Operational Amplifiers
Lesson #5
Chapter 2
BME 372 Electronics I –
J.Schesser
138
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
input
Zero output impedance Inverting
v1
Infinite Bandwidth
v2
vo
+
Non-Inverting input
BME 372 Electronics I –
J.Schesser
139
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.
BME 372 Electronics I –
J.Schesser
140
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 (Summingpoint Constraint
3. Apply standard circuit analyses techniques
such as Kirchhoff’s Laws, Nodal or Mesh
Analysis to solve for the quantities of interest.
BME 372 Electronics I –
J.Schesser
141
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
R2
i2
i1=vin/R1
+
vin
0
R1
vi
-
-
+
+
+
vo
-
-
RL
vin  i1 R1  0 since vi is zero due to the summing - point constraint
i1  i2 due to the summing - point constraint
v0  i2 R2  0 since vi is zero
-
Z in 
vin i1 R1

 R1
i1
i1
R2
vin which is independent of RL (note that the output is opposite to the input : inverted)
R1
BME 372 Electronics I –
J.Schesser
142
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.
BME 372 Electronics I –
J.Schesser
143
Another Example: Inverting Amplifier
R4
R2
i2
i1=vin/R1
0
R1
i3
vi
vin
i4
R3
+
vo
RL
1.
Verify Negative Feedback:
2.
Use the summing point constraint.
3.
Use KVL at the inverting input node for the
branch connected to the source and KCL &
KVL at the node where the 3 resistors are
connected
vin  i1 R1  0 since vi is zero due to the summing - point constraint
i1  i2 due to the summing - point constraint
vi  i2 R2  i3 R3  0  i2 R2  i3 R3 since vi is zero
i4  i3  i2
vo   R4i4  R3i 3
BME 372 Electronics I –
J.Schesser
144
Another Example: Continued
R4
R2
i2
i1=vin/R1
0
R1
i3
vi
vin
i2 R2  i3 R3  i3 
i4
vin  i1 R1  0; i1  i2  i2 
R3
i4  i3  i2  (
+
vo
RL
Av 
vin  i1 R1  0
i2 R2  i3 R3
i4  i3  i2
vo   R4i4  R3i 3
vin
R1
1
R2
R
 1)i2  ( 2  )vin
R3
R3 R1 R1
vo   R4i4  R3i 3   R4 (
 vin (
i1  i2
R2
i2
R3
1
R2
R v
 )vin  R3 2 in
R3 R1 R1
R3 R1
R2 R4 R4 R2

 )
R3 R1 R1 R1
vo
RR R R
 ( 2 4  4  2 )
vin
R3 R1 R1 R1
Z in  Rin 
vin
 R1
i1
Why would we use this design over the simpler one?
1.
Same gain but with smaller values of resistance
2.
Higher gain
BME 372 Electronics I –
J.Schesser
145
Non-inverting Amp
1.
2.
3.
First check: negative feedback?
Next apply, summing point constraint
Use circuit analysis
vin  vi  v f  0  v f  v f
Iin= 0
+
vi
--
vin
R1
vf 
vo  vin ;
R1  R2
Av 
+
vf
--
vin

Since iin  0; Z in 
iin
io
+
-
R2
-+
vo R2  R1
R

 1 2
vin
R1
R1
+
RL
vo
R1
--
Note:
1.
The gain is always greater than one
2.
The output has the same sign as the
input
BME 372 Electronics I –
J.Schesser
146
Non-inverting Amp Special Case
What happens if R2 = 0?
Iin= 0
vin  vi  v f  0  v f  v f
vf 
Av 
R1
vo  vo  vin ;
R1  0
+
vi
+
io
+
-
RL
vo
--
vin
+
--
vo 0  R1

