chapter3 op amp(1)80911
TRANSCRIPT
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OPERATIONAL AMPLIFIER(OP AMP )
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1. Introduction2. Understand the general Op-amp circuit
design Components inside an op.amp.
Characteristics of an op.amp. Circuit symbol and pin configuration
Top view Pin-out configuration Block diagram an op-amp differential amplifier more stages of gain amplifier push-pull amplifier
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3. Understand the differential amplifier4. Understand more stages of gain amplifier5. Understand the push pull amplifier6. Understand the ideal operational amplifier
7. Understand the op-amp configurations Inverting amplifier Non inverting amplifier Summing amplifier Subtractor Differentiator amplifier Integrator amplifier Comparator
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OP - AMP is a solid state device capable of
sensing and amplifying dc and ac input signals.
OP AMP is an amplifier with two inputs
(Differential inputs) and a single output.
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OP AMP consists of 20 transistors 11 resistors
and 1 capacitor.
OP - AMP requires a positive and negative powersupply( Dual power supply ).
This allows the output voltage to swing positiveand negative with respect to ground.
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Op Amp Internal Circuit Diagram
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1.VERY HIGH INPUT RESISTANCE OR EVENINFINITY WHICH PRODUCES NEGLIGIBLE
CURRENT AT THE INPUT.
2.VERY HIGH CURRENT GAIN.
3.VERY LOW OUTPUT IMPEDANCE OR EVENZERO,SO AS NOT TO AFFECT THE OUTPUTOF THE AMPLIFIER BY LOADING.
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Block Diagram (Op Amp)
Internally, the typical Op-Amp has adifferential input, a voltage amplifier and apush pull output.
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An OP
AMP is so named, because it wasoriginally designed to perform mathematical
operations such as addition, subtraction,
multiplication, division, integration,
differentiation etc in analog computer.
Nowadays OP - AMPs are used in analog
computer operations and in timing circuits.
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The Op
Amp is represented by a triangularsymbol .
It has two input and one output terminals.
CIRCUIT SYMBOL AND PIN-OUT CONFIGURATION
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-
+
Inverting Input
Noninverting Input
Output
Positive dc
power supply
Negative dc
power supply
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The terminal with negative sign (-) is called
inverting input.
The terminal with positive sign (+) is callednon inverting input.
The input terminals are at the base and theoutput is at the apex of the triangular
symbol.
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The widely used and most popular type is
Op Amp IC 741.
The top pin on the left side of the notch ispin 1.
The pin number 2 is inverting input and pinnumber 3 is non-inverting input terminal.
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The pin number 6 is the output terminal.
A dc voltage or ac signal connected to pin2 will be 180o out of phase at the output.
A dc voltage or ac signal connected to pin
3 will be in phase at the output.
Pin 4(-) and 7(+) are the power supply
terminals.
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The pin 1 and 5 are Null adjustment pins.
Null adjustment pins are used to null the
output voltage , when equal voltages areinput terminals for perfect balance.
Pin number 8 indicates No connection.
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TOP VIEW OF 8 PIN DIP
Notch
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OP AMP Pin- out configuration
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A single package will often contain severalop-amps
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There are 2 types of application in op-amp
Linear application
Non-linear application
Linear application is where the op-amp operate inlinear region:
Assumptions in linear application:
Input current, Ii = 0
Input voltage: V+=V-
Feedback at the inverting input
Application in op-amp
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Non-linear application is where the op-amp operate innon-linear region
By comparing these two input voltages: positive input
voltages, V
+
and negative input voltage, V
-
where:VO = VCC if V
+ > V-
VO = -VEE if V+ < V-
Input current, Ii = 0
Application in op-amp
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Comparator (Pembanding)
Inverter (Penukar)
Audio amplifier (Penguat Audio) Difference Amplifier (Penguat Beza)
