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ECEN 301 Discussion #18 – Operational Amplifiers 1
Date Day ClassNo.
Title Chapters HWDue date
LabDue date
Exam
3 Nov Mon 18 Operational Amplifiers
8.4LAB 6
4 Nov Tue
5 Nov Wed 19 Binary Numbers 13.1 – 13.2
6 Nov Thu
7 Nov Fri Recitation HW 8
8 Nov Sat
9 Nov Sun
10 Nov Mon 20 Exam Review
LAB 7 EXAM 212 Nov Tue
Schedule…
ECEN 301 Discussion #18 – Operational Amplifiers 2
Give to ReceiveAlma 34:28 28 And now behold, my beloved brethren, I say unto you, do not
suppose that this is all; for after ye have done all these things, if ye turn away the needy, and the naked, and visit not the sick and afflicted, and impart of your substance, if ye have, to those who stand in need—I say unto you, if ye do not any of these things, behold, your prayer is vain, and availeth you nothing, and ye are as hypocrites who do deny the faith.
ECEN 301 Discussion #18 – Operational Amplifiers 3
Lecture 18 – Operational Amplifiers
Answer questions from last lectureContinue with Different OpAmp configurations
ECEN 301 Discussion #17 – Operational Amplifiers 4
Op-Amps – Open-Loop Model1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
–
++
v+
–
+
v–
–
+
vo
–
io
i2
i1–vin
+
–
+
+–
RoutRin
i1
AOLvin
+
vo
–
–
vin
+
v–
v+
NB: op-amps have near-infinite input resistance (Rin) and very small output resistance (Rout)
)(
vvA
vAv
OL
inOLoAOL – open-loop voltage gain
i2
ECEN 301 Discussion #17 – Operational Amplifiers 5
Op-Amps – Open-Loop Model1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
–
+
+–
RoutRin
i1
AOLvin
+
vo
–
–
vin
+
v–
v+
vAvv
vvA
vAv
OL
o
OL
inOLo
)(i2
Ideally i1 = i2 = 0(since Rin → ∞)
What happens as AOL → ∞ ?→ v– ≈ v+
ECEN 301 Discussion #17 – Operational Amplifiers 6
Op-Amps – Closed-Loop Mode
RF
–
+ +
vo
–
i1
RS
vS(t) +–
v+
v–iF
iS
OLSFOLSFoS
SOL
o
FOL
o
S
S
S
S
F
OLo
F
o
S
OLo
S
S
F
o
S
S
FS
ARRARRvv
RAv
RAv
Rv
Rv
RAv
Rv
RAv
Rv
Rvv
Rvv
ii
1/
1/1
//
1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
ECEN 301 Discussion #17 – Operational Amplifiers 7
Op-Amps – Closed-Loop Mode
OLSFOLSFoS ARRARRvv 1
/1
/1RF
–
+ +
vo
–
i1
RS
vS(t) +–
v+
v–iF
iSNB: if AOL is very large these terms → 0
S
F
S
o
RRvv
CLA
:Gain Loop-Closed
NB: if AOL is NOT the same thing as ACL
1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
ECEN 301 Discussion #17 – Operational Amplifiers 8
Op-Amps – Closed-Loop Mode
–+
va RR
–+
vb R R
–+
R1R
R2
vo
1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ?
2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
NB: Current flows through R1 and R2
ECEN 301 Discussion #17 – Operational Amplifiers 9
Op-Amps – Closed-Loop Mode
–+
va RR
–+
vb R R
–+
R1R
R2
vo
1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ?
2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
NB: Inverting amplifiers and (RS = RF) → vo = -vi
ECEN 301 Discussion #17 – Operational Amplifiers 10
Op-Amps – Closed-Loop Mode1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ?
