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OperationalAmplifiers
Lecture4
BME372NewSchesser 94
95
Ideal Amplifiers • Parameters:
vs vo vi
Ro
Ri
Rs
RL Avovi
io ii + +
- -
+
-
vovs
=Ri
Ri + RsAvo
RLRL + RO
This states the gain of the amplifier depends on the external components.This is BAD!!!!!
However, if Ri →∞ and RO → 0; then Ri
Ri + Rs→1 and
RLRL + RO
→1.
Therefore, vovs
→ Avo and the gain of the amplifier is independent of the external components.
BME372NewSchesser
96
Operational Amplifiers
• An operational Amplifier is an ideal differential with the following characteristics: – Infinite input impedance – Infinite gain for the differential signal – Zero gain for the common-mode signal – Zero output impedance – Infinite Bandwidth vo-
+v1v2
Inverting input
Non-Inverting input
BME372NewSchesser
97
Operational Amplifier Feedback • Operational Amplifiers are used with negative feedback • Feedback is a way to return a portion of the output of an
amplifier to the input – Negative Feedback: returned output opposes the source signal – Positive Feedback: returned output aids the source signal
• For Negative Feedback – In an Op-amp, the negative feedback returns a fraction of the
output to the inverting input terminal forcing the differential input to zero.
– Since the Op-amp is ideal and has infinite gain, the differential input will exactly be zero. This is called a virtual short circuit
– Since the input impedance is infinite the current flowing into the input is also zero.
– These latter two points are called the summing-point constraint.
BME372NewSchesser
98
Operational Amplifier Analysis Using the Summing Point Constraint
• In order to analyze Op-amps, the following steps should be followed:
1. Verify that negative feedback is present 2. Assume that the voltage and current at the input
of the Op-amp are both zero (Summing-point Constraint
3. Apply standard circuit analyses techniques such as Kirchhoff’s Laws, Nodal or Mesh Analysis to solve for the quantities of interest.
BME372NewSchesser
99
Example: Inverting Amplifier 1. Verify Negative Feedback: Note that a
portion of vo is fed back via R2 to the inverting input. So if vi increases and, therefore, increases vo, the portion of vo fed back will then have the affect of reducing vi (i.e., negative feedback).
2. Use the summing point constraint.
3. Use KVL at the inverting input node for both the branch connected to the source and the branch connected to the output
RLvin
R2
R1
vivo
-+
0 i1=vin/R1 i2
inverted) :input the toopposite isoutput that the(note oft independen is which -
zero is since 0constraintpoint -summing the todue
constraintpoint -summing the todue zero is since 0
1
2
220
21
11
Lin
i
iin
RvRR
vRivii
vRiv
=
+−==
+=1
1
11
1
RiRi
ivZ inin ===
-+ ++
--
BME372NewSchesser
100
Op-amp
• Because we assumed that the Op-amp was ideal, we found that with negative feedback we can achieve a gain which is:
1. Independent of the load 2. Dependent only on values of the circuit
parameter 3. We can choose the gain of our amplifier by
proper selection of resistors.
BME372NewSchesser
101
Non-inverting Amp 1. First check: negative feedback? 2. Next apply, summing point constraint 3. Use circuit analysis
+
-- vo
+
--
RL R2
R1
+
--
vi
+
--
vf
io
Iin= 0
+
-
vin
1
1 2
2 1 2
1 1
0
;
1
Since 0;
in i f f f
f o in
ov
in
inin in
in
v v v v vRv v v
R Rv R R RAv R R
vi Zi
= + = + =
= =+
+= = = +
= = = ∞Note:1. Thegainisalwaysgreaterthanone2. Theoutputhasthesamesignasthe
input
BME372NewSchesser
102
Non-inverting Amp Special Case What happens if R2 = 0?
+
-- vo
+
--
RL
+
--
vi
io
Iin= 0
+
-
vin
∞===
=+==
==+
=
=+=+=
in
ininin
in
o
inoof
fffiin
ivZi
RR
vvAv
vvvRRv
vvvvv
;0 Since
10
;0
0
1
1
1
1
This is a unity gain amplifier and is also called a voltage follower.
