International University
School of Electrical Engineering
Electronics Devices
Lecture # 3
Textbook: Microelectronic Circuit Design, A.S.Sedra &
K.C. Smith, 6th ed., Oxford University Press.
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1
Overview
• Reading
– Sedra & Smith: Ch. 2
• Supplemental Reading
– James W. Nilsson & Susan A. Riedel, Electric Circuits, 9th Edition.: Ch. 5
•

Background
Armed with our circuit analysis tools and basic understanding of
amplifiers, let’s now look at operational amplifiers (op amps). Op amps
were initially constructed out of vacuum tubes, then discrete transistor
components. With the advent of the integrated circuit, op amp ICs came
out in the 60’s (e.g., from Analog Devices Inc.). They are extremely useful
because they are versatile and one can do almost anything with op amps.
We will begin by looking at an ideal version of the op amp and see how
they are useful. Then, we will investigate various non-idealities of real
amplifier designs and how they affect op amp circuits.
2
The Operational Amplifier (Op-Amp)
Although transistor amplifiers made with ‘discrete’ components (i.e. individually
packaged) are still used for some special purposes like high-quality ‘Hi-Fi’,
most modern signal processing systems use Integrated Circuits (ICs). The one
of the oldest, most commonly used – and cheapest! – IC Operational
Amplifiers is the xx741.
3
The Operational Amplifier (Op-Amp)
4
5
The Operational Amplifier (Op-Amp)
The chip: a bipolar IC
6
The Operational Amplifier (Op-Amp)
 Original integrated circuit Op-Amps

Based on bipolar junction transistors

1965 – Fairchild’s A-709

1969 – Fairchild’s A-741
 Present Op-Amps

Both BJT and MOS technologies available

Normally 20 to 30 transistors
7
The Operational Amplifier (Op-Amp)

Usually Called Op Amps

Op-amp is an electronic device that accepts a varying input signal and
produces a similar output signal with a larger amplitude or that amplify
the difference of voltage at its two inputs.

Usually connected so part of the output is fed back to the input.
(Feedback Loop)

Most Op Amps behave like voltage amplifiers. They take an input
voltage and output a scaled version.

They are the basic components used to build analog circuits.

The name “operational amplifier” comes from the fact that they were
originally used to perform mathematical operations such as integration
and differentiation.

Integrated circuit fabrication techniques have made high-performance
operational amplifiers very inexpensive in comparison to older discrete
devices.
8
The Operational Amplifier (Op-Amp)





