Feedback Amplifiers & Operational Amplifiers (OPAMP)

Feedback Amplifiers and Operational
Amplifiers
Unit IV
Darwin Nesakumar A
Assistant Professor/ECE
RMKEC

 Introduction
 Principle of Feedback Amplifiers
 Classification Feedback Amplifiers
 Basic Concept of Feedback
 Block Diagram of Feedback Amplifier
 Properties of Feedback Amplifiers
 Topologies of Feedback Amplifier
 General Characteristics of Negative Feedback Amplifier
 Voltage Gain
Contents

 Input Impedance
 Output Impedance
 Sensitivities
 Bandwidth Stability
 Effect of Positive Feedback
 Condition of Oscillation
 Barkhausen Criteria
 Introduction to Integrated Circuits
Contents

 Application of Operational Amplifier
 Inverting Amplifier
 Non-Inverting Amplifier
 Adders
 Subtractors
 Constant-Gain Multiplier
 Voltage Follower
 Comparator
 Integrator
 Differentiator
Contents

Introduction
Fraction of the amplifier output is fed back to the input circuit
make the operating point of a transistor insensitive to temperature and
other variations
Feedback system is mainly used in amplifiers, control systems, and etc.,
Feedback system has three components
 Amplifier
 Feedback System
 Mixer or Comparator

Principle of Feedback Amplifier
Amplifier Without Feedback

𝑶𝒑𝒆𝒏 𝑳𝒐𝒐𝒑 𝑮𝒂𝒊𝒏 =
𝑽𝒐
𝑽𝒊
Amplifier Without Feedback

Amplifier With Feedback

Classification Feedback Amplifiers
Positive Feedback

Classification Feedback Amplifiers
Negative Feedback

Block Diagram of Feedback Amplifier
Basic Concept of Feedback
𝑉𝑆 −ac signal at the input side
𝑉𝑓 −Feedback signal
𝑉𝑑 −Difference signal
𝑉𝑂 −Output signal

𝐴 − Gain of the basic amplifier =
𝑉𝑂
𝑉𝑑
𝛽 − Feedback Ratio =
𝑉𝑓
𝑉𝑂
𝐴𝑓 − Feedback Amplifier =
𝑉𝑂
𝑉𝑆

𝑉𝑂
𝑉𝑆
Gain of Positive Feedback Amplifier
Here 𝑉𝑆 𝑎𝑛𝑑 𝑉𝑓 𝑎𝑟𝑒 𝑖𝑛𝑝ℎ𝑎𝑠𝑒
𝑉𝑑= 𝑉𝑆 + 𝑉𝑓
Find 𝑽𝒅
∵ 𝐴 =
𝑉𝑂
𝑉𝑑
𝐴𝑉𝑑 = 𝑉𝑂

𝑉𝑑= 𝑉𝑆 + 𝑉𝑓
𝐴 𝑉𝑆 + 𝑉𝑓 = 𝑉𝑂
∵ 𝛽 =
𝑉𝑓
𝑉𝑂
𝛽𝑉𝑂=𝑉𝑓
𝐴 𝑉𝑆 + 𝛽𝑉𝑂 = 𝑉𝑂
𝐴𝑉𝑆 + 𝐴𝛽𝑉𝑂 = 𝑉𝑂
𝐴𝑉𝑆 = 𝑉𝑂 − 𝐴𝛽𝑉𝑂
𝐴𝑉𝑆 = 1 − 𝐴𝛽 𝑉𝑂

𝑨𝒇 =
𝑨
[𝟏−𝑨𝜷]
𝑳𝒐𝒐𝒑 𝒈𝒂𝒊𝒏 = 𝑨𝜷
𝐴𝑉𝑆 = 1 − 𝐴𝛽 𝑉𝑂
𝐴
1−𝐴𝛽
=
𝑉𝑂
𝑉𝑆
Where 𝐴𝑓=
𝑉𝑂
𝑉𝑆

𝑉𝑂
𝑉𝑆
Gain of Negative Feedback Amplifier
Here 𝑉𝑆 𝑎𝑛𝑑 𝑉𝑓 𝑎𝑟𝑒 𝑖𝑛𝑝ℎ𝑎𝑠𝑒
𝑉𝑑= 𝑉𝑆 − 𝑉𝑓
Find 𝑽𝒅
−
∵ 𝐴 =
𝑉𝑂
𝑉𝑑
𝐴𝑉𝑑 = 𝑉𝑂

𝑉𝑑= 𝑉𝑆 − 𝑉𝑓
𝐴 𝑉𝑆 − 𝑉𝑓 = 𝑉𝑂 ∵ 𝛽 =
𝑉𝑓
𝑉𝑂
𝛽𝑉𝑂=𝑉𝑓
𝐴 𝑉𝑆 − 𝛽𝑉𝑂 = 𝑉𝑂
𝐴𝑉𝑆 − 𝐴𝛽𝑉𝑂 = 𝑉𝑂
𝐴𝑉𝑆 = 𝑉𝑂 + 𝐴𝛽𝑉𝑂
𝐴𝑉𝑆 = 1 + 𝐴𝛽 𝑉𝑂

−
𝐴𝑉𝑆 = 1 + 𝐴𝛽 𝑉𝑂
𝐴
1+𝐴𝛽
=
𝑉𝑂
𝑉𝑆
𝑨𝒇 =
𝑨
[𝟏+𝑨𝜷]
𝑳𝒐𝒐𝒑 𝒈𝒂𝒊𝒏 = 𝑨𝜷
Where 𝐴𝑓=
𝑉𝑂
𝑉𝑆

Properties of Negative Feedback
 Gain Desensitivity
 Extend the Bandwidth of the Amplifier
 Reduce Nonlinear Distortion
 Reduce the Effect of Noise
 Control the Input and Output Impedance

