slide008 op-amp intro
TRANSCRIPT
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EE 4623: Analog Integrated Electronics
Operational Amplifier Applications
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Objectives
Students should be able to: Explain basic amplifier terms including:
Gain, Linear, Distortion, Decibel, Saturation, Biasing, Models,Cascade, Coupling Capacitor, Transconductance, Open- andClosed-Circuit Models.
Calculate gains for amplifier configurations. Understand tradeoffs inherent in amplifier designs.
Identify the four amplifier types.
Understand how frequency response relates to
amplifiers.Analyze op-amp simple circuits and effect of a finite
open-loop gain .
Understand the effects of offset voltage and offsetcurrent in an Op-Amp.
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Amplifiers
o iv A v
Amplifiers are 2-portnetworks:
input port
output port
A is called the amplifiergain.
If the gain is constant, we call this a l inearamplifier.
iv ovA
iv ov
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Amplifier Gain in Decibels
Amplifier gain is expressed in decibels (dB) Originally it was expressed as Bels (named afterAlexander Graham Bell), but these proved to be ofinsufficient size so we multiply Bels by 10 decibels.
Decibels are a log-based ratio and are thereforedimensionless.
Purpose: We want to measure the ratio of some valuerelative to another (e.g. sound power in a stereoamplifier).
Derivation of dB
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Derivation of Decibels
Ratio of power of interest (call it p1) to some other
reference power (say,p2):
However, these values are generally quite huge and
tend to be logarithmically related; thus, creation of
the Bel:
1
2
p
p
1
2
logp
Belp
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Derivation of Decibels (Contd.)
However, as mentioned, the Bel is a bit too small,
so lets multiply it by 10 and call it a decibel (10 x
Bel = 1 dB).
Which gives us the decibel expression for power:
1
210logpower
p
decibel p
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In-Class Exercise: dB for Voltage
Given the following equation for expressing
power in decibels, write a similar equation
expressing voltage in decibels:
1
210logpower
p
decibel p
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dB for Voltage
First, lets relate voltage to power:
2
/p vii v r
p v r
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In-Class Exercise: dB for Voltage
Upon substitution:
Which gives us the decibel expression for
vol tage:
22 2
1 1 1
2 2
2 2 2
10log 10log 10logv r v v
v r v v
1
2
20logvoltagevdecibelv
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Lets Draw some Physical
Conclusions
if dB is positive, then v1 > v2, the signal is
amplified.
if dB is negative, then v1 < v2, the signal is
attenuated.
if dB is 0, then v1 = v2.
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Signal Amplification
A is the amplifiergain.
The premise is that this is a l inearamplifier.
Amplifiers that are not linear exhibit nonlinear
distortion.
)()( tAvtv io
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Gain (in Decibels)
Power Gain,Ap
Voltage Gain,Av
Current Gain,Ai
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The Meaning of Active in EEWhat makes an amplifier an act ivecircuit component?
Active components are devices that add intelligence in someway to a signal that passes through it (in contrast to passivecomponents which consume but do not produce energy).Passive components are incapable of power gain.
Active components can also switch the flow of current and arecapable ofgain. Examples: Transistors, diodes, and vacuumtubes.
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Amplifier Saturation
What is the range ofvI?
L+andLare the positive, and
negative saturation levels,
respectively.
In order to avoid output saturation,
the input must be kept within the
linear range of operation:
I
v v
L LvA A
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Nonlinearity and Biasing We want to operate in the linear
region. What have we added tothe amplifier (below) to achievethe transfer characteristic(right)?
By adding a DC offset to the
input, we bias the input voltageto operate in this region.
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Biasing
The bias point is also referredto as the
quiescent point (Q-point).
DC bias point.
operating point.
The time-varying input signal
vi (t)is superimposed on the
DC bias voltage VI to createthe total instantaneous
input,vI(t)
( ) ( )I I iv t V v t
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Coupling Capacitors
You will be using coupling capacitors rather heavily so itis good to get an idea about what they are accomplishing!
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Symbol Convention
Explain the nomenclature on the following graph:
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Closed- and Open-Circuit
Amplifiers
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Open-Circuit Amplifier Model
What is the relationship ofvo to vi in the open-circuit voltageamplifier model? Here, R
iaccounts for the fact that the amp draws an
input current (ii) from ______. R
oaccounts for changes in output
voltage as the amp is called on to supply output current (io) to _____.
Avois the open-circuit (voltage) gain factor (i.e., it is the gain of the
unloaded amplifier)
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Closed-Circuit Amplifier Model
What is the relationship ofvo
to viin the following closed-
circuit
model? What is Rs? Finally, what is the relationship ofv
oto
vs?
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Procedure/Solution
oL
L
si
ivo
s
o
si
isi
oL
Lvo
i
ov
Lo
oL
Livoo
RR
R
RR
RA
v
v
RR
Rvv
RR
RA
v
vA
RRRR
RvAv
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Cascaded Amplifiers
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Cascaded Amplifiers
Calculate the voltage gains of each stage assuming thenext stage is the load of the previous.
