# ECE2262 Electric Circuits Chapter 4: Operational Amplifier (OP ```ECE2262 Electric Circuits
Chapter 4: Operational Amplifier (OP-AMP) Circuits
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4.1 Operational Amplifiers
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4. Voltages and currents in electrical circuits may represent
signals and circuits can perform mathematical operations on
these signals such as:
scaling (amplification)
!
signal follower (buffer)
!
inverting
!
differentiation and integration
Op- Amps provide solutions to all these problems
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An operational amplifier ("op-amp") is a DC-coupled high-gain electronic
voltage amplifier with a differential input and, usually, a single-ended output.
In this configuration, an op-amp produces an output potential (relative to circuit
ground) that is typically hundreds of thousands of times larger than the potential
difference between its input terminals.
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4.2 Op-Amp Terminals
!
i+
= VCC
i"
= "VCC
Terminals of Op Amp
•
•
•
•
•
inverting input ( ! )
noninverting input ( + )
output
positive power supply (V + = VCC )
negative power supply (V ! = !VCC )
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i+
v+
v!
v0
i!
Terminal Voltage Variables
Terminal Current Variables
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i+
v+
i!
v+
v!
v0
v!
The op amp with power supply terminals removed
When the amplifier is operating within its linear region, the dc voltages
±VCC do not enter into the circuit equations
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4.3 Op-Amp Characteristic and Circuit Models
The voltage transfer characteristic of an op amp
v0
v+ ! v!
#
!VCC if A ( v+ ! v! ) < !VCC
%%
v0 = \$ A ( v+ ! v! ) if ! VCC !A ( v+ ! v! ) " VCC
%
VCC if VCC < A ( v+ ! v! )
%&
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v0
v+ ! v!
• typical values for VCC ! 20 V
• the gain A is no less than 10, 000 !
• in the linear region: v+ ! v! " 20 /10 4 = 2mV
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v0
v+ ! v!
• in the linear region: v+ ! v! " 20 /10 4 = 2mV
• node voltages in the circuits we study are much larger than 2mV
• thus, if an op amp is constrained to its linear operation region we can
assume that
v+ = v!
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• An analysis of the op amp integrated circuit reveals that the equivalent
resistance ( Ri ) seen by the input terminals of the op amp is at least 1 M !
This implies that
i+ = i! = 0
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The model of ideal op amp
i+ = 0
v+ = v!
v+
i! = 0
v!
v+ = v!
v0
i+ = i! = 0
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Example (a) Find v0 if va = 1 V, vb = 2 V. Assume ideal op amp
•
!10 < 6 < 10
i!
i+
v!
v+
v0
• v+ = vb = 2V, since v+ = v! ! v! = vb = 2V
va ! v! 1! 2
1
v0 ! v! v0 ! 2
• i25 =
=
=
= ! mA • i100 =
mA
25k
25k
25
100k
100
• KCL at • : i25 + i100 ! i! = 0 ! i100
v0 ! 2 1
= !i25 !
=
! v0 = 6V
100
25
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(b) For va =1.5 specify the range of vb that avoids op amp saturation
• ii
!
+
v!
v+
Since i100 = !i25 !
v0
v0 ! vb
va ! vb
=!
va =1.5
100k
25k
1
vb = ( v0 + 6 )
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The op amp is in the linear region if !10 " v0 " 10 ! !0.8V " vb " 3.2V
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The linear (more realistic) circuit model of the op amp
i+
v+
i!
A (v + ! v ! )
v0
v!
In the linear region of operation this model represents a more realistic
approximation of an op amp. It includes:
(1) a finite input resistance Ri , (2) a finite open-loop gain A , (3) a nonzero
output resistance R0
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4.4 Voltage Follower and Noninverting Amplifier
A. Voltage Follower : Use the ideal op amp approximation
i+
i!
v!
v+
!VS + i+ RS + v+ = 0 ! v+ = VS ! v! = Vs ! negative feedback
! v0 = v!
!
v0 = VS
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• Negative Feedback
v0 = VS
The portion of the output voltage is applied to the inverting input.
The closed loop feedback greatly reduces the gain of the circuit, i.e.,
when negative feedback is used, the circuit overall gain is determined
mostly by the feedback network, rather than by the op-amp characteristic.
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• Ordinary circuits connection
RL
RL
V0 =
VS
RS + RL
Such a connection changes the behavior of the circuits, e.g., the voltage
RL
V0 =
VS is reduced
RS + RL
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• Connection of circuits through a voltage follower
V0 ! VS
RL
RL >> R0 / A
entire source voltage VS appears across the load resistance RL . Hence, RL does
not draw current from the source network (the current is supplied by the opamp).
Generally, the connection of one circuit to another through a voltage follower
allows both circuits to continue to operate as designed.
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B. The Noninverting Amplifier : Use the ideal op amp approximation
i!
i+
Rg
v!
v0
v+
• Since i+ = 0 ! v+ = vg ! v! = vg
Rs + R f
Rs
Rs
• VD: v! =
v 0 ! vg =
v0 ! v0 =
vg
Rs + R f
Rs + R f
Rs
Rf \$
!
v0 = # 1+ & vg
Rs %
"
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Rf \$
!
v0 = # 1 + & v g
Rs %
"
Operation in the linear region requires that
v0 ! VCC
R f VCC
! 1+
!
Rs
vg
Rf
A noninverting amplifier multiplies the input voltage vg by a gain 1+
Rs
that is independent of the source resistance Rg . Hence, the gain remains
unchanged when the circuits are terminated by an external load.
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Example Find the output voltage when Rx = 60k!