 Desensitivity is defined as the reciprocal of sensitivity
 voltage gain has been reduced due to feedback network
Closed-loop gain of the amplifier with negative feedback is
Gain Desensitivity

Gain Desensitivity
𝐴𝑓 =
𝐴
[1+𝐴𝛽]
(1)
Differentiating the above equation with respect to 𝐴

Gain Desensitivity
𝐴𝑓 =
𝐴
[1+𝐴𝛽]
(1)
Differentiating the above equation with respect to 𝐴
𝑑𝐴𝑓
𝑑𝐴
=
1 + 𝐴𝛽 . 1 − 𝐴 𝛽
[1 + 𝐴𝛽]2

Gain Desensitivity
𝑑𝐴𝑓
𝑑𝐴
=
1 + 𝐴𝛽 . 1 − 𝐴 𝛽
[1 + 𝐴𝛽]2
𝑑𝐴𝑓
𝑑𝐴
=
1 + 𝐴𝛽 − 𝐴𝛽
[1 + 𝐴𝛽]2
𝑑𝐴𝑓
𝑑𝐴
=
1
[1+𝐴𝛽]2 (2)
𝑑𝐴𝑓 =
𝑑𝐴
[1 + 𝐴𝛽]2
Dividing both sides by 𝐴𝑓,

Gain Desensitivity
Dividing both sides by 𝐴𝑓,
𝑑𝐴𝑓
𝐴𝑓
=
𝑑𝐴
[1+𝐴𝛽]2 ×
1
𝐴𝑓
(3)
Substitute equation (1) in (3)
𝑑𝐴𝑓
𝐴𝑓
=
𝑑𝐴
[1 + 𝐴𝛽]2 ×
1 + 𝐴𝛽
𝐴
𝑑𝐴𝑓
𝐴𝑓
=
𝑑𝐴
1 + 𝐴𝛽
×
1
𝐴

Gain Desensitivity
𝑑𝐴𝑓
𝐴𝑓
=
𝑑𝐴
1 + 𝐴𝛽
×
1
𝐴
𝑑𝐴𝑓
𝐴𝑓
=
𝑑𝐴
𝐴
1 + 𝐴𝛽
𝒅𝑨𝒇
𝑨𝒇
represents the fractional change in amplifier voltage gain with feedback
𝒅𝑨
𝑨
denotes the fractional change in voltage gain without feedback
𝟏
𝟏+𝑨𝜷
is called sensitivity and 𝟏 + 𝑨𝜷 called as Desensitivity factor

Gain Desensitivity
𝑑𝐴𝑓
𝐴𝑓
=
𝑑𝐴
𝐴
1 + 𝐴𝛽
𝑑𝐴𝑓
𝐴𝑓
𝑑𝐴
𝐴
=
1
𝐷
=
1
1 + 𝐴𝛽
Stability of amplifier increases with increase in Desensitivity.

Extend the Bandwidth of the Amplifier
Amplifier bandwidth is a function of feedback
𝐵𝑊 𝑖𝑠 bandwidth of amplifier without feedback
𝐵𝑊′ is bandwidth of amplifier with feedback
𝛽 is feedback ratio.
𝐵𝑊′

Extend the Bandwidth of the Amplifier
𝐵𝑊′
Product of voltage gain and bandwidth of an amplifier without feedback and with
feedback remains the same
𝐴𝑣
′
× 𝐵𝑊′
= 𝐴𝑣 × 𝐵𝑊
𝑨𝒗
′
𝒇𝟐
′
− 𝒇𝟏
′
= 𝑨𝒗 𝒇𝟐 − 𝒇𝟏
𝑩𝑾𝒇 = 𝐁𝐖(𝟏 +𝛃𝐀)

Reduction of Noise
𝑉𝑂 = 𝐴1 𝑉𝑆 + 𝑉
𝑛
Find output voltage 𝑽𝑶

Reduction of Noise
Apply Superposition Theorem, Find 𝑽𝑶𝟏
𝑉𝑂1 = 𝑉𝑆
𝐴1𝐴2
1 + 𝐴1𝐴2𝛽 𝑉𝑛=0

Reduction of Noise
Apply Superposition Theorem, Find 𝑽𝑶𝟐
𝑉𝑂2 = 𝑉
𝑛
𝐴1
1 + 𝐴1𝐴2𝛽 𝑉𝑆=0

Reduction of Noise
Overall output voltage 𝑉𝑂 = 𝑉𝑂1 + 𝑉𝑂2
𝑽𝑶 = 𝑽𝑺
𝑨𝟏𝑨𝟐
𝟏+𝑨𝟏𝑨𝟐𝜷
+ 𝑽𝒏
𝑨𝟏
𝟏+𝑨𝟏𝑨𝟐𝜷
Overall Signal to Noise ratio is
𝑆
𝑁
=
𝑉𝑆
𝐴1𝐴2
1 + 𝐴1𝐴2𝛽
𝑉
𝑛
𝐴1
1 + 𝐴1𝐴2𝛽

Reduction of Noise
Overall Signal to Noise ratio is
𝑆
𝑁
=
𝑉𝑆
𝐴1𝐴2
1 + 𝐴1𝐴2𝛽
𝑉
𝑛
𝐴1
1 + 𝐴1𝐴2𝛽
𝑺
𝑵
=
𝑽𝑺
𝑽𝒏
𝑨𝟐

