Give to Receive
Alma 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
1
Lecture 18 – Operational Amplifiers
Continue with Different OpAmp configurations
ECEN 301
Discussion #18 – Operational Amplifiers
2
Op-Amps – Open-Loop Model
1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ?
2.
How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
i1
–
+
vin
+
+
v–
v+
i1
–
+
i2
v–
io
–
Rin
+
vin
v+
–
+
Rout
+
vo
–
–
–
+
AOLvin–
+
vo
–
i2
NB: op-amps have near-infinite input resistance (Rin) and very small output resistance (Rout)
vo  AOL vin


 AOL (v  v )
ECEN 301
AOL – open-loop voltage gain
Discussion #17 – Operational Amplifiers
3
Op-Amps – Open-Loop Model
1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ?
2.
How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
i1
v–
vo  AOL vin
 AOL (v   v  )
vo
v 
 v
AOL

–
–
Ideally i1 = i2 = 0
(since Rin → ∞)
Rin
vin
v+
+
Rout
+
+
AOLvin–
+
i2
vo
–
What happens as AOL → ∞ ?
→ v– ≈ v+
ECEN 301
Discussion #17 – Operational Amplifiers
4
Op-Amps – Closed-Loop Mode
1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ?
2.
How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
RF
RS
iS
vS(t) +
–
v–
i S  i F
vS  v 
vo  v 

RS
RF
iF
–
i1
v+
+
+
vo
–
ECEN 301
vS   vo / AOL 
vo   vo / AOL 

 

RS 
RS
RF 
RF


vS
v
v
v
 S  o  o
RS
RS AOL RF AOL RS
 1
1
1 

vS  vo 


 RF / RS AOL RF / RS AOL 
Discussion #17 – Operational Amplifiers
5
Op-Amps – Closed-Loop Mode
1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ?
2.
How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
RF
RS
iS
vS(t)
+
–
v–
 1
1
1 


vS  vo 


 RF / RS AOL RF / RS AOL 
iF
–
NB: if AOL is very large these terms → 0
i1
v+
+
+
A CL 
vo
–
ECEN 301
Closed - LoopGain :
NB: if AOL is
NOT the same
thing as ACL
Discussion #17 – Operational Amplifiers
vo
vS

RF
RS
6
Op-Amps – Closed-Loop Mode
1.
How can v– ≈ v+ when vo is amplifying (v+ - v-) ?
2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
va
R
R
–
+
R1
R
–
+
vo
R2
vb
R
ECEN 301
R
–
+
NB: Current flows through R1 and R2
Discussion #17 – Operational Amplifiers
7
Op-Amps – Closed-Loop Mode
1.
How can v– ≈ v+ when vo is amplifying (v+ - v-) ?
2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
va
R
R
–
+
R
R1
–
+
vo
R2
vb
R
R
–
+
NB: Inverting amplifiers and (RS = RF)
→ vo = -vi
ECEN 301
Discussion #17 – Operational Amplifiers
8
Op-Amps – Closed-Loop Mode
1.
How can v– ≈ v+ when vo is amplifying (v+ - v-) ?
2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?
iF
R1
-va
i1
R2
-vb
i2
i1  i2  iF
R
–
+
vo
 va  v   vb  v 
vo  v 


R1
R2
R
 va  0  vb  0
v 0

 o
R1
R2
R
va vb vo
 
R1 R2 R
 va vb 
vo  R  
 R1 R2 
ECEN 301
Discussion #17 – Operational Amplifiers
9
More OpAmp Configurations
ECEN 301
Discussion #18 – Operational Amplifiers
10
Op-Amps – Closed-Loop Mode
The Differential Amplifier: the signal to be
amplified is the difference of two signals
RF
iF
RS
iS
vS1(t) +
–
vS2(t) +
–
ECEN 301
RS
v–
–
i1
v+ +
+
i1
RF
vo
–
Discussion #18 – Operational Amplifiers
11
Op-Amps – Closed-Loop Mode
The Differential Amplifier: the signal to be
amplified is the difference of two signals
RF
NB: an ideal op-amp with negative
feedback has the properties
iF
RS
+
–
iS
vS1(t)
vS2(t) +
–
RS
v–
–

i1
v+ +
+
i2
RF
vo

v v
i1  i2  0
–
iF  iS
ECEN 301
Discussion #18 – Operational Amplifiers
12
Op-Amps – Closed-Loop Mode
The Differential Amplifier: the signal to be
amplified is the difference of two signals
RF
Voltage Divider :
RF