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vi1/vs = 0.909 (i.e., a lossof gain at the input)
Av1 vi2/vi1= 9.9
Av2 vi3/vi2= 90.9
Av3 vL/vi3 = 0.909 (Voltage gain of output stage)
Av vL/vi1 =Av1 Av2 Av3 = 818 = ______dB (Total gain)
Gain (Source-to-Load): vL/vs = (vL/vi1)(vi1/vs) = 818 0.909 =_____
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about
Cascaded Amplifiers
To avoid losing signal strength at the input (where the signal isusually very small), the first stage has a relatively high inputresistance. The tradeoff is a moderate voltage gain.
The second stage doesnt need to have as high an inputresistance, and performs the bulk of the voltage gain.
The third stage provides no voltage gain, but functions as a
bu ffer amp li f ier, with a large input resistance and very smalloutput resistance (even lower than RL). This allows connectionto a small 100 load.
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Questions
What would the overall voltage gain be without stage 3?Without stage 2? Without stage 1?
IfRL varies from 10 to 1,000 , what is the range of thecorresponding voltage gain?
Answer: Range is 409 Av 810.
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Other Amplifier Types
Input signals of interest can be voltages or currents.
Some transducers have very high output resistances and can bebetter modeled as current sources.
Output signals can be either as well.
The most popular is the voltage amplifier, but it is just onetype of model. Other are:
Current amplifiers Transconductance amplifiers Transresistance amplifier
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These models are unilateral(i.e., signal flow isuniderectional from input to output). Most real amplifiersshow some reverse transmission (generally undesirable ).
Our transistor models are generally of thetransconductance type.
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Operational Amplifiers: Ideal
The Inverting Configuration
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Circuit Symbol
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741 Op Amp (Pin-Out Diagrams)
Ch t i ti /P ti
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Characteristics/Properties
(Ideal Op-Amp)
Property No.1: Infinite input impedance (Zin)
Input impedance is the ratio of input voltage to input current (Zin = Vin/Iin)
As Iin0Zin.
High-grade op-amps can have input impedance in the T range.
Some low-grade op-amps (on the other hand) can have mA input currents.
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Ideal Properties (Contd)
Property No.2: Zero output impedance (Zout0)
The ideal op-amp acts as a perfect internal voltage source with no internalresistance.
Real-life op-amps have some internal resistance which then is in series with
an external load, thus reducing the output voltage available to the load. Example:
Real-life op-amps have output-impedance in the 100-20 range.
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Ideal Properties (Contd)
Property No.3: Infinite Open-Loop Gain (A vo l)
Open-loop gain is the gain of the op-amp without positive or
negative feedback.
Ideally infinite (typical values range from 20,000 to 200,000 inreal-life devices; normally feedback is applied around the op-ampso that the gain of the overall circuit is defined and kept to a figure
which is more usable).
Open-loop gain falls very rapidly with increasing frequency.
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Ideal Properties (Contd)
Property No.4: Zero Common-Mode Gain (what does this mean?!?)
An ideal op-amp only amplifies the voltage di f ferencebetween its two
inputs.
If these inputs were to be shorted together (thus ensuring zero potential
difference between them), there should be no change in output voltage
for any amount of voltage applied between the two shorted inputs and
ground.
Example:
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Common-Mode Gain (Meaning)
Voltage that is common between either of the inputs and ground (as Vcommon-
mode is in this case) is called common-mode voltage.As we vary this
common voltage, the perfect op-amp's output voltage should hold absolutely
steady (i.e., no observable change in output for any arbitrary change in
common-mode input). This translates to a common-mode voltage gain of
zero.
The performance of a real op-amp in this regard is most commonly measured
in terms of its differential voltage gain (i.e., how much it amplifies the
difference between two input voltages) versus its common-mode voltage gain
(i.e., how much it amplifies a common-mode voltage). The ratio of the former
to the latter is called the common-mode rejection ratio (CMRR):
)(
)(
ModeCommonv
alDifferentiv
A
ACMRR
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Common-Mode Rejection Ratio
An ideal op-amp, withAv(Common-Mode) = 0 would have an infinite CMRR.
Real-life op-amps have high CMRRs (the ubiquitous 741 having somethingaround 70 dB, which works out to a little over 3,000 in terms of a ratio).
A note about feedback compensation
Because the CMRR in a typical op-amp is so high, common-mode gain is usuallynot a great concern in circuits where the op-amp is being used with negativefeedback.
If the common-mode input voltage of an op-amp circuit were to suddenly change(thus producing a corresponding change in the output due to common-modegain), that change in output would be quickly corrected as negative feedback andAvolworked to bring the system back to equilibrium.
Conclusion: A change might be seen at the output, but it would be a lot smaller
than what you might expect!
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Characteristics (Contd)
Property No.5: Infinite Bandwidth (What does this mean?!?)
The ideal op-amp will amplify all signals from DC to the highest
AC frequencies.
In real op-amps, the bandwidth is rather limited; this limitation is
specified by the Gain-Bandwidth product (GB), which = the
frequency where the amplifier gain unity.
Some op-amps (such as the 741 family) have very limited
bandwidth of up to a few KHz.
