# Operational amplifier by uTeJnz

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ST12 – Operational amplifier

Operational amplifier

Lecturer:
Smilen Dimitrov

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ST12 – Operational amplifier
Introduction

•   The model that we introduced for ST

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ST12 – Operational amplifier
Introduction

•   We have discussed

– The units of voltage, current and resistance, from both a microscopic
and macroscopic (electric circuits) perspective
– The definition of an elementary electric circuit, Ohm’s law and Kirschoff
Laws
– Solving and measurement of voltage divider circuit and more
complicated circuits - and applications in sensors

– Resistive based sensors
– AC current, capacitors, and capacitive based sensors
– Semiconductor structures – diode and transistor (and sensor
applications)‫‏‬

•   This time we discuss the operational amplifier as an electronic element, and
a basic example of an integrated circuit

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ST12 – Operational amplifier
Differential amplifier

•   If R1=R2 and that the two transistors are identical → when v1=v2 it follows
that I1=I2, hence I1 = I2 = IE/2 (balanced)
•   v1 ↑ => I1 ↑ => VE1 (=VE) ↑ => VBE2 ↓ (since V2=const.) => I2 ↓ (but IE ≈
const)‫‏‬

HI V
CE  BE
H parameter - links voltage and current
(dimension 1/Ohm = Siemens)‫‏‬

1 B
HI V1 E
V            HI B  E
2 V 2 V

1 I  v v
H 2 12
I
E 2  
H I v v
I  2 1 2

Only interested in changes – 'delta' or differentiation,
here with lower case letters

v v
 12
1
i
2
h
2ib

R
v  2 1 
out v v 2                           Output is related to difference between the inputs
h
2ib
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ST12 – Operational amplifier
Differential amplifier

•   Important – the basis of a differential amplifier

•   Amplifies differential signals, rejects common signals
•   When differential signal is applied to the inputs: this will incrementally
increase and decrease the base voltages to VB1 + ΔV and VB2 - ΔV
– Because Q1 conducts a little more and Q2 a little less, IE now splits
unevenly creating IC1 > IC2 → forces the voltage at VC1 to decrease
and VC2 to increase. The result: a voltage change at each output
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ST12 – Operational amplifier
Differential amplifier

•   suppose a common-mode input signal is applied:
it incrementally increases both inputs to
VB1 + ΔV and VB2 + ΔV
– Because the conduction level of neither
transistor has changed (both bases and
emitters moved by the same amount),
the collector currents did not change.
IC1 = IC2 ≈ IE / 2.
– Subsequently, the voltages at VC1 and VC2 remain the same!
Therefore, the circuit has rejected a signal common to both inputs.

– The bias condition assumes equal
voltages at VB1 and VB2 → forcing
the bias current IE (set by RE) to
split equally between the transistors
→ resulting in IC1 = IC2.
– With RC1 = RC2, equal voltages
develop at VC1 and VC2.
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ST12 – Operational amplifier
Differential amplifier

•   Note that a differential amplifier is a more general form of amplifier than one
with a single input; by grounding one input of a differential amplifier, a single-
ended amplifier results
•   The voltage of each output:
– grows when the input in its own branch grows
– drops when the input in the opposite branch grows. .
•   When both Vbe's are equal → the potential of both bases is also equal →
which means that the voltage between the inputs is zero.
– Vbe of a single transistor will typically change in fractions of a volt (ex.
from 0.620V to 0.638V) → for all practical purposes, the voltage
between inputs stays zero, even when the circuit amplifies.
•   The input resistance of a differential amlifier input is resistance seen into a
base of a transistor – resistance of current source (theoretically infinite),
times hFE - (close to) infinity
•   -Vcc is kept on zero potential (ground) → Vi1 or Vi2 must >= VBEON (typically
0.6V); if -Vcc is -5V → Vi1 or Vi2 must >= -4.4 (-5+VBEON); so by using a
negative potential as a power supply, we can amplify both positive and
negative signals
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ST12 – Operational amplifier
Integrated circuits [IC] - operational amplifier construction (741)‫‏‬

