Transcript Block C: Amplifiers - City University of Hong Kong

EE2301: Basic Electronic Circuit
Block D: Amplifiers
Amplifiers - Unit 1: Amplifier Model
1
Some Applications
WiFi Repeater
Audio Amplifier
Kalaok Amplifier
Power Amplifier
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2
Some Applications
Servo Amplifier
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3
Some Applications
TV and Satellite
Antenna Amplifier
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4
Some Applications
Motion Sensor
Speed Sensor
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Temperature Sensor
Location Sensor
Alcohol Sensor
Heart Beat Rate Sensor
5
Amplifiers Need Everywhere
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Block D Unit 1 Outline
 Concepts in amplifiers
> Gain
> Input resistance
> Output resistance
 Introduction to the Operational Amplifier
> Characteristics of the Ideal Op Amp
> Negative feedback
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Gain: Inverting and Non-inverting
 The gain of an amplifier describes the ratio of the output amplitude to that
of the input
 If the output is an inverted version of the input with a larger amplitude, we
say the amplifier is INVERTING
> Gain is represented by a negative sign
 If the output is simply an amplified version of the input but with no shift in
phase, we say the amplifier is NON-INVERTING
Inverting
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Non-inverting
8
The decibel
 The gain of amplifiers are commonly expressed in decibels (written as dB)
 The decibel is a logarithmic unit related to the power gain:
Gain in dB = 10 log10(Pout/Pin)
Gain in dB = 20 log10(Vout/Vin)
 The dB is extremely useful in finding the overall gain when we cascade
amplifiers (connecting the output of one stage to the input of another) like
in the figure below:
> Note that the overall gain = multiplication of individual gains
> Therefore, overall gain (in dB) = sum of individual gains (in dB)
A1
A2
Total gain (in dB) = 20 log10(A1A2)
= 20 log10(A1) + 20 log10(A2)
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Ideal Amplifier
We will first consider the amplifier as a
black box and see how it connects with
the rest of the system. We can see right
away that the amplifier is really two port
network (REF Block A Unit 3).
Input terminals:
The input terminals of the amplifier are
connected to the source, modeled as a
Thevenin circuit.
Output terminals:
The output terminals of the amplifier
are connected to the load.
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Ideal Amplifier
Now we take a look at how the amplifier looks like on the inside
INPUT port:
Amplifier acts as an equivalent
load with respect to the source
OUTPUT port:
Amplifier acts as an equivalent
source with respect to the load
1) Input resistance Rin: This is seen across the input terminals
2) Dependent source Avin: In the circuit model above, the voltage of the source
depends on the voltage drop across Rin. A is known as the open loop gain.
3) Output resistance Rout: This is seen in series with the dependent source and the
positive output terminal.
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Effect of finite resistances
From the source to the input terminals, there
will be some voltage drop across RS due to
the finite value of Rin. Hence Vin < VS since:
vin 
Rin
vS
RS  Rin
vin is then amplified by a factor A through the
dependent source Avin
From the dependent source to the output terminals, there will be some
voltage drop across Rout since Rout is non-zero. Hence VL < Avin since:
RL
vL 
Avin
Rout  RL
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Impedance Dependence
vL 
RL
vL 
Avin
Rout  RL
Rin
RL
AvS
RS  Rin Rout  RL
vin 
Rin
vS
RS  Rin
 Amplification now dependent on both source and load impedances
 Also dependent on input & output resistance of the amplifier
 Therefore, different performance using different load/source for
same amplifier
Rin should be very large (ideally infinite), so that vin ≈ vS
Rout should be very small (ideally zero), so that vL ≈ Avin
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Impedance: Example 1
RS = Rin = Rout = RL = 50Ω, A = 50
Find vL, (1) without RL, (2) with RL
Vin 
Rin
VS
Rin  RS
 0.5VS
Without RL:
VL  AVin
 50 0.5VS
 25VS
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With RL:
RL
VL 
AVin
RL  RO
1
    50 0.5VS
2
 12.5VS
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Impedance: Example 2
RS = Rin = Rout = RL = 50Ω, A = 40
If two of the above amplifiers are cascaded, find VL
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Impedance: Example 2
vs +
-
Rout
Rout
Rs
Rin
+ Av
-
Rin
 R in 
R in



