Chapter 14: Operational

1
Operational Amplifiers

• Ideal op-amps
• Negative feedback
• Applications
• Useful designs
• Integrators, differentiators
• and filters

Chapter 14 Operational Amplifiers
2
Introduction Ideal Operational Amplifier

• Operational amplifier (Op-amp) is made of many
  transistors,
• diodes, resistors and capacitors in integrated
  circuit technology.
• Ideal op-amp is characterized by
• Infinite input impedance
• Infinite gain for differential input
• Zero output impedance
• Infinite frequency bandwidth

3
Ideal Operational Amplifier

• Equivalent circuit of the ideal op-amp can be
  modeled by
• Voltage controlled source with very large gain
  AOL
• known as open loop gain
• Feedback reduces the gain of op-amp
• Ideal op-amp has no nonlinear distortions

4
Ideal Operational Amplifier
A real op-amp must have a DC supply voltage which
is often not shown on the schematics
5
Inverting Amplifier

• Op-amp are almost always used with a negative
  feedback
• Part of the output signal is returned to the
  input with negative sign
• Feedback reduces the gain of op-amp
• Since op-amp has large gain even small input
  produces large output, thus for the limited
  output voltage (lest than VCC) the input voltage
  vx must be very small.
• Practically we set vx to zero when analyzing the
  op-amp circuits.

with vx 0 i1 vin /R1 i2 i1 and vo -i2
R2 -vin R2 /R1 so Avvo /vin -R2 /R1
i2
i1
6
Inverting Amplifier
Since vo -i2 R2 -vin R2 /R1 Then we see that
the output voltage does not depend on the load
resistance and behaves as voltage source. Thus
the output impedance of the inverting amplifier
is zero. The input impedance is R1 as
Zinvin/i1R1
7
Inverting Amplifier with higher gain
Inverting amplifier gain vo -i2 R2 -vin R2
/R1 Is limited due to fact that it is hard to
obtain large resistance ratio. Higher gains can
be obtained in the circuit below where we
have i1vin/R1i2 from KCL at N2 we have
i2 i3 i4
N2
8
Inverting Amplifier with higher gain
Higher gains can be obtained in the circuit below
where we have i1vin/R1i2 from KCL at N2 we
have i2 i3 i4 Also from KVL1
voi2R2i4R4 gt i4(-vo-i2R2)/R4 and from KVL2
i2R2i3R3 gt i3i2R2/R3
KVL2
KVL1
9
Inverting Amplifier with higher gain
Finally using i2 i3 i4 and i4(-vo-i2R2)/R4 i
3i2R2/R3 we have i2i2R2/R3 (-vo-i2R2)/R4 gt
i2(1R2/R3 R2/R4) -vo/R4
10
Inverting Amplifier with higher gain
i2(1R2/R3 R2/R4) -vo/R4 Substitute i2vin/R1
gt vin/R1 (1R2/R3 R2/R4) -vo/R4 to get the
voltage gain vo/vin-R4/R1 (1R2/R3 R2/R4)
11
Inverting Amplifier with higher gain
So if we chose R1R31kW and R2R410 kW then
the voltage gain is Avvo/vin -R4/R1 (1R2/R3
R2/R4) -10(1101) -120
12
Summing Amplifier
The output voltage in summing amplifier
is vo-ifRf since vi0
if
iB
vi -
13
Summing Amplifier
The output voltage in summing amplifier
is vo-ifRf since vi0
iA
if
iB
vi -
14
Summing Amplifier
The output voltage in summing amplifier
is vo-ifRf since vi0 ifiAiBvA/RAvB/RB
gt vo-(vA/RAvB/RB)Rf
iA
if
iB
vi -
15
Summing Amplifier
The output voltage in summing amplifier
is vo-ifRf since vi0 ifiAiBvA/RAvB/RB
gt vo-(vA/RAvB/RB)Rf For n inputs we will
have vo- Rf Si(vi/Ri)
iA
if
iB
vi -
16
Exercise 14.2
Find the currents and voltages in these two
circuits
a) i1vin/R11V/1kW1mA i2i11mA from
KCL vo-i2R2-10V from KVL iovo/RL-10mA from
Ohms law ixio-i2-10mA-1mA-11mA
17
Exercise 14.2
Find the currents and voltages in these two
circuits
b) i1vin/R15mA i2i15mA i21kW i31kW
gt i35mA i4i2i310mA vo- i21kW- i41kW
-10 V
18
Exercise 14.2
Find expression for the output voltage in the
amplifier circuit
i1v1/R1v1/10kW i2i1v1/10mA v3 - i2R2-
v1/10kW 20kW -2v1
V3 -
19
Exercise 14.2
Find expression for the output voltage in the
amplifier circuit
v3 - i2R2- v1/10kW 20kW -2v1 i5i3i4v3/10k
W v2/10kW vo - i5R5-(v3/10kW v2/10kW
)20kW -2v3 -2v2 4v1 -2v2
V3 -
20
Positive Feedback
When we flip the polarization of the op-amp as
shown on the figure we will get a positive
feedback that saturates the amplifier
output. This is not a good idea.
21
Noninverting amplifier
v1vin i1v1/R1 i2-i1 vo v1 - i2R2 v1
i1R2 v1 R2 v1/R1 v1(1 R2 /R1) Thus
the voltage gain of noninverting amplifier
is Av vo / vin 1 R2 /R1
-i1
22
Voltage Follower
Special case of noninverting amplifier is a
voltage follower
Since in the noninverting amplifier
vo v1(1 R2 /R1) vo v1
so when R20
gt
23
Exercise 14.4

