Chapter 2
Operational Amplifiers
Microelectronic Circuits, 8e
Sedra, Smith, Carusone, Gaudet
Copyright © 2020 Oxford University Press
IN THIS CHAPTER YOU WILL LEARN
•
Operational Amplifier (OP-Amp)
•
Ideal OP-Amp
•
Common Mode rejection and Common-mode voltage
•
OP-Amp Circuits
➢
Inverting amplifier
➢
Non-inverting amplifier
➢
Summing amplifier
➢
Difference amplifier
•
Differential Gain (Ad​ )
•
Common-Mode Gain (Acm)
•
Common Mode rejection Ratio (CMRR)
•
The finite Open Loop Gain Problem
➢
A minor estimation
➢
Further estimation
•
Amplifier frequency response
•
Pole of the system
•
Effect of finite Open-Loop Gain and Bandwidth on Circuit Performance
•
Unity-gain frequency
•
Large-signal limitations
•
Full-power bandwidth
•
Op-Amp Integrator
•
Op-amp differentiator
Operational Amplifier (OP-Amp)
• An electronic unit that behaves like a
Voltage-controlled voltage source
__________________________________
• An active element designed to perform mathematical operations, such as
addition, subtraction, multiplication, division, differentiation, integration,
etc.
• An electronic device consisting of a complex arrangement of resistors,
transistors, capacitors, and diodes
Microelectronic Circuits, 8e
Sedra, Smith, Carusone, Gaudet
Press
Copyright © 2020 Oxford University
Operational Amplifier (OP-Amp) Terminals
• Pin 1: Inverting ( the out pout is out of phase with pin 1)
• Pin 2 : Non Inverting ( the out pout is in phase with pin 2)
• Pin 3: Output
• Pin 4: Connected to a positive supply voltage VCC
• Pin 5: Connected to a negative supply voltage -VEE
Microelectronic Circuits, 8e
Sedra, Smith, Carusone, Gaudet
University Press
Copyright © 2020 Oxford
Operational Amplifier (OP-Amp)
• vd = v2 - v1
• vo = A * vd = A * (v2 - v1)
−Vcc  vo  Vcc
• A is called open loop gain
• In linear circuit, we shall operate the op amp in the linear
Microelectronic Circuits, 8e
region
Sedra, Smith, Carusone, Gaudet
Copyright © 2020 Oxford University Press
Ideal OP Amp
• Purpose: simplify op amp circuit without losing the accuracy in analysis
• Ideal op amp
∞
– Open-loop gain A ≈ ______
i1
∞
– Input Resistance Ri ≈ ______
– Output Resistance Ro ≈ 0_____
• If Ri is ideal in an op amp,
i2
– i1 = ____;
i2 = 0
____
0
i2 – i1 ) = v2 – v1 = ____
– vd = Ri*( ______
 v2 ____
0
= v1
Common-mode rejection
Common-mode rejection is the ability of a differential amplifier to reject signals that are common
to both inputs.
For instance a differential amplifier has two inputs, 𝑣1 = 1 𝑉​ and 𝑣2 = 1 𝑉.
Amplify the difference between the two inputs:
𝑣𝑑 = 𝑣1 − 𝑣2 = 0
(1)
Imagine two wires carrying the same unwanted noise, the amplifier subtracts the two signals.
Because the noise is common to both, it is largely canceled.
Common-mode voltage
It is defined as the average of the two input voltages
𝑣 +𝑣2
2
𝑣𝐶𝑚 = 1
(2)
From (1) and (2):
𝑣𝑑
𝑣1 = 𝑣𝑐𝑚 −
2
𝑣𝑑
𝑣2 = 𝑣𝑐𝑚 +
2
So 𝑣𝐶𝑚 ​ represents what is common to both inputs, while 𝑣𝑑 ​ represents
what is different between them.
Extra Credit: Push Your Limits
The internal circuit of a particular op amp can be modeled by the circuit shown
in Fig. E2.3. Express v3 as a function of v1 and v2. For the case Gm = 20
mA/V, R = 5 kΩ, and μ = 50, find the value of the open-loop gain A.