v L  vs 
A


 R in  R S  R in  R out
1 
1
 1 
 v s   40  40 
2 
2
 2 
 200v s
+
Av
-

RL
 A
 R L  R out
RL



Given: RS = Rin = Rout = RL = 50Ω, A = 40
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Impedance: Example 3
RS = Rin = Rout = RL = 50Ω, A = 20
If it is required for vL = 40vS, how many amplifier stages are needed?
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Impedance: Example 3 solution
From previous example 2:
 RL
v L  
 R L  R out

R in
 A
 R in  R out
n

 v s

n
1
 1 
   20   v s
2
 2 
 0.5  10n v s
To achieve VL = 40VS
0.5 10n  40
n  1.9
Need an integer number of stages, so therefore 2 stages required.
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Block D Unit 1 Outline
 Concepts in amplifiers
> Gain
> Input resistance
> Output resistance
 Introduction to the Operational Amplifier
> Characteristics of the Ideal Op Amp
> Negative feedback
EE2301: Block D Unit 1
Much Simpler to Use
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Block D Unit 1 Outline
 Concepts in amplifiers
> Gain
> Input resistance
> Output resistance
 Introduction to the Operational Amplifier
> Characteristics of the Ideal Op Amp
> Negative feedback
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Operational Amplifier
v+
vv S+
v Svout
Non-inverting input
Inverting input
Positive power supply
Negative power supply
Output
v S+
v+
+
vout
_
vvout = AV(OL) (v+ - v-)
v S-
 The operational amplifier (or op-amp for short) behaves much like an ideal
difference amplifier
 It amplifies the difference between two input voltages v+ and v-
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Op-amp Model
iin
+
vin
Rout
Rin
-
v+
iin
+
+
-
Avin
vin
v-
-
Rout
Rin
+
+
-
vout
AV(OL)vin
General amplifier
model
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-
Op-amp model
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Ideal Op-amp Characteristics
v+
iin
+
vin
v-
-
Rout
Rin
+
+
-
vout
AV(OL)vin
But what about the effect
of infinite open loop gain?
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-
There are 3 main assumptions
we make for an ideal op amp:
1) Infinite input resistance (i.e.
Rin → ∞)
2) Zero output resistance (i.e.
Rout = 0)
3) Infinite open loop gain (i.e.
AV(OL) → ∞)
Given that Rin → ∞, this means that
no current flows into or out of any of
the input terminals (inverting as well
as non-inverting)
Given that Rout = 0, this means that
vout = Avin
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Negative feedback
In this course, we will focus on the op amps being used in negative feedback.
What this means is that we introduce a direct electrical path between the output
and the inverting input terminals.
What this does is to take some of the output and feed it back to the input in the
opposite sense. This provides stability to the circuit and is commonly used to
build amplifiers and filters.
We can see that:
Vout = A(Vin – Vout)
Vin
+
Vout
-
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Re-arranging the terms:
A
Vout 
Vin
A 1
Now if A is infinite, this will then force Vout to equal Vin.
In other words, an infinite A has the effect of forcing the
input terminals to be at the same voltage.
V+ = V24
Before we continue,
let’s go through some op-amp
applications
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Inside the Op Amp
Made up of many transistors and circuits…
Model 741
Amplifiers - Unit 2: The operational amplifier
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Circuit Example 1 - Timer
Amplifiers - Unit 2: The operational amplifier
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Circuit Example 2 - Ramp Generator
Amplifiers - Unit 2: The operational amplifier
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Circuit Example 3 – Optical Receiver
Amplifiers - Unit 2: The operational amplifier
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Circuit Example 5 – Active Filter
Amplifiers - Unit 2: The operational amplifier
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Circuit Example 6 – Dual Power Regulator
Amplifiers - Unit 2: The operational amplifier
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Circuit Example 7 – Radio Receiver
Amplifiers - Unit 2: The operational amplifier
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EE2301: Basic Electronic Circuit
Feel Confusion!
If you want to do these designs,
you have to start from the basic first
Amplifiers - Unit 2: The operational amplifier
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Block D Unit 2 Outline
 Op-amp circuits with resistors only
> Inverting amplifier
> Non-inverting amplifier
> Summing amplifier
> Differential amplifier
> Instrumentation amplifier
 Op-amp circuits with reactive components
> Active filters (Low pass, High pass, Band pass)
> Differentiator & Integrator
 Physical limits of practical op-amps
EE2301: Block D Unit 2
34
Source follower
Vin
Rs
+
Vout
-
RL
Find the gain of the above circuit
The key features of the source follower are:
1) Large input resistance
2) Small output resistance
3) Unity gain (i.e. gain of close to one)
It is therefore commonly used as a buffer between a load and source where the
impedances are not well matched
How to prove it?
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Inverting op amp
R2
R1 and R2 are in series
R1
Vin
Note that the inverting input (node X) is at ground
-
Applying KCL at X
Vout
+
Vin Vout

