EHB 211E
Basics of Electrical Circuits
Asst. Prof. Sheida Faraji
Operational Amplifiers
Week 8 (20th Nov 2023)
Fundamentals of Electric Circuits by C. Alexander and M. Sadiku, Seventh Edition (McGraw-Hill Education)
Introduction: Operational Amplifiers
• What is an operational amplifier (op amp)?
Active element designed to perform mathematical operations: addition,
subtraction, multiplication, division, differentiation, and integration.
❑ Sum, amplify, integrate, or differentiate a signal.
❑ Versatile circuit building block.
❑ Electronic unit that behaves like a voltage-controlled voltage source (VCVS)
❑
• Operational amplifiers (op amps) are popular in practical circuit
design:
Versatile
❑ Inexpensive
❑ Easy to use
❑
• Electronic device consisting of a complex arrangement of resistors,
transistors, capacitors, and diodes.
• Only consider external characteristics of op amps in this course.
Operational Amplifiers
• Consist of two inputs and one output
• Minus(-): inverting input.
• Positive (+): noninverting input
• Inputs applied to noninverting terminal appears with same polarity at the output
• Input applied to inverting terminal appears inverted at the output.
• Connect to two power supplies, positive (𝑉 + ) and negative (𝑉 − ).
Typical op amp package
Top view of op amp
Circuit symbol of op amp
To null the offset voltage, a potentiometer is connected to the offset
null pins of the op amp and adjusted until the output is zero.
Operational Amplifiers
• The op amp is powered by a voltage supply.
• Apply KCL: σ 𝑖𝑖𝑛 = σ 𝑖𝑜𝑢𝑡
𝑖0 = 𝑖1 + 𝑖2 + 𝑖+ + 𝑖−
•
•
•
•
•
𝑣1 : inverting terminal
𝑣2 : noninverting terminal
𝑅𝑖 : Thevenin equivalent resistance seen at input
𝑅0 : Thevenin equivalent resistance seen at output
The differential input voltage 𝑣𝑑 is given by
𝑣𝑑 = 𝑣2 − 𝑣1
• The output of the operational amplifier is given by
𝑣0 = 𝐴𝑣𝑑 = 𝐴(𝑣2 − 𝑣1 )
• where A: open loop voltage gain (no external feedback from output to input)
Operational Amplifiers
• The magnitude of 𝑣0 cannot exceed power supply voltage.
• The output voltage is dependent on and is limited by the power supply.
−𝑉𝐶𝐶 ≤ 𝑣0 ≤ 𝑉𝐶𝐶
• Op amp in three modes as follows:
❑
❑
❑
Positive saturation: 𝑣0 = 𝑉𝐶𝐶
Linear region: −𝑉𝐶𝐶 ≤ 𝑣0 = 𝐴𝑉𝑑 ≤ 𝑉𝐶𝐶
Negative saturation: 𝑣0 = −𝑉𝐶𝐶
• Output voltage 𝑣0 is also dependent on
differential input 𝑣𝑑 (𝑣0 = 𝐴𝑣𝑑 )
Example 1
A 741-op amp has an open-loop voltage gain of 2 × 105 , input resistance of 2 𝑀Ω,
and output resistance of 50 Ω. The op amp is used in the circuit as shown below.
𝑣
Find the closed-loop gain 0 . Determine current 𝑖 when 𝑣𝑠 = 2 𝑉.
𝑣𝑠
Solution
• Apply nodal analysis to equivalent circuit.
Equivalent circuit
• Apply KCL to node 1: σ 𝑖𝑖𝑛 = σ 𝑖𝑜𝑢𝑡
• Apply KCL to node 0:
• This is closed-loop gain because the
20 𝑘Ω feedback resistor closes the loop
between the output and input terminals
Recall:
Current flows from higher
potential (+) to lower potential (-)
Ideal Op Amp
• An Op amp is ideal if it has the following characteristics:
❑
❑
❑
Infinite open-loop gain, A≃ ∞
Infinite input resistance, 𝑅𝑖 ≃ ∞
Zero output resistance, 𝑅0 ≃ 0
• An ideal op amp is an amplifier with infinite open-loop gain, infinite input
resistance, and zero output resistance.