Reduce Nonlinear Distortion
 Non-linear distortion occurs due to active devices
 Occurs additional frequency components along with fundamental frequency called
hormonic distortion
 Negative feedback reduces the non linear distortion by the factor of 𝑫 = 𝟏 + 𝑨𝜷
 The distortion with feedback can be written as,
Df =
D
1 + Aβ

Control the Input and Output Impedance
𝑉𝑖
𝜷
𝑉𝑂

𝑉𝑖
𝜷
𝑉𝑂
𝑽𝒊 − 𝜷𝑽𝑶
𝜷𝑽𝑶
Find 𝑽𝑶
𝑉𝑂 = 𝐴𝑣 𝑉𝑖 − 𝛽𝑉𝑂
𝑉𝑖 − 𝛽𝑉𝑂= 𝑖1𝑍𝑖𝑛
𝑉𝑖 = 𝑖1𝑍𝑖𝑛 + 𝛽𝑉𝑂
𝑉𝑖 = 𝑖1𝑍𝑖𝑛 + 𝛽𝐴𝑣 𝑉𝑖 − 𝛽𝑉𝑂
Substitute 𝑉𝑂 in 𝑉𝑖
𝑉𝑖 = 𝑖1𝑍𝑖𝑛 + 𝛽𝐴𝑣𝑖1𝑍𝑖𝑛
𝑉𝑖 = 𝑖1𝑍𝑖𝑛 1 + 𝛽𝐴𝑣
𝑉𝑖
𝑖1
= 𝑍𝑖𝑛 1 + 𝛽𝐴𝑣

𝑉𝑖
𝜷
𝑉𝑂
𝑽𝒊 − 𝜷𝑽𝑶
𝜷𝑽𝑶
𝑉𝑖
𝑖1
= 𝑍𝑖𝑛 1 + 𝛽𝐴𝑣
Here
𝑉𝑖
𝑖1
= 𝑍𝑖𝑛
′
𝒁𝒊𝒏
′
= 𝒁𝒊𝒏 𝟏 + 𝜷𝑨𝒗

𝑉𝑂
𝑉𝑂
′
𝑉𝑂 = 𝑉𝑂
′
+ 𝑖𝑜𝑍𝑜𝑢𝑡
𝑉𝑂 = 𝐴(𝑉𝑖 − 𝛽𝑉𝑂) + 𝑖𝑜𝑍𝑜𝑢𝑡
𝑉𝑂 + 𝐴𝛽𝑉𝑂 = 𝑖𝑜𝑍𝑜𝑢𝑡 ∵ 𝑉𝑖= 0
𝑽𝟎
𝑽𝟎
,
𝑉𝑂 = 𝐴𝑉𝑖 − 𝐴𝛽𝑉𝑂 + 𝑖𝑜𝑍𝑜𝑢𝑡
∵ 𝐴 = 𝑉𝑂
′
/ (𝑉𝑖−𝛽𝑉𝑂)

𝑉𝑂
𝑉𝑂
′
𝑉𝑂 + 𝐴𝛽𝑉𝑂 = 𝑖𝑜𝑍𝑜𝑢𝑡
1 + 𝐴𝛽 𝑉𝑂 = 𝑖𝑜𝑍𝑜𝑢𝑡
𝑉𝑂
𝑖𝑜
=
𝑍𝑜𝑢𝑡
1 + 𝐴𝛽
Here
𝑉𝑂
𝑖𝑂
= 𝑍𝑜𝑢𝑡
′
𝒁𝒐𝒖𝒕
′
=
𝒁𝒐𝒖𝒕
𝟏 + 𝑨𝜷
1 + Aβ > 1,output impedance of amplifier decreases by a factor(1 + Aβ)

Topologies of Feedback Amplifier
 Voltage-series Feedback or series-shunt feedback
 Voltage-shunt Feedback or shunt-shunt feedback
 Current-series Feedback or series-series feedback
 Current-shunt Feedback or shunt-series feedback
Both voltage and current can be fed back to the input either in series or parallel

Topologies of Feedback Amplifier
 Voltage-series Feedback or series-shunt feedback
Both voltage and current can be fed back to the input either in series or parallel
Method of Sampling
Method of Mixing
Series-Shunt feedback
Method of Mixing
Method of Sampling

Voltage-series Feedback
(i) Gain of the Amplifier
(ii) Input Impedance with Feedback
(iii) Output Impedance with Feedback
Parameters

Gain of the Amplifier
𝑰𝒊
𝑰𝟎

If there is no feedback (𝑉𝑓 = 0), the voltage
gain of the amplifier
𝐴 =
𝑉0
𝑉𝑖
If the feedback signal 𝑽𝒇 is connected in series
𝑉𝑖 = 𝑉𝑆 − 𝑉𝑓
Find output from the feedback network
𝑉𝑓 = 𝛽𝑉0

Voltage-series Feedback Gain of the Amplifier
Find output Voltage 𝑉0
𝑉0 = 𝐴𝑉𝑖
𝑉0 = 𝐴 𝑉𝑆 − 𝑉𝑓 ⇒ 𝐴 𝑉𝑆 − 𝛽𝑉0
𝑉0 = 𝐴 𝑉𝑆 − 𝐴𝛽𝑉0 ⇒ 𝑉0 + 𝐴𝛽𝑉0 = 𝐴 𝑉𝑆
𝑉0 1 + 𝐴𝛽 = 𝐴 𝑉𝑆
𝑉0
𝑉𝑆
=
𝐴
1 + 𝐴𝛽
𝑨𝒇 =
𝑨
𝟏 + 𝑨𝜷

