III. Operational Amplifiers Amplifiers are two-port networks in which the output voltage or current is directly propor- tional to either input voltage or current. Four different kinds of amplifiers exit: Voltage amplifier: A v = V o /V i = constant Current amplifier: A i = I o /I i = consant Transconductance amplifier: G m = I o /V i = constant Transresistance amplifier: R m = V o /I i = constant Our focus in this course is on voltage amplifiers (we also see a transconductance amplifier). i V - + I i o I V o + i A V 0 o i - - + Voltage Amplifier Model R R Voltage amplifiers can be accurately modeled with three circuit elements as shown below. These circuit elements are related to transfer functions of two-port networks discussed before. R i and R o are, respectively, input and output resistances. Gain, A 0 , is the open- loop transfer function, H vo (jω). (For amplifiers the convention is to use A instead of H (jω).) A good voltage amplifier has a large input resistance, R i , and a small output resistance, R o . An ideal voltage amplifier has, R i →∞ and R o → 0. 3.1 Feedback Not only a good amplifier should have sufficient gain, its performance should be insensitive to environmental and manufacturing conditions, should have a large R i , a small R o , a sufficiently large bandwidth, etc. It is easy to make an amplifier with a very large gain. A typical transistor circuit can easily have a gain of 100 or more. A three-stage transistor amplifier can easily get gains of 10 6 . Other characteristics of a good amplifier are hard to achieve. For example, the β of a BJT changes with operating temperature making the gain of the three-stage amplifier vary widely. The circuit can be made to be insensitive to environmental and manufacturing conditions by the use of feedback. Principle of feedback: The input to the circuit is modified by “feeding” a signal propor- tional to the output value “back” to the input. There are two types of feedback (remember the example of a car in the freeway discussed in the class): 1. Negative feedback: As the output is increased, the input signal is decreased and vice versa. Negative feedback stabilizes the output to the desired level. Linear system employs negative feedback. 2. Positive feedback: As the output is increased, the input signal is increased and vice versa. The output of an amplifier with a positive feedback is always at its limit (saturation voltages). (Positive feedback has its uses!) ECE65 Lecture Notes (F. Najmabadi), Winter 2008 45
34
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III. Operational Amplifiers
Amplifiers are two-port networks in which the output voltage or current is directly propor-
tional to either input voltage or current. Four different kinds of amplifiers exit:
Voltage amplifier: Av = Vo/Vi = constant
Current amplifier: Ai = Io/Ii = consant
Transconductance amplifier: Gm = Io/Vi = constant
Transresistance amplifier: Rm = Vo/Ii = constant
Our focus in this course is on voltage amplifiers (we also see a transconductance amplifier).
iV−
+
I i oI
Vo
+
iA V0
o
i
− −
+
Voltage Amplifier Model
R
R
Voltage amplifiers can be accurately modeled with
three circuit elements as shown below. These circuit
elements are related to transfer functions of two-port
networks discussed before. Ri and Ro are, respectively,
input and output resistances. Gain, A0, is the open-
loop transfer function, Hvo(jω). (For amplifiers the
convention is to use A instead of H(jω).)
A good voltage amplifier has a large input resistance, Ri, and a small output resistance, Ro.
An ideal voltage amplifier has, Ri → ∞ and Ro → 0.
3.1 Feedback
Not only a good amplifier should have sufficient gain, its performance should be insensitive to
environmental and manufacturing conditions, should have a large Ri, a small Ro, a sufficiently
large bandwidth, etc. It is easy to make an amplifier with a very large gain. A typical
transistor circuit can easily have a gain of 100 or more. A three-stage transistor amplifier
can easily get gains of 106. Other characteristics of a good amplifier are hard to achieve.
For example, the β of a BJT changes with operating temperature making the gain of the
three-stage amplifier vary widely. The circuit can be made to be insensitive to environmental
and manufacturing conditions by the use of feedback.
Principle of feedback: The input to the circuit is modified by “feeding” a signal propor-
tional to the output value “back” to the input. There are two types of feedback (remember
the example of a car in the freeway discussed in the class):
1. Negative feedback: As the output is increased, the input signal is decreased and vice
versa. Negative feedback stabilizes the output to the desired level. Linear system employs
negative feedback.