• Two important characteristics of the ideal op
amp are:
➢ The currents into both input terminals are zero
𝑖1 = 0 , 𝑖2 = 0 (𝑅𝑖 ≃ ∞)
➢ The voltage across the input terminals is equal to
zero
𝑣𝑑 = 𝑣2 − 𝑣1 = 0 ⇒ 𝑣1 = 𝑣2
Example 2
Determine the current 𝑖0 when 𝑣𝑠 = 1 𝑉 in the circuit shown below. Assume the op
amp is ideal.
Solution:
• Input currents are zero
• Two inputs of op amp have the same voltages
𝑣2 = 𝑣𝑠
• Since 𝑣1 = 𝑣2 , 𝑣1 = 𝑣2 = 𝑣𝑠
• Since 𝑖1 = 0 (no current flow into amp),
40 Ω series with 5 Ω.
• Using voltage division:
• Apply KCL to node O:
Inverting Amplifier
• Inverting amplifier reserves the polarity of the input signal while amplifying it.
• Input 𝑣𝑖 is connected to the inverting terminal through 𝑅1 , and feedback resistor
𝑅𝑓 is connected between the inverting input and output.
• Noninverting input is grounded.
• Apply KCL at node 1:
𝑖1 = 𝑖2 (no current flows into amp)
𝑣𝑖 − 𝑣1
𝑖1 =
𝑅1
𝑣𝑖 − 𝑣1 𝑣1 − 𝑣0
=
𝑅1
𝑅𝑓
𝑣1 − 𝑣0
𝑖2 =
𝑅𝑓
𝑣1 = 𝑣2 = 0
𝑅𝑓
𝑣𝑖
𝑣0
Provide negative
= − ⇒ 𝑣0 = − 𝑣𝑖 output voltage
𝑅1
𝑅𝑓
𝑅1
𝑅𝑓
𝑣0
𝐴𝑣 : voltage gain
𝐴𝑣 =
=−
𝑣𝑖
𝑅1
Gain depends only on the external
elements connected to op amp
Inverting Amplifier
Equivalent circuit of the inverting amplifier
Example 3
For the op amp shown below, if 𝑣𝑖 = 0.5 𝑉, calculate a-) the output voltage 𝑣0 , and
b-) the current in the 10 𝑘Ω resistor.
Solution:
𝑅𝑓
𝑣0 = − 𝑣𝑖
a-)
𝑅1
b-)
Example 4
Determine 𝑣0 in the op amp circuit shown below
Solution:
𝑖1 = 𝑖2 (no current flows into amp)
Noninverting Amplifier
• Noninverting amplifier is an op amp circuit designed to provide a positive voltage
gain.
• Input voltage 𝑣𝑖 is directly applied at the noninverting terminal, and the resistor
𝑅1 is connected between the ground and inverting terminal.
Apply KCL at node 1:
𝑖1 = 𝑖2 (no current flows into amp)
𝑣1 − 𝑣0
0 − 𝑣1
𝑖2 =
𝑖1 =
𝑅𝑓
𝑅1
0 − 𝑣1 𝑣1 − 𝑣0
=
𝑅1
𝑅𝑓
𝑣1 = 𝑣2 = 𝑣𝑖
𝑅𝑓
𝑣𝑖 𝑅1 + 𝑣𝑖 𝑅𝑓
𝑣𝑖 𝑣𝑖 − 𝑣0
Provide positive
𝑣
=
1
+
𝑣
⇒
−
=
⇒ 𝑣0 =
0
𝑖 output voltage
𝑅
𝑅1
𝑅𝑓
𝑅1
1
𝑅𝑓
𝑣0
Gain depends only on the external
𝐴𝑣 : voltage gain elements connected to op amp
𝐴𝑣 =
=1+
𝑣𝑖
𝑅1
Inverting Amplifier: Special Case
• If feedback resistor 𝑅𝑓 = 0 (short circuit) and 𝑅1 = ∞
(open circuit), we will have the following circuit.
• The gain becomes 1.
• It is called voltage follower (or unity gain amplifier)
because output follows input.
