EEG 317: Instrumentation and Measurment I
Lecture 1: Theory of Errors
1.1
Introduction
All branches of engineering perform two main functions: the design of equipment and processes, and the proper operation and maintenance of equipment and processes. These functions require measurement. Hence, the role of measurement cannot be ignored in the engineering field. The measurement of a quantity is fundamentally an act or outcome of comparison between a quantity (whose magnitude is unknown) and a predefined standard. In
order that the results of the measurement are meaningful, there are two basic requirements:
1. The standard used for comparison purposes must be accurately defined and should be
commonly accepted.
2. The apparatus used and the method adopted must be provable
The methods of measurement can be categorized into two, namely, direct methods and
indirect methods. The latter is adopted in engineering measurement systems due to the
inaccuracies in the direct methods as a result of human factors coupled with their lower
sensitivity. Measurements generally involve the use of an instrument as a physical means of
determining a quantity or variable. Therefore, we can define an instrument as a device for
determining the value or magnitude of a quantity or variable. The historical development
of instruments encompasses three phases, namely, mechanical instruments, electrical instruments and electronic instruments. It is important to note that no instrument is perfectly
accurate; thus, no measurement is free from error.
Assignment 1a - Highlight the main features, advantages, and disadvantages of the aforementioned categories of instruments.
1.2
Types of Errors
No measurement, whether conducted in a laboratory or elsewhere, is perfectly accurate. All
have some error or inaccuracy. Hence, it is pertinent to determine how different errors have
entered into measurement, how they are specified, and how they combine to create even
a greater error in measurement systems. Errors may come from different sources and are
usually classified under three main headings:
1-1
1-2
Lecture 1: Theory of Errors
ˆ Gross errors: This class of errors is attributed to human errors, such as misreading of
instruments, incorrect adjustment and improper application of instruments, and computational mistakes.
ˆ Systematic errors: These errors can be further categorized into three, namely instrumental
errors, environmental errors, and observational errors.
– Instrumental errors: these types of errors occur due to the shortcomings in the instrument, misuse of the instruments, and loading effects of instruments.
– Environmental errors: These errors are due to conditions external to the measuring
device, including conditions in the area surrounding the instrument. This may be
effects of temperature, humidity, dust, vibrations or of external magnetic or electrostatic fields.
– Observational errors: There are many sources of observation error. A typical example
is the parallax error, which is incurred when reading a measurement from an angle
rather than in line with the vision of the observer.
ˆ Random errors: Sometimes called residual errors are those due to the cause that cannot
be directly established because of the random variations in the parameter or the system
of measurement.
Assignment 1b - Give an elaborate discussion on the aforementioned types of errors and
how they can be avoided.
In most instruments, the accuracy is guaranteed to be within a certain percentage of fullscale reading. Components are guaranteed to be within a certain percentage of the rated
value. Thus, the manufacturer has to specify the deviations from the true or nominal value
of a particular quantity. These limits lead us to the theory of absolute and relative treatment
of errors.
1.3
Absolute Errors and Relative Errors
Given the nominal value of a resistor as 500 Ω with a tolerance of ±5% as depicted in Figure
1.1(a), this means the resistor will have a value between the limits 475 Ω and 525 Ω. Hence,
we can say the resistor has a possible error of ±25 Ω. The ±25 Ω is referred to as the
absolute error. When the error is expressed as a percentage or as a fraction of the total
1
resistance, it becomes a relative error. Thus the ±25 Ω is ±5%, relative to 500 Ω, or ± 20
of 500 Ω. Note that percentages are usually employed to express errors in resistance and
other electrical quantities. The terms accuracy and tolerance are also used. A resistor with
a possible error of ±5% is said to be accurate to ±5%, or have a tolerance of ±5%.
