The basic operational amplifier or Op Amp is a very important circuit to study because it is so widely used.

1
Introduction

• The basic operational amplifier or Op Amp is a
  very important circuit to study because it is so
  widely used.
• Op Amps were initially made from discrete
  components like vacuum tubes and resistors with
  transistors eventually replacing vacuum tubes
• In the mid 1960s the first integrated circuit op
  amps were produced (mA 709)
• It had quite a few transistors and resistors
  compared to the discrete versions it replaced
• Its characteristics were poor by todays
  standards
• It was expensive, but
• It was all on a single silicon chip and it
  ushered in a new era in analog electronic circuit
  design
• It was more reliable than discrete op amps
• Engineers started using them in large quantities
  and the price dropped dramatically from over ten
  dollars each to 30 cents each and the reliability
  continued to increase
• The op amp is a very versatile circuit and can be
  used in numerous other circuits. The integrated
  circuit op amp characteristics come pretty close
  to that of an ideal amplifier. The near ideal
  behavior makes it much easier to design with op
  amps
• By the end of this group of lecture sessions the
  reader should be able to design some fairly
  complex circuits using op amps
• We will start by treating a op amp as a building
  block and describe its terminal characteristics
  and save a detailed discussion about what is
  inside an op amp for a later course (EGRE 307)

2
The Op Amp Terminals

• The terminals of a circuit can be divided up into
  different groups based on their functions, for
  example the power supply terminals, the signal
  terminals, etc.
• The op amp circuit has three signal terminals,
  two inputs (1 , 2) and one output (3) as shown
  on the figure below
• The amplifier requires dc power to operate. Most
  op amps require two dc power supplies to operate.
  Most of the time the two supplies are equal in
  magnitude but opposite in polarity, for example a
  positive voltage supply V (relative to ground)
  and a negative voltage supply -V (relative to
  ground) connected to terminals 4 and 5 as shown.

1
3
-
inputs
output
2
V
1
4
-
3
inputs
output
2
5
-V
3
Op Amp Terminals, continued

• An op amp can use a single power supply (10V and
  ground for example) but the signal reference of
  the circuit for symmetrical behavior will then be
  at one half the upper supply voltage or 5V in
  this case.
• The reference grounding point in op-amp circuits
  is just the common terminal of the two power
  supplies.
• That is, no terminal of the op amp package is
  physically connected to ground
• In many cases the power supply connections will
  not be explicitly shown on the schematic. They
  will be omitted to reduce the clutter but the
  connections are still implied when you draw the
  triangular op amp symbol.
• An op amp may have other terminals for frequency
  compensation and for removing (or nulling) out
  offset voltages.

5V
10V
1
4
1
4
Output Signal ground is at same level as real
ground
Output signal ground at 5V relative to real
ground
3
-
3
-
inputs
inputs
2
2
5
5
or
-5V
10V
0V
4
Exercise

• What is the minimum number of terminals required
  for a single op amp?
• Five - Two Inputs, Two power connections (either
  /-, symmetric or and Ground, non-symmetric)
  and one signal output referenced to ground
  (symmetric) or 1/2 the positive supply in a
  non-symmetric configuration
• What is the minimum number of terminals required
  for an integrated circuit package which contains
  four op amps (assume all four share power
  supplies)?
• Fourteen, four pairs of inputs, two power
  connections and four outputs

5
The Ideal Op Amp

• The op amp is designed to sense the difference
  between the voltage signals applied to its two
  inputs (v2-v1), and multiply this by a number A,
  and cause the resulting voltage A (v2-v1) to
  appear at the output terminal relative to ground.
  Note that v1 and v2 are also implied to be
  relative to ground as shown in the figure below.
• The ideal op amp is not supposed to draw any
  current from the signal source (it will draw
  current from the power supply terminal which are
  not shown).
• The input impedance is supposed to be infinite
• The output is supposed to act as an ideal voltage
  source independent of the current drawn by any
  load placed on the op amp.
• The output impedance is supposed to be zero

1

• The output has the same sign (is in phase with)
  the input v2 and is out of phase (opposite sign)
  as v1 . For this reason input terminal 1 is
  called the inverting input (distinguished by a
  - sign) and input terminal 2 is called the
  non-inverting input (distinguished by a
  sign).
• The op amp only responds to the difference signal
  v2-v1 and ignores any signal common to both
  inputs.