BME372NewSchesser
103
Special Amplifiers
• Summer (Homework Problem) • Instrumentation Amplifier – Uses 3 Op-amps – One as a differential amplifier – Two Non-inverting Amps using for providing gain
BME372NewSchesser
104
Medical Instrumentation Amplifier
Differential Amplifier
+
--
R2
R1
+
-
v2
+
--
R2
R1
+
-
v1
R3
R4 +
-
vo
R5
R6
Non-inverting Amplifier
Non-inverting Amplifier
BME372NewSchesser
105
Medical Instrumentation Amplifier
+
--
R2
R1
+
-
v2
+
--
R2
R1
+
-
v1
R3
R4
vo
R5
R6
Differential Amplifier
Non-inverting Amplifier
Non-inverting Amplifier
v2D
v1D
+
-
BME372NewSchesser
106
Medical Instrumentation Amplifier 2
6 5
2
6 6 5 5
5 62
6 6 5 5
41
3 4
5 62 41
6 3 4 6 5 5
5 5 6 41 2
6 5 3 4
5 6 4
5 3 4
51 2
6
1 1( )
( )
( )
( )
Chose 1
( )
D x x o
oDx
oDx
y D x i x
oDD
o D D
o D D
v v v vR R
vv vR R R R
R R vv vR R R R
Rv v v v vR R
R R vv R vR R R R R R
R R R Rv v vR R R RR R RR R R
Rv v vR
− −=
−− + =
+ −− =
= = − =+
+ −− =+
+= −+
+ =+
= −
R3
R4
vo
R5
R6
Differential Amplifier
v2D
v1D
vx
+
-
4
3
5
6
3564
45356454
43
4
5
65
1 Chose
RR
RR
RRRRRRRRRRRR
RRR
RRR
=
=+=+
=+
+
vi=0
vy
BME372NewSchesser
107
Medical Instrumentation Amplifier
+
--
R2
R1
+
-
v2
+
--
R2
R1
+
-
v1
R3
R4 +
-
vo
R5
R6
Differential Amplifier
Non-inverting Amplifier
Non-inverting Amplifier
v2D
v1D
BME372NewSchesser
108
Medical Instrumentation Amplifier
+
--
R2
R1
+
-
v2
+
--
R2
R1
+
-
v1 Non-inverting Amplifier
Non-inverting Amplifier
v2D
v1D vA
AD
AD
AD
AD
vRRv
RRRv
vRRv
RRRv
vRRv
RRRv
Rvv
Rvv
1
21
1
211
1
22
1
212
1
22
12
22
1
2
2
22
Likewise
)11(
−+=
−+=
−+=
−=−
BME372NewSchesser
109
Medical Instrumentation Amplifier
))(1())((
)]([
)(
21
1
2
6
521
1
21
6
5
1
22
1
21
1
21
1
21
6
5
1
21
1
211
1
22
1
212
21
6
5
vvRR
RRvv
RRR
RRv
vRRv
RRRv
RRv
RRR
RRv
vRRv
RRRv
vRRv
RRRv
vvRRv
o
AAo
AD
AD
DDo
−+=−+=
−+−−+=
−+=
−+=
−=
+
-- R2
R1
+
-
v2
+
--
R2
R1 + -
v1
R3
R4
vo
R5
R6
Differential Amplifier
Non-inverting Amplifier
Non-inverting Amplifier
+
-
-
+
BME372NewSchesser
110
Uses of the Differential Amplifier
BME372NewSchesser
111
Integrators and Differentiators
∫∫ −=−=
==
t
in
t
o
in
dxxvRC
dxxiC
v
tiRtvti
002
21
)(1)(1
)()()(
vin
C
Rvi
vo
-+
0 i2 i1
vin
R
Cvi
vo
-+
0 i2 i1
dttdvRCRtiv
tidttCdvti
ino
in
)()(
)()()(
2
21
−=−=
==
BME372NewSchesser
112
Frequency Analysis
vinvi
vo
-+
0 i1=vin/Z1
i2
1in
o
Lin1
o
1
11in
ZZ
VV
ZVZZ
ZIVII
ZIV
2
2
22
2
- )()(
oft independen is which )(-
zero )(virtually is since 0)()(constraintpoint -summing the todue )()(
zero )(virtually is since 0)()()(
=
=
+−==
+=
ωω
ω
ωωωω
ωωω
jj
j
vjjjj
vjjj
i
i
ZLvin
C
Rvi vo
-+
0i1=vin/Z1 i2
ZLvin
R
Cvi vo
-+
0i1=vin/Z1 i2
integratoran 1- )()( 2
RCjjj
ωωω −==
1in
o
ZZ
VV tordifferenia a - )(
)( 2 RCjjj ωωω −==
1in
o
ZZ
VV
Z2
Z1
BME372NewSchesser
113
Frequency Response
vin
CRvi
vo
-+
0 i2 i1
vin
RCvi
vo
-+
0 i2 i1
1- )()( 2
RCjjj
ωωω −==
1in
o
ZZ
VV
RCjjj ωωω −==
1in
o
ZZ
VV 2-
)()(
ω
ω
BME372NewSchesser
114
Frequency Response
vin
C2R1vi
vo
-+
0 i2 i1
vin
R2
C1vi
vo
-+
0 i2
i1
R2 VoutVin
= −Z2Z1
= −R2R1
1(1+ jωC2R2 )
=R2R1
1
1+ (ωC2R2 )2∠π − tan−1(ωC2R2 )
0
500
1,000
1,500
2,000
2,500
1.E+00 1.E+02 1.E+04 1.E+06 1.E+08HZ
R151C110mfR2100k
R1
12 2 1 1 2 1 11 12
1 1 1 1 1 1 1
tan ( )(1 ) 21 ( )
out
in
V Z R j C R R C R C RV Z R j C R R C R
ω ω π ωω ω
−= − = − = ∠− −+ +
0
500
1,000
1,500
2,000
2,500
1.E+00 1.E+02 1.E+04 1.E+06 1.E+08HZ
R151C110mfR2100k
BME372NewSchesser
115
Homework • Probs 2.2, 2.5, 2.6, 2.10, 2.22, 2.24, 2.25,
2.28 • Calculate and plot the output vs frequency
for these circuits. R1=1k, R2=3k, C=1µf Use Matlab to calculate the Bode plot
vin
C
R1vi
vo-+
R2
vinCR1
vivo
-+
R2
vin
C
R1vi
vo-+
R2R2
vinC
R1
vivo
-+
BME372NewSchesser