Mathematically,
ρm = ρt ± εa ,
(1.1)
Lecture 1: Theory of Errors
1-3
Figure 1.1: Resistor tolerance is represented by a gold color band.
where ρm is the actual or measured value, ρt is the nominal value, and εa is the absolute
error. The relative error in fractional and percentage form can be expressed as given in (1.2)
and (1.3), respectively.
εa
(1.2)
εr = ,
ρt
Percentage relative error %εr = εr × 100.
(1.3)
Also, from (1.1), limiting values in terms of relative errors can be expressed as
ρm = ρt ± εa = ρt ± εr ρt = ρt (1 ± εr ).
(1.4)
In addition, when the deviation limits from the nominal value are given, we can take the
maximum deviation from the ρm to be the erroneous quantity. Thus we have
εa = ρm − ρt .
(1.5)
Illustration: A component manufacturer constructs a certain resistance to be anywhere
between 1.14 kΩ and 1.26 kΩ and classifies them to be 1.2 kΩ resistors. What tolerance
should be stated?
Solution: From (1.5)
εa = 1.26 kΩ − 1.2 kΩ = 0.06 kΩ.
(1.6)
Thus, εa = ±0.06 kΩ. From (1.2)
±0.06 kΩ
= 0.05
1.2 kΩ
% εr = 0.05 × 100 = 5%.
εr =
Take note of the extension of the question in Example 2-1, page 16 of reference [1].
(1.7)
(1.8)
1-4
Lecture 1: Theory of Errors
1.4
Accuracy, Precision, Resolution, and Significant
Figures
1.4.1
Accuracy and Precision
Accuracy refers to the degree of closeness or conformity to the true value of a quantity under
measurement. Precision refers to the degree of agreement within a group of measurements
or instrument. The term ’Precise’ means sharply defined. Consider the digital voltmeter
Figure 1.2: Digital voltmeter with a precision of 1 mV.
indication shown in Figure 1.2. For the 8.135 V quantity indicated, the last (right-side)
numeral refers to millivolts. If the measured quantity increases or decreases by 1 mV, the
reading becomes 8.136 V or 8.134 V, respectively. Therefore, the voltage is measured with
precision of 1 mV. Thus we can infer that the measurement precision depends on the smallest
change that can be observed in a measured quantity. Suppose the digital voltmeter has an
accuracy of ±0.2%. The measured voltage is 8.135 V ±0.2%, or 8.135 ±16 mV, meaning
that the actual voltage is somewhere between 8.119 V and 8.151 V. Thus we find that the
quantity is measured with a precision of 1 mV, the measurement accuracy is ±16 mV
1.4.2
Resolution
The measurement precision of an instrument defines the smallest change in measured quantity that can be observed. This smallest change in measured value to which an instrument
will respond is referred to as the resolution of the instrument. In the case of the digital
voltmeter depicted in Figure 1.2, the meter can read to a precision of 1 mV, 1 mV is the
smallest voltage change that can be observed. Thus, the measurement resolution is 1 mV.
Lecture 1: Theory of Errors
1.4.3
1-5
Significant figures
The number of significant figures used in a measured quantity indicates the precision of the
measurement. For the 8.135 V measurement in Figure 1.2, the four significant figures show
that the measurement precision is 0.001 V, or 1 mV. Assuming the precision of 10 mV or 0.01
V, the display would be 8.13 V or 8.14 V; i.e., there would be only three significant figures.
The former, with more significant figures, expresses a measurement of greater precision than
the latter.
Consider the result of using an electronic calculator to determine a resistance value from
digital measurement of voltage and current.
R=
8.14 V
V
=
= 3.493 562 232 kΩ.
I
2.33 mA
(1.9)
It will be unreasonable to have an answer containing 10 significant figures. The only reasonable approach is to use the same number of significant figures in the answer as in the original
quantities. So the calculation becomes
R=
1.5
V
8.14 V
=
= 3.49 kΩ.
