BIE-601_Unit-1

BIE-601
Electronic Instrumentation-II
Lecture-1
Mohd. Umar Rehman
umar.ee.amu@gmail.com
January 21, 2021
BIE-601 Lecture-1 January 21, 2021 1 / 68

Unit-I: Signal Conditioning
Signal conditioning- ac and dc signal conditioning, comparators,
current-to-voltage and voltage-to-current converter, attenuators,
A to D and D to A converters, instrumentation amplifiers(IA)-single
op-amp and three op-amp configuration, IA specifications, application
of instrumentation amplifier using transducer bridge as temperature
indicator.

Quick Review
What is measurement?
What is an instrument?
What is instrumentation?

Quick Review
What is measurement?
To report the magnitude of a physical quantity in numerical terms.
What is an instrument?
The device that measures.
What is instrumentation?
Art and Science of measurement.

Contd...
BEE-401: Electrical Measurements → Basic Principles of
measurement, electrical measurement devices
BIE-401, BLE-303: Electronic Instrumentation-I → Electronic
measurement techniques
BIE-601: Electronic Instrumentation-II → Data (Acquisition,
Conditioning, Transmission)

Block Diagram of an Instrumentation System
Sensor/
Transducer
Physical
Quantity
Signal
Conditioning
Sensor
O/P
Measurement
Measurable
Qty Display/
Further
Processing
Measured
Qty

Block Diagram...
Sensor/
Transducer
Physical
Quantity
Signal
Conditioning
Sensor
O/P
Measurement
Measurable
Qty Display/
Further
Processing
Measured
Qty
Unit-1

Block Diagram...
Sensor/
Transducer
Physical
Quantity
Signal
Conditioning
Sensor
O/P
Measurement
Measurable
Qty Display/
Further
Processing
Measured
Qty
Unit-1
Unit-2

Block Diagram...
Sensor/
Transducer
Physical
Quantity
Signal
Conditioning
Sensor
O/P
Measurement
Measurable
Qty Display/
Further
Processing
Measured
Qty
Unit-1
Unit-2 Unit-3

Block Diagram...
Sensor/
Transducer
Physical
Quantity
Signal
Conditioning
Sensor
O/P
Measurement
Measurable
Qty Display/
Further
Processing
Measured
Qty
Unit-1
Unit-2 Unit-3
Unit-4

Contd...
The first stage of any instrumentation system is the
sensor/transducer stage.
Transducer converts one form of energy to another form.
We are concerned with only electrical transducers e.g. photodiode.
Transducer at the very first stage is known as sensor.

Signal Conditioning
Signal: Any useful information contained in a time function
v(t), i(t) etc.
Conditioning: To make a signal suitable for measurement
(/transmission/display) etc.
E. g. The output of a thermocouple is a very small voltage which
needs to be amplified in order to successfully measure it.
Definition: Signal Conditioning can be defined as the process of
modifying the output signals of a transducer into a usable form.
It may include amplification, filtering, attenuation, linearization
etc.

Signal Conditioning Requirements of
various transducers

Why Signal Conditioning?
Signal conditioning is necessary due to the following reasons:
1. The signal may be too small and may not be sufficient to produce
a reading.
2. The signal may contain noise due to nearby em interference.
3. The signal may be non-linear in nature and may require
linearization.
4. The signal may require A/D conversion or vice-versa.
5. The signal may require any other processing to make it suitable
for acquisition, measurement, transmission and display.

Advantages of Signal Conditioning
In many cases, mechanical processes may be used for signal condition-
ing. However, mechanical signal conditioning suffers from serious dis-
advantages like undesirable loading of the instrument, backlash, poor
frequency response etc.
In such cases, electrical signal conditioning is a better choice due to the
following reasons:
(i) Any physical quantity can be converted into electrical form which
is most suitable form for measurement.
(ii) Unwanted signals can be filtered using suitable filter circuits.
(iii) Power amplification is easy.

DC Signal Conditioning System

DC Signal Conditioning System
The above figure shows a DC signal conditioning system
DC signal conditioning system is used in DC measurements
where transducers like RTD, strain gauge etc. are being used and
bridge is supplied with DC.
The main part of a DC signal conditioning system is the DC
Amplifier.

DC Signal ... Contd
The DC amplifier has the following characteristics
(i) Good thermal and long-term stability.
(ii) High CMRR
(iii) Can be calibrated at low frequencies
(iv) Can recover from overload conditions unlike its AC counterpart.
(v) The major disadvantage of DC amplifier is that it suffers from the
problem of drift. As a result, some undesirable low frequency
signals are also amplified and appear in the output.
(vi) This problem can be overcome by the use drift reduction circuits
and filters.