Amplifiers provide gains in voltage or current.
Op amps can convert current to voltage.
Op amps can provide a buffer between two
circuits.
Op amps can be used to implement
integrators and differentiators.
Lowpass and bandpass filters.
9
Op-Amp Terminals
Terminals of primary interest:
• inverting input
• noninverting input
• output
• positive power supply (+Vcc)
• negative power supply (-Vcc)
Offset null terminals may be
used to compensate for a
degradation in performance
because of aging and
imperfections.
“The op amp is a differential-input, single-ended-output amplifier”
This means that at the output, the voltage is obtained
with respect to a reference, usually called ground.
10
Ideal Op-Amp
The op amp is designed to sense the
difference between the voltage signals
applied to the two input terminals and
then multiply it by some gain factor A
such that the voltage at the output
terminal is A(v2-v1).
– One of the input terminals (1) is called an inverting input terminal
denoted by ‘-’
– The other input terminal (2) is called a non-inverting input terminal
denoted by ‘+’
– The gain A is often referred to as the differential gain or open-loop gain
– We can model an ideal amplifier as a voltage-controlled voltage source
(VCVS)
11
Ideal Op-Amp
Inverting
+VS
_
i(-)
RO
vid
Noninverting
i(+)
Ri
Output
vO = Advid
A
+
-VS
• i(+), i(-) : Currents into the amplifier on the inverting and noninverting lines
respectively
• vid : The input voltage from inverting to non-inverting inputs
• +VS , -VS : DC source voltages, usually +15V and –15V
• Ri : The input resistance, ideally infinity
• A : gain of the amplifier. Gain A is called the differential gain or open-loop gain
• RO: The output resistance, ideally zero
• vO: The output voltage; vO = AOLvid where AOL is the open-loop voltage gain
12
Ideal Op Amps Characteristics
 Ideal op amp characteristics:
– The input impedance is infinite (i.e., i1 = 0 and i2 = 0)
– The output terminal can supply an arbitrary amount of current (ideal VCVS)
and the output impedance is zero so output voltage is connected directly to
dependent voltage source.
– The op amp only responds to the voltage difference between the signals @
2 input terminals and ignores any voltages common to both inputs. In other
words, an ideal op amp has infinite common-mode rejection.
– The frequency response of an ideal op amp is flat for all frequency. In other
words, it amplifies signals of any and all frequencies by the same amount A.
– Lastly, A is or can be treated as being infinite.
 We will see later that real op amps do not have the characteristics
above, but we strive to make them behave as close to an ideal op amp
as possible.
13
Ideal Op Amps Characteristics
14
Typical vs. Ideal Op Amps
Typical Op Amp:
 The input resistance
(impedance) Rin is very
large (practically infinite).
 The voltage gain A is very
large (practically infinite).
Ideal Op Amp:
 The input resistance is
infinite.
 The gain is infinite.
 The op amp is in a
negative feedback
configuration.
15
Terminal Voltages and Currents
Terminal voltage variables
All voltages are considered as
voltages rises from the common
node.
Terminal current variables
All current reference directions
are into the terminal of the opamp.
16
Terminal Voltages and Currents
The terminal behavior of the
op amp as linear circuit
element is characterized by
constraints on the input
voltages and input currents.
Voltage transfer
characteristic:
 VCC

v0   A v p  v n

 VCC




 VCC  Av p  v n   VCC
Av p  v n   VCC
A v p  v n   VCC
When the magnitude of the input voltage difference (|vp – vn|) is
small, the op amp behaves as a linear device, as the output voltage
is a linear function of the input voltages (the output voltage is equal
to the difference in its input voltages times the gain, A.
17
18
DC off-set at the output of an Op-Amp
DC off-set: In any practical Op Amp, a very small differential input,
vIN1-vIN2, is require to make the voltage on this node (and VOUT) zero.
In a practice, an Op Amp will be used in a
feed-back circuit like the example shown to
the left, and the value of vOUT with vIN = 0 will
be quite small. For this example (in which Avd
= -2 x 106, and VOFFSET = 0.1 V) vOUT is only
0.1μV.
19
Terminal Voltages and Currents
For ideal op amp:
Input voltage constraint:
v p  vn
Input current constraint: i p  i n  0
Apply Kirchhoff’s current law
i p  in  i0  ic  ic  0