Voltage-series Feedback Input Impedance with Feedback
𝑰𝒊
𝑰𝟎

Apply KVL to the input side
𝑉
𝑠 = 𝑍𝑖𝐼𝑖 + 𝑉𝑓
Substitute 𝑉𝑓 = 𝛽𝑉0
𝑉
𝑠 = 𝑍𝑖𝐼𝑖 + 𝛽𝑉0
𝑉
𝑠 = 𝑍𝑖𝐼𝑖 + 𝛽𝐴𝑉𝑖
∵ 𝑉0 = 𝐴𝑉𝑖
∵ 𝑉𝑖 = 𝐼𝑖𝑍𝑖
𝑉
𝑠 = 𝑍𝑖𝐼𝑖 + 𝛽𝐴𝐼𝑖𝑍𝑖
1 + 𝛽𝐴 𝐼𝑖𝑍𝑖 = 𝑉𝑆
𝒁𝒊
𝒁𝒐
𝑉𝑆
𝐼𝑖
= 𝑍𝑖𝑓 𝑍𝑜𝑓
𝑍𝑜𝑓
′

𝒁𝒊
𝒁𝒐
1 + 𝛽𝐴 𝐼𝑖𝑍𝑖 = 𝑉𝑆
1 + 𝛽𝐴 𝑍𝑖 =
𝑉𝑆
𝐼𝑖
∵
𝑉𝑆
𝐼𝑖
= 𝑍𝑖𝑓
𝒁𝒊𝒇 = 𝟏 + 𝜷𝑨 𝒁𝒊
Voltage-series feedback, the input resistance increases by a factor (1 + 𝛽𝐴V)
𝑉𝑆
𝐼𝑖
= 𝑍𝑖𝑓

Voltage-series Feedback Output Impedance with Feedback
Applying KVL to the output loop
𝑭𝒊𝒏𝒅 𝑽𝟎
𝑉0 = 𝐴𝑉𝑖 + 𝐼0𝑍0
Substitute 𝑉𝑖
𝑉𝑖 = 𝑉𝑆 − 𝑉𝑓
To find output impedance, short circuit 𝑉
𝑠
𝑉𝑖 = −𝑉𝑓
𝑉0 = −𝐴𝑉𝑓 + 𝐼0𝑍0
𝑉
𝑜
𝒁𝒐𝒖𝒕

Voltage-series Feedback Output Impedance with Feedback
𝑉0 = −𝐴𝑉𝑓 + 𝐼0𝑍0
𝑉0 = −𝐴𝛽𝑉0 + 𝐼0𝑍0
𝑉0 + 𝐴𝛽𝑉0 = 𝐼0𝑍0
1 + 𝐴𝛽 𝑉0 = 𝐼0𝑍0
𝑉0
𝐼0
=
𝑍0
1 + 𝐴𝛽
𝒁𝟎𝒇 =
𝒁𝟎
𝟏+𝑨𝜷
𝑉
𝑜
𝒁𝒐𝒖𝒕

Voltage-shunt Feedback
Parameters

Voltage-shunt Feedback
𝐴 =
𝑉0
𝐼𝑖
If the feedback signal 𝑰𝒇 is connected in parallel
Find the input current 𝐼𝑠
Then the overall gain of the amplifier is
𝐴𝑓 =
𝑉0
𝐼𝑠

Voltage-Shunt Feedback
𝐼𝑓 = 𝛽𝑉0
𝐼𝑠 = 𝐼𝑖 + 𝐼𝑓
overall gain of the amplifier is,
𝐴𝑓 =
𝑉0
𝐼𝑠
⇒
𝑉0
𝐼𝑖+𝐼𝑓
From the circuit we know that ,
𝐼𝑖 =
𝑉0
𝐴

Voltage-Shunt Feedback Gain of the Amplifier
𝐴𝑓 =
𝑉0
𝐼𝑖+𝐼𝑓
𝐴𝑓 =
𝑉0
𝑉0
𝐴
+𝛽𝑉0
𝐴𝑓 =
𝐴𝑉0
𝑉0+𝐴𝛽𝑉0
𝐴𝑓 =
𝐴 𝑉0
(1+𝐴𝛽)𝑉0
𝑨𝒇 =
𝑨
(𝟏+𝑨𝜷)

Voltage-shunt Feedback Input Impedance with Feedback

𝑍𝑖𝑓 =
𝑉𝑖
𝐼𝑠
Substitute 𝐼𝑠
𝑍𝑖𝑓 =
𝑉𝑖
𝐼𝑖 + 𝐼𝑓
𝑍𝑖𝑓 =
𝑉𝑖
𝐼𝑖 + 𝛽𝑉0

𝑍𝑖𝑓 =
𝑉𝑖
𝐼𝑖 + 𝛽𝑉0
Input impedance can be obtained by dividin
equation by 𝐼𝑖
𝑍𝑖𝑓 =
𝑉𝑖
𝐼𝑖
𝐼𝑖
𝐼𝑖
+
𝛽𝑉0
𝐼𝑖
𝒁𝒊𝒇 =
𝒁𝒊
𝟏 + 𝑨𝜷

Voltage-shunt Feedback Output Impedance with Feedback
Applying KVL to the output loop
𝑭𝒊𝒏𝒅 𝑽𝟎
𝑉0 = 𝐴𝐼𝑖 + 𝐼0𝑍0
𝑰𝟎 =
𝑉0 − 𝐴𝐼𝑖
𝑍0
To find output impedance, open circuit 𝐼𝑠
𝐼𝑖 = −𝐼𝑓
𝐼0
∵ 𝐼𝑖 =
𝑉0
𝐴
∵ 𝐼𝑠 = 𝐼𝑖 + 𝐼𝑓
From the circuit we know that,