1
vin
R1
Since iin  0; Z in 
--
vin

iin
This is a unity gain amplifier and is also called a voltage follower.
BME 372 Electronics I –
J.Schesser
147
Some Practical Issues when Designing Op-amps
• Since ratios of resistor values determines the gain,
choosing the proper resistor values is crucial
– Too small means large currents drawn
– Too large yields another set of problems
• Open Loop Gain is not constant but a function of
frequency
• Non-linearities of the amplifier
– Voltage clipping
– Slew rate
• DC imperfections
– Offsets
– Bias Currents
BME 372 Electronics I –
J.Schesser
148
Selecting Resistor Values
• Let say we want a gain of 10. This
means that R2 = 9R1.
• If we chose R1=1Ω, then for a 10 volt
output, there will be 1 A flowing threw
R1 and R2.
vin
• THIS IS DANGEROUS!!!!
• On the other hand if R1=10 MΩ, then
there may be unwanted effects due to
pickup of induced signals
• Therefore choosing values between
100Ω and 1MΩ is optimum
BME 372 Electronics I –
J.Schesser
Iin= 0
+
vi
io
+
+
-
--
R2
+
vo
-+
vf
R1
--
Av  1 
--
R2
R1
149
Frequency Issues
• Let’s assume that the open loop gain of our Op-amp
is a function of frequency.
20 log |AOL( f )| dB
AOL ( f ) 
AoOL
1  j ( f / f BOL )
20 log |A0OL|
where AoOL is the open-loop gain for f  0,
f BOL is called the open-loop break
frequency since when f  f BOL
then AOL ( f )  AoOL / 2 or at the half power point.
Note that when f  AoOL f BOL , AOL ( f )  1
fBOL
This is will define an important relationship for the amplifier
when feedback is used
BME 372 Electronics I –
J.Schesser
150
Frequency Issues
AOL ( f ) 
+
+
AoOL
1  j ( f / f BOL )
Vi
Vin
R1
R1
 VO   VO ; where  
R1  R2
R1  R2
R2
+
-+
Vf
VO
AOL

Vin 1   AOL
(Note for the ideal Op-amp AOL  
and ACL 
1


R1  R2
R
 1 2
R1
R1
which what we expected.)
Vo
R1
--
V
Vin  V i   VO  O   VO ; since V O  V i AOL
AOL
 ACL 
io
+
-
--
Using phasors:
Vf 
Iin= 0
--
AoOL
1  j ( f / f BOL )
AOL ( f )
AOL ( f )


ACL ( f ) 
 AoOL
1   AOL ( f ) 1   AOL ( f ) 1 
1  j ( f / f BOL )
AoOL
AoOL
1  j ( f / f BOL )


 AoOL
1  j ( f / f BOL )
1   AoOL  j ( f / f BOL )

1  j ( f / f BOL ) 1  j ( f / f BOL )
AoOL
AoCL
1   AoOL


f
f
1  j(
) 1  j(
)
f BOL (1   AoOL )
f BCL
BME 372 Electronics I –
J.Schesser
151
Iin= 0
Frequency Issues
20 log |AOL( f )| dB
20 log |A0OL|
+
Vi
--
Vin
+
Vf
20 log |A0CL|
fBOL
R2
Vo
R1
--
fBOL(1+AoOL)
A interesting factor now comes to light:
AoCL f BCL 
--
io
+
-
-+
20 log |AOL( f )| dB
+
AoOL
f BOL (1   AoOL )  AoOL f BOL
1   AoOL
Where:
AoCL 
AoOL
and f BCL  f BOL (1   AoOL )
1   AoOL
The bandwidth product is constant!!!
BME 372 Electronics I –
J.Schesser
152
Example
• For an amplifier with Open-loop dc gain of 100
dB with Open-loop breakpoint of 40 Hz the
Bandwidth product for β=1, 0.1, 0.01
β
1
AOCL
0.9999
AOCL(dB) fBCL
0
4 MHz
0.1
9.9990
20
400 kHz
0.01
99.90
40
40 kHz
BME 372 Electronics I –
J.Schesser
153
20
10
0
0
5
10
15
20
Voltage clipping
Iin= 0
-10
+
Vi
-20
•
•
--
The Op-amp has a limit as to how
large an output voltage and current
can be produced. (See figure 2.28)
For example for the circuit shown it
has the following limitations: ±12 V
and ±25 mA
a)
b)
If the load resistor is 10k Ω, what is
the maximum input voltage which can
be handled without clipping
Repeat for 100 Ω
AvCL  1  3 / 1  4
a ) I o max 
12
12
Vo max
V
 o max  4 
RL
R1  R2 10 3000  1000
Vin
+
Vf
--
io
+
-
3k
-+
b) I o max 
+
Vo
1k
--
Vo max
V
12
12
 o max 