•   The operational amplifier is based on the differential amplifier - it is a whole
circuit, and we cannot really treat it as a 'basic' electronic element (of the
likes of a resistor or a capacitor).
•   However, it is quite commonly produced as an 8-pin integrated circuit (IC),
with a standardized layout for the pins; and in such a form, the op-amp finds
many uses - so it is an 'element', in the sense of being a basic building block
found in many circuits.
– Opamp is a DC-coupled high-
gain electronic voltage amplifier
with differential inputs and,
usually, a single output.
– the output of the op-amp is
controlled by negative feedback
(external elements), which
almost completely determines
the output voltage for any given
input.
– originally designed to perform
mathematical operations                                                      8
ST12 – Operational amplifier
Integrated circuits [IC] - operational amplifier construction (741)‫‏‬

•   A monolithic integrated circuit (also known as IC,
microcircuit, microchip, silicon chip, or chip) is a
miniaturized electronic circuit that has been
manufactured in the surface of a thin substrate
of semiconductor material.

•   The first integrated op-amp to become widely
available, in the late 1960s, was the bipolar Fairchild μA709, created by Bob
Widlar in 1965;
•   it was rapidly superseded by the 741, which has better performance and is
more stable and easier to use.
•   The μA741 is still in production, and has become ubiquitous in electronics —
many manufacturers produce a version of this classic chip, recognisable by
part numbers containing '741.'
•   Better designs have since been introduced, some based on the FET (late
1970s) and MOSFET (early 1980s). Many of these more modern devices
can be substituted into an older 741-based circuit and work with no other
changes, to give better performance.
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ST12 – Operational amplifier
Integrated circuits [IC] - operational amplifier construction (741)‫‏‬

•   741

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ST12 – Operational amplifier
Opamp

•   Op-Amp is nothing more than a differential amplifier that amplifies the
difference between two inputs. One input has a positive effect on the output
signal, the other input has a negative effect on the output.
•   The Op-Amp requires two power supplies; a positive voltage supply and a
negative voltage supply, both with respect to our circuit
ground/Earth/chassis connection.
• The theoretically perfect Op-Amp has an infinite voltage gain, an infinite
bandwidth and infinite input impedances. In this way it just senses an input
voltage level without actually interfering with that voltage in any way.
•   The perfect Op-Amp also has a zero-Ohm output impedance. It may
therefore be used to drive heavy (in electronic terms) circuits.
•   A typical operational amplifier has input impedances of 100M-ohms, 1-ohm
output impedance and will drive up to 20mA of output current. Supply
voltages may be as high as +20V and -20V (total 40 volts from +ve supply to
-ve supply).
•   The typical voltage gain/bandwidth of a normal Op-Amp is about 1 000 000.
This is to say that at DC it will have a voltage gain of one Million, but at
1MHz it will only have a voltage gain of 1.
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ST12 – Operational amplifier
Opamp

•   The basic function of the Op-Amp is to multiply a voltage level by the gain of
the amplifier. If you were to couple a DC level of +1v into the + input of our
Op-Amp then the output would be 1v X 1000000 or one Million volts. The
output, however, cannot exceed the supply voltage, so the output will be
+20v DC.
•   If you were to couple a DC level of +1v into the - input of our Op-Amp then
the output would be -1v X 1000000 or MINUS one Million volts. The output,
however, still cannot exceed the supply voltage, so the output will be -20v
DC.
•   If you were to couple a DC level of +1v into both the - and + inputs of our Op-
Amp then the output would be (-1v X 1000000) plus (1v X 1000000) = 0v.
•   In other words, both inputs act on the output simultaneously and the output
is the sum of both input functions. If both inputs are identical then the output
should always be zero.
– This is a good test for an Op-Amp. If you connected both inputs to the
same input, then the output SHOULD be zero volts. In reality, there are
small differences in the circuit's characteristics and components, this will
result in a small 'offset' voltage.
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ST12 – Operational amplifier
Operational amplifier as an electronic element
linear amplifier – the output is
•   Symbol                                                             directly proportional to the
amplitude of input signal.
open-loop gain, A - the
voltage gain without feedback
(~106)‫‏‬
closed-loop gain, G - the
voltage gain with negative
feedback
negative feedback - the
output is connected to the
inverting input forming a
feedback loop (usually
through a feedback resistor).
•   Ideal opamp
– large forward transfer function,
– virtually nonexistent reverse transfer function,
– large input impedance (any signal can be supplied to the op-amp without loading problems)
RIN → ∞
– small output impedance, (the power supplied by the op-amp is not limited), ROUT → 0
– wide bandwidth, and
– infinite (open-loop) gain . A → ∞           V openloop
OUT   V V        G A      V V     
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ST12 – Operational amplifier
Operational amplifier as an electronic element