v  vs 2
RF  RS
iF
RS
+
–
iS
vS1(t)
vS2(t) +
–
ECEN 301
RS
v–
–
i1
v+ +
+
i2
RF
vo
–
v   v   vS 1  iS RS
vS 1  v 
iS 
RS
Discussion #18 – Operational Amplifiers
vo  v 
iF 
RF
13
Op-Amps – Closed-Loop Mode
The Differential Amplifier: the signal to be
amplified is the difference of two signals
RF
vo  iF RF  v 
iF
RS
+
–
iS
vS1(t)
vS2(t) +
–
ECEN 301
RS
v–
 iS RF  v 
–
i1
v+ +
+
i2
RF
vo
–
  vS 1
vS 2
vS 2 RF 
vo  RF 





R
R

R
R
R

R
S
F
S
S
F 
 S
RF
vS 2  vS1 

RS
Discussion #18 – Operational Amplifiers
14
Op-Amps – Level Shifter
Level Shifter: can add or subtract a DC offset from a
signal based on the values of RS and/or Vref
RF
Vsensor
RS
AC voltage with DC offset
v–
–
v+ +
Vref
+
–
DC voltage
ECEN 301
Discussion #18 – Operational Amplifiers
+
vo
–
15
Op-Amps – Level Shifter
Example1: 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)
RF
Vsensor
RS
v–
–
v+ +
Vref
ECEN 301
+
–
+
vo
–
Discussion #18 – Operational Amplifiers
16
Op-Amps – Level Shifter
Example1: 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)
RF
Vsensor
RS
–
v+ +
Vref
ECEN 301
+
–
Find the Closed-Loop voltage gain by
using the principle of superposition on
each of the DC voltages
v–
+
vo
–
Discussion #18 – Operational Amplifiers
17
Op-Amps – Level Shifter
Example1: 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)
RF
RS
Vsensor +
–
DC from sensor:
Inverting amplifier
v–
–
v+ +
+
vo
–
ECEN 301
vosen  ACL vsenDC
RF

vsenDC
RS
Discussion #18 – Operational Amplifiers
18
Op-Amps – Level Shifter
Example1: 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)
RF
DC from reference:
Noninverting amplifier
v–
RS
–
v+ +
Vref
+
–
+
vo
–
ECEN 301
voref  ACL vref
 RF
 1 
 RS
Discussion #18 – Operational Amplifiers

vref

19
Op-Amps – Level Shifter
Example1: 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)
RF
Vsensor
RS
v–
v+ +
Vref
ECEN 301
+
–
voDC  vosen  voref
–
+
vo
 RF 
RF
vref

vsenDC  1 
RS
 RS 
–
Discussion #18 – Operational Amplifiers
20
Op-Amps – Level Shifter
Example1: 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)
RF
Vsensor
RS
v–
–
v+ +
Vref
ECEN 301
+
–
Since the desire is to remove all
DC from the output we require:
 R
RF
vsDC  1  F
RS
 RS
RF / RS
 vsDC
1  RF / RS
220/ 10
 (1.8)
1  220/ 10
 1.714V
0
+
vo
–
vref
Discussion #18 – Operational Amplifiers

vref

21
Op-Amps – Ideal Integrator
The Ideal Integrator: the output signal is the integral
of the input signal (over a period of time)
CF
The input signal is AC, but
not necessarily sinusoidal
RS
iS(t)
vS(t)
+
–
iF(t)
v–
–
i1
v+
+
+
vo(t)
–
ECEN 301
Discussion #18 – Operational Amplifiers
NB: Inverting
amplifier setup
with RF replaced
with a capacitor
22
Op-Amps – Ideal Integrator
The Ideal Integrator: the output signal is the integral
of the input signal (over a period of time)
CF
iS (t )  iF (t )
RS
iS(t)
vS(t) +
iF(t)
v–
–
i1
v+

vS (t )
d vo (t )  v  (t )
 C F
RS
dt
+
+
–
vo(t)
–

dvo (t )
v (t )
 S
dt
RS C F
1
vo (t )  
RS C F

t

vS ( )d
dv (t )
Recall i(t )  C dt
ECEN 301
Discussion #18 – Operational Amplifiers
23
Op-Amps – Ideal Integrator
Example2: find the output voltage if the input is a square wave of amplitude
+/–A with period T
T = 10ms, CF = 1uF, RS = 10kΩ
CF
iS(t)
vS(t) +
–
v–
–
vs(t)
RS
A
iF(t)
i1
v+
+
T/2
+
vo(t)
-A
–
ECEN 301
T
Discussion #18 – Operational Amplifiers
time
24
Op-Amps – Ideal Integrator
Example2: find the output voltage if the input is a square wave of amplitude
+/–A with period T
T = 10ms, CF = 1uF, RS = 10kΩ
CF
RS
iS(t)
vS(t) +
–
iF(t)
v–
1
vo (t )  
RS CF
–
i1
v+
+
+
vo(t)
–
ECEN 301