 123mA
RL
R1  R2 100 3000  1000
For this case, maximum output current will occur
before maximum output voltage is reached.
I o max 
V
Vo max
Vo max
V
 o max  25mA  o max 
100 3000  1000
RL
R1  R2
Vo max  2.44V
 4.2mA
For this case, maximum output voltage will occur Vin max  2.44 4  0.61V
before maximum output current is reached.
BME 372 Electronics I –
Vin max  12 4  3V
J.Schesser
154
Slew Rate
• Slew Rate is a phenomenon which occurs
when the Op-Amp can not keep up the
change in the input.
• Therefore, we identify the maximum rate of
change of the Op-amp as the Slew Rate SR
BME 372 Electronics I –
J.Schesser
155
DC imperfections
• We saw that we have to provide DC voltages to an
amplifier in order to provide it with the power to support
amplification.
– For a differential amplifier which must handle both positive and
negative voltages
• The process of designing this DC circuitry is called biasing
• As a result, biasing currents flow through the amplifier
which affects its performance.
• In particular, a voltage during to the biasing will appear at
the output without any input signal.
• These extra voltage can be due to:
– Bias currents flowing in the feedback circuitry
– Bias current differentials
– Voltage offsets due to the fact that the Op-amp circuitry is not
ideal
BME 372 Electronics I –
J.Schesser
156
Special Amplifiers
• Summer (Homework Problem)
• Instrumentation Amplifier
– Uses 3 Op-amps
– One as a differential amplifier
– Two Non-inverting Amps using for providing
gain
BME 372 Electronics I –
J.Schesser
157
Medical Instrumentation Amplifier
Non-inverting Amplifier
R5
+
-
R6
-
--
+
v2
R2
+
R1
R3
R1
vo
R4
R2
+
v1
Differential Amplifier
+
--
Non-inverting Amplifier
BME 372 Electronics I –
J.Schesser
158
Medical Instrumentation Amplifier
Non-inverting Amplifier
R5
+
-
v2D
R6
-
--
+
v2
R2
+
R1
v1D
R3
R1
vo
R4
R2
+
v1
Differential Amplifier
+
--
Non-inverting Amplifier
BME 372 Electronics I –
J.Schesser
159
Medical Instrumentation Amplifier
Differential Amplifier
R5
vx
v1D
+
vi=0
vy
R4
Chose
R5  R6 R4
1
R5 R3  R4
R4 R5  R4 R6  R5 R3  R5 R4
R4 R6  R5 R3
R6 R3

R5 R4
v
v2 D
1
1
 vx (  )  o
R6
R6 R5
R5
-
v2D R6
R3
v2 D  vx vx  vo

R6
R5
vo
R  R6
v
v2 D
 vx ( 5
) o
R6
R6 R5
R5
vy 
R4
v1D  vx  vi  vx
R3  R4
v
R  R6
v2 D
R4

v1D ( 5
) o
R6 R3  R4
R6 R5
R5
vo 
R5
R  R6 R4
 v2 D )
(v1D 5
R6
R5 R3  R4
Chose
vo 
R5  R6 R4
1
R5 R3  R4
R5
(v1D  v2 D )
R6
BME 372 Electronics I –
J.Schesser
160
Medical Instrumentation Amplifier
Non-inverting Amplifier
R5
+
-
R6
-
--
+
v2
R2
+
R1
R3
R1
vo
R4
R2
+
v1
Differential Amplifier
+
--
Non-inverting Amplifier
BME 372 Electronics I –
J.Schesser
161
Medical Instrumentation Amplifier
Non-inverting Amplifier
+
-
v2
R2
+
v2D
--
R1
vA
v1D
R1
R2
+
+
--
R2
1 1
v2 D  R2 (  )v2  v A
R2 R1
R1
R1  R2
R2
v2 D 
v2  v A
R1
R1
Likewise
-
v1
v2 D  v2 v2  v A