0
R1 R2
Closed-loop gain
Vout
R2

Vin
R1
Negative sign indicates a 1800
shift in the phase
Gain is set by the resistor values
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Non-inverting amplifier
iin
Note that since iin = 0:
R
+
1) v+ = vin (no voltage drop across R)
2) We can apply voltage divider rule to
RS & RF
+
vin _
RS
vRS 
vout
RS  RF
+
_
vout
iin
+
RS
vRS
-
RF
-
But we also note that A is infinite, so:
1) vin = vRS
2) Hence we then obtain:
EE2301: Block D Unit 2
vout
RF
G
 1
vin
RS
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Summing Amplifier
RS1
Apply NVA at the inverting input
terminal:
vS1 vS 2 vout


0
RS1 RS 2 RF
v
vout
v 
  S 1  S 2 
RF
 RS 1 RS 2 
+
vS1 _
EE2301: Block D Unit 2
-
RS2
+
+
vS2 _
Hence the form of the gain relation can be
described by:
Vout = -(A1vS1 + A2vS2)
A1 and A2 are set by the resistor values
chosen.
RF
+
vout
-
We can extend this result to write a
general expression for the gain:
N
RF
vSn
n 1 RSn
vout  
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Differential amplifier
R2
V1
V2
One way of analyzing this circuit is to apply
superposition (find Vout for V1 or V2 only)
R1
R1
Vout
+
If we short V2 first, we obtain:
V1
R2
+
R1
EE2301: Block D Unit 2
Vout1
vout1  
[R2/(R1+R2)]V2
Vout2
R2
+
If we short V1 now, we obtain:
-
R2
R1
vout 2 
R2
v2
R1
R2
v1
R1
Hence, finally: Vout = (R2/R1)(v2 – v1)
Output is the difference between the
inputs amplified by a factor set by the
resistor values.
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Difference amplifier
In the case whereby all the resistors are different, as shown in the circuit below, while
vout1 remains unchanged, vout2 now becomes:
vout 2
 R2  R4 
v2
 1  
 R1  R3  R4 
R2
Hence the overall gain expression is given by:
 R2  R4 
R
v2  2 v1
vout  1  
R1
 R1  R3  R4 
This is similar to the form of the summing
amplifier except that we take the difference
between the two inputs:
V1
V2
R1
R3
+
Vout
R4
Vout = A2V2 – A1V1
A2 and A1 are set by the resistor values
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Diff Amp: Example 1
Find vout if R2 = 10kΩ and R1 = 250Ω
R2
V1
V2
R1
R1
+
Vout
R2
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Diff Amp: Example 2
For the same circuit in the previous example, given v2 = v1:
Find vout if v1 and v2 have internal resistances of 500Ω and 250Ω respectively
 R2  R4 
R2
v2  v1
vout  1  
R1
 R1  R3  R4 
R2
500Ω
V1
V2
R1
R1
+
Vout
Now: R2 = 10kΩ, R1 = 750Ω, R3
= 500Ω, R4 = 10kΩ
250Ω
R2
vout = 13.65 v2 – 13.33v1
We can see that as a result of the source resistances:
1)
Differential gain has changed
2)
vout is not zero for v2 = v1
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EE2301: Basic Electronic Circuit
Recap in last lecture
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Block D Unit 2 Outline
 Ideal Op-Amp Characteristic
> Three basic assumptions
1. Infinitive input impedance
2. Zero output impedance
3. Infinitive open-loop gain
 V+ = V-
 Op-amp circuits with resistors only
> Source Follower
> Inverting amplifier
> Non-inverting amplifier
> Summing amplifier
> Differential amplifier (Difference amplifier)
> Instrumentation amplifier
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Source follower
Vin
Rs
+
Vout
-
RL
Find the gain of the above circuit
The key features of the source follower are:
1) Large input resistance
2) Small output resistance
3) Unity gain (i.e. gain of close to one)
It is therefore commonly used as a buffer between a load and source where the
impedances are not well matched
EE2301: Block D Unit 2
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Inverting op amp
R2
R1
Vin
Vout
Vout
R2