• Find voltage gain Avvo/vin and input impedance
• With the switch open
• With the switch closed
• a.
• From KVL vini1Ri1Rvo
• i20 and i1Ri2R gt i10
• so vinvo and Avvo/vin 1

24
Exercise 14.4

• Find voltage gain Avvo/vin and input impedance
• With the switch open
• With the switch closed
• a.
• Input impedance Zinvin/iin vin/0 inf

25
Exercise 14.4

• Find voltage gain Avvo/vin and input impedance
• With the switch open
• With the switch closed
• b. for closed switch i2vin/R
• and i1Ri2R gt i1i2 gt
  vini1Ri1Rvo
• so vinvin/RRvin/RRvo gt -vinvo
• and Avvo/vin -1

26
Exercise 14.4

• Find voltage gain Avvo/vin and input impedance
• With the switch open
• With the switch closed
• b. i2vin/R
• Input impedance Zinvin/iin vin/(i1i2)
• and i1i2 gt
• Zinvin/iin vin/(2vin/R)R/2

27
Voltage to Current Converter
Find the output current io as a function of vin
28
Voltage to Current Converter
Find the output current io as a function of vin
vin ioRf so iovin/Rf
io
29
Exercise 14.6

1. Find the voltage gain vo/vin
2. Calculate the voltage gain vo/vin for R110 kW,
   R2 100 kW
3. Find the input resistance

30
Exercise 14.6

• Find the voltage gain vo/vin
• Calculate the voltage gain vo/vin for R110 kW,
  R2 100 kW
• Find the input resistance
• From KCL1 vin/R1 (v2-vin)/R2 gt v2/R2
  vin(1/R21/R1)
• From KCL2 (v2-vin)/R2v2/R1(v2-v0)/R2 0 gt

v2
31
Exercise 14.6

• Find the voltage gain vo/vin
• Calculate the voltage gain vo/vin for R110 kW,
  R2 100 kW
• Find the input resistance
• From KCL1 vin/R1(v2-vin)/R2 gt v2/R2
  vin(1/R21/R1) ()
• From KCL2 (v2-vin)/R2v2/R1(v2-v0)/R2 0 gt
• v2(2/R21/R1) (vinv0)/R2
• () v2 vin(1R2/R1) gt vin(1R2/R1)(2/R21/R1)
  (vinv0)/R2
• vin (R2 (1R2/R1)(2/R21/R1)-1)v0