t

vS ( )d
t
1  0


v
(

)
d


v
(

)
d

S
S
0

RS CF  
1
 vo (0) 
RS CF
Discussion #18 – Operational Amplifiers

0
t
vS ( )d
25
Op-Amps – Ideal Integrator
Example2: find the output voltage if the input is a square wave of amplitude
+/–A with period T
T = 10ms, CF = 1uF, RS = 10kΩ
NB: since the vs(t) is periodic, we can find
vo(t) over a single period – and repeat
CF
RS
iS(t)
vS(t) +
–
0t T /2
iF(t)
v–
–
i1
v+
+
+
1
vo (t )  vo (0) 
RS C F

0
t
vS ( )d
t
vo(t)
–
 0  100 Ad
0
  100A
t
0
 100At
ECEN 301
Discussion #18 – Operational Amplifiers
26
Op-Amps – Ideal Integrator
Example2: find the output voltage if the input is a square wave of amplitude
+/–A with period T
T = 10ms, CF = 1uF, RS = 10kΩ
NB: since the vs(t) is periodic, we can find
vo(t) over a single period – and repeat
CF
RS
iS(t)
vS(t) +
–
T /2t T
iF(t)
v–
–
i1
v+
+
+
vo(t)
–
ECEN 301
1  t
T 

vo (t )  vo   
v
(

)
d

S


 2  RS C F  T / 2
t
T
 100A  100 ( A)d
T /2
2
T
t
 100A  100A T / 2
2
 100A(T  t )
Discussion #18 – Operational Amplifiers
27
Op-Amps – Ideal Integrator
Example2: find the output voltage if the input is a square wave of amplitude
+/–A with period T
T = 10ms, CF = 1uF, RS = 10kΩ
CF
T/2
iS(t)
vS(t) +
–
iF(t)
v–
–
i1
vo(t)
RS
T
v+
+
+
vo(t)
-50AT
–
time
ECEN 301
Discussion #18 – Operational Amplifiers
28
Op-Amps – Ideal Differentiator
The Ideal Differentiator: the output signal is the
derivative of the input signal (over a period of time)
RF
The input signal is AC, but
not necessarily sinusoidal
iF(t)
v–
CS
–
iS(t)
vS(t)
+
–
i1
v+
+
+
vo(t)
–
ECEN 301
Discussion #18 – Operational Amplifiers
NB: Inverting
amplifier setup
with RS replaced
with a capacitor
29
Op-Amps – Ideal Differentiator
The Ideal Differentiator: the output signal is the
derivative of the input signal (over a period of time)
RF

iF(t)
–
iS(t)
vS(t) +
i1
v+

d vS (t )  v  (t )
vo (t )
CS

dt
RF
v–
CS
iS (t )  iF (t )
+
+
–
vo(t)
–
dv (t )
dvS (t )
vo (t )   RF CS
dt
NB: this type of differentiator is rarely
used in practice since it amplifies noise
Recall i(t )  C dt
ECEN 301
Discussion #18 – Operational Amplifiers
30
Op-Amps – Closed-Loop Mode
Circuit Diagram
RS
Inverting
Amplifier
+
–
vS
RS1
Summing
Amplifier
+
–
vS1
RS2
+
–
vS2
vo  ACL vS
RF
–
+
–
+
ACL
RF

vS
RS
+
vo
–
N
RF
+
vo
–
vo   AOLn vSn
RSn
+
–
ECEN 301
vSn
Discussion #18 – Operational Amplifiers
n 1
N
RF
 
