Ajal op amp

Op-Amp
ByBy
AJAL.A.J ( ASSISTANT PROFESSOR )AJAL.A.J ( ASSISTANT PROFESSOR )
ECE DEPARTMENTECE DEPARTMENT
MAIL:MAIL: ec2reach@gmail.comec2reach@gmail.com
MOB: 8907305642MOB: 8907305642

WHAT ?
• An Operational Amplifier (Op-Amp) is an integrated
circuit that uses external voltage to amplify the input
through a very high gain.
Key WordsKey Words:
Op Amp Model
Ideal Op Amp
Op Amp transfer characteristic
Feedback
Virtual short

History of the Op-Amp
• 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

1950’s-1960’s
The end of Vacuum Tubes was built up
during the 1950’s-1960’s to the advent of
solid-state electronics
1.The Transistor
2.The Integrated Circuit
3.The Planar Process

1960s:
• 1960s: beginning of the Solid State Op-Amp
• Example: GAP/R P45 (1961 – 1971)
– Runs on ± 15 V, but costs $118 for 1 – 4
• The GAP/R PP65 (1962) makes the Op-Amp into a
circuit component as a potted module

@ 1963
• The solid-state decade saw a proliferation of Op-Amps
– Model 121, High Speed FET family, etc.
• Robert J. Widlar develops the μA702 Monolithic IC Op-
Amp (1963) and shortly after the μA709
• Fairchild Semiconductor vs. National Semiconductor
– National: The LM101 (1967) and then the LM101A (1968) (both by
Widlar)
– Fairchild: The “famous” μA741 (by Dave Fullager 1968)
and then the μA748 (1969)

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

Ideal Op-Amp Characteristics
• Open-loop gain G is infinite
• Rin is infinite
• Zero input current
•Rout is zero

Ideal Op-Amp Analysis
To analyze an op-amp feedback circuit:
• Assume no current flows into either input terminal
• Assume no current flows out of the output terminal
• Constrain: V+ = V-

11
Ideal Vs Practical Op-Amp
Ideal Practical
Open Loop gain A ∝ 105
Bandwidth BW ∝ 10-100Hz
Input Impedance Zin ∝ >1MΩ
Output Impedance Zout 0 Ω 10-100 Ω
Output Voltage Vout
Depends only
on Vd =
(V+−V−)
Differential
mode signal
Depends slightly
on average input
Vc = (V++V−)/2
Common-Mode
signal
CMRR ∝ 10-100dB
+
−
~
AVin
Vin Vout
Zout=0
Ideal op-amp
+
− AVin
Vin Vout
Zout
~
Zin
Practical op-amp

@ Practical opamp
3 categories are considering
• 3 categories are considering
 Close-Loop Voltage Gain
 Input impedance
 Output impedance

Ref:080114HKN Operational Amplifier 13
Ideal Op-Amp Applications
Analysis Method :
Two ideal Op-Amp Properties:
(1) The voltage between V+ and V− is zero V+ = V−
(2) The current into both V+ and V− termainals is zero
For ideal Op-Amp circuit:
(1) Write the kirchhoff node equation at the noninverting
terminal V+
(2) Write the kirchhoff node eqaution at the inverting
terminal V−
(3) Set V+ = V− and solve for the desired closed-loop gain

2 TYPES
1. Inverting Amplifier
2. Non-Inverting Amplifier

1. Inverting Amplifier Analysis
virtual ground

2. Non-Inverting Amplifier Analysis

17
Op-Amp Properties
(1) Infinite Open Loop gain
- The gain without feedback
- Equal to differential gain
- Zero common-mode gain
- Pratically, Gd = 20,000 to 200,000
(2) Infinite Input impedance
- Input current ii ~0A
- T-Ω in high-grade op-amp
- m-A input current in low-grade op-
amp
(3) Zero Output Impedance
- act as perfect internal voltage source
- No internal resistance
- Output impedance in series with load
- Reducing output voltage to the load
- Practically, Rout ~ 20-100 Ω
+
−
V1
V2
Vo
+
−
Vo
i1~0
i2~0
+
−
Rout
Vo'
Rload
outload
load
oload
RR
R
VV
+
′=

Op-Amp Buffer
Vout = Vin
Isolates loading effects
A
High output
impedance
B
Low input
impedance

20
Op-Amp Differentiator
RC
dt
dV
v i
o 





−=
+
−
R
C
VoVi
0
to t1 t2
0
to t1 t2

Op-Amp Differential Amplifier
If R1 = R2 and Rf = Rg:

Closed-loop gain
2
21
21
2
1
R
RR
v
RR
AR
Av
v
IN
IN
o
+
≈
+
+
=
Closed-loop gain
Af=vo/vin
Suppose A=106
, R1=9R, R2=R,
10INo vv ≈
Gain
INf vA=
)1(
2
1
R
R
Af +=
Closed-loop gain: determined by resistor ratio
insensitive to A, temperature

Negative feedback
�
)1(
2
1
R
R
v
v
A
i
o
f +==
1
2
R
R
v
v
A
i
o
f −==
• We can adjust the closed-loop gain by changing the ratio of R2 and R1.
• The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is
large enough) and we can make it arbitrarily large or small and with the desired
accuracy depending on the accuracy of the resistors.

