0

V

0

V

15.00V

-15.00V-14.80V

R4

1k5.000V5.000V

0

U1

AD741

3

2

74

6

1

5+

-

V+

V-

OUT

OS1

OS2

0

C20.01u

0V

V3

15Vdc

V2

15Vdc

18. Explain the working of Integrator [AUC NOV 2013]

19. INTEGRATOR

The circuit in which the output voltage waveform is the integration of the input voltage waveform is called as integrator.

CIRCUIT DIAGRAM


Analysis

Applying KCL at node V2 we get I1= Ib + if Due to high input Z , Ib<< If I1~If Relation between current and voltage across a capacitor is given by Ic = c.dvc / dt In case of an integrator if = icf I1=cf dvc/dt-----------------1 But i1=vin-v2/r1 and vc=v2-vo Equating i1 in 1 Vin-v2 /r1 =n cf dvc/dt =cf d/dt(v2-vo) V1-v2=0 Vin/r1=cf d/dt(-vo) Vo=-1/r1cf ∫vin dt +c Applications

Triangular wave or ramp generators.

A/D converters.

Integral type controllers.

In analog computers to solve differential equations.


20. Design an OP-AMP based second order active low pass filter with cut off

Frequency 1kHz. (8) [AUC MAY 2012]

Choose C=0.01µf

unit ii: applications of op-amp part -a (2 marks) 1....

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