vSn
n 1 Rn
31
Op-Amps – Closed-Loop Mode
Circuit Diagram
Noninverting
Amplifier
Voltage
Follower
ECEN 301
–
+
R
vS
+
–
–
+
+
–
vo  ACL vS
RF
RS
ACL
 RF 
vS
 1 
 RS 
+
vo
–
+
vo
–
vo  ACL vs
Discussion #18 – Operational Amplifiers
 vs
32
Op-Amps – Closed-Loop Mode
Circuit Diagram
RF
RS
Differential
Amplifier
ECEN 301
+
–
–
+
RS
v1
+
–
v2
RF
+
vo
–
ACL
RF
vS 2  vS1 
vo 
RS
Discussion #18 – Operational Amplifiers
33
Op-Amps – Closed-Loop Mode
Circuit Diagram
RS
Ideal
Integrator
+
–
vS
CF
–
+
+
vo(t)
–
ACL
1
vo  
RF CS

t

vS ( )d
RF
CS
–
Ideal
Differentiator
ECEN 301
+
–
vS
+
+
vo(t)
–
dv S (t )
vo   RF CS
dt
Discussion #18 – Operational Amplifiers
34
Op-Amps
Example 3: find an expression for the gain if vs(t) is sinusoidal
CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
CF
vs(t)
R1
R2
i1(t)
i2(t)
iS(t)
ECEN 301
iF(t)
v+
+
vo(t)
i1
CS
v–
–
Discussion #18 – Operational Amplifiers
35
Op-Amps
Example 3: find an expression for the gain
CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
ZF=1/jωCF
1.
2.
Node b
Vs(jω) Z1
I1(jω)
Node a
IF(jω)
Z2
I2(jω)
v+
Iin
ZS
v–
IS(jω)
NB: v+ = v– = vo
and Iin = 0
ECEN 301
+
–
Transfer to frequency domain
Apply KCL at nodes a and b
KCL at a :
I1  I F  I 2  0
Vo(jω)
VS  Va Vo  Va Va  V 


0
Z1
ZF
Z2
1
 1
1
1
1 
1 
  VS  
Va  
   Vo  
 Z1 Z F Z 2 
 Z2 ZF 
 Z1 
 1 j 1 
 1 j 
1
Va  
   Vo  
  VS  
3 6 2
2 6 
3
Discussion #18 – Operational Amplifiers
36
Op-Amps
Example 3: find an expression for the gain
CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
ZF=1/jωCF
1.
2.
Node b
Vs(jω) Z1
I1(jω)
Node a
ECEN 301
IF(jω)
Z2
I2(jω)
v+
+
Iin
ZS
IS(jω)
v–
–
Transfer to frequency domain
Apply KCL at nodes a and b
KCL at b :
I 2  I S  I in  0
Vo(jω)
Va  Vo Vo  0

0
Z2
ZS
 1
 1 
1 
Va    Vo 
   0
 Z2 
 ZS Z2 
1
 j 1 
Va    Vo 
 0
2
 6 2
Discussion #18 – Operational Amplifiers
37
Op-Amps
Example 3: find an expression for the gain
CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
ZF=1/jωCF
Vs(jω) Z1
I1(jω)
IF(jω)
Z2
I2(jω)
v+
+
Iin
ZS
IS(jω)
v–
–
1.
2.
3.
Vo(jω)
Transfer to frequency domain
Apply KCL at nodes a and b
Express Vo in terms of Vs
Va 5  j   Vo 3  j   2VS
3Va  Vo 3  j   0
6VS
Vo   2
  j5  6
ECEN 301
Discussion #18 – Operational Amplifiers
38
Op-Amps
Example 3: find an expression for the gain
CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
ZF=1/jωCF
Vs(jω) Z1
I1(jω)
IF(jω)
Z2
I2(jω)
+
Vo(jω)
Transfer to frequency domain
Apply KCL at nodes a and b
Express Vo in terms of VS
Find the gain (Vo/VS)
Iin
ZS
IS(jω)
ECEN 301
v+
1.
2.
3.
4.
v–
–
Vo
6
 2
VS
  j5  6
Discussion #18 – Operational Amplifiers
39
Op-Amps
Example 3: find an expression for the gain
CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
Frequency Response
ZF=1/jωCF
Vs(jω) Z1
I1(jω)
IF(jω)
Z2
I2(jω)
v+
+
Vo(jω)
Iin
ZS
IS(jω)
v–
–
2nd order
Lowpass filter
ECEN 301
Discussion #18 – Operational Amplifiers
40
Instrumentation Op-Amp
• Special configuration used for common-mode
voltage rejection

2R 
(V2  V1 )
Vout  1 
Rgain


Advantages:
- Common-mode voltage rejection
- Very high input impedance for V1 and V2
- Set gain with one resistor
Connect sensor with twist pair in
differential configuration to minimize
external noise pickup. Using grounded
shield also helps.