Vin
R1
+
Negative sign indicates a 1800
shift in the phase
Gain is set by the resistor values
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Non-inverting amplifier
iin
RS
vRS 
vout
RS  RF
R
+
+
vin _
vout
iin
+
vout
RF
G
 1
vin
RS
EE2301: Block D Unit 2
+
_
RS
vRS
-
RF
-
47
Summing Amplifier
RS1
 vS 1 vS 2 
vout

 

RF
 RS 1 RS 2 
+
vS1 _
RF
-
RS2
+
vS2 _
+
+
vout
-
We can extend this result to write a
general expression for the gain:
N
vout
RF
 
vSn
n 1 RSn
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Differential amplifier
R2
V1
V2
R1
R1
Vout
+
R2
 R2 
vout   v2  v1 
 R1 
R2
V1
V2
R1
R3
+
 R2  R4 
R2
v2  v1
vout  1  
R1
 R1  R3  R4 
R4
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Diff Amp: Example 2
For the same circuit in the previous example, given v2 = v1:
Find vout if v1 and v2 have internal resistances of 500Ω and 250Ω respectively
4  R2
 
 2 R
R2 R
 
v2v2 v1v1
vvoutout 
1 
R R

R1
 1  1R3  R4 
R2
500Ω
V1
V2
R1
R1
+
Vout
Now: R2 = 10kΩ, R1 = 750Ω, R3
= 500Ω, R4 = 10kΩ
250Ω
R2
vout = 13.65 v2 – 13.33v1
We can see that as a result of the source resistances:
1)
Differential gain has changed
2)
vout is not zero for v2 = v1
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Instrumentation Amplifier
V1
+
-
R3
R1
R2
vo1
RX
-
V2
R1
vo2
+
-
Vout
R2
+
R3
The instrumentation amplifier takes care of this problem by including a noninverting amplifier (which possesses an infinite input resistance) between the
differential amplifier an the inputs.
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Instrumentation Amplifier
V1
+
Vo1
-
R3
Vo1
R1
+
RX/2
-
1st stage comprises a pair of non-inverting
amplifiers:
Vout
Vo2
R3
Vo1 = (2R1/Rx +1)V1
Closed-loop gain:
2nd stage is a differential amplifier:
Vout = (R3/R2)(2R1/Rx + 1)(V2 - V1)
Vout = (R3/R2) (Vo2 - Vo1)
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Op amp example 1
Problem 8.5
Find v1 in the following 2 circuits
What is the function of the source follower?
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Op amp example 1 solution
Fig (a):
6Ω || 3Ω = 2Ω
v1 = {2 / (2+6)}*Vg = 0.25Vg
Fig (b):
Voltage at the non-inverting input = 0.5Vg
Voltage at output = Voltage at non-inverting input = 0.5Vg
This slide is meant to be blank
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Op amp example 2
Problem 8.7
Find the voltage v0 in the following circuit
Transform to Thevenin
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Op amp example 2 solution
Thevenin equivalent circuit:
Rth = 6 + 2||4 = 22/3 kΩ
Vth = {4/(2+4)}*11 = 22/3 V
Closed loop gain expression:
A = - 12kΩ / Rth (With Vth as input source)
Vout = 12/(22/3) * (22/3) = 12V
This slide is meant to be blank
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Op amp example 3
R2
Find the closed-loop gain
R1
vin
-
Current through RB = 0A (Infinite input
resistance of op amp)
+
Consider:
Voltage across RB = 0V
RB
+
vout
-
Voltage at inverting input = 0V