I10
v1
A(v2-v1)
3
2
I20
v2
Common terminal of the power supplies
6
Common Signal vs. Differential Signal

• The part of the signal that is the same for both
  inputs is called the common mode component and is
  not amplified
• If both inputs see the same signal the DIFFERENCE
  (v2-v1) is zero and there is nothing to amplify
  and the output is zero. This means that the
  amplifier has rejected (not amplified) what the
  two signals had in common
• An ideal opamp has infinite common-mode rejection
• For the time being we will consider the opamp to
  be a differential input single - ended - output
  (output from terminal 3 to ground)

A(v2 - v1)
v2
0
v1
Vp-p
1 2
Vp-p
on each side out of phase
Vp-p
1 2
Vp-p
on each side in phase
0
7
The Ideal Op Amp continued

• The gain, A is called the differential gain
  (based on the difference between terminals 2 and
  1)
• The gain, A is also sometimes called the open
  loop gain as opposed to later on when we add a
  feedback path from output to input and look at
  the closed loop gain
• Opamps are direct-coupled (DC) or direct current
  (DC) amplifiers.
• DC amplifiers have many applications but can also
  pose some practical problems
• IDEAL opamps amplify from zero frequency to
  infinite frequency, they are said to have
  infinite bandwidth

1

• The gain of an ideal opamp should approach
  infinity, but then how could we make use of an
  ideal opamp since any difference times infinity
  would be infinity?
• We usually dont use opamps in an open loop
  configuration
• Feedback lowers the gain to practical levels
• The value for A of a real opamp might be on the
  order of a million

I10
v1
A(v2-v1)
3
2
I20
v2
Common terminal of the power supplies
8
Exercise 2.2

• Consider an op amp that is ideal except that its
  open-loop gain A103. The op amp is used in a
  feedback circuit and the voltages appearing at
  two of its three signal terminals are measured.
  In each of the following cases, use the measured
  values to find the expected value of the voltage
  at the third terminal.
• A) v2 0V, v3 2V
• v1-(v3/A)-v2 -(2/1000)-0 -0.002V
• B) v2 5V, v3 -10V
• v1-(v3/A)-v2 -(-10/1000)-5 5.01V
• C) v1 1.002V, v2 0.998V
• v3A(v2 - v1) 1000(0.998-1.002) 4.0V
• D) v1 -3.6V, v3 -3.6V
• v2(v3/A)v1 (-3.6/1000)3.6 3.6036V

9
Exercise 2.3

• The internal circuit of a particular op amp can
  be modeled by the circuit shown to the right.
• Express v3 as a function of v1 and v2.
• For the case when Gm 10 mA/V, R 10kW and m
  100, find the value of the open-loop gain A.

1
v1
Gmv1
3
mvd
vd
2
R
v2
Gmv2
10
The Analysis of Circuits Containing Ideal Op Amps

• The Inverting Configuration
• Resistor R2 is connected from the output
  terminal, 3, back to the inverting or negative
  terminal input terminal, 1. We say that R2
  applies negative feedback, if R2 were connected
  between the output, 3, and the other input, 2, we
  would say that it was positive feedback.
• We have grounded input number 2 and applied the
  input signal between terminal one and ground.
  This is known as a single sided input since
  terminal 2 does not contribute any information to
  the signal (since it is grounded).
• The output is taken at terminal 3 relative to
  ground and is called a single sided output. Note
  that the impedance at terminal 3 (output) of an
  ideal opamp is infinite, thus the voltage vo will
  not depend on the value of the current that might
  be supplied to a load placed across vo.