Filter (Penapis)
Summing Amplifier (Penguat Jumlahan)
Application in op-amp
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Inverting Amplifier
Non-Inverting Amplifier
Summing Amplifier
Unity Follower
Difference Amplifier
Integrators
Differentiators
Op-amp Circuit Application
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Vo = A(V+ - V)
Vo/A = V+ - V
Let A --- infinitythen,
V+ - V--- 0
Summary of op-amp behavior
V+ = V
I+ = I = 0
Seems strange, butthe input terminalsto an op-amp act asa short and open atthe same time
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Write node equations at + and - terminals
(Ii=I+ = I-= 0)
Set V+ = V-
Solve for Vo
To analyze an op-amp circuit for linear
operation
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IDEAL OPERATIONAL AMPLIFIERS
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Differential Amplifier circuits In differential amplifier, there are two inputs Vi1 and Vi2 andthere are three outputs (1) at Vo1, (2) at Vo2, and (3) across
between Vo1 and Vo2 Dual supply +VCC and VEE are used so that Vi1 and Vi2 can beconnected to the BJT directly without coupling capacitor
IC1 + IC2 = 2Ic flows
through REthen VRE = 2Icx REVi1
RC
-VEE
+VCC
Vo1
Vo2
RC
RE
Vi2
Q1 Q2AC
AC
VE = -0.7V
VB = 0V dc
+
-
Ce
E
EEC
EEEEREEC
I
mV26r
R2
7.0VI
7.0VV7.0VRI2
CCCC2C1C RIVVV
DC analysis
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Example
In differential amplifier shown,find (1) VB1= VB2 (2) VE1= VE2 (3) IC1 = IC2 and (4) VC1= VC2
VB = 0V dc
VE = -0.7V
Vi1
-9V
+9V
Vo1
Vo2
RC
3.3kVi2
Q1
Q2
AC
AC
3.9k
+-
VC
IC
mA26.1k3.32
7.09
R2
7.0VI
7.0VV7.0VRI2
E
EEC
EEEEEEC
V1.4k9.3mA26.19RIVV CCCCC
V7.0V,V0V EB
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Due to biasing requirements or leakage, asmall amount of current (typically ~10nanoamperes for bipolar op-amps, tens ofpicoamperes for JFET input stages, and only a
few pA for MOSFET input stages) flows into theinputs. When large resistors or sources withhigh output impedances are used in the circuit,these small currents can produce large
unmodeled voltage drops. If the input currentsare matched, and the impedancelooking out of both inputs are matched, thenthe voltages produced at each input will bee ual.
Input Bias Current
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-The output voltage of an op-amp when thedifferential input is zero should be also zero.
- However, due to unavoidable internalimbalances and due to non-zero bias currents, asmall voltage, V
IO, is seen between the
terminals.- ICs provide a means to compensate for this.- This is generally done by connecting anexternal potentiometer to pins designated withOffset Null.- With zero input voltage, the output is set to zeroby adjusting the potentiometer- The pinout for the 741 op-amp (the most
common op-amp IC) is shown next.
Input Offset Voltages
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Input Offset Current
- Input offset current refers to thedifference between the bias currents ofthe amplifier. Again, ideally the two
currents should be equal to obtain azero output voltage. However, there hasto be a difference between the two bias
currents to set the output to zero. Thisdifference is referred to as input offsetcurrent.
- The difference in IBias between the
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Common-mode Gain
Common mode gain is the undesiredgain when the same signal is appliedto both inputs. Ideally it should be
zero.
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Common-mode Rejection Ratio
The ability of a differential amplifier to notpass (reject) the portion of the signalcommon to both the + and - inputs.
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Multistages Amplifier
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Push Pull Amplifier
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Operation (Push Pull)
Most audio power amplifiers use a Class Bconfiguration which employs two commoncollector (emitter-follower) stages where onetransistor provides power to the load during one-
half of the waveform cycle (it pushes) and asecond transistor provides power to the load forthe other half of the cycle (it pulls).
Neither transistor remains on for the entire cycle,giving each transistor time to rest and coolduring the waveform cycle. This makes for amore power-efficient amplifier circuit, but leads
to a distinct type of nonlinearity known as
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Operation (Push Pull)
Distortion occurs because there is a delaybetween the time one transistor turns off and theother transistor turns on.
There will be no output signal until Vin 0.6V .
