THEORY AND EXPLANATION OF OPERATIONAL AMPLIFIER AND ITS APPLICATION IN FORMING NON-INVERTING AMPLIFIERS, DIFFERENTIAL AMPLIFIERS, INSTRUMENTATIONAL AMPIFIERS ALONG WITH FEW SOLVED NUMERICAL PROBLEMS FOR BETTER UNDERSTANDING, THE CONTENT IS PRECISE AND SUFFICIENT TO UNDERSTAND THE THEORY RELATED TO THE TOPICS

1. 1)Topics=> Non-inverting amplifiers,
Differential amplifiers, Instrumentation
amplifiers

2.  Infinite input impedance
 Zero output impedance
 Zero common-mode gain or, equivalently, infinite common-
mode rejection
 Infinite open-loop gain A
 Infinite bandwidth
Since, the value of A is ideally infinite therefore the Op-Amp
will be used with other components to apply feedback to
close the loop.
CLOSED LOOP
CONFIGURATION
INVERTING
CONFIGURATION
NON-INVERTING
CONFIGURATION
CHARACTERISTICS OF AN IDEAL OP-
AMP

3. It consists of one Op-Amp and 2 resistors R1 & R2, R2 is connected
from output terminal back to the inverting terminal, hace applying
negative feedback. The input signal source is connected to non-
inverting terminal while R1 is connected to terminal 1 other end being
connected to the ground.
The closed loop gain for an ideal Op-Amp in non-inverting configuration is 1+R2/R1.
Also, current through Op-Amp is zero, therefore, the voltage Vo is fed back to resisitors such that
V1=Vo(R1/R1+R2)
Now, as Vi is increased, Vd tend to increase and therefore Vo also increases(A being large) and
as a result V1 also increases which drives back Vd to zero. This action of negative feedback is
called degeneration feedback.
Effect of Finite Open Loop Gain
Assuming the Op-Amp to be ideal except for
having finite open loop Gain.As A becomes
infinitely large, G becomes equal to the closed
loop gain. The % error in G resulting from the
finite Op-Amp A is
% gain error= -((1+(R2/R1))/(1+A+(R2/R1))) * 100
The input impedance is infinte while while output
impedance is zero.
Q1. Explain the design of non-inverting configuration of an Op-Amp
along with the effect of finite loop gain
NON-INVERTING AMPLIFIER

4. Q1. Use the superposition principle to find the output voltage of the circuit
Q2. Design a non inverting
amplifier with a gain of 2. At
the maximum output voltage
of 10V the current in the
Voltage divider is to be 10 μA.
Examples on Non-Inverting amplifier

5. Q3. For the circuit, find the values of iI, v1, i1, i2, vO, iL, and iO. Also find
the voltage gain vO/vI, the current gain iL/iI, and the power gain PL/PI.

6. A difference amplifier is one that responds to the difference between the two signals
applied at its input and ideally rejects signals that are common to the two inputs. For
practical circuits,Vo=AdVd+ AcmVcm
The efficacy of a differential amplifier is measured by the degree of its rejection of
common mode signal in preference to differential signal. This is quantified as a by
a measure CMRR(common mode rejection ratio), CMRR= 20* log(Ad/Acm)
A differential amplifier is introduced since an Op- Amp due to its ideally infinte gain
makes to use by itself therefore an appropriate feedback network is connected to the
op-amp to create a circuit whose closed loop gain is finite, stable and predicatable.
A single Op-Amp difference amplifier
The combination of inverting and non-inverting configuration such that the gain by each configuration is
equal which will reject common-mode signal. The attenuation is required at the positive input path so that
1+R2/R1 can become equal to R2/R1.
A DIFFERENTIAL AMPLIFIER is usually required to have a
High input resistance. The input resistance between the two
input terminals is called The Differential input
resistance Rid ,
Rid=Vid/I1, Vid=I1R1+ I2R3(Since, I1=I2 AND R1=R3),
Vid=2I1R1, Rid= 2R1.
 If the amplifier is required to have a large differential
gain (R2/R1), then R1 of necessity will be relatively
small and the input resistance will be correspondingly
low.
 Another drawback of the circuit is that it is not easy to
vary the differential gain of the amplifier.
DIFFERENTIAL AMPLIFIER

7. Examples on “Differential amplifiers”
Q1.Consider the difference-amplifier circuit
for the caseR1=R3=2kOhm andR2=R4=200kOhm. ANS1.
If the resistors have 1% tolerance (i.e., each can be
within±1% of its nominal value), to find the worst-case
common-mode gain Acm and hence the corresponding
value of CMRR.
Q2. Find values for the resistances in the circuit of Fig.
2.16 so that the circuit behaves as a difference amplifier
with an input resistance of 20 k and a gain of 10.
ANS2.

9. Instrumentation amplifier overcomes all the drawbacks of differential amplifier. The difference amplifier here is
connected at its each input terminal to the non-inverting configuration which acts as a voltage follower with gain such
that high input resistance can be achieved along with gain. The gain if achieved in the the very first stage i.e.through
the non-inverting configuration at the input terminals of the differential amplifier eases the burden on the latter to
further acheive the gain rather the task of rejecting common mode signals becomes its only priority.
1) The gain achieved by non-inverting
configuration can be so much so that that
the op-amp can saturate and if not saturate
then the differential amplifier in either case
has to deal with such large magnitudes of
signal leading to reduction in the value of
CMRR.
1) The two amplifier channels in the first
stage have to be perfectly matched,
otherwise a spurious signal may
appear between their two outputs.
Such a signal would get amplified by
the difference amplifier in the second
stage.
1) To vary the differential gain Ad,
two resistors have to be varied simultaneously, here these two resistors labelled as R1.
In the figure given the removal of node at X can make the amplifier overcome all these drawbacks.
Due to virtual short circuit at terminal 1 and 2 of each non-inverting op-amp the volatge across 2R1 becomes
V2-V1 i.e.Vd and hence a current flows equal to Vd/2R1. Hence, the voltage across differential amplifier
becomes Vo2-Vo1=(2R2Vd/2R1)+2R1Vd/2R1=Vd(1+R2/R1) and this voltage after passing through
differential amplifier becomes (R4/R3)(1+R2/R1)Vd i.e. Ad=(R4/R3)(1+(R2/R1))
THIS SHOWS THAT THE GAIN CAN BE VARIED ONLY BY VARYING THE VALUE OF 2R1.
INSTRUMENTATIONAL AMPLIFIER

10. EXAMPLES ON INSTRUMENTATION AMPLIFIER
Consider the instrumentation amplifier with a common-mode input voltage of +5V (dc) and a
differential input signal of 10-mV-peak sine wave. Let (2R1)=1k, R2 =0.5 M, and R3 =R4 =10 k.
Find the voltage at every node

11. EXAMPLES ON INSTRUMENTATION AMPLIFIER