R2
input
R1
1
4
-
3
vI
vo -
output
2
11
Closed-Loop Gain (G) Analysis

• G is defined as
• We can draw the equivalent circuit for the opamp.
  Assume the opamp is powered up and working and
  the output is finite. With a finite output and
  infinite gain the difference between the inputs
  is negligibly small. Since terminal 2 is held at
  ground then terminal one also has to be at ground
  (forced there by the circuit NOT connected to
  ground).
• Using our definition we write

i2
We say that the two input terminals are tracking
each other in potential
R2
R2
i1
R1
R1
1
1
-
3
vI
-
3
vo -
vI
vo -
v2-v1
0
A(v2-v1)
2
2
12
Virtual Ground

• The circuit forces the two input terminals to be
  at virtually the same potential, they are
  virtually shorted together. A physical
  connection for charge movement does not exist but
  since they track each other in potential is like
  there is a connection (a virtual short circuit)
• If terminal 2 is grounded then we can refer to
  terminal 1 as a virtual ground. It is forced to
  be at zero volts even though it is not directly
  connected to ground it acts like it is.

13
Back to the analysis

• Since we now know v1 we can use Ohms law and
  find the current flowing through the resistor R1.
• Since this current can not flow into the input
  terminal of an ideal op amp (infinite input
  resistance) it must flow through the resistor R2
  to the low impedance terminal, 3 (output). We can
  now apply Ohms law to R2 and determine vo.
• The negative sign for the closed loop gain
  indicates that the amplifier provides signal
  inversion
• If we apply a 50mV peak-to-peak sine-wave input
  signal the output will be an amplified sine-wave
  with a 180 degree phase shift (inversion).

14
The Effect of Finite Open-Loop Gain

• How do our results change if we let the open loop
  gain, A, be finite?
• If we let the output be denoted by vO, then the
  voltage difference between the inputs is vO/A.
• Since the positive input is grounded the negative
  input must be at - vO/A.
• We can now find the current through resistor R1
• The infinite input resistance of the op amp
  forces all of the current to flow through R2.
• We can write another expression which contains
  vO.
• Collecting terms and solving for the closed-loop
  gain G, we get

Note As A approaches infinity, G approaches the
ideal value that we previously derived
15
Input and Output Resistance

• As A approaches infinity the voltage at the
  inverting terminal approaches zero (our virtual
  ground)
• In order to minimize the effect of A on G we
  should make
• If we assume an ideal op amp with an infinite
  input resistance the overall input resistance of
  the closed loop circuit (containing the ideal op
  amp) will be
• The value of the open loop gain, A, has very
  little effect on the input resistance
• For a high input resistance we need R1 high, but
  if we want the gain to also be high we might
  require R2 to be very large (maybe impracticably
  large). Can you think of a way to increase the
  input resistance?
• The output of the inverting configuration is
  taken at the terminal of an ideal voltage source
  (whose value is (v2-v1)A) which can supply
  infinite current, RoutV/I is zero.

Ro0

• Putting the previous results together we have the
  following model of the inverting closed loop
  amplifier circuit (using an ideal opamp)

vI _
RiR1
16
Example 2.1

• Consider the inverting amplifier configuration
  shown below with R11kW and R2100kW.
• A) Find the closed-loop gain for the cases
  A103, 104, 105. In each case determine the
  percentage error in the magnitude of G relative
  to the ideal value of R2/R1 (obtained with A that
  is infinite). Also determine the voltage v1 that
  appears at the inverting input terminal when vI
  0.1 V.
• We can use the equation for the closed loop
  gain, Greal in terms of the open loop gain, A and
  compare the results to the ideal closed loop gain
  Gideal.

17
Example 2.1 continued

• B) If the open-loop gain A is cut in half, from
  100,000 to 50,000, what is the corresponding
  percentage change in the magnitude of the
  closed-loop gain G?
• For A100,000 G was 99.9, so cutting the open
  loop gain in half from 100,000 to 50,000 only
  changed the closed loop gain by -0.1 percent
• Using negative feedback in a low gain
  configuration makes the circuit much more
  impervious to possible variations in the
  open-loop gain of the op amp.