R2
R1
Non-inverting Amplifier
R1  R2
R2
v1 D 
v1  v A
R1
R1
BME 372 Electronics I –
J.Schesser
162
Medical Instrumentation Amplifier
Non-inverting Amplifier
R5
+
-
--
R2
R1
v2 D 
R1  R2
R
v2  2 v A
R1
R1
v1D 
R1  R2
R
v1  2 v A
R1
R1
vo 
R5 R1  R2
R
R  R2
R
v1  2 v A  ( 1
v2  2 v A )]
[
R6
R1
R1
R1
R1
vo 
R5 R1  R2
R
R
(
)(v1  v2 )  5 (1  2 )(v1  v2 )
R6
R1
R6
R1
vo
R3
-
R1
R4
+
R2
-
R5
(v1 D  v2 D )
R6
R6
+
+
-
v2
vo 
Differential Amplifier
+
v1
+
--
Non-inverting Amplifier
BME 372 Electronics I –
J.Schesser
163
Wheatstone Bridge as a sensor
R2
V
rD
-
rC
v2D R1
+
rB
rA
v1D
R3
vo
R4
BME 372 Electronics I –
J.Schesser
164
Wheatstone Bridge
V
V
rC
V2 r
B
V
rC
V2 r
B
V2
rA
rD
rB
rA
V1
V1
rD
rA
rC
rD
Using Thevinin's Theorem on the Wheatstone Bridge
Left side
r
rr
V2  B V;r2  B C
rB  rC
rB  rC
V1
+
V2
--
r2
Right side
r
rr
V1  A V;r1  A D
rA  rD
rA  rD
+
V1
r1
--
BME 372 Electronics I –
J.Schesser
165
Wheatstone Bridge
V
VBridge  V2  V1  (
rC
rD
V2 r
B
rA
V1
rB
r
 A )V
rB  rC rA  rD
When bridge is balanced
rB
r
r
r
 A  rB rD  rA rC  A  B
rB  rC rA  rD
rD rC
(nominally, rA  rB  rC  rD )
and VBridge  0
BME 372 Electronics I –
J.Schesser
166
Wheatstone Bridge
V2  v x v x  vo

R1  r2
R2
R2
v
V2
1
1
 vx (
 ) o
R1  r2 R2
R1  r2
R2
vx
--
R1
r2
+
V1
--
r1
+
V2
-
+
R  R2  r2
v
V2
 vx ( 1
) o
R1  r2
(R1  r2 )R2
R2
R3
vo
vx  v y 
R4
V
R3  r1  R4 1
V2
R4
R  R2  r2
v

V1 ( 1
) o
R1  r2 R3  r1  R4
(R1  r2 )R2
R2
R4
R4
R  R2  r2
V1
v
V2

( 1
)
 o
R1  r2 R3  r1  R4
R2
(R1  r2 ) R2
vo 
R2
R4
R  R2  r2
[(
)( 1
)V1 V2 ]
R1  r2 R3  r1  R4
R2
Chose
vo 
BME 372 Electronics I –
J.Schesser
R1  R2  r2
R4
1
R2
R3  r1  R4
R2
R2
r
r
(V1 V2 ) 
( A  B )V
R1  r2
R1  r2 rA  rD rB  rC
167
Wheatstone Bridge
R2
vx
--
R1
r2
r1
+
V1
+
V2
-
+
Chose
R3
vo
R4
--
R1  R2  r2
R4
1
R2
R3  r1  R4
R1  R2  r2 R3  r1  R4

R2
R4
R1  R2 
R2
rB rC
rB  rC

R3 
rArD
 R4
rA  rD
R4
Hard to deal with; then let's make sure that r2  R1  R2 and r1  R3  R4
R1  R2 R3  R4

 R2 R3  R2 R4  R4 R1  R4 R2  R2 R3  R4 R1
R2
R4
R4 R2

R3 R1
vo 
R2
r
r
( A  B )V
R1 rA  rD rB  rC
Note that the output depends on the unbalance of the bridge and when the bridge is balanced vo  0
Gain is still
R2
R1
BME 372 Electronics I –
J.Schesser
168
Integrators and Differentiators
C
i1
R
i2
i1 (t ) 
0
1t
1 t
vo    i2 ( x)dx  
 vin ( x)dx
C0
RC 0
-
vi
+
vin
vin (t )
 i2 (t )
R
vo
R
i1
C
i2
0
vi
vin
i1 (t ) 
+
Cdvin (t )
 i2 (t )
dt
vo
vo  i2 (t ) R   RC
BME 372 Electronics I –
J.Schesser
dvin (t )
dt
169
Frequency Analysis
i1=vin/Z1
Z1
Vin ( j )  I 1 ( j )Z1 ( j )  0 since vi is (virtually) zero
Z2
i2
0
vi
I 1 ( j )  I 2 ( j ) due to the summing - point constraint
-
Vo ( j )  I 2 ( j )Z 2  0 since vi is (virtually) zero
vo
+
vin
-
Z2
V ( j ) which is independent of Z L
Z1 in
Vo ( j )
Z
- 2
Vin ( j )
Z1
C
i1=vin/Z1
R
i2
0
vi
vin
R
i1=vin/Z1
C
+
vo
ZL
Vo ( j )
Z
 - 2   1 an integrator
Vin ( j )
Z1
jRC
i2
0
vi
vin
+
vo
ZL
Vo ( j )
Z
 - 2   jRC a differeniator
Vin ( j )
Z1
BME 372 Electronics I –
J.Schesser
170
Frequency Response
i1
R
i2
Vo ( j )
Z
1
- 2 
Vin ( j )
Z1
jRC
C
0
-
vi
+
vin
vo
ω
i1
C
i2
vin
R
0
vi
Vo ( j )
Z
 - 2   jRC
Vin ( j )
Z1
+
vo
ω
BME 372 Electronics I –
J.Schesser
171
Frequency Response
Vout Z 2
1
1
R
R