I
2.33 mA
(1.10)
Measurement Error Combinations
When using two or more instruments to calculate a quantity, it must be assumed that the
errors due to the instrument inaccuracy combine in the worst possible way. The resulting
error is then larger than the error in any one instrument. This can be easily found by
considering the relative increment of the measured value if the final result is in the form of
an algebraic equation.
1.5.1
Sum of Quantities
Where a quantity is determined as the sum of two measurements, the total error is the sum
of absolute errors in each measurement. As illustrated in Figure 1.3(a),
E = V1 + V2
= (V1 ± ∆V1 ) + (V2 ± ∆V2 )
= (V1 + V2 ) ± (∆V1 + ∆V2 )
(1.11)
Illustration: Three resistors have a nominal resistance of 330 Ω each and are connected in
series. One has a ±5% tolerance, and the other two are ±10%. Calculate the maximum and
minimum values of the total resistance.
1-6
Lecture 1: Theory of Errors
Figure 1.3: Measurement error combinations. (a) Error in sum of quantities equal sum of errors. (b)
errors in difference of quantities equals sum of errors. (c) Percentage error in product or quotient
of quantities equals sum of percentage errors
Solution: Let
R1 = 330 Ω ± 5%
= 330 Ω ± 16.5
(1.12)
R2 = R3 = 330 Ω ± 10%
= 330 Ω ± 33
(1.13)
RT otal = (330 Ω ± 16.5) + 2(330 Ω ± 33)
= 990 ± 82.5
(1.14)
max
Thus, we have RTmin
otal = 907.5 and RT otal = 1072.5
1.5.2
Difference of Quantities
Figure 1.3(b) depicts a scenario in which a potential difference is determined as the difference
between two measured voltages. Here again, the errors are additive.
E = V1 − V2
= (V1 ± ∆V1 ) − (V2 ± ∆V2 )
= (V1 − V2 ) ± (∆V1 + ∆V2 )
(1.15)
Lecture 1: Theory of Errors
1.5.3
1-7
Product of Quantities
Consider Figure 1.3(c). When a calculated quantity is the product of two quantities, the
percentage error is the sum of the percentage errors in each quantity.
P = EI
= (E ± ∆E)(l ± ∆I)
= EI ± E∆I ± I∆E ± ∆E∆I
(1.16)
P = EI ≃ ±(E∆I + I∆E)
(1.17)
Since ∆E∆I is very small,
E∆I + I∆E
× 100%
EI
!
∆I ∆E
=
+
× 100%
I
E
Percentage error =
% error in P = (% error in I) + (% error in E)
1.5.4
(1.18)
(1.19)
Quotient of Quantities
Following the illustration in Figure 1.3(c), it can be shown that the percentage error is the
sum of the percentage errors in each quality.
%error in
1.5.5
E
= (% error in I) + (% error in E)
I
(1.20)
Quantity Raised to Power
% error in AB = B(% error in A)
Illustration: An 820 Ω resistor with an accuracy of ±10% carries a current of 10 mA. The
current was measured by an analog ammeter on a 25 mA range with an accuracy of ±2% of
full scale. Calculate the power dissipated in the resistor, and determine the accuracy of the
result.
Solution:
P = I 2R
= (10 mA)2 × 820 Ω
= 82 mW
Error in R = ±10%
(1.21)
1-8
Lecture 1: Theory of Errors
The absolute error of the instrument can be computed as
εa = εr ρ t
= ±2% × 25 = ±0.5 mA
(1.22)
The magnitude of the current being measured is 10 mA, and the corresponding relative error
at this current can be computed as, that is,
±0.5 mA
% = ±5%
10 mA
(1.23)
error in I 2 = 2(±5%) = ±10%
(1.24)
error in P = (% error in I 2 ) + (% error in R)
= ±(10% + 10%) = ±20%
(1.25)
error in I =
1.6
Basics of Statistic
Revise arithmetic mean, deviation, standard deviation, and probable error in Section 2.5 of
reference [1]
References
[1] David A. Bell, Electronic Instrumentation and Measurements, Second Edition.