i0   ic  ic

Even though the current at the input
terminal is negligible, there are still
appreciable current at the output
terminal.
20
Terminal Voltages and Currents
Input Signal modes
The input signal can be applied to an op-amp in differentialmode or in common-mode.
21
Terminal Voltages and Currents
Signal modes
The input signal can be applied to an op-amp in differential-mode
or in common-mode.
Common-mode signals are applied
to both sides with the same phase
on both.
Usually, common-mode signals are
from unwanted sources, and affect
both inputs in the same way. The
result is that they are essentially
cancelled at the output.
22
Common-Mode Rejection Ratio
The ability of an amplifier to amplify differential signals and reject
common-mode signals is called the common-mode rejection
ratio (CMRR).
Acm is zero in ideal op-amp and much less than 1 is practical
op-amps.
Aol ranges up to 200,000 (106 dB)
CMRR = 100,000 means that desired signal is amplified 100,000
times more than un wanted noise signal.
CMRR can also be expressed in decibels as
23
Common-Mode Rejection Ratio
What is CMRR in decibels for a typical 741C op-amp?
The typical open-loop differential gain for the 741C is 200,000 and
the typical common-mode gain is 6.3.
24
Voltage and Current Parameters
VO(p-p): The maximum output voltage swing is determined by the
op-amp and the power supply voltages
VOS: The input offset voltage is the differential dc voltage required
between the inputs to force the output to zero volts
IBIAS: The input bias current is the average of the two dc currents
required to bias the differential amplifier
IOS: The input offset current is the difference between the two dc
bias currents
25
Impedance Parameters
ZIN(d) : The differential input
impedance is the total resistance
between the inputs
ZIN(cm) : The common-mode input
impedance is the resistance between
each input and ground
Zout: The output impedance is the
resistance viewed from the output of
the circuit.
26
Slew Rate
Slew rate is the maximum rate of change of the output voltage in
response to a step input voltage
Determine the slew rate for the output
response to a step input.
27
Negative Feedback
Most of the time, this feedback path is
provided by using a resistor.
In general the rule is
this: If, when the output
voltage increases, the
voltage at the inverting
input also increases
immediately, then we
have negative feedback.
For ideal op amps, we can assume
that the op amp has negative
feedback if there is a signal path
from the output to the inverting
input of the op amp.
Feedback Path
Inverting Input
Output
Noninverting Input
+
28
Negative Feedback
Negative feedback is the process of returning a portion of the output
signal to the input with a phase angle that opposes the input signal.
Ex.: Open loop gain is in order of 100,000.
Even an extremely small input saturates the output.
Vin * Aol = (1mv * 100,1000) = 100V
 It is not well-controlled parameter.
The advantage of negative
feedback is that precise values
of amplifier gain can be set. In
addition, bandwidth and input
and output impedances can be
controlled.
29
Negative Feedback
Negative feedback is used in op-amp circuits to stabilize the gain
and increase frequency response.
o Controlled Gain
o Increased bandwidth
o Increased input impedance
o Reduced output impedance
The closed loop gain, Acl is the voltage gain of op-amp with external
feedback
30
Op Amps in the Inverting Configuration
v0
G
vI
v0
v2  v1 
0
A
31
Op Amps in the Inverting Configuration
We can adjust the closed-loop gain by changing the ratio of R2 and R1
If the input is a sine wave, then the output is a sign wave phase-shifted by 180.