�
The terminal 1 is a virtual ground since
terminal 2 is grounded.
Inverting configuration,
This is a classic example of what negative feedback does. It takes an amplifier with
very large gain and through negative feedback, obtain a gain that is smaller, stable,
and predictable. In effect, we have traded gain for accuracy. This kind of trade off is
common in electronic circuit design… as we will see more later.
Negative feedback

�
Inverting configuration,
Input Resistance:
Assuming an ideal op amp (open-loop gain A = infinity), in the closed-loop inverting
configuration, the input resistance is R1.
1
11 /
R
Rv
v
i
v
R
i
ii
in ===
Negative feedback

1
2
R
R
v
v
A
i
o
f −==
Inverting configuration,01
1
=== o
i
in RR
i
v
R
We can model the closed-loop
inverting amplifier (with A =
infinite) with the following
equivalent circuit using a
voltage-controlled voltage
source…
Negative feedback �

29
Frequency-Gain Relation
• Ideally, signals are amplified
from DC to the highest AC
frequency
• Practically, bandwidth is limited
• 741 family op-amp have an limit
bandwidth of few KHz.
• Unity Gain frequency f1: the gain at
unity
• Cutoff frequency fc: the gain drop by
3dB from dc gain Gd
(Voltage Gain)
(frequency)
f1
Gd
0.707Gd
fc0
1
GB Product : f1 = Gd fc
20log(0.707)=3dB

30
GB Product
Example: Determine the cutoff frequency of an op-amp having a unit gain
frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV
Sol:
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 / Gd = 10 MHz / 20 V/mV
= 10 × 106
/ 20 × 103
= 500 Hz
(Voltage Gain)
(frequency)
f1
Gd
0.707Gd
fc0
1
10MHz
? Hz

Op Amp transfer characteristic curve
saturation
)( −+ −== vvAvAv io
active region

So far, we have been looking at
the amplification that can be
achieved for relatively small
(amplitude) signals.
For a fixed gain, as we increase
the input signal amplitude, there
is a limit to how large the output
signal can be. The output
saturates as it approaches the
positive and
negative power supply voltages.
In other words, there is limited
range across which the gain is
linear. −+ > vv += Vvo
−+ < vv −= Vvo
Op Amp transfer characteristic curve

Applications of Op-Amps
Filters
Types:
•Low pass filter
•High pass filter
•Band pass filter
•Cascading (2 or more filters connected
together) R2
+
-
+
V0
__
+ Vcc
- Vcc
-
+
R1
C
Low pass filter
Low pass filter Cutoff
frequency 
Low pass filter transfer
function

• Electrocardiogram EKG Amplification
– Need to measure difference in voltage from lead 1 and lead 2
– 60 Hz interference from electrical equipment

• 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”

Strain Gauge
Use a Wheatstone bridge to
determine the strain of an
element by measuring the
change in resistance of a
strain gauge
(No strain) Balanced Bridge
R #1 = R #2
(Strain) Unbalanced Bridge
R #1 ≠ R #2

Strain Gauge
Half-Bridge Arrangement
Using KCL at the inverting and non-inverting
terminals of the op amp we find that  ε ~ Vo = 2ΔR(Rf /R2
)
R + ΔR Rf
+
- +
V0
__
+ Vcc
- Vcc
-
+
Rf
Vref
R
R - ΔR
R
Op amp used to amplify
output from strain gauge

• Piezoelectric Transducer
– Used to measure force, pressure, acceleration
– Piezoelectric crystal generates an electric charge in
response to deformation
• Use Charge Amplifier
– Just an integrator op-amp circuit

•Goal is to have VSET = VOUT
•Remember that VERROR = VSET – VSENSOR
•Output Process uses VERROR from the PID controller to adjust
Vout such that it is ~VSET
P
I
D
Output
Process
Sensor
VERRORVSET
VOUT
VSENSOR
PID Controller – System Block
Diagram

Applications
PID Controller – System Circuit Diagram
Source:
http://www.ecircuitcenter.com/Circuits/op_pid/op_pid.htm
Calculates VERROR = -(VSET + VSENSOR)
Signal conditioning allows you to
introduce a time delay which could
account for things like inertia
System to control
-VSENSOR

Applications
PID Controller – PID Controller Circuit Diagram
VERR
Adjust Change
Kp RP1, RP2
Ki RI, CI
Kd RD, CD
VERR PID

• Example of PI Control:
Temperature Control
• Thermal System we
wish to automatically
control the temperature
of:
• Block Diagram of
Control System:

• Voltage
Error
Circuit:
• Proportiona
l-Integral
Control
Circuit:
• Example of PI Control: Temperature Control

44
Multiple Inputs
(1) Kirchhoff node equation at V+
yields,
(2) Kirchhoff node equation at V−
yields,
(3) Setting V+
= V–
yields
0=+V
0
_
=
−
+
−
+
−
+
− −−−
c
c
b
b
a
a
f
o
R
VV
R
VV
R
VV
R
VV
∑=
−=





++−=
c
aj j
j
f
c
c
b
b
a
a
fo
R
V
R
R
V
R
V
R
V
RV
+
−
Rf
Va
Vo
Rb
Ra
Rc
Vb
Vc

Homework
1) Design a circuit to 5.0==
i
o
f
V
V
A
2) Find the vo=?

Review: Two fundamental Op Amp
Structure Af Input voltage
( )terminal
Feed back
( )terminal
Inverting
Amp
_ _
Non
inverting
Amp
+ _

Ajal op amp

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