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Operational Amplifier

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Output Impedance of an Inverting Amplifiers with Finite AOL, rin, and r OL ... This amplifier has very high input impedance = Zin ... – PowerPoint PPT presentation

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Title: Operational Amplifier


1
Operational Amplifier
VCC
-VEE
VCC and VCC are the dc power supply
connections. Usually VCC -VEE .
Normally they are not shown in schematics.
2
Common Mode and Differential Voltages
vicm Common Mode Voltage 0.5(v1 v2) Thus
Common Mode Voltage is the average of voltages at
the non-inverting and inverting inputs vid
Differential Voltage (v1- v2) Thus Differential
Voltage is the difference of voltages at the
non-inverting and inverting inputs
3
Zout
Zin
4
Equivalent Circuit of an Op-Amp
  • Ideally
  • Zin ?
  • Zo 0
  • AOL ?
  • No parasitic capacitance or inductance
  • (bandwidth ?)
  • Zero gain for common mode voltages.

5
Using Op-Amps
  • For Stability Op-Amps use negative feedback. This
    can be
  • achieved by feeding back the output to the
    inverting input. Some
  • Examples Inverting, Non-inverting Amplifiers,
    Integrators.
  • This does not mean positive feedback is not
    useful. Comparators,
  • Schmitt Triggers are some examples where positive
    feedback is
  • used.

6
Inverting Amplifiers
7
Analysis of Inverting Amplifiers
Since Zin ?, we can safely
assume that current entering the inverting
terminal is 0. This means the inverting input is
at zero potential but no current can flow through
it. Thus we say that the inverting input is at
virtual ground
8
Analysis of Inverting Amplifiers (2)
Thus we can say that
LP81
and
LP82
Applying KCL at the inverting input i1- i2 0
LP83
or
LP84
LP85
or
9
Analysis of Inverting Amplifiers (3)
Input impedance of inverting amplifier
Zin
LP86
Output impedance of inverting amplifier
Since output voltage expression (LP85) is
independent of load resistance, Zo 0
10
Inverting Adder
11
Inverting Adder(2)
Applying KCL at the inverting input
LP87
LP88
LP89
Let R1R2R3Rn R0 R
LP810
LP811
?
LP812
?
12
Example of inverting Adder
If v1 2 V, v2 -3 V, R1 100k ?, R2 33 k?, R0
47 k? in the above circuit calculate v0.
13
Example of inverting Adder(2)
Applying KCL at the inverting input
?
or
or
14
Output Impedance of an Inverting Amplifiers with
Finite AOL, rin, and r OL
Worked out on Greenboard
15
Non-Inverting Amplifiers
16
Non-Inverting Amplifiers(2)
Since AOL ?, for proper
amplification the voltage at the inverting input
(v-) must be equal to the voltage at the
non-inverting input (v). Otherwise, vo (v-
v-)AOL will be saturated.
? v- vin
LP813 Since Zin
?, ii 0
LP814

LP815
Equating LP813 and LP815
LP816
17
Voltage Follower
A voltage follower is a special case of a
non-inverting amplifier. In this the output is
directly connected to the inverting input.
Thus, v0 vin

LP817 This amplifier has very high
input impedance Zin Thus it is used as an input
buffer in many applications.
18
Differential Amplifier
  • Uses concepts of both inverting and non-inverting
    amplifier
  • Used as rear-end of Instrumentation amplifier
    (Voltage followers
  • form the front-end)
  • Needed in applications with high input impedance
    and CMRR
  • (CMRR Common Mode Rejection Ratio)

Ad Differential gain Acm Common mode gain
19
Differential Amplifier Analysis
Since for linear operation input voltages at the
inverting and non-inverting input are equal
Rearranging
Choosing R2 R4 and R1 R3
  • The configuration rejects any common mode signal
    existing
  • between v1 and v2.

