chapter 5 op-amp_stdnt note
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8/8/2019 Chapter 5 Op-Amp_stdnt Note
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CHAPTER 5
Operational Amplifier
OPERATIONALAMPLIFIER ( OP-AMP )
( Chapter 5 )
Industrial Electronics
DEK 3113
Sumaiya Binti Mashori
October 05, 2010
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_ _
++
1) Op-amp ( Operational Amplifier )high gain DC amplifier to increase signal strength
V+
V -
Non-inverting input
Inverting input
Vs -
Vs+
Vout
The Operational Amplifier The Operational Amplifier
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i(+), i(-) : Currents into the amplifier on the inverting and non-inverting lines
respectivelyvid : The input voltage from inverting to non-inverting inputs
+VS , -VS : DC source voltages, usually +15V and ±15V
Ri : The input resistance, ideally infinity
A : The gain of the amplifier. Ideally very high, in the 1x1010 range.
RO: The output resistance, ideally zero
vO: The output voltage; vO = AOLvid where AOL is the open-loop voltage gain
Ideal OpIdeal Op--Amp Amp+VS
-VS
vid
Inverting
Non-inverting
Output
+
_ i(-)
i(+)
vO = Advid
RO
ARi
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Characteristic of Ideal OpCharacteristic of Ideal Op--Amp Amp
1) High input impedance
Zi = Ri = 2) Low output impedance
Zo = Ro = 0
3) Ri = ( assume open cct )
i(-)
i(+)V+
V -
o/c
i(+) = i(-) = 0
4) Ro = 0 ( assume short cct )
V+
V -
s/c
Vout
RO = 0
Vdiff AOL .Vdiff
V diff =V + - V - = 0
V +=V -
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Types Of OpTypes Of Op--Amp Amp
1)Inverting Amplifier2)Non-Inverting Amplifier
3)Inverting Summing Amplifier4)Non-Inverting Summing Amplifier5)Integrator
6)Differentiator7)Voltage Comparator8)Voltage Follower
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Amplify the input,Vs and invert the value to be negative
value.
+
_
RL
vin
+
_
Rii1
RFiF
Vout
Inverting Amplifier Inverting Amplifier (Vin at(Vin at ± ±veve terminal)terminal)
Derivation: hand written
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Inverting Amplifier Inverting Amplifier (Vin at(Vin at ± ±veve terminal)terminal)
+
_
500
+
_
i1
iF
VoutV+
V-
i-
i+
Example1: Find a) Vo
b) i1
c) Av
0.5V
1k
20k
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Inverting Amplifier Inverting Amplifier (Vin at(Vin at ± ±veve terminal)terminal)
Solution Example1:
Phase shift 180o
Find a) Vo Find b) i1
Find c) Av
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Inverting Amplifier Inverting Amplifier (Vin at(Vin at --veve terminal)terminal)
Exercise1: If we need an amplifier of gain =10, Designthe amplifier using inverting amplifier
1) Assume
+
_
RL
0.5V+
_
10ki1
iF
Vout
100k
2) Design
V+
V-
i-
i+
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Inverting Amplifier Inverting Amplifier (Vin at(Vin at --veve terminal)terminal)
3) Find Vo 4) Find i1
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NonNon--Inverting Amplifier Inverting Amplifier (Vin at +(Vin at +veve
terminal)terminal)
+
_
RL
vin
+
_
Rii1
iF
Vout
RF
Derivation: hand written
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NonNon--Inverting Amplifier Inverting Amplifier (Vin at +(Vin at +veve
terminal)terminal)
Example1: Vin = 0.5V ; Ri = 1k ; RF = 20k ; RLD = 500
Find a) Vo
b) i1c) Av
Diff polarity
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NonNon--Inverting Amplifier Inverting Amplifier (Vin at +(Vin at +veve
terminal)terminal)
Exercise1: Design using non-inverting amplifier with thegain= 10
1) Assume
+
_
RL+
_
i1
iF
Vout
90k
10k
0.5V
V+
V-
i-
i+
2) Design
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NonNon--Inverting Amplifier Inverting Amplifier (Vin at +(Vin at +veve
terminal)terminal)
3) Find Vo 4) Find i1
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+
_
R3
i1
iF
Vout
Inverting Summing Amplifier Inverting Summing Amplifier
i2V2
V1
R1
R2
RF
i-
i+
Derivation: hand written
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+
_
i1
iF
Vout
NonNon--Inverting Summing Amplifier Inverting Summing Amplifier
i2V2
V1
R1
R2
RF
i-
i+
R3i3
Derivation: hand written
2 input to +ve terminal
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+
_
i1
iF
Vout
NonNon--Inverting Summing Amplifier Inverting Summing Amplifier
i2V2
V1
R1
R2
RF
i-
i+
R4i4
Derivation: try on your own
i3V3
R3 3 input to +ve terminal
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Change square wave to triangle.
+
_ RiR
iC
Vout
C
Integrator Integrator (phase shifter )(phase shifter )
Vi (t)
Derivation: hand written
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Change triangle to square wave.
+
_
R
iC
iR
Vout
C
Differentiator Differentiator (phase shifter )(phase shifter )
Vi (t)
Derivation: hand written
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Has 2 independent input - output based on relation between
2 input
Function: as Switch
Voltage Comparator Voltage Comparator
Example: hand written
Vout
_
+
Vin
Vref
+Vcc
-Vcc
Vout
_
+Vin
Vref
+Vcc
-Vcc
(1) (2)
Concept:
V-
V+
V-
V+
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For impedance matching.
Negative feedback-for stability
+
_
Vout
Voltage Follower Voltage Follower (unity gain)(unity gain)
Vin
Example: hand written
i-
i+