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

7. 

10. Ideal Op-Amp Characteristics
Internal differential gain Aod is infinite.
Differential input voltage (v2-v1) is zero.
Effective input resistance is infinite (i1 and i2 are zero).
Output resistance is zero so output voltage is connected directly to dependent voltage source.

13. V0=V1-i2R2=0 - (VI/R1)R2
i1 = i2

14. Ex 9.1
For the circuit given in Fig., design an inverting op-amp such that Av = -15 and the input resistance Ri = 20 kW. Assume an ideal voltage source.

19. Op-Amp Integrator

21. Noninverting Op-Amp
Between (1) and (2) there is a virtual short circuit (v1 = v2)

22. Voltage Follower
Voltage follower is used as an impedance transformer or a buffer, where Zin  ∞, Zout  0

23. Voltage Follower
Sever loading effect
Almost zero loading effect

24. Applications and Design of Integrated Circuits
Chapter Fifteen
Applications and Design of Integrated Circuits

26. 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.

27. Active filters

28. 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.

31. 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.

34. (cont.)

38. (cont.)

39. (cont.)

40. There are number of ways to solve this problem; perhaps
the easiest is using the principle of superposition and virtual short concept.
To apply superposition we first reduce VI2=0

41. (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 of
VO1 and VO2.

42. 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

43. Differential input resistance
We know the condition for difference amplifier
V2-V1=
 R1=R3 and R2=R4
Since two input of the op amp track each other in potential,
we may write a loop equation

45. EX 9.7
For the difference amplifier shown in Fig. if R1 = R3 = 10 kW, R2 = 20 kW, and R4 = 21 kW, determine vo when (a) vI1 = +1 V, vI2 = -1 V, (b) vI1 = vI2 = +1 V,