𝒁𝒐𝒇 =
𝒁𝟎
𝟏+𝑨𝜷
𝐼0
𝐼𝑖 = −𝐼𝑓 ⇒ −𝛽𝑉0
𝐼0 =
𝑉0 − (−𝐴𝛽𝑉0)
𝑍0
𝐼0 =
𝑉0 + 𝐴𝛽𝑉0
𝑍0
𝐼0 =
(𝟏 + 𝐴𝛽)𝑉0
𝑍0

𝒁𝒐𝒇 =
𝒁𝟎
𝟏+𝑨𝜷
𝐼0
𝐼0 =
(𝟏 + 𝐴𝛽)𝑉0
𝑍0
𝐼0
𝑉0
=
(𝟏 + 𝐴𝛽)
𝑍0
∵ 𝑍0𝑓 =
𝑉0
𝐼0

Current-series Feedback
Parameters

𝐴 =
𝐼0
𝑉𝑖
If the feedback signal 𝑽𝒇 is connected in series
Find the input voltage at the amplifier 𝑉𝑖
Then the overall gain of the amplifier is
𝐴𝑓 =
𝐼0
𝑉
𝑠

Current-Series Feedback
𝑉𝑓 = 𝛽𝐼0
𝑉𝑖 = 𝑉
𝑠 − 𝑉𝑓
From the circuit we know that ,
𝑉𝑖 =
𝐼0
𝐴
𝑉𝑖 = 𝑉
𝑠 − 𝛽𝐼0
𝐼0
𝐴
= 𝑉
𝑠 − 𝛽𝐼0

Current-Series Feedback Gain of the Amplifier
𝐼0
𝐴
= 𝑉
𝑠 − 𝛽𝐼0
𝐼0
𝐴
+ 𝛽𝐼0 = 𝑉
𝑠
𝐼0 + 𝐴𝛽𝐼0
𝐴
= 𝑉
𝑠
1 + 𝐴𝛽 𝐼0
𝐴
= 𝑉
𝑠
1 + 𝐴𝛽
𝐴
=
𝑉
𝑠
𝐼0

Current-Series Feedback Gain of the Amplifier
𝐴𝑓 =
𝐼0
𝑉𝑠
=
𝐴
(1+𝐴𝛽)
1 + 𝐴𝛽
𝐴
=
𝑉
𝑠
𝐼0
𝑨𝒇 =
𝑨
(𝟏+𝑨𝜷)

Current-Series Feedback Input Impedance with Feedback

Input impedance of the Amplifier
𝒁𝑰𝑭 =
𝑽𝒔
𝑰𝒊
𝑰𝟎
𝑰

Input Impedance with feedback is given as
𝑍𝑖𝑓 =
𝑉
𝑠
𝐼𝑖
𝑉
𝑠 − 𝐼𝑖𝑍𝑖 − 𝑉𝑓 = 0
Apply KVL to the input loop,
𝑉
𝑠 = 𝐼𝑖𝑍𝑖 + 𝑉𝑓
𝑉
𝑠 = 𝐼𝑖𝑍𝑖 + 𝛽𝐼0
The output current is given by
∵ 𝑉𝑓 = 𝛽𝐼0
𝐼0 = 𝐴𝑉𝑖
𝒁𝑰𝑭 =
𝑽𝒔
𝑰𝒊
𝑰𝟎
𝑰

𝑉
𝑠 = 𝐼𝑖𝑍𝑖 + 𝛽𝐼0
Substitute 𝐼0 value
𝑉
𝑠 = 𝐼𝑖𝑍𝑖 + 𝛽𝐴𝑉𝑖
𝑉
𝑠 = 𝐼𝑖𝑍𝑖 + 𝛽𝐴𝐼𝑖𝑍𝑖
∵ 𝑉𝑖 = 𝐼𝑖𝑍𝑖
𝑉
𝑠 = 𝐼𝑖𝑍𝑖 1 + 𝛽𝐴
𝑉
𝑠
𝐼𝑖
= 𝑍𝑖 1 + 𝛽𝐴
𝑍𝑖𝑓 = 𝑍𝑖 1 + 𝛽𝐴
𝒁𝑰𝑭 =
𝑽𝒔
𝑰𝒊
𝑰𝟎
𝑰

𝒁𝒊𝒇 = 𝒁𝒊 𝟏 + 𝜷𝑨
Input resistance increases by a factor of (1 + 𝑨𝛽)
𝑉
𝑠
𝐼𝑖
= 𝑍𝑖 1 + 𝛽𝐴
𝒁𝑰𝑭 =
𝑽𝒔
𝑰𝒊
𝑰𝟎
𝑰

Output impedance of the Amplifie
𝒁𝑰𝑭 =
𝑽𝒔
𝑰𝒊
𝑰𝟎
𝑰
Applying KCL to the output node
𝐴𝑉𝑖 + 𝐼 = 𝐼0
Input voltage is given by,
𝑉𝑠 − 𝑉𝑖 − 𝑉𝑓 = 0 ∵ 𝑉
𝑠 = 0
𝑉𝑖 = 𝑉𝑓 ∵ 𝑉𝑓 = 𝛽𝐼0
𝑉𝑖 = 𝛽𝐼0 or 𝑉𝑖 = 𝛽𝐼
Substitute 𝑉𝑖 value
𝐼𝐴𝛽 + 𝐼 = 𝐼0

Output impedance of the Amplifie
𝒁𝑰𝑭 =
𝑽𝒔
𝑰𝒊
𝑰𝟎
𝑰
𝐼𝐴𝛽 + 𝐼 = 𝐼0
𝐼 𝐴𝛽 + 1 = 𝐼0
𝐼 1 + 𝐴𝛽 =
𝑉
𝑍0
𝑍0 1 + 𝐴𝛽 =
𝑉
𝐼
𝒁𝒐 𝟏 + 𝑨𝜷 = 𝒁𝒐𝒇
Output resistance increase by a factor (𝟏 + 𝑨𝜷).