•
i!
i+
v
v!!
v+
! 1
v+
VS ! v+
1 \$ VS
• KCL at • : ! ! i+ +
= 0 ! v+ #
+ &=
! v+ = 320mV
Rx
15k
" 15k Rx % 15k
4.5k
67.5k
• VD: v! =
v0 ( v! = v+ ) ! v0 =
v+ = 4.8V
63k + 4.5k
4.5k
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4.5 Inverting Amplifier : assume an ideal op amp
•
i!
v! v+
• KCL at • : is + i f ! i! = 0
vs
v0 ! v! v0
• v+ = 0 ! v! = v+ = 0 ! is =
and i f =
=
Rs
Rf
Rf
• Since i! = 0 ! i f = !is !
Rf
v0 = ! vs
Rs
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Inverting-amplifier equation
Rf
v0 = ! vs
Rs
Rf
The upper limit on the gain
is determined by the power supply voltage VCC
Rs
and the value of the signal voltage vs :
v0 ! VCC
Rf
R f VCC
!
vs ! VCC !
!
Rs
Rs
vs
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If R f is removed, the negative feedback path is opened and the amplifier is
operating in open loop.
v!
Opening the feedback path drastically changes the behavior of the circuit.
It can be shown (use the realistic model of op-amp) that the output
voltage is
v0 = !A v!
Hence, we can call A the open-loop gain of the op amp.
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4.6 The Summing and Difference-Amplifier Circuits
A. The Summing-Amplifier Circuit
i!
•
v!
• KCL at the inverting terminal:
v! ! va v! ! vb v! ! vc v! ! v0
+
+
+
+ i! = 0
Ra
Rb
Rc
Rf
• Since v+ = 0 ! v! = 0 and i! = 0 ! solving the above w.r.t. v0 we obtain
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Rf
Rf %
" Rf
v0 = ! \$
va +
vb +
vc '
Rb
Rc &
# Ra
Inverting-summing amplifier equation
Rf
• If Ra = Rb = Rc = Rs ! v0 = ! ( va + vb + vc )
Rs
• The scaling factors in summing-amplifier circuits are entirely determined by
the external resistors: R f , Ra , Rb ,……, Rn .
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Example (a) Find v0 if va = 0.1 V, vb = 0.25 V.
v!
•
v+
v! ! va v! ! vb v! ! v0
+
+
= 0 ( i! = 0, v! = 0)
• KCL at v! :
5k
25k
250k
• v0 = ! {50va + 10vb } = -7.5 V
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(b) If vb = 0.25 V, how large can va be before the op-amp saturates ?
v!
•
v+
• From the previous in (a) solution we have v0 = ! {50va + 10vb }
• v0 = ! {50va + 2.5} ! !10 " v0 " 15 is satisfied if
!0.35V " va " 0.15V
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B. The Difference -Amplifier Circuit
•
i!
v!
v+
va ! v! v0 ! v!
va ! v! v0 ! v!
• KCL at v! :
+
! i! = 0 !
+
=0
Ra
Rb
Ra
Rb
v !v
v
v !v
v
v !v
v
• KCL at v+ : b + ! + ! i+ = 0 ! b + ! + = 0 ! b ! ! ! = 0
Rc
Rd
Rc
Rd
Rc
Rd
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" va ! v! v0 ! v!
\$\$ R + R = 0
a
b
#
\$ vb ! v! ! v! = 0
Rd
\$% Rc
Rb !
Ra \$ Rd
Rb
v0 = # 1+ &
vb ' va
Ra "
Rb % Rc + Rd
Ra
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If Rd = Rb and Rc = Ra !
Rb !
Ra \$ Rd
Rb
Rb !
Ra \$ Rb
Rb
v0 = # 1+ &
vb ' va =
1+ &
vb ' va
#
Ra "
Rb % Rc + Rd
Ra
Ra "
Rb % Ra + Rb
Ra
•
i!
Ra v!
v0 =
Rb v
+
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Rb
vb ! va )
(
Ra
•
i!
Ra v!
v0 =
Rb
vb ! va )
(
Ra
Rb v
+
The scaling is controlled by the external resistors. The relationship between the
output voltage and the input voltage is not affected by connecting a nonzero
load resistance across the output of the amplifier.
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Example
200 !
R
Since v! = v+ and i! = i+ = 0 then VD yields
R
v0 6k + R
v1 =
v0 ! Gain =
=
= 31 if R = 200 !
6k + R
v1
R
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Example
R1
R1 + R2
v! = v+ = Vin , VD: Vin =
V0 ! V0 =
Vin
R1 + R2
R1
V0
R
20k
= 1+ 2 = 1+
= 7.06
Vin
R1
3.3k
V0
14.12
V0 = Gain !Vin = 14.12 V ! I 0 =
=
= 606 µ A
23.3k 23.3k
! Gain =
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i! = 0
Example
V
V
RI
v2 ! V V
v2 ! V V
• KCL at v+ :
! ! i+ = 0 !
=
! V = R + R v2
I
2
R2
RI
R2
RI
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v1 ! V v0 ! V
v1 ! V V ! v0
+
! i! = 0 !
=
!
• KCL at v! :
R1
RF
R1
RF
"
RF
RF %
! v0 = !
v1 + \$ 1+ ' V
R1
R1 &
#
"
RF
RF % RI
v0 = ! v1 + \$ 1+ '
v2
R1
R1 & RI + R2
#
Note: If
RF RI
=
= ! ! v0 = ! ( v2 " v1 )
R1 R2
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RI
V=
v2
RI + R2
Summary: Signal Processing Circuits
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If
Rf
R1 R2
then Vout =
=
V2 ! V1 )
(
R f Rg
R1
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4.7 Problems ! Set 5
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