Lecture 5: Operational Amplifiers
Dr. Jin Zhang
E-mail: jzhang_gzy@126.com
21 April 2022
广西师范大学2021届硕士毕业答辩
CONTENTS
01
Introduction of Operational Amplifiers
02
Bias Circuits
03
Electrical Characteristics
04
Op-Amp Applications
01
Introduction of Operational Amplifiers
μA741 OP-AMP
Schematic
Symbol
Lecture 5: Operational Amplifiers
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01
Introduction of Operational Amplifiers
Bias circuit: Determining the quiescent point
Input stage: Differential pre-amplifying circuit
Intermediate stage: Main amplifier (common-emitter
amplifier)
Output stage: Power stage
Lecture 5: Operational Amplifiers
44
02
Bias Circuits—— Current Source
➢ Mirror current source
• T0 and T1 are the same.
I R = (VCC − U BE ) R
U BE1 = U BE0，I B1 = I B0
Reference current
I C1 = I C0 = I C
I R = I C0 + I B0 + I B1 = I C +
IC =

 +2
2I C

 IR
• When β >> 2, IC ≈ IR.
Lecture 5: Operational Amplifiers
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02
Bias Circuits—— Current Source
➢ Microcurrent source
• It is required to provide a small static current without using a large resistance.
I E1 = (U BE0 − U BE1 ) Re
UBE0
IR = βIs·e
UT
UBE1
IE1 = βIs·e
Re
IR
=e
IE1
UT
UBE0−UBE1
IE1Re
UT
UT
=e
• Firstly, determine IE0 and IE1. Then select R and Re.
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02
Bias Circuits——Current Source
➢ Multi-stage current source: Based on proportion current sources.
UBE0+IE0Re0
=UBE1+IE1Re1
=UBE2+IE2Re2
=UBE3+IE3Re4
Since UBE is almost constant for every stage，
IE0Re0≈ IE1Re1 ≈ IE2Re2 ≈ IE3Re3
• The resistances (Re ) are selected according to the needed static currents.
Lecture 5: Operational Amplifiers
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02
Bias Circuits——Current Source
➢ Multi-stage current source: Based on a multi-emitter transistor.
The areas of emitter layers are S0、S1、S2，
I C1 S1 I C2 S 2
= ， =
I C0 S0 I C0 S0
• The areas of emitter layers are determined according to the needed static currents.
Lecture 5: Operational Amplifiers
88
03
Electrical Characteristics
Parameters
uO
(uP − u N )
Open-loop differential-mode gain Aod
20 lg Aod = 20 lg
Differential-mode input resistance rid
(uP − u N )
rid =
iP
Common mode rejection ratio KCMR
K CMR = 20 lg
Lecture 5: Operational Amplifiers
Aod
Ac
99
03
Electrical Characteristics
Parameters
Input offset current IIO
Since the two input transistors are never exactly matched, each will
operate at a slightly different current.
Input bias current
Input offset current
Lecture 5: Operational Amplifiers
10
10
03
Electrical Characteristics
Parameters
Input offset voltage UIO
Lecture 5: Operational Amplifiers
Although the op-amp output should be 0 V when the input is 0 V, in actual
operation there is some offset voltage at the output. An output offset voltage
will also result due to any difference in dc bias currents at both inputs.
11
11
03
Electrical Characteristics
Parameters
Common-mode input
voltage range Uicm
Maximum peak output
voltage swing Uopp
Lecture 5: Operational Amplifiers
This parameter lists the range over which the input voltage may vary. Inputs
larger in amplitude than this value will probably result in output distortion
and should be avoided.
This parameter lists the largest amount the output may vary.
12
12
03
Electrical Characteristics
μA741 Electrical Characteristics: VCC =±15 V, TA =25°C
Lecture 5: Operational Amplifiers
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13
01
Op-Amp Applications ——Voltage Transfer Characteristic
In linear region:
Saturation region
uo＝Aod(uP－uN)
where Aod is open-loop differential-common voltage
gain.
Since Aod is very large, and the output is a limited
vale, (uP－uN) is very small.
Linear region
Saturation region
Lecture 5: Operational Amplifiers
When (uP－uN) is larger than a threshold voltage, the
Op-Amp output is +UOM or － UOM. The Op-Amp is
operated in amplitude limitation region.
14
14
01
Op-Amp Applications ——Voltage Transfer Characteristic
Feature:
• A loop of voltage negative feedback is introduced.
Linear region
Since uO is a finite value, and Aod＝∞,
uN－uP＝0
i.e.
uN＝uP…………virtual short
In linear region:
uo＝Aod(uP－uN)
Lecture 5: Operational Amplifiers
Since rid＝∞,
iN＝iP＝0………virtual break
15
15
04
Op-Amp Applications—— Constant-Gain Multiplier
uN＝uP…………virtual short
iN＝iP＝0………virtual break
iF
Inverting input
iN
+
iR
Noninverting input
iP
_
uI
For N-point： iF = iR =
R
Rf
uO = −iF Rf = −  uI
R
u N = uP = uI
Rf
uO = (1 +
)  uN
R
Rf
uO = (1 +
)  uI
R
Lecture 5: Operational Amplifiers
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16
04
Op-Amp Applications—— Voltage Summing
uN＝uP…………virtual short
iN＝iP＝0………virtual break
Summing
u N = uP = 0
iF = iR1 + iR 2 + iR 3
u I1 u I2 u I3
=
+
+
R1 R2 R3
uO = −iF Rf = − Rf (
uI1 uI2 uI3
+
+ )
R1 R2 R3
How to realize the subtraction function?
Lecture 5: Operational Amplifiers
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Thank you！
Lecture 5: Operational Amplifiers
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