AC Signal Conditioning System

AC Signal Conditioning System
The above figure shows an AC signal conditioning system which
overcomes the problems encountered in a DC signal conditioning
system.
Sometimes, transducer parameters can have variations, which can
cause inaccurate measurement.
This problem is taken care of by using a modulator.
The modulator uses amplitude modulation and the carrier
oscillator provides oscillating signal with certain frequency to run
the AC bridge.
This carrier signal is also provided to phase sensitive
demodulator circuit to multiply the signal coming from the bridge
via AC amplification.
Due to the use of a phase sensitive demodulator the polarity of
the output indicates the direction of parameter change in the
output of the bridge.

Some Features of AC Signal Conditioning System
1. Sometimes AC mains frequency may disturb the measurement. In
such cases, active filters are used to reject this frequency and also
prevent the overloading of the AC amplifier
2. The carrier frequency components of the data signal are filtered
out by the phase sensitive demodulator
3. The utility of AC signal conditioning systems lies where the
variable reactance transducers are employed and also where
signals are transmitted via long cables connecting transducers to
signal conditioning equipment

Comparators
+
−

Comparators...contd

Comparators...contd
A comparator is a circuit with two inputs and a single output. The
two inputs can be compared with each other i.e. one of them can
be considered as a reference terminal.
When the non-inverting input voltage is higher or greater than the
inverting input voltage, the output of the comparator is high.
When the non-inverting input voltage is lower or less than the
inverting input voltage, the output of the comparator is low.
The comparator circuit shown in above figure consists of a fixed
reference voltage Vref applied to the inverting input and a
sinusoidal signal applied to the non-inverting input Vin
When Vin > Vref, the output goes to the positive saturation
Vout = +Vsat = +VCC
When Vin < Vref, the output goes to the negative saturation
Vout = −Vsat = −VEE

Comparators...contd
Hence, the output changes from one saturation level to another
whenever Vin = Vref
Diodes D1 and D2 are used to protect the op-amp from damage
due to excessive input voltage.
The differential input voltage of the op-amp is either +0.7 V or
−0.7 V because of the diodes D1 and D2

Operational Amplifier (Review)
It is the type of amplifier which is widely used in instrumentation
systems
As the name sugegests, this amplifier can perform a number of
mathematical functions like +, −,
R
,
d
dt
, log, antilog etc.
It is a high gain directly coupled differential amplifier having very
high input impedance and very low input impedance
Op-amp consists of a number of transistors, diodes etc.
Symbolic representation is:
−
+
v1
vo = A(v2−v1)
v2

Characteristics of an Ideal Op-Amp
1. Infinite voltage gain
2. Infinite input impedance
3. Zero output impedance
4. Zero output voltage, when input is zero
5. Infinite Bandwidth
6. Infinite CMRR
7. Infinite slew rate so that the output voltage changes
simultaneously with the input voltage

Inverting Amplifier
−
+
vin
+
−
R1
R2
RL
vout
vout
vin
= −
R2
R1

Non-Inverting Amplifier
−
+
vin
+
−
R1
R2
RL
vout
vout
vin
= 1 +
R2
R1

Summing Amplifier
−
+
v3
R
RL
vout
R1
R3
R2
v1
v2
vout = −

R
R1
V1 +
R
R2
V2 +
R
R3
V3


Integrating Amplifier
−
+
R
vout
vin
C
vout = −
1
RC
Z t
0
vin(τ) dτ

Differentiating Amplifier
−
+
R
vout
vin
C
vout = −RC
dvin
dt

See Yourself
Logarithmic amplifier
Antilog amplifier

Try Yourself
Problem 1: Suggest a method to provide double of the input voltage
at the output of an op-amp.
Problem 2: Suggest a method to provide the following output of an
op-amp based summer vo = v1 + 2v2 + 3v3

Current to Voltage Converter
−
+
Is
RL
RS
vout

Current to Voltage Converter...Contd
A device that produces a voltage proportional to an input current
signal is called as I to V converter
As shown in the above figure, a current source
(photo-cell/photomutliplier tube etc.) provides a current
proportional to the luminous flux. However, this current is
independent of the load impedance
This current is provided to the inverting terminal of the op-amp.
RS is the shunt source resistance
As there is virtual ground at the inverting terminal, therefore
current through RS is zero.
Hence, all the source current flows through the load resistance RL
and the resulting output voltage is vout = −RLIS

Current to Voltage Converter...Contd
The above equation clearly indicates that the output voltage is
directly proportional to the source current.
However, in this circuit the lower limit on the current
measurement is set by biasing the current of the inverting terminal
Often a capacitor is connected in parallel with the load resistance
to reduce the HF noise.
I to V converter is commonly used in sensing current from
photodetectors and D/A converters

Voltage to Current Converter
IL
I
Figure: Floating Load Configuration

Voltage to Current Converter...Contd
Often in electronic measurement systems, such as CRT we need to
convert a voltage to a proportional current.
The requirement of such a converter is that the output current
should be independent of the load resistance.
The above figure shows a floating load configuration of the V to I
converter
This circuit is similar to a positive scaler circuit except that the
feedback resistance is replaced by the load resistance.
The output equation assuming virtual ground is:
IL = I =
Vin
R
Thus, the output current is proportional to the input voltage and
independent of the load resistance. Such a circuit is used in A/D
converters.