2. Positive feedback: As the output is increased, the input signal is increased and vice
versa. The output of an amplifier with a positive feedback is always at its limit (saturation
voltages). (Positive feedback has its uses!)
ECE65 Lecture Notes (F. Najmabadi), Winter 2008 45
For feedback to work well, the amplifier gain should be large. In this limit, the response of
the overall circuit is set by the feedback loop (and not by the gain of amplifier). As such, a
voltage amplifier can be made into many other useful circuits such as active filter, integrator,
etc. We will first explore the concept of feedback through operational amplifier circuits.
3.2 Operational Amplifiers
Operational amplifiers (OpAmps) are general purpose voltage amplifiers employed in a va-
riety of circuits. OpAmps are “DC” amplifiers with a very large gain, A0 (105 to 106), high
input impedance (> 10 − 100 MΩ), and low output resistance (< 100 Ω). They are con-
structed as a “difference” amplifier, i.e., the output signal is proportional to the difference
between the two input signals.
s+V
−Vs
Vd
VoVp
Vn
−+−
+
−+ +
−
+
−
vo = A0vd = A0(vp − vn)
+ and − terminals of the OpAmp are called, respectively, non-
inverting and inverting terminals. vs and −vs are power supply
attachments. An OpAmp chip should be powered for it to work,
i.e., power supply attachments are necessary. These connections,
however, are not usually shown in the circuit diagram.
3.2.1 OpAmp Models
i
o
−+
p
n
dd0
o
+
+
−+
−
Linear Model
RR
v
v
vA v
v−+
p
n
d
p
n
d0
o
+
+
−+
−
Ideal Model
v
v
v
i
i
A v
v
Because Ri is very large and Ro is very small, ideal model of the OpAmp assumes Ri → ∞and Ro → 0. Ideal model is usually a very good model for OpAmp circuits. Very large input
resistance also means that the input current into an OpAmp is very small:
First Golden Rule of OpAmps: ip ≈ in ≈ 0 (Also called “Virtual Open Principle”)
As discussed before, concepts of very large and very small employed above require a frame of
reference. A “rule of thumb” for ignoring Ri and Ro and employing the ideal model for the
OpAmp (and using the first golden rule) is to ensure that all impedances connected to the
OpAmp circuit are much smaller than Ri and much larger than Ro (for a “typical” OpAmp,
all impedances should between 1 kΩ and 1 MΩ)
ECE65 Lecture Notes (F. Najmabadi), Winter 2008 46
Another important feature of OpAmp is that the OpAmp will be in saturation without
negative feedback because its gain is very high. For example, take an OpAmp with a gain
of 105 and vsat = 15 V. Then, for OpAmp to be in linear region, vi ≤ 15× 10−5 = 150 µV (a
very small value). As such, OpAmps are rarely used by themselves. They are always part
of a circuit which employ either negative feedback (e.g., linear amplifiers, active filters) or
positive feedback (e.g., comparators). Examples below shows several OpAmp circuits with
negative feedback.
3.2.2 Inverting Amplifier
Li
i
o
1
2
Rv
i
v
R
R
++
−−
+
−
1
L
i
o
d
2
n
p
−
d0
−+
+
−
−
+
R
R
v
v
R
v
v
v
A v
The first step in solving OpAmp circuits is to replace the OpAmp with its circuit model
(ideal model is usually very good).
vp = 0, vo = A0vd = A0(vp − vn) = −A0vn
Using node-voltage method and noting in ≈ 0:
vn − vi
R1
+vn − vo
R2
= 0
Substituting for vn = −vo/A0 and multiplying the equation by R2, we have:
− R2
A0R1
vo −R2
R1
vi −vo
A0
− vo = 0 → vo
[
1 +1
A0
+R2
A0R1
]
= −R2
R1
vi
vo
vi
= − R2/R1
1 +1
A0
(
1 +R2
R1
)
If R2 and R1 are chosen such that their ratio is not very large, 1 + R2/R1 ≪ A0, then the
voltage transfer function of the OpAmp is
vo
vi
≈ − R2
R1
ECE65 Lecture Notes (F. Najmabadi), Winter 2008 47
This circuit is called an inverting amplifier because the voltage transfer function is “nega-
tive.” (A “negative” sinusoidal function looks inverted.) The negative sign actually means
that there is 180 phase shift between input and output signals.