Same as inverting op amp:
A = -R2 / R1
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Op amp example 4
Find the closed-loop gain
R2
Consider:
-
Current through R2 = 0A (Infinite
input resistance of op amp)
Voltage across R2 = 0V
Vout = VVout = Vin (Infinite open loop gain of
op amp)
EE2301: Block D Unit 2
+
vin
-
+
+
vout
-
58
Common and differential mode
A differential amplifier should ideally amplify ONLY DIFFERENCES between the inputs.
That is to say identical inputs should give an output of zero.
Ideal differential amplifier: Vout = A(V2 – V1)
In reality this is not the case as we have seen in a previous example. The output of the
amplifier is more accurately described by:
Vout = A2V2 + A1V1
A2: Gain when V1 = 0 (V2 is the only input)
A1: Gain when V2 = 0 (V1 is the only input)
We would like to express
A1 and A2 by Acm and Adm
It is then useful to describe the performance of a differential amplifier in terms of the
gain when both input are identical (common mode) and when the inputs are equal and
out-of-phase (differential)
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Common mode rejection ratio
Common mode gain (Acm): Gain when both inputs are exactly the same (V1 = V2
= Vin)
Vocm = (A2 + A1)Vin Voltage output for common input
Acm  Vocm Vin  A2  A1
Differential mode gain (Adm): Gain when both inputs are equal but out-of-phase (V2 =
-V1 = Vin)
V = (A - A )V Voltage output for differential input
odm
2
1
in
Adm  Vodm 2Vin   A2  A1  2
Divide by 2 since the difference of a pair of differential inputs is twice that of each input
 v  v1 
vout  Adm v2  v1   Acm  2

2 

The common mode rejection ratio (CMRR) is simply the ratio of the differential mode
gain (Adm) over the common mode gain (Acm):
CMRR = Adm/Acm
A large CMRR is therefore desirable for a differential amplifier
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Block D Unit 2 Outline
 Op-amp circuits with resistors only
> Inverting amplifier
> Non-inverting amplifier
> Summing amplifier
> Differential amplifier
> Instrumentation amplifier
 Op-amp circuits with reactive components
> Active filters (Low pass, High pass, Band pass)
> Differentiator & Integrator
 Physical limits of practical op-amps
EE2301: Block D Unit 2
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EE2301: Basic Electronic Circuit
Let’s con’t in this lecture
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Active filters
Range of applications is greatly expanded if reactive components are used
Addition of reactive components allows us to shape the frequency response
Active filters: Op-amp provides amplification (gain) in addition to filtering effects
Substitute R with Z now for our analysis:
Vout
ZF
 j   
VS
ZS
ZF and ZS can be arbitrary (i.e.
any) complex impedance
Vout
ZF
 j   1 
VS
ZS
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Active low pass filter
Vout
ZF
 j   
VS
ZS
Z S = RS
Z F  RF || C F
 R  
1 

  F   RF 
jC F 
 jC F  
RF

Amplification
1  jC F RF
Vout
 j    RF RS
VS
1  jCF RF
EE2301: Block D Unit 2
Shaping of frequency
response
64
Active high pass filter
Vout
ZF
 j   
VS
ZS
Z F = RF
Z S  RS  1 jCS
 1  jCS RS  jCS
Amplification
Vout
 j    jCs RF
VS
1  jCS RS
ω→∞, Vout/VS → -RF/RS
EE2301: Block D Unit 2
Shaping of frequency response
65
Active band pass filter
Vout
ZF
 j   
VS
ZS
Z S  RS  1 jCS
 1  jCS RS  jCS
Z F  RF || C F

RF
1  jC F RF
Vout
jCs RF
 j   
1  jCS RS 1  jCF RF 
VS
ZS: CS blocks low frequency inputs but lets high frequency inputs through
High pass filter
ZF: CF shorts RF at high frequency (reducing the gain), but otherwise looks just like a high
pass filter at lower frequencies
Low pass filter
EE2301: Block D Unit 2
66
Active band pass filter
Vout
jCs RF
 j   
1  jCS RS 1  jCF RF 
VS
ωHP = 1/(CSRS) – Lower cut-off frequency
ωLP = 1/(CFRF) – Upper cut-off frequency
For ωHP < ωLP: Frequency response curve is shown below
EE2301: Block D Unit 2
67
Second-order Low Pass Filter
Vout
ZF
 j   
VS
ZS
Z F  R2 || C

Z S  R1  jL
R2
1  jCR2
Vout
R2
 j   
1  jCR2 R1  jL 
VS
R1, R2, C and L are specially chosen so that: ω0 = 1/(CR2) = R1/L.
The frequency response function then simplifies to:
Vout
R2
 j   
2
VS
R1 1  j  0 
EE2301: Block D Unit 2
68
Second-order Low Pass Filter
R1, R2, C and L are specially chosen so that:
ω0 = 1/(CR2) = R1/L.
H v  j 
Vout
R2
 j   
2
VS
R1 1  j  0 
Above ω0, Hv is reduced by a factor of 100 for a ten fold
increase in ω (40dB drop per decade)
First-order filter: Hv is reduced by a factor of 10 for a ten fold
increase in ω (20dB drop per decade)
EE2301: Block D Unit 2
69
Ideal integrator
KCL at inverting input: iS = -iF
Virtual ground at inverting input: iS = VS/RS
For a capacitor: iF = CF[dVout/dt]
VS
dVout
 C
RS
dt
Output is the integral of
the input
dVout
VS

dt
CF RS
1
Vout (t )  
VS (t )dt

CF RS
EE2301: Block D Unit 2
70
Ideal differentiator
KCL at inverting input: iS = -iF
Virtual ground at inverting input: iF = Vout/RF
For a capacitor: iS = CS[dVS/dt]
Vout
dVS
 CS
RF
dt
Vout   RF CS
Output is the timedifferential of the input
EE2301: Block D Unit 2
dVS
dt
71
Physical limit: Voltage supply limit
The effect of limiting supply voltages is that
amplifiers are capable of amplifying signals
only within the range of their supply voltages.
RS = 1kΩ, RF = 10kΩ, RL = 1kΩ;
VS+ = 15V, VS- = -15V; VS(t) = 2sin(1000t)
Gain = -RF/RS = -10
Vout(t) = 10*2sin(1000t) = 20sin(1000t)
But since the supply is limited to +15V and -15V,
the op-amp output voltage will saturate before
reaching the theoretical peak of 20V.
EE2301: Block D Unit 2
72
Physical limit: Frequency response limit
So far we have assumed in our ideal op-amp model that the open loop gain AV(OL) is
infinite or at most a large constant value. In reality, AV(OL) varies with a frequency
response like a low pass filter:
ω0: frequency when the
response starts to drop off
The consequence of a finite bandwidth is a fixed gain-bandwidth product
If closed loop gain is increased, -3dB bandwidth is reduced
Increasing the closed loop gain further results in a bandwidth reduction till the
gain-bandwidth produce equals the open-loop gain
Gain bandwidth product = A0ω0
EE2301: Block D Unit 2
73