•An understanding of the complex circuitry within the op amp is not necessary to use this amplifying circuit in the construction of an amplifier.

Ideal Op-Amp Characteristics

1. Internal differential gain Aod is infinite.

2. Differential input voltage (v2-v1) is

zero.3. Effective input resistance is infinite

(i1 and i2 are zero).

4. Output resistance is zero so output voltage is connected directly to dependent voltage source.

i1 = i2

V0=V1-i2R2=0 - (VI/R1)R2

Ex 9.1

For the circuit given in Fig., design an inverting op-amp such that Av = -15 and the input resistance Ri = 20 k. Assume an ideal voltage source.

Op-Amp Integrator

t

ICodt)t(v

CRVv

021

1

Noninverting Op-Amp

1

21R

R

v

vA

i

o

v Between (1) and (2) there is a

virtual short circuit (v1 = v2)

Voltage Follower

1

1

1

1

2

v

i

o

v

ARIf

R

R

v

vA

Voltage follower is used as an impedance transformer or a buffer, where Zin ∞, Zout 0

Voltage Follower

010.RR

R

v

v

SL

L

I

O

1I

O

v

v

Sever loading effect Almost zero loading effect

Chapter Fifteen

Applications and Design of

Integrated Circuits

Passive and active filters

• Filters are building blocks of communication and instrumentations.

• The oldest technology based on inductors and capacitors are called passive LC filters, which are incompatible with any of the modern techniques for assembling electronic systems.

• Active-RC filters utilize op amp together with resistors and capacitors.

• At present, the most viable approach for realizing fully integrated monolithic filters is the switch capacitance techniques.

Active filters

Uses of the filters• Some of the uses are listed below:

• in electronic power supply ripple smoothing

• as tuned circuits in RF stages

• for selection of particular sideband as in SSB transmitters

• for harmonic frequency suppression in transmitters

• As wave trap for eliminating undesirable signal from radio and TV receivers• for improving high and low frequency response in video amplifier

• for restricting audio frequency band pass in SSB transmitters

• For eliminating undesirable frequency from motors, generators and other electrical and electronic equipment.

One pole active filters• Usually, passive filters suffers from loading effects, substantially

reducing the maximum gain from the unity.• Loading effect can be reduced by using active filters.

(cont.)

(cont.)

(cont.)

There are number of ways to solve this problem; perhapsthe easiest is using the principle of superposition and virtual short concept.

To apply superposition we first reduce VI2=0

(cont.)

Next, we reduce the VI1=0 and evaluate the corresponding output voltageVO2

R3 and R4 form a voltage divider. Therefore,

or

The superposition principle tells us that the output voltage VO is equal to the sum ofVO1 and VO2.

What is the condition under which this circuit will act as a difference amplifier?

Let VI1=VI2 VO=0

Which is clearly that of a difference amplifier with a gain of R2/R1

Differential input resistanceWe know the condition for difference amplifier

R1=R3 and R2=R4

V2-V1=

Since two input of the op amp track each other in potential, we may write a loop equation

EX 9.7

For the difference amplifier shown in Fig. if R1 = R3 = 10 kR2 = 20 kand R4 = 21 kdetermine vo when (a) vI1 = +1 V, vI2 = -1 V, (b) vI1 = vI2 = +1 V,