Current-shunt Feedback
Parameters

𝑰𝒇

Gain of the Amplifier with feedback
If there is no feedback (𝑉𝑓 = 0), the voltage gain
of the amplifier
𝐴 =
𝐼0
𝐼𝑖
If the feedback network is connected in parallel wi
th input, overall gain of the feedback amplifier,
𝐴𝑓 =
𝐼0
𝐼𝑠
Find 𝑰𝒔

Find 𝑰𝒔
𝐼𝑠 = 𝐼𝑖 + 𝐼𝑓
𝐴𝑓 =
𝐼0
𝐼𝑖 + 𝐼𝑓
Find output from the feedback networks
𝐼𝑓 = 𝛽𝐼0
Find Input current to the amplifier
𝐼𝑖 =
𝐼0
𝐴

𝐴𝑓 =
𝐼0
𝐼0
𝐴
+ 𝛽𝐼0
Substitute 𝐼𝑖 𝑎𝑛𝑑 𝐼𝑓 in 𝐴𝑓
𝐴𝑓 =
𝐴𝐼0
𝐼0 + 𝐴𝛽𝐼0
⇒
𝐴𝐼0
𝐼0 1 + 𝐴𝛽
𝑨𝒇 =
𝑨
𝟏 + 𝑨𝜷

Current-Shunt Feedback
Input Impedance with Feedback
𝑰𝒇
Input impedance with feedback is
𝑽𝒊
𝑍𝑖𝑓 =
𝑉𝑖
𝐼𝑠

𝑰𝒇
Applying KCL to the input node
𝐼𝑠 = 𝐼𝑓 + 𝐼𝑖
Substitute 𝐼𝑓 in 𝐼𝑠
𝐼𝑠 = 𝛽𝐼0 + 𝐼𝑖
𝑽𝒊

𝑰𝒇
Applying KCL to the input node
𝐼𝑠 = 𝐼𝑓 + 𝐼𝑖
Substitute 𝐼𝑓 in 𝐼𝑠
𝐼𝑠 = 𝛽𝐼0 + 𝐼𝑖
𝑽𝒊
Find output current 𝑰𝟎 with respect to amplifier

𝑰𝒇
𝑽𝒊
Find output current 𝑰𝟎 with respect to amplifier
𝐼0 = 𝐴𝑖𝐼𝑖
𝐼𝑠 = 𝛽𝐴𝑖𝐼𝑖 + 𝐼𝑖
𝐼𝑠 = 𝐼𝑖 𝛽𝐴𝑖 + 1

𝑰𝒇
𝑽𝒊
𝐼𝑠 = 𝐼𝑖 𝛽𝐴𝑖 + 1
Input impedance with feedback is
𝑍𝑖𝑓 =
𝑉𝑖
𝐼𝑠
𝑍𝑖𝑓 =
𝑉𝑖
𝐼𝑖 𝛽𝐴𝑖 + 1

𝑰𝒇
𝑽𝒊
𝑍𝑖𝑓 =
𝑉𝑖
𝐼𝑖 𝛽𝐴𝑖 + 1 ∵ 𝑉𝑖 = 𝑍𝑖𝐼𝑖
𝑍𝑖𝑓 =
𝑍𝑖𝐼𝑖
𝐼𝑖 𝛽𝐴𝑖 + 1
𝒁𝒊𝒇 =
𝒁𝒊
𝜷𝑨𝒊 + 𝟏

Current-Shunt Feedback Output Impedance with Feedback
𝑰𝒇
𝑽𝒊 𝑽𝒐
Output impedance of the circuit is,
𝑍𝑜𝑓 =
𝑉
𝑜
𝐼𝑜
Applying KCL to the output node
𝐴𝑖𝐼𝑖 +
𝑉
𝑜
𝑍𝑜
= 𝐼𝑜

𝑰𝒇
𝑽𝒊 𝑽𝒐
𝐴𝑖𝐼𝑖 +
𝑉
𝑜
𝑍𝑜
= 𝐼𝑜
we know that, 𝐼𝑠= 𝐼𝑓 + 𝐼𝑖
𝐼𝑓 = −𝐼𝑖 and 𝐼𝑖 = −𝛽𝐼𝑜
𝐴𝑖 −𝛽𝐼𝑜 +
𝑉
𝑜
𝑍𝑜
= 𝐼𝑜