2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
–+
R1R
R2
vo
-va
-vb
21
21
21
21
21
000
Rv
RvRv
Rv
Rv
Rv
Rv
Rv
Rv
Rvv
Rvv
Rvv
iii
bao
oba
oba
oba
F
iF
i1
i2
ECEN 301 Discussion #18 – Operational Amplifiers 11
More OpAmp Configurations
ECEN 301 Discussion #18 – Operational Amplifiers 12
Op-Amps – Closed-Loop ModeThe Differential Amplifier: the signal to be amplified
is the difference of two signals
–
+ +
vo
–
i1
RF
vS2(t) +–
v+
v–
iFRS
iS
i1vS1(t) +
–
RF
RS
ECEN 301 Discussion #18 – Operational Amplifiers 13
Op-Amps – Closed-Loop ModeThe Differential Amplifier: the signal to be amplified
is the difference of two signals
021
iivv
RF
–
+ +
vo
–
i1
vS2(t) +–
v+
v–
iFRS
iS
i2
vS1(t)+–
RF
RS
NB: an ideal op-amp with negative feedback has the properties
SF ii
ECEN 301 Discussion #18 – Operational Amplifiers 14
Op-Amps – Closed-Loop ModeThe Differential Amplifier: the signal to be amplified
is the difference of two signals
SF
Fs RR
Rvv
2
:Divider VoltageRF
–
+ +
vo
–
i1
vS2(t) +–
v+
v–
iFRS
iS
i2
vS1(t)+–
RF
RS
S
SS
SSS
Rvvi
Rivvv
1
1
F
oF R
vvi
ECEN 301 Discussion #18 – Operational Amplifiers 15
Op-Amps – Closed-Loop ModeThe Differential Amplifier: the signal to be amplified
is the difference of two signals
RF
–
+ +
vo
–
i1
vS2(t) +–
v+
v–
iFRS
iS
i2
vS1(t)+–
RF
RS
12
221
SSS
F
FSS
FS
FS
S
S
SFo
FS
FFo
vvRR
RRRRv
RRv
RvRv
vRi
vRiv
ECEN 301 Discussion #18 – Operational Amplifiers 16
Op-Amps – Level ShifterLevel Shifter: can add or subtract a DC offset from a
signal based on the values of RS and/or Vref
–
+ +
vo
–
RF
Vref+–
v+
v–RSVsensor
AC voltage with DC offset
DC voltage
ECEN 301 Discussion #18 – Operational Amplifiers 17
Op-Amps – Level ShifterExample1: design a level shifter such that it can remove
a 1.8V DC offset from the sensor signal (Find Vref)RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt)
–
+ +
vo
–
RF
Vref+–
v+
v–RSVsensor
ECEN 301 Discussion #18 – Operational Amplifiers 18
Op-Amps – Level ShifterExample1: design a level shifter such that it can remove
a 1.8V DC offset from the sensor signal (Find Vref)RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt)
–
+ +
vo
–
RF
Vref+–
v+
v–RSVsensor
Find the Closed-Loop voltage gain by using the principle of superposition on each of the DC voltages
ECEN 301 Discussion #18 – Operational Amplifiers 19
Op-Amps – Level ShifterExample1: design a level shifter such that it can remove
a 1.8V DC offset from the sensor signal (Find Vref)RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt)
–
+ +
vo
–
RF
v+
v–RS
Vsensor+–
DC from sensor:Inverting amplifier
senDCS
F
senDCCLosen
vRRvAv
ECEN 301 Discussion #18 – Operational Amplifiers 20
Op-Amps – Level ShifterExample1: design a level shifter such that it can remove
a 1.8V DC offset from the sensor signal (Find Vref)RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt)
–
+ +
vo
–
RF
v+
v–RS
Vref+–
DC from reference:Noninverting amplifier
refS
F
refCLoref
vRR
vAv
1
ECEN 301 Discussion #18 – Operational Amplifiers 21
Op-Amps – Level ShifterExample1: design a level shifter such that it can remove
a 1.8V DC offset from the sensor signal (Find Vref)RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt)