•   Two rules can be used to analyze op-amp circuits:
– Rule 1: The input currents I+ and I- are zero, ( I+ = I- = 0 {RIN→∞} )‫‏‬
– Rule 2: The voltages (input potentials) V+ and V- are equal, ( V+ = V- {A→∞} )‫‏‬
•   To apply these rules requires negative feedback.

•   Real op-amps can only approximate to the ideal – but in analysis we can take
them as ideal in first approximation

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ST12 – Operational amplifier
Measuring (testing) opamps

•   We can't measure to find out the pin function – must read datasheet

•   The input impedance of an OP-AMP is very high and probing either input
with a multimeter or CRO [cathode ray oscilloscope] will change the voltage
on the input and alter the state of the output. The reason is this: The voltage
on either input is extremely critical.
•   It is also impossible to measure the difference in potential between the
inverting input and non-inverting input. Thus the normal method of probing
and testing an OP-AMP with a multimeter or CRO DOES NOT WORK!
•   Don't be tricked by a CRO. It puts a load on the OP-AMP and if the line
under investigation is HIGH IMPEDANCE, the CRO will affect the amplitude
of the signal. The amplitude on the display will be reduced (attenuated) as
the frequency increases.
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ST12 – Operational amplifier
Measuring (testing) opamps

•   If you suspect an op-amp is faulty, check it by substitution but first:

– Is each pin of the op-amp connected? - Check visually that none of its pins have become
wrapped under its body instead of being inserted into the board.

– Is the output finite? - If the output is within a volt or so of either power rail then either it has failed,
or there is an excessive voltage at its inputs.

– Is the input consistent with the output? - Measure the voltage between
the inverting and non-inverting inputs, it should be within millivolts of
zero

•   There is actually very little that can go wrong in an opamp based circuit.
Opamps usually work or they don't - intermittent states can occur, but are
very uncommon.

•   Almost all faults with a newly built opamp based circuit will be the result of
wiring mistakes.

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ST12 – Operational amplifier
Symmetric (dual, split) power supply

•   Type of a power supply - having a ground, positive
and negative terminal is known as a symmetric,
split, bipolar or dual voltage (power) supply.
– On the other hand, using a single battery would
be considered using a single or unipolar supply.

•   Positive and negative terminals of a symmetric source are also known as
supply rails.

•   Opamp needs a symmetric power supply - to account for amplifying negative
voltages, as well as for compensating for the transistor turn-on voltage
(0.6V) for input voltages close to 0.

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ST12 – Operational amplifier
Symmetric (dual, split) power supply

•   In opamp design, this issue is adressed in two ways - either a design of a
symmetric power section; or designing the opamp circuit to work with a
single supply

– Voltage divider could be used
as a symmetric converter, but
isn't stable

– Redesigning for single supply means changing the
operating point, and the values of the signals involved

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ST12 – Operational amplifier
Symmetric (dual, split) power supply
– Split supply                  – Single supply

•   A converter using an opamp (based on voltage divider, but more stable)‫‏‬

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ST12 – Operational amplifier
Basic circuits - comparator

•   The simplest use of an operational amplifier is as a comparator - Without
external components the op-amp functions as a comparator.