20
A Voltage to Current converter
Show that when Q1 and Q2 have high ? and working
in active region
where IL is the load current.
21
Solution to the Voltage to Current Converter
Circuit
U1 works like a voltage follower and thus its
inverting input is at vin. Thus the collector
voltage of Q1 ? Vcc-vin. U2 also works like a
voltage follower and thus its inverting input is
as at Vcc-vin.This implies that the emitter
current of Q2 ? (Vcc-( Vcc-vin))/R vin/R ?
IL
22
Positive Feedback
QuestionWhy wont the above circuit work as an
amplifier?
23
Positive Feedback(2)
Answer Because of positive feedback
To start with, v- v vin and
LP818
Let us say due to some disturbance v vin ?v
LP819
Then
Subsequently,
LP820
24
Positive Feedback(3)
Thus due to positive feedback, the disturbance
gets amplified and drives the op-amp to either
positive or negative saturation ( the output
goes near VCC or VEE).
25
Positive Feedback(4)
Example In the above circuit VCC 15V and
-VEE -15V. R1 R2 10 k?,Vin 5V. Initially v0
10V and v 5V. What happens to v0 if due
to some disturbance v becomes i) 5.01 V, ii)
4.99 V. The output Saturation voltage is ? 13V.
AOL 105.
26
Positive Feedback(4)
  • In this case ?v v - v- 5.01-5 0.01V.
  • Hence AOL ?v 1050.01 103. Since the output
    can increase only
  • as far as 13V, v0 saturates to 13V.
  • In this case ?v v - v- 4.99-5 -0.01V.
  • Hence AOL ?v -1050.01 -103. Since the output
    can increase only
  • as far as -13V, v0 saturates to -13V.

Negative feedback ,on the other hand, opposes any
disturbance and stabilizes amplifiers.
27
Op-Amp Integrator
28
Op-Amp Integrator (2)
Since the inverting input is at virtual ground
LP821
LP822
Applying KCL at the inverting input
i1i2 0
LP823
LP824
LP825
29
Example on Op-Amp Integrator
V
A rectangular waveform is applied to the input of
an integrator. If R 10 k? and C 0.1 ?F
sketch vo. Assume zero initial value for vo.
30
Example on Op-Amp Integrator (2)
RC 1040.110-6 10-3 sec
Thus at t 0.5 ms, vo -2(0.5-0)10-3/10-3
0 -1.0V
31
Example on Op-Amp Integrator (2)
Thus at t 1.5 ms, vo 2(1.5-0.5)10-3)/10-3
-1.0V 1V . etc.
32
Op-Amp Differentiator
33
Op-Amp Differentiator (2)
Since the inverting input is at virtual ground
LP826
LP827
Applying KCL at the inverting input
i1i2 0
LP828
LP829
LP830
Differentiators are avoided in practice as they
amplify noise
34
Imperfections and Limitations of Op-Amps
  • Bias and Offset Currents
  • Due to inherent manufacturing imperfections,
    undesirable base or
  • gate currents (dc) enter the inverting and the
    non-inverting inputs of the
  • op-amps. They are in the order of nA for BJT
    based op-amps and pA
  • for FET based op-amps. If IB- is the magnitude of
    this current entering
  • the inverting input and IB the current entering
    the non-inverting input

Then bias current, IB
and offset current, Ioff
Normally,
  • Offset voltage
  • The output of an op-amp will usually not be zero
    with zero input
  • voltage. This is because of input offset voltage.

35
Imperfections and Limitations of Op-Amps(2)
  • Slew-rate (SR) limitations
  • The magnitude of the rate of change of the
    output voltage is limited. This
  • is called the slew-rate limitation

Example for a 741, SR 0.5 V/?s
36
Imperfections and Limitations of Op-Amps(3)
  • Output voltage swing
  • For a specified supply voltage, the output
    voltage range is normally 2-3
  • volts less. For example, when a 741 is supplied
    with ? 15V, the
  • guaranteed output range is ? 12V.

37
Imperfections and Limitations of Op-Amps(4)
  • Output current limit
  • An Op-Amp can supply a specified maximum value of
    current. For a 741
  • it is ? 40 mA.

38
Imperfections and Limitations of Op-Amps(5)
  • Full Power Bandwidth
  • Frequency range (0-fFP) for which the Op-Amp can
    produce an
  • undistorted sinusoidal output with its peak
    amplitude equal to the guaranteed maximum output.

Relationship between SR and fFP
Let
for undistorted output
Clearly
At the upper limit
39
Imperfections and Limitations of Op-Amps(6)
or
or
Example for a LF 351, Vom 10V and SR 12V/
?s. What is the fFP for this amplifier?
fFP
40
Imperfections and Limitations of Op-Amps(7)
  • Gain-bandwidth product
  • Because of internal capacitances used in the
    op-amps, there exists an upper limit on gain with
    respect to the frequency of input signal. To be
    more precise the gain-bandwidth product is
    constant above a certain frequency of operation.
    Under closed loop operation we normally trade-off
    gain with bandwidth.(? maximum bandwidth at
  • unity gain.
  • Example If for the following op-amp we can get
    an open loop gain
  • of 100 dB (105) upto 40Hz, find its maximum band
    width for a
  • closed loop gain of 40 (100) dB. (Ans 40 kHz).
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