Note that the voltage transfer function of the circuit is “independent” of the OpAmp gain,
A0, and is set by the values of the resistors R1 and R2. While A0 is quite sensitive to
environmental and manufacturing conditions (can very by a factor of 10 to 100), the resistor
values are quite insensitive and, thus, the gain of the system is quite stable.
2
1
i o
Feedback CircuitR
R
v v+
+
−
−
+
−
This stability is achieved by a negative feedback
(see figure). The output voltage is sampled via R2
and is applied in parallel to the input signal (via
R1) to the input terminals of the OpAmp (called
“parallel-parallel” feedback). If vo increases, this re-
sistor forces vn to increase, reducing vd = vp−vn and
vo = A0vd, and stabilizes the OpAmp output.
This is a negative feedback as vo is connected to the inverting terminal of OpAmp. Also, it
is obvious that R1 is needed, otherwise feedback would not work as vn = vi is a fixed value
and the input to OpAmp chip would not change when vo changes.
Feedback can be applied in a different manner. The output voltage is sampled and is applied
in series with vi to the input terminals of the OpAmp (called “parallel-series” feedback, i.e.,
parallel to the output, in series with in the input). We explore this feedback configuration
below.
3.2.3 Non-inverting Amplifier
i
21
i o
I
RR
v v+
−
⇒2
1
i
o
+
d
+
−
n
p
d0
−
p
−+
R
R
v
v
v
v
v
A v
i
21
i
i
o
FeedbackCircuit
RR
v
i
v
+
− ++
− −
The left circuit above is a non-inverting amplifier. It employs a “parallel-series” feedback
(see figure to the right above) as the output is sampled through a combination of R2 and R1
resistors and is applied in series to vi to the input terminals of the OpAmp. Furthermore,
as the output vo is connected to the inverting terminal, it is a negative feedback.
ECE65 Lecture Notes (F. Najmabadi), Winter 2008 48
Replacing the OpAmp with its ideal model (middle circuit above) and noting that in ≈ 0
and ip ≈ 0, we get:
vp = vi
Voltage divider: vn = vo
R1
R1 + R2
vd = vp − vn = vi − vo
R1
R1 + R2
Substituting for vd = vo/A0 in the last equation above, we get:
vo
A0
= vi − vo
R1
R1 + R2
→ vo
(
1
A0
+R1
R1 + R2
)
= vi
vo
vi
=1
1
A0
+R1
R1 + R2
=R1 + R2
R1
× 1
1 +R1 + R2
A0R1
=(
1 +R2
R1
)
× 1
1 +1
A0
(
1 +R2
R1
)
If R2 and R1 are chosen such that their ratio is not very large, 1 + R2/R1 ≪ A0, then the
voltage transfer function of the OpAmp is
vo
vi
= 1 +R2
R1
This circuit is called a non-inverting amplifier because its voltage transfer function is “posi-
tive.” (as opposed to the inverting amplifier we a negative voltage transfer function.) Note
that the voltage transfer function is “independent” of the OpAmp gain, A0, and is only set
by the values of the resistors R1 and R2.
3.2.4 OpAmp Circuits with Negative Feedback
An important feature of OpAmp circuits with negative feedback is that because the OpAmp
is NOT saturated, vd = vo/A0 is very small (because A0 is very large). As a result,
Negative Feedback → vd ≈ 0 → vn ≈ vp
Second Golden Rule of OpAmps: For OpAmps circuits with negative feedback, the
OpAmp adjusts its output voltage such that vd ≈ 0 or vn ≈ vp (also called “virtual Short
Principle”). This rule is derived by assuming A → ∞. Thus, vo cannot be found from
ECE65 Lecture Notes (F. Najmabadi), Winter 2008 49
vo = A0vd = ∞ × 0 = indefinite value. The virtual short principle replace vo = A0vd
expression with vd ≈ 0.