V  V V
 V ; i i
o  CC 1   2

V  ;1 V
  CCi i
o VV 2

•   Equality of input values is very difficult to achieve in practice.
•   The speed at which the change in output results from a change in input
(often called the slew rate in operational amplifiers) is typically in the order of
10ns to 100ns, but can be as slow as a few tens of μs.
•   A dedicated voltage comparator chip, such as the LM339, is designed to
interface directly to digital logic (for example TTL or CMOS).
– Circuits specially built for use as comparators, such as the LM311 and
the LF311, are not op-amps, and cannot be used as op-amps. Among
other things, they would not be stable.
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ST12 – Operational amplifier
Basic circuits - comparator

•   Note that you cannot compare with
ground, in a single supply opamp
comparator

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ST12 – Operational amplifier
Voltage follower (buffer)‫‏‬

•   Voltage follower implements a negative feedback by connecting the output
to the inverting input.
•   This is a 'total' feedback, the amplification is 1 (also known as unity gain) -
that is, the input is duplicated (followed) at the output.
•   the benefit is that the original source of the input voltage now 'sees' the near
infinite input resistance of the opamp

I  I 0                 Vi  V
U d  V  V              Vo  V 
V  V  U d  0          Vo  Vi

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ST12 – Operational amplifier
Voltage follower (buffer)‫‏‬

•   The circuit works because
Vo  A  U d  A  V  V   A  Vi  V0 

•   Growth of Vi as amplified as AV+, which tends to infinity, but
– as whole of the output is brought to the inverting input, which figures with
a minus sign - the infinite gain is compensated (as Vd tends to zero).

•   However, if the inverting and non-inverting input switch places, there is no
more voltage following
–   growth of input Vi goes to inverting input, and shows up as -AVi in the output expression.
–   That negative voltage then comes back on the non-inverting input, so again we have -AVi in the output
expression.
–   So it turns out that Vo = -2AVi, which basically means that the opamp will immediately show the negative
supply voltage as output

– you cannot make an inverting follower, by simply switching the inverting
and non-inverting inputs in the regular voltage follower.

•   in general true for all opamp circuits - the non-inverting and inverting input
are not reversable.
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ST12 – Operational amplifier
Voltage follower (buffer)‫‏‬

•   Most sensor interfaces take a range of voltages as input, and convert them into a digital
representation.
•   If you make your own sensors, or use sensors not perfectly matched to your system, you can
usually improve the quality of your data by scaling the voltages into the precise range needed by
•   The voltage follower is an extremely simple circuit that simply outputs a low impedance voltage
that is identical to the input.
•   This would be fairly useless, except that it changes high impedance inputs to low impedance, and
makes the signal stronger. Used in conjunction with an inverting op amp, it can be a simple way to
•   A buffer amplifier (sometimes simply called a buffer) is one that provides electrical impedance
transformation from one circuit to another. If the voltage is transferred unchanged (the voltage gain
is 1), the amplifier is a unity gain buffer; also known as a voltage follower.

a noticeable distortion for
single supply: the output
(red) stays at around 0.2V,
when the input is below
0.2V ; the reason is the
base-emitter turn-on
voltages of the differential
section.                                                      24
ST12 – Operational amplifier
Opamp amplifiers - inverting

•   The inverting amplifier:
I  I  0
U d  V  V
V  V  U d  0
–     Since I+ = 0, then I1 = I2 ( KCL: +I2 -I- -I1 = 0 )‫‏‬
–     Since V+ is connected to ground, and the differential voltage Vd
is zero, then V- must also keep the ground potential (zero).

V  Vi 0  Vi   V                    V  V  Vo  0 Vo                 Vi V0           R
I1                   i               I2  o                                   V0  Vi 2
R1      R1     R1                      R2     R2    R2                R1 R2           R1

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ST12 – Operational amplifier
Opamp amplifiers - non-inverting

•   The non-inverting amplifier:
I  I  0
U d  V  V
V  V  U d  0
–   Since I+ = 0, then I1 = I2 ( KCL: +I2 -I- -I1 = 0 )‫‏‬
–   Since Vi is connected to V+, and the differential voltage Vd
is zero, then V- must also keep the potential Vi.
V  V0  V   V      R  R1            R 
V  0 Vi                  V  V Vo  Vi                             0  Vi 2      V0  Vi 1  2 