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Characteristics (Contd)
Property No.6: Zero Output Offset (What does this mean?!?)
The output offset is the output voltage of an amplifier when bothinputs are grounded, i.e.;
Ideal op-amp: Zero output offset.
Real op-amps: Finite amount of output offset voltage
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The Inverting Configuration
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Closed-Loop Gain
Determine the closed-loop voltage gain vo/vI
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(Solution)
1
2
21211
111
11
12
0
0
0
R
R
v
v
RiRivv
R
v
R
v
R
vvi
A
v
vve
evA
I
o
o
III
vol
o
in
in
ovol
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Synthesis of the Feedback Resistor
Design an inverting op-amp where the closed-loop gain is =10 and the total resistance is
150K.
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(Solution)
2
1
2 1
2 1
1 1
1
2
10
10
150
10 150
13.63
136.3
o
I
v R
v R
R R
R R K
R R K
R K
R K
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Synthesis of the Feedback Resistor
Design an inverting op-amp where the closed-loop gain is =10 and the total resistance is
150K.
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(Solution)
k3.136
;k63.13
k15010
k150:But
10
10
2
1
11
12
12
1
2
R
R
RR
RR
RR
R
R
v
v
I
o
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Closed-Loop Gain II
Determine the closed-loop gain vo/vi
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(Solution)
1 2
1
2 2
2 2 23
3 3 1 3
24 2 3
1 1 3
4 4
22 2 4 4 2 4
1 1 1 3
2 4 2 4 2 4 2
1 1 1 3 1 1 3
2 4 2
1 1
0
1
1
I
x
xI
II
o x
I Io I
o I I
o
i
vi iR
v i R
v i R Ri v
R R R R
v Ri i i v
R R R
v v i R
v v Rv i R i R R v R
R R R R
R R R R R R Rv v v
R R R R R R R
v R R R
v R R
3
R
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The Very Real Effects of Finite
Open-Loop Gain
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Recall the Ideal Op-Amp Model:
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In Reality, v1 0
The Effect of Finite Open Loop
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The Effect of Finite Open-Loop
Gain
012 vol
oin
A
vvve
111
11
R
Av ol
vv
R
Av ol
vv
R
vvi
oI
oI
I
21
:
RiA
vv
KVL
vol
oo
21
RR
A
vv
A
v vol
oI
vol
o
vol
volI
o
AR
R
ARR
RR
v
v
1
2
12
12
1
make...thereforeshouldThus,/)/1(1
/
???Why
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What is a Weighted Summer?
Simply put, it is an op-amp configured to produce the weighted sum
of two or more voltage inputs
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Analysis of a Weighted Summer
Use superposition to analyze the following weighted summer (i.e.,
write an expression forvo in terms of the input voltages and input andfeedback resistors).
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Solution
n
n
fff
o vR
Rv
R
Rv
R
Rv ...2
2
1
1
n
nn
R
vi
R
vi
R
vi ...,,
2
2
2
1
1
1
nf iiiii ...21
f
o
R
vi
0:Where
fo iRv
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The Non-Inverting Configuration
How to derive the closed-loop gain (vo/vI) for this
configuration:
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Closed-Loop Gain Derivation
111
11
0
:lawsOhm'
R
v
R
v
R
vvi III
2
1
21 0
:KVL
RRvvv
Rivv
IIo
Io
vol
I
ovcl
A
RR
vvA
:Observe
1
:Thus
1
2
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Design Exercise
Design a non-inverting op-amp where the gain
is +10 and the total resistance is 150k. Sketch
the final circuit.
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Solution
1 2
1
2
1
2
150k
1 10
15k
135k
R R
RR
R
R
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The Difference Amplifier
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What is a Difference Amplifier?
A Difference amplifier combines features of the inverting amplifier
and the non-inverting amplifier. It is the complement of the summing
amplifier and allows the subtraction of two voltages or, as a special
case, the cancellation of a signal common to the two inputs.
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Use superposition
set v1= 0, solve forvo (i.e.,
this is a non-invertingamp)
set v2= 0, solve forvo (i.e.,
this is an inverting amp)
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Result of setting v1= 0 and then solving forvo (non-
inverting configuration):
2
43
4
1
22 1 v
RR
R
R
Rvo
43
42
RR
Rv
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Result of setting v2= 0 and then solving forvo (inverting
configuration):
1
1
21 v
R
Rvo
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Add the two results
vo = vo1 +vo2
1
21
43
4
1
22 1
R
Rv
RR
R
R
Rvvo
Hint on the Shunt Resistor
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Hint on the Shunt Resistor
Observation
WithoutR2, what
happens if there is a
tiny dc voltage?
WithoutR2, what
happens if there is atiny dc input offset
current?
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Examples and Problems
Examples: 1.2, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 1.10, 1,11, 1.12, 1,13,
Problems: 1.8, 1.11, 1.16, 1.18, 1.19, 1.25, 1.26 1.31, 1.43, 1.44
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Assignment No: 04
Problems: 1.16, 1.18, 1.19, 1.25, 1.26 1.31, 1.43, 1.44
Last date of submission: 07-05-2013