The second golden rule of OpAmps allows us to solve OpAmp circuits in a much simpler
manner following the “recipe” below.
Recipe for solving OpAmp circuits:
1) Replace the OpAmp with its circuit model.
2) Check for negative feedback, if so, vp ≈ vn.
3) Solve. Best method is usually node-voltage method. You can solve simple circuits with
KVL and KCLs. Do not use mesh-current method.
4) Find problem unknowns in term of node voltages.
Li
i
o
1
2
Rv
i
v
R
R
++
−−
+
−
For example, for the inverting amplifier we will have:
Negative Feedback → vn ≈ vp ≈ 0
vn − vi
R1
+vn − vo
R2
= 0 → vi
R1
+vo
R2
= 0
Av =vo
vi
= − R2
R1
Note that you should not write a node equation at OpAmp output as its a node attached to
a voltage source. The value of vo is Avd and is indefinite. Instead of using this equation, we
used vd ≈ 0.
Input and Output resistances of inverting amplifier configuration can now be found.
From the circuit,
ii =vi − 0
R1
→ Ri =vi
ii= R1
The input impedance of the inverting amplifier circuit is R1 (although input impedance of
OpAmp is very large).
The voltage transfer function for this circuit, Av (= Hv) is independent of RL, i.e., it
would be the same in the limit of RL → ∞. Thus, Hv = Hv0 for this circuit. Since
Hv = [ZL/(Zo + ZL)]Hv0 (See page 13), condition of Hv = Hv0 leads to ZL = RL = 0.
ECE65 Lecture Notes (F. Najmabadi), Winter 2008 50
i
21
i o
I
RR
v v+
−
Similarly, for the non-inverting amplifier circuit above,
we will have:
vp = vi
Negative Feedback: vn ≈ vp = vi
vn − 0
R1
+vn − vo
R2
= 0
Note that again we do not write a node equation at OpAmp output as its a node attached
to a voltage source. Substituting for vn = vi, we get
R2
R1
vi + vi − vo = 0 ⇒ vo
vi
= 1 +R2
R1
Input Resistance: ii = ip = 0. Therefore, Ri → ∞.
Output Resistance: Av is independent of RL, so Ro = 0.
Note that Ri → ∞ and Ro = 0 should be taken in the context that we are using an “ideal”
OpAmp model. In reality, the above circuit will have input and output resistances equal to
that of the OpAmp itself.
3.3 OpAmps as linear amplifiers
3.3.1 Voltage Follower
In some cases, we have two-terminal networks which do not match well, i.e, the input
impedance of the later stage is not large enough, or the output impedance of preceding stage
is not low enough. A “buffer” circuit can be used in between these two circuits to solve the
matching problem. A “buffer” circuits has a gain of 1 but has a large input impedance and a
small output impedance. Because the gain of buffer is 1, it is also called a “voltage follower.”
io
+
−v
vThe non-inverting amplifier above has Ri → ∞ and Ro =
0 and, therefore, can be turned into a voltage follower
(buffer) by adjusting R1 and R2 such that the gain is 1.
vo
vi
= 1 +R2
R1
= 1 → R1 → ∞
So by setting R1 → ∞ (replace R1 with an open circuit), we have a gain of unity. We note
that this expression is valid for any value of R2. As we want to minimize the number of
components in a circuit as a rule (cheaper circuits!) we set R2 = 0 (short circuit) and remove
R2 from the circuit.
ECE65 Lecture Notes (F. Najmabadi), Winter 2008 51
3.3.2 Non-Inverting Summer
fs
1
2
2
1
po
n
RR
R
R
−
+v
v
vv
v
Negative Feedback: vn ≈ vp
vn − 0
Rs
+vn − vo
Rf
= 0 → vo =(
1 +Rf
Rs
)
vn
vp − v1
R1
+vp − v2
R2
= 0 → vp
(
1
R1
+1
R2
)
=v1
R1
+v2
R2
Substituting for vn in the 1st equation from the second (noting vp = vn):
vo =1 + Rf/Rs
1/R1 + 1/R2
(
v1
R1
+v2
R2
)
So, this circuit also signal adds (sums) two signals. It does not, however, inverts the signals.