refS
FsenDC
S
F
orefosenoDC
vRRv
RR
vvv
1
–
+ +
vo
–
RF
Vref+–
v+
v–RSVsensor
ECEN 301 Discussion #18 – Operational Amplifiers 22
Op-Amps – Level ShifterExample1: design a level shifter such that it can remove
a 1.8V DC offset from the sensor signal (Find Vref)RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt)
V
RRRRvv
vRRv
RR
SF
SFsDCref
refS
FsDC
S
F
714.110/2201
10/220)8.1(
/1/
10
–
+ +
vo
–
RF
Vref+–
v+
v–RSVsensor
Since the desire is to remove all DC from the output we require:
ECEN 301 Discussion #18 – Operational Amplifiers 23
Op-Amps – Ideal IntegratorThe Ideal Integrator: the output signal is the integral
of the input signal (over a period of time)
–
+ +
vo(t)
–
i1
CF
RS
vS(t) +–
v+
v–iF(t)
iS(t)
The input signal is AC, but not necessarily sinusoidal
NB: Inverting amplifier setup with RF replaced with a capacitor
ECEN 301 Discussion #18 – Operational Amplifiers 24
Op-Amps – Ideal IntegratorThe Ideal Integrator: the output signal is the integral
of the input signal (over a period of time)
–
+ +
vo(t)
–
i1
CF
RS
vS(t) +–
v+
v–iF(t)
iS(t)
t
SFS
o
FS
So
oF
S
S
FS
dvCR
tv
CRtv
dttdv
dttvtvdC
Rtv
titi
)(1)(
)()(
)()()(
)()(
ECEN 301 Discussion #18 – Operational Amplifiers 25
Op-Amps – Ideal IntegratorExample2: find the output voltage if the input is a square wave of amplitude
+/–A with period TT = 10ms, CF = 1uF, RS = 10kΩ
–
+ +
vo(t)
–
i1
CF
RS
vS(t) +–
v+
v–iF(t)
iS(t)
time
vs(t)
A
-A
T/2 T
ECEN 301 Discussion #18 – Operational Amplifiers 26
Op-Amps – Ideal IntegratorExample2: find the output voltage if the input is a square wave of
amplitude +/–A with period TT = 10ms, CF = 1uF, RS = 10kΩ
–
+ +
vo(t)
–
i1
CF
RS
vS(t) +–
v+
v–iF(t)
iS(t)
t
SFS
o
t
SSFS
t
SFS
o
dvCR
v
dvdvCR
dvCR
tv
0
0
0
)(1)0(
)()(1
)(1)(
ECEN 301 Discussion #18 – Operational Amplifiers 27
Op-Amps – Ideal IntegratorExample2: find the output voltage if the input is a square wave of
amplitude +/–A with period TT = 10ms, CF = 1uF, RS = 10kΩ
–
+ +
vo(t)
–
i1
CF
RS
vS(t) +–
v+
v–iF(t)
iS(t)
At
A
Ad
dvCR
vtv
t
t
t
SFS
oo
100
100
1000
)(1)0()(
0
0
0
NB: since the vs(t) is periodic, we can find vo(t) over a single period – and repeat
2/0 Tt
ECEN 301 Discussion #18 – Operational Amplifiers 28
Op-Amps – Ideal IntegratorExample2: find the output voltage if the input is a square wave of
amplitude +/–A with period TT = 10ms, CF = 1uF, RS = 10kΩ
–
+ +
vo(t)
–
i1
CF
RS
vS(t) +–
v+
v–iF(t)
iS(t)
)(100
1002
100
)(1002
100
)(12
)(
2/
2/
2/
tTA
ATA
dATA
dvCR
Tvtv
t
T
t
T
t
T SFS
oo
NB: since the vs(t) is periodic, we can find vo(t) over a single period – and repeat
TtT 2/
ECEN 301 Discussion #18 – Operational Amplifiers 29
Op-Amps – Ideal IntegratorExample2: find the output voltage if the input is a square wave of
amplitude +/–A with period TT = 10ms, CF = 1uF, RS = 10kΩ
–
+ +
vo(t)
–
i1
CF
RS
vS(t) +–
v+
v–iF(t)
iS(t)
timevo(t)
T/2 T
-50AT
ECEN 301 Discussion #18 – Operational Amplifiers 30
Op-Amps – Ideal DifferentiatorThe Ideal Differentiator: the output signal is the
derivative of the input signal (over a period of time)
–
+ +
vo(t)
–
i1
CS
RF
vS(t) +–
v+
v–
iF(t)