v2
v0 / vin 131
32
Exercise 14.6

• Find the voltage gain vo/vin
• Calculate the voltage gain vo/vin for R110 kW,
  R2 100 kW
• Find the input resistance
• From KCL1 vin/R1(v2-vin)/R2 gt v2/R2
  vin(1/R21/R1) ()
• From KCL2 (v2-vin)/R2v2/R1(v2-v0)/R2 0 gt
• v2(2/R21/R1) (vinv0)/R2
• () v2 vin(1R2/R1) gt vin(1R2/R1)(2/R21/R1)
  (vinv0)/R2
• vin (R2 (1R2/R1)(2/R21/R1)-1)v0 gt v0 / vin
  100(110)0.12-1

v2
v0 / vin 131
33
Matrix equations for op-amp circuits
Example
34
Example
35
Design of Simple Amplifiers
Practical amplifiers can be designed using op-amp
with feedback.
We know that for noninverting amplifier Av vo /
vin 1 R2 /R1 so to obtain Av10 we could use
R11W and R29W But such low output resistance
will draw too much current from the power supply
36
Design of Simple Amplifiers
The same gain can be obtained with large
resistance values. But for high output
resistance are sensitive to bias current and we
must use a filtering output capacitor to remove
the noise.
Av vo / vin 1 R2 /R1
37
Op-Amp Imperfections in a Linear Mode

• We consider the following op-amp imperfections
• Nonideal linear operation,
• Nonlinear characteristics
• Dc offset values.

• Input and output impedances
• Ideal opamp have Rin0,
• Real op-amp has

38
Op-Amp Imperfections in a Linear Mode

• We consider the following op-amp imperfections
• Nonideal linear operation,
• Nonlinear characteristics
• Dc offset values.

• Voltage gain
• Ideal op-amp has infinite gain and bandwidth,
• Real op-amp has the gain that changes with
  frequency.
• Open loop gain

Open loop bandwidth
Terminal frequency ft gain drops to 1
39
Op-Amp Imperfections in a Linear Mode
Negative feedback is used to lower the gain and
extend the bandwidth. Open loop gain From KVL
So the closed loop gain
where
40
Op-Amp Imperfections in a Linear Mode
Using open loop gain We get
So we will get closed loop dc gain
closed loop voltage gain
closed loop bandwidth
41
Op-Amp Imperfections in a Linear Mode
Comparing to open loop, the closed loop gain is
reduced And closed loop bandwidth is larger
The gainbandwidth product stays the same
42
Nonlinear Limitations

• Nonlinear limitations
• Output voltage swing is limited and depend on
  power supply voltage
• for
• Maximum output current is limited

43
Nonlinear Limitations
When voltage or current limits are exceeded,
clipping of the output signal occurs causing
large nonlinear distortions
44
Nonlinear Limitations
Another nonlinear limitation is limited rate of
change of the output signal known as the
slew-rate limit SR
Using slew rate we can find maximum frequency
known as full-power bandwidth. Assuming
So the full-power bandwidth
45
Dc offset values

• There are three dc offset values related to
  op-amp design
• Bias currents IB, IB- related to
  differential inputs
• Offset current ideally zero value
• Offset voltage results in nonzero output for
  zero input
• They can be represented as additional dc sources
  in the
• op-amp model

46
Industrial op-amp
741 Amplifier is the most popular amplifier it
has AOL100000
47
Industrial op-amp
741 Amplifier BJT transistor level schematic
48
(No Transcript)
49
(No Transcript)
50
(No Transcript)
51
(No Transcript)
52
(No Transcript)
53
(No Transcript)
54
(No Transcript)
55
(No Transcript)
56
(No Transcript)
57
(No Transcript)
58
(No Transcript)
59
(No Transcript)
60
(No Transcript)
61
(No Transcript)