3.3.3 Inverting Summer
1
2
2
1
n
p
o
f
R
R
−
+
v
v
v
v
v
R
Negative Feedback: vn ≈ vp
vp = 0 → vn = 0
vn − v1
R1
+vn − v2
R2
+vn − vo
Rf
= 0
vo = −Rf
R1
v1 −Rf
R2
v2
So, this circuit adds (sums) two signals and invert them.
3.3.4 Difference Amplifier
1
2
s
f
1
2n
op
R
R
R
R
v
vv
v−
+
v
Negative Feedback: vn ≈ vp
vp − v2
R2
+vp − 0
R3
= 0 −→ vn ≈ vp =R3
R2 + R3
v2
vn − v1
R1
+vn − vo
Rf
= 0
Substituting for vn in the 2nd equation, one can get:
vo = − Rf
R1
v1 +(
1 +Rf
R1
) (
R3
R2 + R3
)
v2
ECE65 Lecture Notes (F. Najmabadi), Winter 2008 52
If one choose the resistors such thatR3
R2
=Rf
R1
, then
vo =Rf
R1
(v2 − v1)
3.3.5 Transconductance Amplifier or Current Source
L
1
i L
R
R
v i+
−Negative Feedback: vn ≈ vp = vi
−io = iL =vn
R1
=vi
R1
iovi
= − 1
R1
= Gm = const
So, this circuit is a “transconductance” amplifier: the output current is proportional to the
input voltage (see page 45). Alternatively, if a fixed value of vi is applied to the circuit, the
current iL is independent of value of RL and output voltage vo. As such, this circuit is also
an independent current source. Note that when the output of a circuit is the current, a load
is necessary for the circuit to function properly.
This circuit is similar to the non-inverting amplifier circuit. It is an example of circuits
which perform different functions depending on the location of the output terminals. For
this circuit, if the output terminals are between the output of the OpAmp chip and the
ground, it is a non-inverting amplifier while if the output terminals are taken across RL it
would be a current-source. Another example of this type of circuits is RC first-order filters
that we examined before (see pages 27 and 31). If the output is taken across the resistor, it
would be a high-pass filter while if the output is taken across the capacitor, it would be a
low-pass filter.
3.3.6 Grounded Current Source
L
1
f
3
2
i
L
R
R
R
R
R
v
i
−
+
The problem with the above current source is that the
load is not grounded. This may not be desirable in
some cases. This circuit is also a current source with
a grounded load if Rf/R1 = R3/R2.
Exercise: Compute iL and show that it is independent
of RL.
ECE65 Lecture Notes (F. Najmabadi), Winter 2008 53
3.4 Active Filters, Integrators & Differentiators
io
1
2
VV
Z
Z
+
−
Consider the circuit shown. This is an inverting ampli-
fier with impedances instead of resistors. Following the
inverting amplifier solution, we find:
H(jω) =Vo
Vi
= − Z2
Z1
Various filter circuits can be made with different choices
for Z1 and Z2.
3.4.1 1st Order Low-Pass Filter:
io
2
21
2V
V
C
RR
Z
+
−
Z1 = R1
Z2 = R2 ‖ C2 =R2
1 + jωC2R2
H(jω) = − Z2
Z1
= − R2/R1
1 + jωR2C2
Comparing the above with the generalized transfer function of a 1st order low-pass filter:
H(jω) =K
1 + jω/ωc
We find that the above circuit is a low pass filter with
K = − R2
R1
and ωc =1
R2C2
The minus sign in front of K indicates an additional −180 phase shift. A low-pass RC or
RL filter has a phase shift of 0 at low frequencies and −90 at high frequencies. The above
amplifier has a phase shift of −180 at low frequencies and −270 at high frequencies (or
alternatively +180 at low frequencies and +90 at high frequencies as we can add 360 to the
phase angle). Another difference with passive RC or RL filters is that the gain, |K| = R2/R1
can be set to be larger than one (i.e., amplify the signals in the pass band). As such this