The closed-loop gain is (ideally) independent of op amp open-loop gain A (if A is large
enough) and we can make it arbitrarily large or small and of desired accuracy
depending on the accuracy of the resistors.
This is a classic example of what negative feedback does. It takes an amplifier with
very large gain and through negative feedback, obtain a gain that is smaller, stable,
and predictable. In effect, we have traded gain for accuracy. This kind of trade off is
common in electronic circuit design.
32
Inverting Amplifier Example
Determine the gain of the inverting amplifier shown
33
Finite Open-Loop Gain
(*)
34
Example 2.1
Consider the inverting configuration with R1 = 1 kΩ and R2 = 100 kΩ.
(a) Find the closed-loop gain for the cases A = 103, 104, and 105. In each case
determine the percentage error in the magnitude of G relative to the ideal
value of R2/R1 (obtained with A = ∞). Also determine the voltage v1 that
appears at the inverting input terminal when vI = 0.1 V.
(b) If the open-loop gain A changes from 100,000 to 50,000 (i.e., drops by
50%), what is the corresponding percentage change in the magnitude of the
closed-loop gain G?
35
Example 2.1 - Solution
(a) Substituting the given values in Eq. (*), we obtain the values given
in the following table, where the percentage error ε is defined as
The values of v1 are obtained from v1 = –vO ⁄ A = GvI ⁄ A with vI = 0.1V
(b) Using Eq. (2.5), we find that for A = 50,000, |G| = 99.80. Thus a
−50% change in the open-loop gain results in a change of only −0.1%
in the closed-loop gain!
36
Input Resistance
37
Output Resistance
38
Example 2.2
Assuming the op amp to be ideal, derive an expression for the closed-loop gain
v0/vI of the circuit shown in Fig.. Use this circuit to design an inverting amplifier
with a gain of 100 and an input resistance of 1 MΩ. Assume that for practical
reasons it is required not to use resistors greater than 1 MΩ.
Fig. Circuit for Example 2.2. The
circled numbers indicate the
sequence of the steps in the
analysis.
39
Example 2.2 - Solution
At the inverting input terminal of the op amp, the voltage is
Here we have assumed that the circuit is “working” and producing a finite output
voltage vO. Knowing v1, we can determine the current i1 as follows:
Since zero current flows into the inverting input
terminal, all of i1 will flow through R2, and thus
Now we can determine the
voltage at node x:
Find the current i3:
40
Example 2.2 - Solution
Next, a node equation at x yields i4:
Finally, we can determine vO from
Thus the voltage gain is given by
which can be written in the form
Select R1 = 1 MΩ  with the limitation of using resistors no greater than 1 MΩ, the
maximum value possible for the first factor in the gain expression is 1 and is obtained by
selecting R2 = 1 MΩ. To obtain a gain of −100, R3 & R4 must be selected so that the second
factor in the gain expression is 100. If we select the maximum allowed value of 1 MΩ for R 4,
then the required value of R3 can be calculated to be 10.2 kΩ.  this circuit utilizes three 1MΩ resistors and a 10.2-kΩ resistor. In comparison, if the inverting configuration were used
with R1 = 1 MΩ we would have required a feedback resistor of 100 MΩ, an impractically
large value!
41
Model of Closed-Loop Inverting Amplifier
We can model the closed-loop inverting amplifier (with A = ∞)
with the following equivalent circuit using a voltage-controlled
voltage source…
42
Inverting Configuration with General
Impedances
43
Inverting Integrator
We replace Z2 (the negative feedback impedance) with a capacitor and Z1 is a resistor.
V0 ( s)
V ( j )
1 / sC
1
1
1
1
1