𝑰𝒇
𝑽𝒊 𝑽𝒐
𝐴𝑖 −𝛽𝐼𝑜 +
𝑉
𝑜
𝑍𝑜
= 𝐼𝑜
𝑉𝑜
𝑍𝑜
= 𝐼𝑜 + 𝐴𝑖𝛽𝐼𝑜
𝑉
𝑜
𝑍𝑜
= 1 + 𝐴𝑖𝛽 𝐼𝑜

𝑰𝒇
𝑽𝒊 𝑽𝒐
𝑉
𝑜
𝑍𝑜
= 1 + 𝐴𝑖𝛽 𝐼𝑜
𝑉𝑜
𝐼𝑜
= 1 + 𝐴𝑖𝛽 𝑍𝑜
𝒁𝒐𝒇 = 𝟏 + 𝑨𝒊𝜷 𝒁𝒐

Operational Amplifier(OP-AMP)
Introduction
 Low cost integrating circuit consisting of
Transistors
Resistors
Capacitors
 Able to amplify a signal due to an external power supply
 Op-Amp is a very high gain differential amplifier with very high input
impedance (in terms of Mega Ω) and a low output impedance (less
than 100Ω)

 used in analog computers to perform mathematical operations to solve
differential and integral equations.
 Op-amps are linear integrated circuits (ICs) that use relatively low dc
supply voltages and are reliable and inexpensive
Operational Amplifier(OP-AMP)
Introduction

Applications of Op-Amps
 Simple Amplifiers
 Sign Changer
 Summers
 Comparators
 Integrators
 Differentiators
 Analog to Digital Converters

Pin Configuration of Op-Amps
8 Pin DIP (Dual Inline Package) Package

Pin Configuration of Op-Amps
8 Metal CAN package

Characteristics of an Ideal Op-Amp
 Infinite voltage gain
 Infinite input resistance
 Zero output resistance
 Zero output voltage when input is zero
 Infinite Common Mode Rejection Ratio (CMRR)
 Gain is independent of input frequency
 Infinite slew rate (Slew rate is defined as the maximum rate of change of an op amps
output voltage, and is given in units of volts per microsecond.)

Input Signal Modes
Single-ended differential mode
Double-ended differential mode

Input Signal Modes
Common-Mode operation

Op-Amp Parameters
Common-Mode Rejection Ratio
Common-Mode Rejection Ratio (CMRR) =
𝑨𝒅
𝑨𝑪𝑴
Common-Mode Rejection Ratio (CMRR) = 𝟐𝟎𝒍𝒐𝒈
𝑨𝒅
𝑨𝑪𝑴
in dB

Inverting Amplifier
If there is no current at the inverting input, then 𝐼𝑓 = 𝐼𝑖𝑛
Since inverting terminal is virtual ground, then
𝐼𝑖𝑛 =
𝑉𝑖𝑛−𝑉𝐴
𝑅𝑖
∵ 𝑉𝐴 = 0

Inverting Amplifier
𝐼𝑖𝑛 =
𝑉𝑖𝑛
𝑅𝑖
Find current through feedback 𝑅𝑓
𝐼𝑓 =
𝑉𝐴 − 𝑉𝑜𝑢𝑡
𝑅𝑓
Since 𝐼𝑓 = 𝐼𝑖𝑛
−𝑉𝑜𝑢𝑡
𝑅𝑓
=
𝑉𝑖𝑛
𝑅𝑖
𝑽𝒐𝒖𝒕
𝑽𝒊𝒏
=
−𝑹𝒇
𝑹𝒊
Closed Loop Gain 𝑨𝒄𝒍(𝑰) =
−𝑹𝒇
𝑹𝒊
𝐼𝑓 =
−𝑉𝑜𝑢𝑡
𝑅𝑓
𝐼𝑖𝑛 =
𝑉𝑖𝑛−𝑉𝐴
𝑅𝑖
∵ 𝑉𝐴 = 0

Inverting Amplifier- Input & Output Impedance
Input Impedance 𝑍𝑖 = 𝑅𝑖

Sign Changer
𝐴𝑐𝑙(𝐼) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
=
−𝑅𝑓
𝑅𝑖
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= −1
𝑉𝑜𝑢𝑡 = − 𝑉𝑖𝑛
Inverter or Sign Changer

Non-Inverting Amplifier
Current through 𝑅𝑖 = Current through 𝑅𝑓
𝑉𝑖𝑛 − 0
𝑅𝑖
=
𝑉𝑜𝑢𝑡 − 𝑉𝑖𝑛
𝑅𝑓
𝑉𝑖𝑛𝑅𝑓 = 𝑉𝑜𝑢𝑡 − 𝑉𝑖𝑛 𝑅𝑖
𝑉𝑖𝑛𝑅𝑓 = 𝑉𝑜𝑢𝑡𝑅𝑖 − 𝑉𝑖𝑛𝑅𝑖
𝑉𝑖𝑛 𝑅𝑓 + 𝑅𝑖 = 𝑉𝑜𝑢𝑡𝑅𝑖

Non-Inverting Amplifier
𝑉𝑖𝑛 𝑅𝑓 + 𝑅𝑖 = 𝑉𝑜𝑢𝑡𝑅𝑖
𝑅𝑓 + 𝑅𝑖
𝑅𝑖
=
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
= 1 +
𝑅𝑓
𝑅𝑖
Closed Loop Gain 𝑨𝒄𝒍 = 𝟏 +
𝑹𝒇
𝑹𝒊

Voltage Follower
𝑨𝒄𝒍 = 𝟏 +
𝟎
𝑹𝒊
𝑨𝒄𝒍 = 𝟏
𝐴𝑐𝑙 = 1 +
𝑅𝑓
𝑅𝑖

Adder or Summing Amplifier
Infinite impedance and virtual ground
𝐼𝑓 = 𝐼1 + 𝐼2 + 𝐼3
When all the three inputs are applied the output
voltage is
𝑉𝑜𝑢𝑡 = −𝐼𝑓𝑅𝑓
𝑉𝑜𝑢𝑡 = − 𝐼1 + 𝐼2 + 𝐼3 𝑅𝑓
= −
𝑉1
𝑅1
+
𝑉2
𝑅2
+
𝑉3
𝑅3
𝑅𝑓
If 𝑅1 = 𝑅2 = 𝑅3 = 𝑅, then