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Infinite gain for the differential input signal
Zero gain for the common-mode inputsignal
Infinite input impedance
Zero output impedance
Infinite bandwidth
Characteristics of Ideal Op Amps
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(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- 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
+
V1
V2 Vo
+
Vo
i1~0
i2~0
+
Rout
Vo'Rload
outload
load
oloadRR
RVV
Ideal characteristics of Op-Amp
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Input offset voltage
0
when the input terminals are shorted so that , theoutput is a Virtual Ground or vout = 0
Offset current 0
there is assumed to be no leakage or bias currentinto the device
Bandwidth -
the frequency magnitude response is considered tobe flat everywhere with zero phase shift)
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Ideal Practical
Open Loop gainA 105
BandwidthBW 10-100Hz
Input ImpedanceZin >1M
Output ImpedanceZout 0 10-100
Output Voltage Vout Depends onlyon Vd = (V+V)
Differential
mode signal
Depends slightly
on average input
Vc = (V++V)/2Common-Mode
signal
CMRR 10-100dB
+
~
AVin
Vin Vout
Zout=0
Ideal op-amp
+
AVinVin VoutZ
out
~
Zin
Practical op-amp
Ideal Vs Practical Op-Amp
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We are using the two
ideal op ampproperties discussedabove to analyze thiscircuit.
Since the amplifierhas infinite gain, it willdevelop its outputvoltage, Vout, with zeroinput voltage.
INVERTING AMPLIFIER
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INVERTING AMPLIFIER
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Since the differentialinput is zero, the full
input voltage mustappear across Ri, makingthe current in Ri,
Iin = Vin / Ri .
since there is no current
flow into either inputterminal because theinput impedance isinfinite, the current Iin
must also flow in Rf.
INVERTING AMPLIFIER
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INVERTING AMPLIFIER
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INVERTING AMPLIFIER
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V 0 Ii 0
I1 If Ii
Vs V
R1V Vo
Rf
V V 0
Vo
Vs
Rf
R1
Vo Rf
R1
Vs
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THE NON INVERTING AMPLIFIER
The input signal is appliedto the non inverting input
terminal.
A resistor Ra, is connected
from the inverting inputto the ground.
The feedback resistor Rfis
connected between the
output and the invertinginput.
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+Vin
Vo
RaRf
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THE NON INVERTING AMPLIFIER
Since the potential at theinverting input and that at the
non inverting input are same .
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+Vin
Vo
RaRf
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THE NON INVERTING AMPLIFIER
(R1 + R2 / R1) = VOUT /VA
1 + ( R2 / R1) = VOUT /VIN
A v =1 + ( R2 / R1).
The output and input
voltages are in phase.
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Non Inverting Amplifier
(1) Kirchhoff node equation at V+yields,
(2) Kirchhoff node equation at V
yields,
(3) Setting V+ = V yields
or
+VinVo
RaRf
iVV
00
f
o
a RVV
RV
0
f
oi
a
i
R
VV
R
V
a
f
i
o
R
R
V
V1
The output and input voltages are in phase.
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The Summing Amplifier
The summingamplifier provides
an output equal to
the sum of theinput voltages.
Here we have an
inverting amplifier,
used to sum two
input voltages.
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The Summing Amplifier
The input voltagesV1 and V2 are
connected to the
inverting inputthrough the
resistors R1, R2 and
RF
is the feedback
resistance.
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The Summing Amplifier
S i A lifi
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Summing Amplifier
This circuit is called
a weighted summer
3
3
2
2
1
1
3
3
2
2
1
1
3
3
2
2
1
1
321
;0
:
;0
:
0
R
V
R
V
R
V
RV
R
V
R
V
R
V
R
VVinsert
R
VV
R
VV
R
VV
R
VVso
Iwhile
IIIII
KCLuse
VV
fo
f
o
f
o
i
RfiRRR
V1
V2
V3
R1
R2
R3
Rf
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Subtrator Amplifier
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43
21
RR
RR
)(21
2
4VV
R
RVO
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dttvRC
tv
dt
tdvC
R
tv
III
io
i
Ci
)(1
)(
)(0
)( 0
sC
1
Cj
1X
:impedancecetanCapaci
C
Integrator Amplifier
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dt
tdvRCtv
R
tvV
dt
tdv
C
II
i
o
oi
RC
)()(
)()(
Differentiator Amplifier
C t
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Comparator
Vo(V)
10
-5
t
VS(V)
t
Compare V+ and V-
V+=0V-=VS
When:VS>0,V
+>V- so Vo=10VVS
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1VV
O
Unity Follower(Pengikut Voltan)
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Instrumentation Amplifier
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Non-linear application is where the op-ampoperate in non-linear region
By comparing these two input voltages: positive
input voltages, V+
and negative input voltage, V-
where:
VO = VCC if V+ > V-
VO = -VCC if V+ < V-
Input current, Ii = 0
Non-linear application in op-amp
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