 0


1 
 
Vi ( s)
R
sRC
Vi ( j )
jRC RC
RC
RC
How about in the time domain?
t
1
vC  vC (0)   iC ( t )dt
C0
t
1
v0 ( t )  
v I ( t )dt  vC (0)
RC 0
vR = vi
CR the integrator time constant. This
integrator circuit is said to be an inverting
integrator. It is also known as a Miller
integrator
44
Inverting Lossy Integrator
While the DC gain in the previous integrator circuit is infinite, the
amplifier itself will saturate. To limit the low-frequency gain to a known
and reliable value, add a parallel resistor to the capacitor.
1

RC
Frequency ω is known as the integrator
frequency and is simply the inverse of the
integrator time constant.
45
Differentiator
46
Weighted Summer
We can also building a summer
All these currents sum i = i1 + i2 +… + in
vO = 0 – iRf
That is, the output voltage is a weighted sum of the input signals v1,
v2, . . . , vn. This circuit is therefore called a weighted summer.
47
Summing Amplifier Circuit Example 1
Find the output voltage of
the following Summing
Amplifier circuit
Solution
we can now substitute the values of
the resistors in the circuit as follows,
48
Summing Amplifier Circuit Example 2
(a) Find v0 with va =0.1 V and vb = 0.25 V.
(b) If vb = 0.25 V, how large can va be before the op amp
saturates.
(c) If va = 0.1 V, how large can vb be before the op amp
saturates.
(d) Repeat (a), (b), and (c) with the polarity of vb reversed.
49
Summing Amplifier Circuit Example 2
Solution
50
Summing Amplifier Circuit Example 2
Solution
51
Summing Amplifier Applications
Summing Amplifier Audio Mixer
If the input resistances of a summing amplifier are connected to
potentiometers the individual input signals can be mixed together by
varying amounts. For example, measuring temperature, you could add a
negative offset voltage to make the display read "0" at the freezing point or
produce an audio mixer for adding or mixing together individual waveforms
(sounds) from different source channels (vocals, instruments, etc) before
sending them combined to an audio amplifier.
52
Summing Amplifier Applications
Digital to Analogue Converter (DAC summing amplifier circuit)
Another useful application of a Summing Amplifier is as a weighted sum
digital-to-analogue converter. If the input resistors, Rin of the summing
amplifier double in value for each input, for example, 1kΩ, 2kΩ, 4kΩ, 8kΩ,
16kΩ, etc, then a digital logical voltage, either a logic level "0" or a logic
level "1" on these inputs will produce an output which is the weighted sum
of the digital inputs. Consider the circuit below.
53
Summing Amplifier Applications
54
Characteristic of 4-Bit DAC
55
Bias Current and Offset voltage with
compensation techniques
Transistors within op-amp need bias current.
Practical op-amp has small input bias currents.
Small imbalances in transistors produce a small offset voltage between the inputs.
For op-amps with a BJT input stage, bias current can create a small output error
voltage. To compensate for this, a resistor equal to Ri||Rf is added to one of the inputs.
56
Non-Inverting Configuration
57
Non-Inverting Example 1
Determine the gain of the noninverting amplifier shown
58
Non-Inverting Example 2
(a) Calculate v0 if va = 1 V and vb = 0 V.
(b) Calculate v0 if va = 1 V and vb = 2 V.
(c) If va = 1.5 V , specify the range of vb that avoids amplifier
saturation.
59
Non-Inverting Example 2 – Sol.
a) A negative feedback path exists from the op amp's output to its inverting
input through the 100 k resistor,  assume the op amp is working in
linear operating region.  write a node-voltage equation at the inverting
input terminal. The voltage at the inverting input terminal is 0, as vp = vb = 0
from the connected voltage source, and vn = vp from the voltage constraint.
The node-voltage equation at vn is: i25 = i100 = in
i25 = (va - vn)/25 = 1/25 mA
i100 = (vo - vn)/100 = vo/100 mA
The current constraint requires in = 0. Substituting the values for the three
currents into the node-voltage equation, we obtain
vo = -4 V
b) Vp = vb = vn = 2 V
i25 = - i100
vo = 6 V
60
Non-Inverting Example 2 – Sol.
c) vn = vp = vb, and i25 = -i100
va = 1.5 V
Solving for vb as a function of vo gives
Now, if the amplifier is to be within the linear region of operation,
-10 V  vo  10 V.
Substituting these limits on vo into the expression for vb, we see that vb
is limited to
-0.8 V  vb  3.2 V.
61
Non-Inverting Example 3
Assume that the op amp in the circuit shown is ideal.
• Calculate v0 for the following values of if vs : 0.4, 2.0, 3.5, -0.6, -1.6 and 2.4 V.
• Specify the range of vs required to avoid amplifier saturation.
62
Non-Inverting Example 3 – Sol.
63
Non-Inverting Configuration
A special case of the inverting amplifier is when Rf =0 and Ri = ∞.
This forms a voltage follower or unity gain buffer with a gain of 1.
This configuration offers very high input impedance and its very low
output impedance. These features make it a nearly ideal buffer
amplifier for interfacing high-impedance sources and lowimpedance loads.
vin
It produces an excellent circuit for
isolating one circuit from another,
which avoids "loading" effects.
+
vout
-
64
Difference Amplifier
65
Finite Open-Loop Gain and BW
So far, we have assumed infinite gain and infinite bandwidth (BW) for the
amplifier, but that is not reality. Amplifiers have finite gain and BW. Here’s
an example of the open-loop gain vs. frequency plot of an amplifier.
Notice that the gain can be very high at low
frequency, but starts to roll off at a low
frequency also. They are also “frequency
compensated” to roll off at -20dB/dec (or a
single pole) to guarantee that op
amp circuits will be stable (more on
this later in the semester when we
talk about the guts of building
amplifiers and feedback stability).
66
Finite Open-Loop Gain and BW
We can represent frequency response characteristics of this amplifier as we
did for a single time constant low-pass filter.
A( s) 
A0
;
1  s / b
A(j  ) 
A0
1  j / b
For frequencies much greater than ωb (ω >> ωb) we can approximate the
gain as…
A
A( j ) 
A( j ) 
0
b
j
A0b