Fig. Grounded Load Configuration

Another configuration is shown in the above figure.
In this circuit the load is grounded and the input voltage controls
the load current.
Used in low voltage AC/DC voltmeters, LED and Zener Diode
testers.

Circuit Analysis
Applying KCL at node 1, we get
I1 + I2 = IL + IB(= 0) (1)
Vin − V1
R
+
Vout − V1
R
= IL
Vin + Vout − 2V1 = ILR
V1 =
Vin + Vout − ILR
2
(2)

Circuit Analysis...Contd
Since, the op-amp is in non-inverting mode, therefore,
Gain = 1 +
R
R
= 2
Vout = 2V1
Using this value of Vout in equation (2), we get
Vout = 2V1 = Vin + Vout − ILR

Vout = Vin +

Vout − ILR
IL =
Vin
R
IL ∝ Vin
Independent of load resistance.

D/A Converter

D/A Converter Truth Table

D/A Converter...Contd
Most of the D/A and A/D converters utilize the basic D/A
conversion process. Hence usually, D/A conversion is studied
first.
D/A conversion is the process of taking a value represented in
digital code such as Binary Coded Decimal (BCD) or straight
binary and converting it to a voltage or current that is
proportional to the digital value.
The above diagram shows a typical 4-bit D/A converter. Let us
focus only on the relationship between input and output without
going into internal details.
The four digital inputs represented by A, B, C, D are obtained
from the output of a register in a digital system.
We know that these four inputs will produce 24 = 16 different
binary numbers.

D/A Converter...Contd
For these various combinations of input bits, corresponding
output voltage is obtained.
In general, we have the relationship
Analog Output = K × Digital Input
The constant of proportionality K is constant for a given DAC. It
has unit of voltage if the output of the DAC is also voltage.
For the above shown DAC, the value of K is 1 because the analog
output directly corresponds to the digital input.

A/D Converter

A/D Converter...Contd
It involves the process of taking an analog voltage and converting
it to its equivalent digital code.
The A/D conversion process is generally more complex than the
D/A conversion.
Many different methods have been developed in the literature for
A/D conversion process.
We will see a basic function of an A/D converter.

The above diagram shows a basic A/D converter.
It consists of an op-amp comparator, a control unit, register and a
DAC.
The timing for the operation is provided by the input CLK signal.
The control unit contains the logic circuit for generating the
proper sequence of operations in response to start command which
initiates the conversion process.
The comparator has two analog inputs and a digital output. The
digital o/p switches the states depending upon which analog i/p
is greater.

The basic operation can be summarized as follows:
1. The start command pulse initiates the operation.
2. The control unit modifies the binary number stored in the register. The
modification rate is determined by the clock input.
3. The binary number in the register is converted into an analog voltage
VAX by DAC.
4. The comparator compares VAX with the analog input VA. As long as
VAX VA, the comparator o/p stays high. When VAX exceeds VA by
some minimum threshold voltage, the comparator o/p becomes low.
And the modification process in the register is stopped.
At this point, VAX is a close approximation of VA. The digital number in
the register which is digital equivalent of VAX is also digital equivalent
of the VA within the resolution and accuracy of the DAC.
5. The control logic activates the end of conversion signal when the
conversion is complete.

Try Yourself
Q. The input voltages to the comparator of a 4-bit ADC at a particular
time instant are: VA = 5.02 volts, VAX = 5 volts.
Determine the output of ADC.
Answer: 0101

Instrumentation Amplifier (IA/INA)
The simplest form of signal conditioning is the amplification of
the sensor output signal.
For this purpose, a special type of amplifier known as
instrumentation amplifier is used.
It is basically a differential amplifier with characteristics close to
that of an ideal amplifier.
The important features of an IA are as follows:
1. Selectable gain with high gain accuracy and linearity
2. Very high CMRR
3. Low thermal and time drifts
4. Low power consumption
5. High i/p impedance and low o/p impedance

IA versus Ordinary Op-Amp
The main differences between an ordinary op-amp and IA are as
follows:
1. The IA is often a package consisting of op-amps wired up with
accurate and stable feedback to give the desired gain which can be
varied by a single resistance.
2. The IA can be used directly for amplification since it works in
closed loop configuration.
3. Depending upon the configuration used, the IA gives very high
CMRR even when the source impedance exceeds 1 MΩ
See Yourself: Datasheet of any commercial IA. µA 725, LH0036

3 Op-Amp based IA

3 Op-Amp based IA
The above circuit shows a 3 op-amp based IA.
It consists of a basic differential amplifier A3 whose inputs are
supplied from two non-inverting amplifiers A1 A2.
The o/p of A1 is Vo1 and that of A2 is Vo1.
The overall output is:
Vo = −
R2
R1
(Vo1 − Vo2)
Let us now find out the values of Vo1 and Vo1 for the circuit shown
above.