𝑣0 = 𝑣𝑖
Voltage follower does not provide
any amplification to the signal
• The voltage follower is used as an
intermediate-stage (or buffer) amplifier to
isolate one circuit from another. It
minimizes the interaction between two
stages and eliminates inter-stage loading.
Example 5
For the op amp circuit shown below, calculate the output voltage 𝑣0 .
Solution:
Method 1: Using superposition
Solution
Method 2: Using nodal analysis
Summing Amplifier
• Op amp can perform addition besides amplification.
• A summing amplifier (aka summer) is an op amp circuit that combines several
inputs and produces an output that is the weighted sum of the inputs.
• Current entering each op amp input is zero.
• Apply KCL at node a:
𝑖 = 𝑖1 + 𝑖2 + 𝑖3
𝑣1 − 𝑣𝑎
𝑖1 =
𝑅1
𝑣2 − 𝑣𝑎
𝑖2 =
𝑅2
𝑣3 − 𝑣𝑎
𝑖3 =
𝑅3
𝑣𝑎 − 𝑣0
𝑖=
𝑅𝑓
𝑣0 𝑣1 𝑣2 𝑣3
− =
+
+
𝑅𝑓 𝑅1 𝑅2 𝑅3
𝑣𝑎 = 0
𝑅𝑓
𝑅𝑓
𝑅𝑓
𝑣0 = −
𝑣1 + 𝑣2 + 𝑣3
𝑅1
𝑅2
𝑅3
Example 6
• Calculate 𝑣0 and 𝑖0 in the op amp circuit shown below.
Solution:
Difference Amplifier
• Difference (aka differential) amplifiers are used in various applications where
there is a need to amplify the difference between two input signals.
• A difference amplifier is a device that amplifies the difference between two
inputs but rejects any signals common to the two inputs.
• Current entering each op amp input is zero.
• Apply KCL at node a:
𝑣1 − 𝑣𝑎 𝑣𝑎 − 𝑣0
=
𝑅1
𝑅2
𝑅2
𝑅2
𝑣0 =
+ 1 𝑣𝑎 − 𝑣1
𝑅1
𝑅1
Eq 1
Difference Amplifier
• Apply KCL at node b:
𝑅4
𝑣2 − 𝑣𝑏 𝑣𝑏 − 0
𝑣2
=
⇒ 𝑣𝑏 =
𝑅3 + 𝑅4
𝑅3
𝑅4
𝑣𝑎 = 𝑣𝑏
• Substitute 𝑣𝑏 into 1st Eq
𝑅2
𝑣0 =
𝑅1
𝑅1
1+𝑅
𝑅2
2
𝑣 − 𝑣
𝑅3 2 𝑅1 1
1+
𝑅4
• If 𝑅1 = 𝑅2 and 𝑅3 = 𝑅4 , the
difference amplifier becomes a
subtractor with output
𝑣0 = 𝑣2 − 𝑣1
𝑅1 𝑅3
𝑅2
• When
, 𝑣0 =
=
𝑣2 − 𝑣1
𝑅2 𝑅4
𝑅1
• In this case, the difference amplifier
rejects a signal common to the two
inputs, i.e., 𝑣0 = 0 when 𝑣1 = 𝑣2
Example 7
Design an op amp circuit with inputs 𝑣1 and 𝑣2 such that 𝑣0 = −5𝑣1 + 3𝑣2
Difference Amplifier
Solution:
• This circuit can be designed in two ways.
• Design 1: design it using only one op amp.
• Two inputs since 𝑣0 = 3𝑣2 − 5𝑣1
𝑅2
𝑣0 =
𝑅1
𝑅1
1+𝑅
𝑅2
2
𝑣 − 𝑣
𝑅3 2 𝑅1 1
1+
𝑅4
5
3
𝑅2
= 5 ⇒ 𝑅2 = 5𝑅1
𝑅1
1ൗ
5
𝑅1
1+𝑅
2
5
= 3 ⇒ 𝑅3 = 𝑅4
𝑅3
1+
𝑅4
We may choose:
𝑅1 1
=
𝑅2 5
𝑅1 = 10 𝑘Ω
𝑅3 = 20 𝑘Ω
𝑅2 = 50 𝑘Ω
𝑅4 = 20 𝑘Ω
Solution
• Design 2: design it using two op amps.
