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Operational Amplifiers:
• An operational amplifier (op-amp) are used in analog electronic design for– Amplification– Analog filtering– Buffering– Threshold detection
Basic Electric CircuitsOperational Amplifiers
inverting input
noninverting inputoutput
V-
V+
The basic op amp with supply voltage included is shownin the diagram below.
-V Vout V
Basic Electric CircuitsOperational Amplifiers
In most cases only the two inputs and the output areshown for the op amp. However, one should keep inmind that supply voltage is required, and a ground.The basic op amp without a ground is shown below.
Outer op amp diagram.
Basic Electric CircuitsOperational Amplifiers
A model of the op amp, with respect to the symbol, isshown below.
V1
V2
_
+
Vd Ri
Ro
AVd
Vo
Op Amp Model.
Basic Electric CircuitsOperational Amplifiers
The previous model is usually shown as follows:
Ri
Ro
AVd
_
+
Vd
V1
V2
Vo
+
_
Basic Electric CircuitsOperational Amplifiers
Application: As an application of the previous model,consider the following configuration. Find Vo as a function of Vin and the resistors R1 and R2.
+
_
R2
R1
+
_
+
_
Vin Vo
Basic Electric CircuitsOperational Amplifiers
In terms of the circuit model we have the following:
Ri
Ro
AVi
_
+
ViVin Vo
+
_
+
_
R1
R2
ab
Total op amp schematic for voltagegain configuration.
Basic Electric CircuitsOperational Amplifiers
Ri
Ro
AVi
_
+
ViVin Vo
+
_
+
_
R1
R2
ab
Circuit values are:R1 = 10 k R2 = 40 k Ro = 50 A = 100,000 Ri = 1 meg
Basic Electric CircuitsOperational Amplifiers
We can write the following equations for nodes a and b.
(1)
(2)
( )10 1 40
( )5040
i oin i i
i oo i
V VV V Vk meg k
V VV AVk
Basic Electric CircuitsOperational Amplifiers
Simplifying 1
inio VVV 10012625 (3)
Simplifying 2
010410005.4 95 io VxVx (4)
Basic Electric CircuitsOperational Amplifiers
From Equations 3 and 4 we find;
ino VV 99.3
Fortunately, we are not required to do elaborate circuitanalysis, as above, to find the relationship between theoutput and input of an op amp. Simplifying the analysisis our next consideration.
(5)
Basic Electric CircuitsOperational Amplifiers
For most all operational amplifiers, Ri is 1 meg orlarger and Ro is around 50 or less. The open-loop gain, A, is greater than 100,000.
Ideal Op Amp:The following assumptions are made for the ideal op amp.
i
o
RohmsinputInfinite
RohmsoutputZero
AgainloopopenInfinite
;.3
0;.2
;.1
Basic Electric CircuitsIdeal Op Amp:
_
+ ++
++
_
__ _
Vi
V1
V2 = V1Vo
i1
i2
= 0
= 0
(a) i1 = i2 = 0: Due to infinite input resistance.
(b) Vi is negligibly small; V1 = V2.
Ideal op amp.
Basic Electric CircuitsIdeal Op Amp:
Find Vo in terms of Vin for the following configuration.
+
_
R2
R1
+
_
+
_
Vin Vo
Gain amplifier op amp set-up.
Basic Electric CircuitsIdeal Op Amp:
+
_
R2
R1
+
_
+
_
Vin Vo
a
Vi
Writing a nodal equation at (a) gives;
21
)(RVV
RVV oiiin
(6)
Basic Electric CircuitsIdeal Op Amp:
21
)(RVV
RVV oiiin
With Vi = 0 we have;
With R2 = 4 k and R1 = 1 k, we have
ino VV 4 Earlierwe got ino VV 99.3
(7)
inVRRV1
20
Basic Electric CircuitsIdeal Op Amp:
Example : Consider the op amp configuration below. Find V0
+
+
+
_
__
3 VVin
6 k
1 k
V0
a
Assume Vin = 5 V
Basic Electric CircuitsOperational Amplifiers
+
+
+
_
__
3 VVin
6 k
1 k
V0
a
At node “a” we can write;
kV
k)V( in
63
13 0
From which; V0 = -51 V (op amp will saturate)
solution
Operational Amplifiers
Summing Amplifier
Example : Given the following, Find V0:Rfb
R1
R2
V2
V1V0
a
fbRV
RV
RV 0
2
2
1
1 (11)
Example continued
Equation 11 can be expressed as;
2
21
10 V
RR
VRR
V fbfb (12)
If R1 = R2 = Rfb then,
210 VVV (13)
Therefore, we can add signals with an op amp.
Basic Electric CircuitsOperational Amplifiers
Example : The non-inverting op amp. Consider the following:
R0
Rfb
V0V2_
+
+
_
a
Basic Electric CircuitsOperational Amplifiers
Writing a node equation at “a” gives;
20
0
02
0
02
0
2
1
,
11
0)(
VRR
V
giveswhich
RRV
RV
so
RVV
RV
fb
fbfb
fb
Basic Electric CircuitsOperational Amplifiers
Example: Non-inverting Input.Find V0 for the following op amp configuration.
+
_
+
+
_
_ 4 V
2 k
6 k
5 k
10 k
V0
a
Vx
33
Basic Electric CircuitsOperational Amplifiers
The voltage at Vx is found to be 3 V.
Writing a node equation at “a” gives;
0105
0
k
)VV(k
V xx
orVVV x 930