iS(t)
The input signal is AC, but not necessarily sinusoidal
NB: Inverting amplifier setup with RS replaced with a capacitor
ECEN 301 Discussion #18 – Operational Amplifiers 31
Op-Amps – Ideal Differentiator
dttdvCRtv
Rtv
dttvtvdC
titi
SSFo
F
oSS
FS
)()(
)()()(
)()(
NB: this type of differentiator is rarely used in practice since it amplifies noise
The Ideal Differentiator: the output signal is the derivative of the input signal (over a period of time)
–
+ +
vo(t)
–
i1
CS
RF
vS(t) +–
v+
v–
iF(t)
iS(t)
ECEN 301 Discussion #18 – Operational Amplifiers 32
Op-Amps – Closed-Loop ModeCircuit Diagram ACL
Inverting Amplifier
Summing Amplifier
N
nSn
n
F
N
nSnOLno
vRR
vAv
1
1
–+ +
vo
–
+–
vS
RS
RF
SS
F
SCLo
vRRvAv
–+ +
vo
–
+–
+–
+–
RSn
RS2
RS1
vSn
vS2
vS1
RF
ECEN 301 Discussion #18 – Operational Amplifiers 33
Op-Amps – Closed-Loop ModeCircuit Diagram ACL
Noninverting Amplifier
Voltage Follower
s
sCLo
vvAv
SS
F
SCLo
vRR
vAv
1
–+ +
vo
–
+–
R
RS
RF
vS
–+ +
vo
–+–
ECEN 301 Discussion #18 – Operational Amplifiers 34
Op-Amps – Closed-Loop ModeCircuit Diagram ACL
Differential Amplifier
12 SSS
Fo vv
RRv
–+ +
vo
–+–
+–
RS
RS
RF
RF
v1
v2
ECEN 301 Discussion #18 – Operational Amplifiers 35
Op-Amps – Closed-Loop ModeCircuit Diagram ACL
Ideal Integrator
Ideal Differentiator
dttdvCRv S
SFo)(
t
SSF
o dvCR
v )(1
–
+ +vo(t)
–
+–
vS
CS
RF
–+ +
vo(t)–
+–
vS
RS
CF
ECEN 301 Discussion #18 – Operational Amplifiers 36
Op-Amps Example3: find an expression for the gain if vs(t) is sinusoidal
CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
+
–
vo(t)i1
CF
R2 v+
v–
iF(t)
i2(t)
R1
i1(t) CS
iS(t)
vs(t)
ECEN 301 Discussion #18 – Operational Amplifiers 37
Op-Amps
+
–
Vo(jω)Iin
ZF=1/jωCF
Z2 v+
v–
IF(jω)
I2(jω)
Z1
I1(jω)ZS
IS(jω)
Vs(jω)
Node a
Node b
31
621
21
631
111111
0
0:aat KCL
1221
21
21
Soa
SF
oF
a
a
F
aoaS
F
VjVjV
ZV
ZZV
ZZZV
ZVV
ZVV
ZVV
III
1. Transfer to frequency domain2. Apply KCL at nodes a and b
NB: v+ = v– = vo
and Iin = 0
Example3: find an expression for the gainCF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
ECEN 301 Discussion #18 – Operational Amplifiers 38
Op-Amps
+
–
Vo(jω)Iin
ZF=1/jωCF
Z2 v+
v–
IF(jω)
I2(jω)
Z1
I1(jω)ZS
IS(jω)
Vs(jω)
Node a
Node b
021
621
0111
000
:bat KCL
22
2
2
jVV
ZZV
ZV
ZV
ZVV
III
oa
Soa
S
ooa
inS
1. Transfer to frequency domain2. Apply KCL at nodes a and b
Example3: find an expression for the gainCF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
ECEN 301 Discussion #18 – Operational Amplifiers 39
Op-Amps
033 jVV oa
+
–
Vo(jω)Iin
ZF=1/jωCF
Z2 v+
v–
IF(jω)
I2(jω)
Z1
I1(jω)ZS
IS(jω)
Vs(jω)
1. Transfer to frequency domain2. Apply KCL at nodes a and b3. Express Vo in terms of Vs
Soa VjVjV 235
656
2
jVV S
o
Example3: find an expression for the gainCF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
ECEN 301 Discussion #18 – Operational Amplifiers 40
Op-Amps
+
–
Vo(jω)Iin
ZF=1/jωCF
Z2 v+
v–
IF(jω)
I2(jω)
Z1
I1(jω)ZS
IS(jω)
Vs(jω)
1. Transfer to frequency domain2. Apply KCL at nodes a and b3. Express Vo in terms of VS
4. Find the gain (Vo/VS)
656
2
jVV
S
o
Example3: find an expression for the gainCF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F