In this script, we will discuss the most important topic of Operational Amplifier, Now we will discuss

  • What is Ideal Operational Amplifier?
  • Characteristics of Ideal Operational Amplifier
  • Two important properties of the ideal op-amp 
  • Two important properties of the ideal op-amp 
  • Math

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What is Ideal Operational Amplifier?

An ideal op amp is an amplifier with infinity open-loop gain, infinity input resistance, and zero output resistance. An ideal op amp provides only an approximate analysis, maximum modern amplifiers have such large gains and input impedance that approximately analysis is a good one.

 



Characteristics of Ideal Operational Amplifier

1. Infinite open-loop gain, A ≃ ∞

2. Infinite input resistance, R1 ≃ ∞

3. Zero output Resistance ≃ 0

4. Infinite Bandwidth of Operation ≃ ∞

5. Zero Offset Voltage ≃ 0


Two important properties of the ideal op-amp 

1. The currents into both input terminals are zero.

    i1=0, i2 = 0

When infinite input resistance. An infinite resistance between the input terminals implies that an open circuit exits there and current cannot enter the op-amp. But the output current is not necessarily zero.

2. The voltage across the input terminals is equal to zero i.e

     Vd=V2-V1= 0 or

     V1 = V2

When an ideal op amp has zero current in its two input terminals and the voltage between the two input terminals is equal to zero.

Two important properties of the ideal op-amp 

  • voltage follower
  • selective inversion circuit
  • a current-to-voltage converter
  • active rectifier
  • integrator
  • voltage comparator
Problem: 1 Find out the Output current.



 V2 = Vs
Here i1 =  0, and the 40-k and 5-k resistors are in series; the same current flows through them. v1 is the voltage across the 5-k resistor.

Hence, using the voltage division principle,

v1 = 5/5+40 Vo = Vo/9

We know that V2=V1...................(1)

Put the value v1 equation 1 

  V2= Vo/9

 Vs= Vo/9 [V2=Vs]

 Vo/Vs = 9

At node O,
 io = Vo/40+5 +Vo/20mA ...............................(2)

When vs  = 1 V, 
 vo =  9 V.
 Substituting for vo = 9 V in Equation (2)
  io = Vo/(40+5) +Vo/20 mA 
      = 9/45 + 9/20 mA
      = o,65mA

This is close to the value of 0.657 mA obtained