18
Example 2.2

• We are going to analyze a different inverting op
  amp circuit (see below)
• Part A) Derive an expression for the closed-loop
  gain vO/vI
• We can now write an expression for the voltage at
  node X

R2
R4
i2
i4
X
R3
i3
R1
0
-
vI
vo -
vI
i1
Ideal
19
Exercise 2.4 (design)

• Using the circuit shown below design and
  inverting amplifier having a gain of -10 and an
  input resistance of 100kW. Give the values of R1
  and R2.

R2
input
R1
1
4
3
vI
-
vo -
output
2
20
Exercise 2.5 Transresistance Amplifier

• The circuit shown below can be used to implement
  a transresistance amplifier. Find the value of
  the input resistance Ri, the transresistance Rm,
  and the output resistance Ro.

10kW
1
4
input
3
-
vo -
output
2
21
Exercise 2.5 Transresistance Amplifier continued

• Part b) If a signal source of 0.5 mA in parallel
  with a 10kW resistor (see below) is connected to
  the input of this amplifier find its output
  voltage.

RF 10kW
1
4
3
-
vo -
RS 10kW
0.5mA
output
2
Since terminal 2 is grounded terminal 1 is forced
to a virtual ground this means that both sides of
the source resistance Rs are at ground and no
current flows through it. The entire 0.5 mA must
flow through RF the feedback resistor since the
input impedance of the op amp is
infinitely large. Therefore
22
The Inverting Configuration with General
Impedances

• The Inverting Configuration
• Replace resistors R1 and R2 with impedances Z1
  and Z2 as shown below.

Z2
input
Z1
1
4
3
vI(s)
-
output
vo(s) -
2
23
Example 2.3

• Derive an expression for the transfer function of
  the circuit shown below. Express the transfer
  function in a standard form for Bode Plots (see
  page 32 of Sedra and Smith 4th edition).

C2
R2
R1
1
-
3
vI
vo -
2
In Bode Plot Form
The dc gain is
The corner frequency is
24
Example 2.3 continued

• We could arrive at the same results regarding the
  circuit with impedances by using our knowledge of
  the frequency dependent behavior of the
  capacitor.
• The capacitor behaves as an open circuit at low
  frequencies. With the capacitor essentially
  removed the circuit is the same as the resistive
  configuration we have already analyzed and the
  gain is just -(R2/R1).
• At high frequencies the capacitor acts like a
  short-circuit and resistor R2 will become shorted
  out reducing the gain to zero at some point
  (frequency).
• Design the circuit so that the dc gain is 40 dB
  with a corner (-3dB) frequency of 1kHz and an
  input resistance of 1kW.
• The input resistance is simply R1 1kW since the
  inverting input is at ground
• We can now find the capacitance value which
  causes the corner frequency (f0) to be at 1kHz.

25
Example 2.3 continued

• Recall that the gain of a low pass single time
  constant circuit will fall off at -20dB per
  decade, so that if the gain is 40 db at f01kHz
  then the gain will be zero dB two decades higher
  at f100kHz.
• We also know that there is a -90 degree phase
  shift for frequencies more than ten times the
  corner frequency, but we have to recall that in
  the case of the inverting configuration there is
  already a -180 degree (inversion) shift so that
  the total phase shift is -270 degrees.