Adder or Summing Amplifier
If 𝑅1 = 𝑅2 = 𝑅3 = 𝑅, then
𝑽𝒐𝒖𝒕 = −
𝑹𝒇
𝑹
𝑽𝟏 + 𝑽𝟐 + 𝑽𝟑
If the gain of the amplifier is unity then,
𝑅1 = 𝑅2 = 𝑅3 = 𝑅𝑓
𝑽𝒐𝒖𝒕 = − 𝑽𝟏 + 𝑽𝟐 + 𝑽𝟑

Subtractor
When V2 is zero
𝑉𝑜1 =
−𝑅𝑓
𝑅1
𝑉1 (1)
When 𝑽𝟏 is zero
𝑉𝐵 = 𝑉2
𝑅𝑓
𝑅2+𝑅𝑓
(2)
Potential of node A is same as B i.e. VA = VB
𝐼 =
𝑉𝐴 − 𝑉1
𝑅1
⇒ 𝐼 =
𝑉𝐴
𝑅1
𝐼 =
𝑉𝐴
𝑅1
=
𝑉𝐵
𝑅1
(3)
Ref : Superposition Theorem

Subtractor 𝐼 =
𝑉𝐴
𝑅1
=
𝑉𝐵
𝑅1
(3)
Find the current at feedback path
𝐼 =
𝑉𝑜2−𝑉𝐴
𝑅𝑓
=
𝑉𝑜2−𝑉𝐵
𝑅𝑓
(4)
Equating the equations (3) and (4),
𝑉𝑜2−𝑉𝐵
𝑅𝑓
=
𝑉𝐵
𝑅1
𝑉𝑜2
𝑅𝑓
=
𝑉𝐵
𝑅1
+
𝑉𝐵
𝑅𝑓
𝑉𝑜2 = 𝑉𝐵
1
𝑅1
+
1
𝑅𝑓
𝑅𝑓
𝑅𝑓
𝑅1
+ 1

Subtractor
1
𝑅1
+
1
𝑅𝑓
𝑅𝑓
𝑅𝑓
𝑅1
+ 1
Substitute equation (2) in (5)
𝑉𝑜2 = 𝑉2
𝑅𝑓
𝑅2+𝑅𝑓
𝑅𝑓
𝑅1
+ 1
Hence using Superposition principle
𝑉
𝑜 = 𝑉𝑜1 + 𝑉𝑜2
𝑉
𝑜 =
−𝑅𝑓
𝑅1
𝑉1 + 𝑉2
𝑅𝑓
𝑅2+𝑅𝑓
𝑅𝑓
𝑅1
+ 1

Subtractor
𝑉
𝑜 = 𝑉𝑜1 + 𝑉𝑜2
𝑉
𝑜 =
−𝑅𝑓
𝑅1
𝑉1 + 𝑉2
𝑅𝑓
𝑅2+𝑅𝑓
𝑅𝑓
𝑅1
+ 1
If the resistances are selected as R1 = R2
𝑉
𝑜 =
−𝑅𝑓
𝑅1
𝑉1 + 𝑉2
𝑅𝑓
𝑅1+𝑅𝑓
𝑅𝑓
𝑅1
+ 1
𝑉
𝑜 =
−𝑅𝑓
𝑅1
𝑉1 + 𝑉2
𝑅𝑓
𝑅1+𝑅𝑓
𝑅1+𝑅𝑓
𝑅1
𝑉
𝑜 =
−𝑅𝑓
𝑅1
𝑉1 + 𝑉2
𝑅𝑓
𝑅1
𝑽𝒐 =
𝑹𝒇
𝑹𝟏
𝑽𝟐 − 𝑽𝟏

Subtractor
𝑽𝒐 =
𝑹𝒇
𝑹𝟏
𝑽𝟐 − 𝑽𝟏
Output voltage is proportional to the difference
between the two input voltages
If 𝑅1 = 𝑅2 = 𝑅𝐹 is selected
𝑽𝒐 = 𝑽𝟐 − 𝑽𝟏

Integrator
Because of virtual ground and infinite impedance
𝑖 = 𝑖𝑐
𝑖 =
𝑉𝑖−0
𝑅
=
𝑉𝑖
𝑅
(1)

Integrator
voltage across capacitor is
𝑉
𝑐 = 0 − 𝑉
𝑜
𝑉
𝑐 = −𝑉
𝑜
𝑖𝑐 =
𝐶𝑑𝑉
𝑐
𝑑𝑡

Integrator
𝑖𝑐 =
𝐶𝑑𝑉
𝑐
𝑑𝑡
𝑖𝑐 = −
𝐶𝑑𝑉𝑜
𝑑𝑡
(2)
From equations (1) and (2)
𝑉𝑖
𝑅
= −
𝐶𝑑𝑉
𝑜
𝑑𝑡

Integrator
𝑉𝑖
𝑅
= −
𝐶𝑑𝑉
𝑜
𝑑𝑡
𝑑𝑉𝑜
𝑑𝑡
= −
𝑉𝑖
𝑅𝐶
Take integration on both sides
𝑽𝒐 = −
𝟏
𝑹𝑪
𝟎
𝒕
𝑽𝒕 𝒅𝒕 scale multiplier
𝟏
𝑹𝑪

Feedback Amplifiers & Operational Amplifiers (OPAMP)

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