I1  
R1

R1
I2  o
R2

R2                   R1    R2      R2      R2 R1              R1 


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ST12 – Operational amplifier
Analogy for divided feedback in opamps

•   no hydraulic analogy for opamps as such - there is still another physical
analogy for inverting opamp amplifier
– If we draw a lever diagram next
to the amplifier schematic, with
the distance between fulcrum
and lever ends representative
of resistor values,
• the motion of the lever will
signify changes in voltage
at the input and output
terminals of the amplifier.)‫‏‬

•   Changing the resistor ratio R2/R1 changes the gain of the amplifier circuit,
just as
– changing the fulcrum position on the lever changes its mechanical
displacement 'gain.'

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ST12 – Operational amplifier
Summing amplifier (summer, mixer)‫‏‬

•   The summing amplifier:
I  I  0
U d  V  V
V  V  U d  0
–   Since I- = 0, then I2 = I11+I12+I13+… ( KCL: +I2 -I- -I11 -I12 -I13 - ...
= 0 )‫‏‬
–   Since V+ is connected to ground, and the differential voltage Vd is
zero, then V- must also keep the ground potential (zero).
V  V  Vo  0 Vo                V  Vi1 0  Vi1
I2  o                       I 11       
V
  i1                         V0    V    V          
R2      R2     R2                R1     R1       R1                               i1  i 2  ... 
R               
I12 
V  Vi 2 0  Vi 2

V
  i2
R2     1 R1           
R1       R1        R1

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ST12 – Operational amplifier
Differential opamp amplifier

•   Differential opamp amplifier:
I  I  0
U d  V  V
V  V  U d  0
–      (KCL node V-)‫ ‏‬I f  I 1  I   0  I f   I 1
–      (KCL node V+)‫ ‏‬I 2  I g  I   0  I 2  I g
Vo  V              V  0                     V1  V                     V2  V
If                  Ig                        I1                         I2 
Rf                  Rg                           R1                         R2

Rg
– V+ is set by a voltage divider from V2                                                 V 
R2  R g
V2

Vo  V    V  V                              Rf                                    R f  R1          Rf
– since If = -I1                       1                        Vo             V1  V   V         Vo               V         V1
Rf          R1                               R1                                      R1              R1

R f  R1      Rg             Rf
– since V- = V+
V                                           V2         V1
R2  R g
o
R1                         R1

– When R1=R2 and Rf=Rg, we have
amplified difference:      Rf
Vo     V2  V1 
R1                                                                                                               29
ST12 – Operational amplifier
Instrumentation amplifier

•   Instrumentation amplifier:
–   points Vo1 and Vo2 are inputs of the
differential amplifier, so write directly the
expression for the output voltage
R3
Vo        Vo 2  Vo1 
–                             R2
I+ and I- must be zero, single current Ix flows
through R1, Rgain and second R1. We can
write three expressions for this current
Vo1  V1 V2  Vo 2   V  Vo 2
Ix                       o1
R1       R1      2 R1  R gain

– we can express Vo through Ix                     Vo 2  Vo1   I x 2 R1  R gain       Vo   I x
R3
2R1  Rgain 
R2

Vo1  Vo 2  V2  V1 I x 2 R1  R gain   V2  V1
– we can express V2-V1 through Ix                         2I x 
R1

R1
V2  V1   I x  R gain

– replace this Ix, into the
V2  V1 R3
expression for Vo; and we obtain                          Vo                  2R1  Rgain 
dependency directly between Vo                                       R gain R2
and V2-V1

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ST12 – Operational amplifier
Sensing application

•   The operational amplifier does not have a 'sensor' conterpart, it is almost
unavoidable for sensor circuits.

– High impedance transducers
such as piezoelectric sensors,
hydrophones, and some
accelerometers require an
amplifier that converts a transfer
of charge into a voltage change.
– when interfacing strain gauges, commonly a Wheatstone bridge is used – it
must be amplified with a differential amplifier; more commonly, an
instrumentation amplifier is used with a strain gauge Wheatstone bridge.

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