26
The Inverting Integrator

• Consider the inverting configuration with an
  impedance due to the capacitor used to provide
  the feedback
• We apply a time-varying input signal vI(t). The
  virtual ground at terminal 1 causes the input
  signal to appear across the resistor and a
  current iI(t) to flow. This current flow through
  the capacitor C since the input impedance of the
  ideal op amp is infinitely high.
• A charge will begin to accumulate on capacitor C.
  If we assume that the circuit began operating at
  time t0 then at some arbitrary time t, the
  charge on the capacitor will be given by
• The capacitor voltage will be where

• The output voltage is
• The output is proportional to the time integral
  of the input. The product RC is the integrator
  time constant

C
R
1
3
-
vo(t) -
vI(t)
2
27
Another way to analyze the inverting integrator

• Looking at the frequency domain
• And
• The integrator transfer function has a magnitude
  of
• The phase angle (f) is 90 degrees
• Constructing a Bode plot of the transfer function
  shows that as w doubles the magnitude of the
  response is cut in half. You can prove to
  yourself that this is a -6 dB decrease for each
  octave (eight-fold) increase in frequency and a
  -20 dB decrease for each decade increase in
  frequency.
• The frequency at which the integrator gain
  becomes zero dB is known as the integrator
  frequency and is simply the reverse of the
  integrator time constant.

-20 dB per decade
1/CR
w (log scale)
28
Some additional comments about the integrator

• As seen on the integrator behaves like a low-pass
  filter with a corner frequency of zero
• At zero frequency (dc) the gain is infinite (open
  loop gain of an ideal opamp)
• The feedback element is a capacitor and at dc it
  would be an open circuit (no feedback or an open
  loop)
• Any tiny dc level in the input will theoretically
  produce a very large output. What really happens
  is that the op amps output usually saturates at
  a level very close to the positive or negative
  power supply levels of the op amp depending on
  the polarity of the tiny dc level.
• We can reduce the gain of the dc signal by
  placing a high valued resistor in parallel with
  the capacitor. The high value should have little
  effect (no longer ideal though) on the integrator
  but it will provide a dc feedback path.

29
Example 2.4

• Find the output produced by a Miller integrator
  in response to an input pulse of 1-V height and
  1-ms width. Let R10kW and C10nf.
• From our earlier work we know the response will
  be of the form
• Substituting in the given values we find

Assume that the initial voltage on the capacitor
is zero
1V
t
0
1ms
t
-10V
30
Example 2.4 continued

• The 1V signal through a 10kW resistor produces a
  constant 0.1mA current through the capacitor
  which causes the voltage to increase linearly
• If the integrator capacitor is shunted by a 1-MW
  resistor, how will the response be modified?
• The current will now be supplied into a single
  time constant network of Rf in parallel with C.
• Appendix F gives a review of single time constant
  circuits and the resulting output responses. The
  solution to the differential equation that arises
  is
• Where is the final value of the output
• and is the initial value which is zero
• Therfore

0
1ms
t
-9.5V
31
The Differentiator (noise magnifier, rarely used)
R
C
1
3
-
vo(t) -
vI(t)
2
CR is the differentiator time constant
32

• Like a single time constant high-pass filter
• Spike at the output every time there is a sharp
  change in the input

1/CR
20 dB per decade
w (log scale)
Improvement, but makes it non-ideal
R
C
1
3
-
vI(t)
vo(t) -
2
33
The Weighted Summer
R1
v1
i
Rf
i1
R2
i
1
3
-
v2
vo -
0V
i2
2
R3
v3
i3
34
Exercise 2.6

• Consider a symmetrical square wave of 20-V
  peak-to-peak, zero average, and two milli-second
  period applied to a Miller integrator. Find the
  value of the time constant RC such that the
  triangular waveform at the output has a 20 Volt
  peak-to-peak amplitude.
• Refer back to example 2.4

35
The Non-Inverting Configuration

• The Non-Inverting Configuration
• The input signal vi is applied directly to the
  positive input terminal while one terminal of R1
  is grounded
• Analysis,
• Assuming that the op amp is ideal with infinite
  gain, a virtual short circuit exists between the
  two input terminals and the difference input
  signal is
• since the voltage at the inverting terminal (is
  equal to that at the non-inverting terminal) is
  equal to vI we can determine the current through
  resistor R1.

R2
R1
1
4
3
-
output
vo(s) -
Virtual short
input
2
vI(s)