 1  t  A0b
t is called the unity-gain BW. So the gain can be represented as: A(s) = t /s
– assuming ωb is very small (low)
• So given this equation, we can find the gain at any frequency (assuming a
single-pole magnitude response)
67
Frequency Response of Closed-Loop Amplifiers
Let’s look at the closed-loop gain equation we derived earlier for for an
amplifier with finite op-amp open-loop gain A.
if A0 >> 1+R2/R1, then we can approximate the equation as…
v0 ( s)
 R2 / R1
 R2 / R1
 R2 / R1
t



where 3dB 
vI ( s) 1  s(1  R2 / R1 ) 1  s(1  R2 / R1 ) 1  s
1  R2 / R1
A0b
t
3dB
Therefore, the closed-loop gain has a response that rolls off at –20dB/dec at a
frequency, ω-3dB, that is a function of the gain set by the input and feedback
resistors.
68
Gain-Bandwidth Tradeoff
69
Gain BW Product (GBW)
The product of gain and BW is a very useful value when designing
amplifiers and amplifier circuits
– Provides a measure of how “good” you amplifier is (want higher GBW)
– GBW is constant anywhere along the plot above for a particular design
70
BW for Multi-Stage Amps
We define the bandwidth of an amplifier to be
BW = fupper cutoff – flower cutoff ≈ fupper cutoff
Now, consider multiple amplifier stages (iterative stage amp)
Assume we use identical stages and we can write the expression for
gain of each stage as:
AMB
A1  A2  ....  An 
1  s / p
71
BW for Multi-Stage Amps (2)
72
Optimizing BW
73
Optimizing BW (2)
74
Input Offset Voltage Compensation
Most ICs provide a mean of compensation.
An external potentiometer to the offset null pins of IC package
External potentiometer
Adjust for zeros output
75
EXERCISES 1
Consider the noninverting amplifier circuit shown in Fig. (a). As shown, the circuit is
designed for a nominal gain (1 + R2/R1) = 10V/V. It is fed with a low-frequency sinewave signal of peak voltage Vp and is connected to a load resistor RL. The op amp
is specified to have output saturation voltages of ±13 V and output current limits of
±20 mA.
(a) For Vp = 1 V and RL = 1 kΩ, specify the signal resulting at the output of the amplifier.
(b) For Vp = 1.5 V and RL = 1 kΩ, specify the signal resulting at the output of the amplifier.
(c) For RL = 1 kΩ, what is the maximum value of Vp for which an undistorted sine-wave
output is obtained?
(d) For Vp = 1 V, what is the lowest value of RL for which an undistorted sine-wave output
is obtained?
(a)
EXERCISES 1 – Sol.
(a) For Vp = 1 V and RL = 1 kΩ, the output will be a sine wave
with peak value of 10 V. This is lower than output saturation
levels of ±13 V, and thus the amplifier is not limited that way.
Also, when the output is at its peak (10 V), the current in the load
will be 10V/1kΩ = 10 mA, and the current in the feedback
network will be 10V/(9+1) kΩ = 1 mA, for a total op-amp
output current of 11 mA, well under its limit of 20 mA.
EXERCISES 1 – Sol.
(b) Now if Vp is increased to 1.5 V, ideally the output would be a
sine wave of 15-V peak. The op amp, however, will saturate at ±13
V, thus clipping the sine-wave output at these levels.
Let’s next check on the op-amp output current: At 13-V output and
RL = 1 kΩ, iL = 13 mA and iF = 1.3 mA; thus iO = 14.3 mA, again
under the 20-mA limit. Thus the output will be a sine wave with its
peaks clipped off at ±13 V, as shown in Fig. (b).
(b)
EXERCISES 1 – Sol.
(c) For RL = 1 kΩ, the maximum value of Vp for undistorted sinewave output is 1.3 V. The output will be a 13-V peak sine wave,
and the op-amp output current at the peaks will be 14.3 mA.
(d) For Vp = 1 V and RL reduced, the lowest value possible for RL
while the output is remaining an undistorted sine wave of 10-V
peak can be found from