3 Op-Amp based IA...contd
Current from E to F:
IEF =
Vo1 − Vo2
2R + RG
(1)
IGH =
V1 − V2
RG
(2)
Equating (1) (2), we get
Vo1 − Vo2
2R + RG
=
V1 − V2
RG
Vo1 − Vo2 =

2R + RG
RG

(V1 − V2) (3)

3 Op-Amp based IA...contd
Substitute the above value of differential voltage in the output voltage
equation, then:
Vo = −
R2
R1

2R + RG
RG

(V1 − V2)
Vo =
R2
R1

1 +
2R
RG

(V2 − V1)
Gain =
R2
R1

1 +
2R
RG

Hence, we have shown that the gain of an IA is dependent only on RG
and can be varied as per requirement.

Try Yourself
Find the gain of the IA with the following parameters:
RG (in Ω ) = last 3 digits of your faculty no.,
rest all resistors are of value 10 kΩ
[Mid-Sem Exam 2019]

IA using Transducer Bridge

IA using...Contd
The above circuit shows a simplified circuit of an IA using a
transducer bridge.
One of the arm of the bridge consists of a resistive transducer
whose resistance changes as a function of some other physical
quantity like temperature, pressure, light intensity (LDR) etc.
The condition of balance of the above bridge is Vb = Va or in terms
of resistance
RC
RB
=
RT
RA
The bridges is balanced at a desired reference condition, which
depends upon the specific value of the physical quantity to be
measured.
Under this condition, resistors RA, RB, RC are selected in such a
way that they are all equal to the transducer resistance RT

IA using...Contd
As the physical quantity to be measured changes,RT also changes
causing the bridge to become unbalanced (Vb 6= Va).
Hence, the o/p voltage of the bridge is a function of change in
resistance of the transducer.
Let us now derive the expression for Vo in terms of ∆R.

To prove that Vo = f (∆R)
Using voltage division rule
Vb =
RB
RB + RC
E
Va =
RA
RA + RT + ∆R
E
Vab = Va − Vb
=

RA
RA + RT + ∆R
−
RB
RB + RC

E
If RA = RB = RC = RT = R, then
Vab = E

R
2R + ∆R
−
1
2


Proof...Contd
Vab = E

2R − 2R − ∆R
2(2R + ∆R)

Vab = E

−
∆R
2(2R + ∆R)

Vo =
RF
R1
Vab
Vo = −
RF
R1

∆R
2(2R + ∆R)

E
Vo = f (∆R)

Contd...
It can be seen from the above equation that Vo is a function of
change in the resistance ∆R of the transducer.
Since this change is caused by a physical quantity to be measured,
a meter connected at the o/p can be calibrated suitably.
Other applications of IA with transducer bridge include:
temperature indicator using thermistor, temperature controller,
light intensity meter, analog weighing scale.

Temperature Indicator
Thermistor is a very sensitive temperature transducer. Most
thermistors have negative temperature coefficient (NTC)
characteristics.
The above circuit of IA with transducer bridge can be used to
indicate the temperature by connecting a thermistor in the
transducer arm.
The output meter then can be calibrated in terms of ◦C or ◦F
By selecting the gain of the IA, the meter can be calibrated to read
a desired temperature.
In this circuit, the deflection of the meter depends upon the
amount of unbalance in the bridge, which in turn depends on ∆R
The change ∆R of the thermistor can be determined as follows:
∆R = temperature coefft. of resistance [final temp. − initial temp.]

Ex. 14.3
In the above circuit,
R1 = 2.2 kΩ, RF = 10 kΩ, RA = RB = RC = 120 kΩ, E = +5 V, and
op-amp supply voltage = ±15 V. The transducer is a thermistor with
the following specifications.
RT = 120 kΩ at a reference temperature of 25 ◦C.
(a) Determine the output voltage at 0 ◦C and 100 ◦C
(b) Now, suppose that a meter is connected at the output which is
calibrated directly to give reading in ◦C

Announcements
1. Mid-Semester Exam Syllabus will be all syllabus covered
before mid-sem, and it will be notified once again.
2. Pattern will remain same as previous years’.
3. Home Assignment is posted on the webpage:
tinyurl.com/murehman

BIE-601_Unit-1

More Related Content

What's hot

Similar to BIE-601_Unit-1

More from Mohammad Umar Rehman

Recently uploaded

BIE-601_Unit-1