 2
 2
  tan 1 ( C1R1 )
2
Vin Z1
R1 (1  j C2 R2 )
R1 1 ( C1 R1 )
R2
i1
R1
i2
C2
0
2,500
2,000
-
vi
1,500
+
vin
vo
1,000
500
0
1.E+00
i2
i1
R1
vin
C1
vi
1.E+02
1.E+04
HZ
1.E+06
1.E+08
Z
R
R
Vout
j C1 R1
 C1R1

 2  2
 2
   tan 1 ( C1R1 )
2
Vin
Z1
R1 (1 j C1R1 )
R1 1  ( C1 R1 )
2
R2
0
R1 51 C1 10mf R2 100k
R1 51 C1 10mf R2 100k
2,500
+
vo
2,000
1,500
1,000
500
0
1.E+00
BME 372 Electronics I –
J.Schesser
1.E+02
1.E+04
HZ
1.E+06
1.E+08
172
Integrators and Differentiators
• Integrators and Differentiators are used in analog
computers
• An analog computer solves a differential equations
• By using Integrators and Differentiators one can
“program” a particular differential equation to be solved
• Usually only integrators are used since the gain of a
differentiator occurs at high frequencies while the opposite
is true for the integrator.
• Since the frequency response of an real Op-amp attenuates
high frequencies, using a differentiator conflicts with the
characteristics with a real Op-amp
• Also noise (high frequencies) are amplified by
differentiators
BME 372 Electronics I –
J.Schesser
173
R-Wave Detector
BME 372 Electronics I –
J.Schesser
174
Positive Feedback
• What happens to our Op-amp circuit if positive
feedback is used?
Using KCL at the input :
R
i2
i1
iin  0; since Amplifier has infinite input impedance
iin
R
vi
vin
i1  i2  iin  0
+
-
vo
RL
vi  vin vi  vo

0
R
R
vo  AOL vi where AOL is the open loop gain
vi 
1
1
(vin  vo )  (vin  AOL vi )
2
2
• Even if vin = 0 and there is a slight voltage at vi then
– vo = AOLvi will increase vi
– vo will grow even larger
– Eventually this will reach an extreme since there is not
infinite energy in the circuit
BME 372 Electronics I –
J.Schesser
175
Positive Feedback
• Then our positive feedback design will operate
between its positive and negative extremes, say ±5V
R
i2
i1
1
vi  (vin  vo )
2
For vo to be at  5V , vi  0
iin
R
vi
vin
+
vo
-
1
1
vi  (vin  vo )  (vin  5)  0
2
2
This is true as long as vin  -5V
Vo
For vo to be at  5V , vi  0
+5
1
1
vi  (vin  vo )  (vin  5)  0
2
2
This is true as long as vin  5V
+5
-5
RL
Vin
-5
BME 372 Electronics I –
J.Schesser
176
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 perform the plot
R2
C
vin
-
vi
R1 C
+
R2
vo
+
vin
-
vi
R1
R2
vo
R2
R1
C
vin
R1
vi
C
+
vo
vin
BME 372 Electronics I –
J.Schesser
vi
+
vo
177
Homework
rC =
R-ΔR
V
V2 r =
B
R+ΔR
rD=R+ΔR
V
rA = 1
R-ΔR
Wheatstone Bridge - Strain Gauge
•
•
A strain gauge shown here is used with a difference amplifier. Calculate the
amplifier output signal as a function of ΔR and difference amplifier resistors.
Assume that ΔR<<R.
For the strain gauge calculate the value of each of the difference amplifier
resistors for a value of R=10Ω and a percent change in R of ±10% if the
output of the amplifier is less than ±5 volts. Assume that the bridge is
powered with 5 volts.
BME 372 Electronics I –
J.Schesser
178
Homework
Calculate and plot the output vs frequency for
this circuits. R1=1k, R2=3k, C=1μf. Use
Matlab to perform the plot.
C
R1
R2
+
vi
vin
C
-
BME 372 Electronics I –
J.Schesser
vo
179