• In this case, cascade an inverting amp
and two input inverting summer amp as
shown in the figure.
• For the summer:
𝑅𝑓
𝑅𝑓
𝑅𝑓
𝑣0 = −
𝑣 + 𝑣 + 𝑣
𝑅1 1 𝑅2 2 𝑅3 3
𝑣0 = −
Inverting amp
5𝑅1
5𝑅1
𝑣𝑎 +
𝑣1 ⇒ 𝑣0 = −𝑣𝑎 − 5𝑣1
5𝑅1
𝑅1
Summer amp
• For the inverter amp:
𝑅𝑓
𝑣0 = − 𝑣𝑖
𝑅1
3𝑅3
𝑣𝑎 = −
𝑣 ⇒ 𝑣𝑎 = −3𝑣2
𝑅3 2
𝑣0 = 3𝑣2 − 5𝑣1
We may choose: 𝑅1 = 𝑅3 = 10 𝑘Ω
Example 8
An instrumental amplifier shown below is an amplifier of low-level signals used in
process control or measurement applications and commercially available in singlepackage units. Show that
𝑅2
2𝑅3
𝑣0 =
1+
𝑅1
𝑅4
𝑣2 − 𝑣1
Difference
amplifier
Solution
• Amplifier A3 is a difference amplifier.
Eq 1
Eq 2
Eq 3
• Substitute Eq 3 into Eq 2 and then substitute into Eq 1
Instrumentation Amplifiers
• One of the most useful and versatile
op amp circuit which can be used for
precision measurement and process
control.
• Typical applications of IAs: isolation
amplifiers, data acquisition system,
etc.
• Extension of the difference amplifier.
It amplifies the difference between
its input signal.
• Typically consists of three op amps
and several resistors.
• RG: external gain-setting resistor.
𝑅2
2𝑅3
𝑣0 =
1+
𝑅1
𝑅4
𝐴𝑣
2𝑅
𝐴𝑣 = 1 +
𝑅𝐺
𝑣2 − 𝑣1
Voltage gain
𝑣0 = 𝐴𝑣 𝑣2 − 𝑣1
Instrumentation Amplifiers
• The instrumentation amplifier has three major characteristic as follows:
❑
❑
❑
The voltage gain is adjusted by one external resistor 𝑅𝐺 .
The input impedance of both inputs is very high and does not vary as the gain is adjusted.
The output 𝑣0 depends on the difference between the inputs 𝑣1 and 𝑣2 , not the voltage
common to them (common-mode voltage).
• For IA, very small changes or differences in the input will result in very large
output. How big the output signal also depends on how big its gain is.
𝑣0 = 𝐴𝑣 𝑣2 − 𝑣1
Schematic symbol of IA
Cascaded Op Amp Circuit
• To obtain a large overall gain, connect op amp circuits in cascaded way.
• Cascaded connection: head-to-tail arrangement of two or more op amp circuits
such that the output of one is the input of the next
• Cascaded connection is connecting block in a head-to-tail way.
• In cascaded connection, each circuit in the string is called a “stage”.
• Input signal increased by the gain of the individual stage.
• Overall gain: 𝐴 = 𝐴1 𝐴2 𝐴3
Example 9
For the circuit shown below, find 𝑣0 and 𝑖0 .
Solution:
Example 10
If 𝑣1 = 1 𝑉 and 𝑣2 = 2 𝑉, find 𝑣0 in the op amp circuit shown below.
Solution:
• The op amp circuit is
composed of three circuits
• Two inverting amps and one
summing (summer) amp
• A and B: inverting amp
• C: Summing amp
𝑅𝑓
6
𝑣𝑎 = − 𝑣𝑖
𝑣𝑎 = − 1 = −3 𝑉
𝑅1
2
8
𝑣𝑏 = − 2 = −4 𝑉
4
𝑅𝑓
𝑅𝑓
𝑅𝑓
𝑣0 = −
𝑣 + 𝑣 + 𝑣
𝑅1 1 𝑅2 2 𝑅3 3
10
10
𝑣0 = −
−3 +
−4
5
15
⇒ 𝑣0 = 8.667 𝑉
Summary: