Download Operational Amplifiers

Shantilal Shah Government
Engineering College
Bhavnagar
 Electrical Engg. Department
Branch :- ELECTRICAL
Semester :Batch:- B1
rd
3
SUBJECT:
ANALOG ELECTRONICS
SUBJECTED BY:
PROF. A. M.UPADHYAY
Group members
NAME
1) Gamit sujal J.
ENROLL.NO
(130430109018)
2) Chaudhari piyush U. (130430109008)
What is an Op-Amp? – The
Surface
 An Operational Amplifier (Op-Amp) is an
integrated circuit that uses external voltage
to amplify the input through a very high
gain.
 We recognize an Op-Amp as a massproduced component found in countless
electronics.
What an Op-Amp looks
like to a lay-person
What an Op-Amp looks
like to an engineer
History of the Op-Amp – The Dawn
 The Vacuum Tube Age
 The First Op-Amp: (1930 – 1940) Designed by Karl Swartzel for the
Bell Labs M9 gun director
 Uses 3 vacuum tubes, only one input, and ± 350 V to attain a gain of
90 dB
 Loebe Julie then develops an Op-Amp with two inputs: Inverting
and Non-inverting
Mathematics of the Op-Amp
 The gain of the Op-Amp itself is calculated as:
G = Vout/(V+ – V-)
 The maximum output is the power supply voltage
 When used in a circuit, the gain of the circuit (as opposed
to the op-amp component) is:
Av = Vout/Vin
Op-Amp Characteristics
• Open-loop gain G is typically over 9000
• But closed-loop gain is much smaller
• Rin is very large (MΩ or larger)
• Rout is small (75Ω or smaller)
• Effective output impedance in closed loop is very small
Symbols for Ideal and Real Op Amps
OpAmp
uA741
LM111
LM324
Terminals on an Op Amp
Positive power supply
(Positive rail)
Non-inverting
Input terminal
Output terminal
Inverting input
terminal
Negative power supply
(Negative rail)
Open Circuit Output Voltage
vo = A vd
 Ideal Op Amp
vo = ∞ (vd)
Open Circuit Output Voltage
 Real Op Amp
Voltage
Range
Positive Saturation A vd > V+
Linear Region
V- < A vd < V+
Negative
Saturation
A vd < V-
Output
Voltage
vo ~ V+
vo = A vd
vo ~ V-
The voltage produced by the dependent voltage source inside the op amp is
limited by the voltage applied to the positive and negative rails.
Voltage Transfer Characteristic
Range where
we operate
the op amp as
an amplifier.
vd
Ideal Op Amp
v2
i2 = 0
v1 = v2
vd = 0 V
i1 = 0
v1
Because Ri is equal to ∞W,
the voltage across Ri is 0V.
Ideal Operational Amplifier
 The “ideal” op amp is a special case of the ideal differential amplifier
with infinite gain, infinite Rid and zero Ro .
v
v  o
id A
and
lim vid  0
A 
 If A is infinite, vid is zero for any finite output voltage.
 Infinite input resistance Rid forces input currents i+ and i- to be
zero.
 The ideal op amp operates with the following assumptions:
 It has infinite common-mode rejection, power supply rejection,
open-loop bandwidth, output voltage range, output current
capability and slew rate
 It also has zero output resistance, input-bias currents, inputoffset current, and input-offset voltage.
Almost Ideal Op Amp
 Ri = ∞ W
 Therefore, i1 = i2 = 0A
 Ro = 0 W
 Usually, vd = 0V so v1 = v2
 The op amp forces the voltage at the inverting input terminal
to be equal to the voltage at the noninverting input terminal
if there is some component connecting the output terminal to
the inverting input terminal.
 Rarely is the op amp limited to V- < vo < V+.
 The output voltage is allowed to be as positive or as negative
as needed to force vd = 0V.
Inverting Amplifier: Input and
Output Resistances
Rout is found by applying a test
current (or voltage) source to the
amplifier output and determining the
voltage (or current) after turning off
all independent sources. Hence, vs
=0
vx  i R  i R
2 2 11
But i1=i2
vx  i ( R  R )
1 2 1
vs
R   R since v  0
in i
1
s

Since v- = 0, i1=0. Therefore vx
= 0 irrespective of the value of ix
.
Rout  0
Inverting Amplifier:Voltage Gain
 The negative voltage gain


vs  isR  i R  vo  0
1 2 2

But is= i2 and v- = 0 (since vid= v+ - v-=
0)
R
vs
vo
is 
 2
and Av 
R
vs
R
1
1
implies that there is a 1800
phase shift between both dc
and sinusoidal input and
output signals.
The gain magnitude can be
greater than 1 if R2 > R1
The gain magnitude can be less
than 1 if R1 > R2
The inverting input of the op
amp is at ground potential
(although it is not connected
directly to ground) and is said
to be at virtual ground.
The Non-inverting Amplifier:
Configuration
• The input signal is applied to the non-inverting input terminal.
• A portion of the output signal is fed back to the negative input
terminal.
• Analysis is done by relating the voltage at v1 to input voltage vs and
output voltage vo .
Non-inverting Amplifier: Voltage Gain, Input
Resistance and Output Resistance
R
vs  v  v
1
and
v  vo
id 1
1
R R
1 2
vs  v
But vid =0
1
R R
vo  vs 1 2
R
1
R
v o R1  R2
 Av 

 1 2
R
vs
R
1
1
vs
R  
Since i+=0
in i

Rout is found by applying a test current source to the amplifier
output after setting vs = 0. It is identical to the output resistance of
the inverting amplifier i.e. Rout = 0.
Since i-=0
Non-inverting Amplifier:
Example
 Problem: Determine the output voltage and current for the given
non-inverting amplifier.
 Given Data: R1= 3kW, R2 = 43kW, vs= +0.1 V
 Assumptions: Ideal op amp
 Analysis:
R
43kW
Av 1 2 1
15.3
R
3kW
1
vo  Avvs (15.3)(0.1V)1.53V
Since i-=0,

vo
1.53V
io 

 33.3A
R  R 43kW  3kW
2 1
Types of Gain
if
is
i1
io
i
i2
Gain Error
 Gain Error is given by
GE = (ideal gain) - (actual gain)
For the non-inverting amplifier,
GE 
1
A
1

 1 A  (1 A )

 Gain error is also expressed as a fractional or
percentage error.1 
A
1
1
FGE   1 A 

1
1 A A

PGE 

1
100%
A
Types of Closed Loop Gain
Gain
Variable
Name
Equation
Units
Voltage Gain
Current Gain
AV
AI
vo/vs
io/is
None or V/V
None or A/A
Transresistance Gain
Transconductance
Gain
AR
AG
vo/is
io/vs
V/A or W
A/V or W1
Finite Common-Mode Rejection
Ratio (CMRR)
A(or Adm) = differential-mode gain
Acm = common-mode gain
vid = differential-mode input voltage
vic = common-mode input voltage
v
v
v  v  id
v  v  id
1 ic 2
2 ic 2
A real amplifier responds to
signal common to both inputs,
called the common-mode input
voltage (vic). In general,
v v 
vo  A (v v ) Acm 1 2 
dm 1 2

2 

vo  A (v ) Acm(v )
ic
dm id
An ideal amplifier has Acm = 0, but for a
real amplifier,




Acm v 
v


ic  A v  ic 
vo  A v 

dm id
dm id CMRR 
A


dm 
A
CMRR  dm
Acm
and CMRR(dB) 20log (CMRR)
10
Op-Amp Summing Amplifier
Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg:
Op Amps Applications
 Audio amplifiers
 Speakers and microphone circuits in cell phones,
computers, mpg players, boom boxes, etc.
 Instrumentation amplifiers
 Biomedical systems including heart monitors and
oxygen sensors.
 Power amplifiers
 Analog computers
 Combination of integrators, differentiators, summing
amplifiers, and multipliers
Applications of Op-Amps
 Simple EKG circuit
 Uses differential
amplifier to cancel
common mode signal
and amplify differential
mode signal
 Realistic EKG circuit
 Uses two non-inverting
amplifiers to first
amplify voltage from
each lead, followed by
differential amplifier
 Forms an
“instrumentation
amplifier”
Applications of Op-Amps
Filters
Low pass filter
Types:
•Low pass filter
•High pass filter
•Band pass filter
•Cascading (2 or more filters connected
together)
R2
R1
Low pass filter transfer
function
+
Low pass filter Cutoff
frequency 
C
+ Vcc
+
- Vcc
+
V0
__
Summary
 The output of an ideal op amp is a voltage from a dependent
voltage source that attempts to force the voltage at the inverting
input terminal to equal the voltage at the non-inverting input
terminal.
 Almost ideal op amp: Output voltage limited to the range between V+
and V-.
 Ideal op amp is assumed to have Ri = ∞ W and Ro = 0 W.
 Almost ideal op amp: vd = 0 V and the current flowing into the
output terminal of the op amp is as much as required to force v1 = v2
when V+< vo< V-.
 Operation of an op amp was used in the analysis of voltage
comparator and inverting amplifier circuits.
 Effect of Ri < ∞ W and Ro > 0 Wwas shown.
THANK YOU