Microelectronic Circuits, 8e
Sedra, Smith, Carusone, Gaudet
Press
Copyright © 2020 Oxford University
Inverting Amplifier
The key idea is that the input signal is applied to the inverting terminal (−), while the
noninverting terminal (+) is connected to ground
𝑣
𝑣𝑖
Goal is to find 𝐺𝑚 = 0
Gm is called closed loop gain
Terminal 1 is being a virtual ground—that is, having zero voltage but not physically
connected to ground
Microelectronic Circuits, 8e
Sedra, Smith, Carusone, Gaudet
Copyright © 2020 Oxford University Press
Inverting Amplifier
• Ideal op amp:
0
0
– vin = ____;
v1 ____
= v2 = _____
– iin = ____
0
iin
vin
– KCL at node 1: i1 ____
= i2
vi − v1
R1
i1 = _________
v1 − vo
Rf
i2 = _________
0
v1 = ____
Rf
vo
= −
R1
vi
vo of an inverting amplifier is an inverted and scaled replica of vi The
Negative sign means output is 180° out of phase with the input.
Extra Credit: Earn It. Master It.
Determine vo and the current in the 10 kΩ resistor in the circuit below if
vi = 0.5 v.
Example
Assuming the op amp to be ideal, derive an expression for the closed-loop gain
vO/vI of the circuit shown in Fig. 2.8. Use this circuit to design an inverting
amplifier with a gain of 100 when R1=1 MΩ. Assume that for practical
reasons it is required not to use resistors greater than 1 MΩ.
Microelectronic Circuits, 8e
Sedra, Smith, Carusone, Gaudet
Press
Copyright © 2020 Oxford University
Noninverting Configuration
The input signal 𝑣𝑖 is applied directly to the positive input terminal of the op amp while
one terminal of 𝑅1 is connected to ground.
𝑣
Goal is to find 𝐺𝑚 = 𝑣0
𝑖
Gm is called closed loop gain
Microelectronic Circuits, 8e
Sedra, Smith, Carusone, Gaudet
Copyright © 2020 Oxford University Press
Non-inverting Amplifier
• Ideal op amp:
vi
= v2 = _____
– v1 ____
– KCL at node v1: i1 ____= i2
0 − v1
R1
i1 = _________
v1 − vo
Rf
i2 = _________
i1 = i2
v1 = vi
vi is connected to the OP-Amp at
the non-inverting terminal
Rf
vo
= 1+
R1
vi
vo of a non-inverting amplifier is a scaled replica of vi
Extra Credit: Test Your Skills!
Design a noninverting amplifier with a gain of 2. At the maximum output
voltage of 10 V the current in the voltage divider is to be 10 µA.
Voltage Follower
A voltage follower is simply a noninverting op-amp with the output connected
directly to the inverting input
Application: buffer amplifier
•
Isolate one circuit from another
•
minimize the interaction between two
stages
•
Eliminate inter-stage loading
Microelectronic Circuits, 8e
Sedra, Smith, Carusone, Gaudet
University Press
Copyright © 2020 Oxford
Extra-Credit Challenge: Put Your Skills to Work!
A measurement application produces an open-circuit voltage of 1 V and has a
source resistance of 1 MΩ. This signal needs to be connected to a load of 1 kΩ
resistance. Find the load voltage if the connection is made (a) directly and (b)
through a unity-gain voltage follower.
Summing Amplifier
• Ideal op amp:
0
– va = ____
i1 + i2 + i3
– KCL at node a: i = __________
v1 − va
R1
i1 = _________
v2 − va
R2
i2 = _________
v3 − va
R3
i3 = _________
va − vo
Rf
i = _________
va = 0
Rf
Rf 
 Rf
vo = −  v1 + v2 + v3 
R2
R3 
 R1
vo of a summing amplifier is an inverted and scaled
sum of the input voltages
Extra Credit: Push Your Limits!
For the summing amplifier shown below,
a) Find vo if va=0.1 v and vb=0.25 v.
b) If vb=0.25 v, how large can va be before the op amp saturates?
c) If va=0.1 v, how large can vb be before the op amp saturates?
Example
Use the superposition principle to find the output voltage of the circuit shown
in Fig. E2.9
Superposition:
• Keep one independent source active and turn off all other independent sources
• Voltage source → short circuit (V=0)
• Current source → open circuit (I=0).
• Find the desired voltage or current due to that source
• Repeat for each independent source.
• Add all individual contributions algebraically to obtain the total response.
Microelectronic Circuits, 8e
Sedra, Smith, Carusone, Gaudet
University Press
Copyright © 2020 Oxford