Theory of Errors
1 / 14
Introduction
The measurement of a quantity is fundamentally an act or outcome of comparison
between a quantity and a predefined standard.
The standard used for comparison purposes must be accurately defined and should be
commonly accepted.
The apparatus used and the method adopted must be provable.
Methods of measurement can be categorized into two, namely, direct methods and
indirect methods.
Measurements generally involve the use of an instrument as a physical means of
determining a quantity.
An instrument can be defined as a device for determining the value or magnitude of a
quantity or variable.
Based on historical development, instruments can be categorized under mechanical
instruments, electrical instruments and electronic instruments. (Assignment)
No instrument is perfectly accurate; thus, no measurement is free from error.
2 / 14
Types of Errors
Gross error: This class of errors is attributed to human errors.
Systematic errors: These errors can be further categorized into three, namely
instrumental errors, environmental errors, and observational errors.
Instrumental errors: these types of errors occur due to the shortcomings in the instrument,
misuse of the instruments, and loading effects of instruments.
▶ Environmental errors: These errors are due to conditions external to the measuring device,
including conditions in the area surrounding the instrument. This may be effects of
temperature, humidity, dust etc.
▶ Observational errors: There are many sources of observation error. A typical example is the
parallax error.
▶
Random errors: are those due to the cause that cannot be directly established because
of the random variations in the parameter or the system of measurement (Assignment).
3 / 14
Absolute Errors and Relative Errors
The diagram above illustrates a resistor with a nominal resistance of 500 Ω having a
tolerance of ±5%
The resistance can also be expressed as 500 ± 25. The ±25 Ω is referred to as the
absolute error.
When the error is expressed as a percentage or as a fraction of the total resistance, it
1
becomes a relative error i.e., ±5% or ± 20
.
4 / 14
Absolute Errors and Relative Errors Contd.
Mathematically,
ρm = ρt ± εa ,
(1)
ρm is the actual or measured value, ρt is the nominal value, and εa is the absolute error.
The relative error in fractional and percentage form can be expressed as given in (2) and (3),
respectively.
εr =
εa
,
ρt
Percentage relative error %εr = εr × 100.
(2)
(3)
The limiting values in terms of relative errors can be expressed as
ρm = ρt ± εa = ρt ± εr ρt = ρt (1 ± εr ).
(4)
5 / 14
Absolute Errors and Relative Errors Contd.
Illustration: A component manufacturer constructs a certain resistance to be anywhere
between 1.14 kΩ and 1.26 kΩ and classifies them to be 1.2 kΩ resistors. What tolerance
should be stated?
Solution:
εa = 1.26 kΩ − 1.2 kΩ = 0.06 kΩ.
(5)
Thus, εa = ±0.06 kΩ. From (2)
±0.06 kΩ
= 0.05
1.2 kΩ
% εr = 0.05 × 100 = 5%.
εr =
(6)
(7)
6 / 14
Accuracy, Precision, Resolution, and Significant Figures
Accuracy refers to the degree of closeness or conformity to the true value of a quantity
under measurement.
Precision refers to the degree of agreement within a group of measurements or
instrument. The term ’Precise’ means sharply defined.
If the measured quantity increases or decreases by 1 mV, the reading becomes 8.136 V or
8.134 V, respectively. Therefore, the voltage is measured with precision of 1 mV.
Suppose the digital voltmeter has an accuracy of ±0.2%, what will be the measurement
accuracy in terms of absolute error?
7 / 14
This smallest change in measured value to which an instrument will respond is referred to
as the resolution of the instrument. The measurement resolution is 1 mV.
The number of significant figures used in a measured quantity indicates the precision of
the measurement.
For the 8.135 V measurement shown in the previous slide, the four significant figures
show that the measurement precision is 0.001 V, or 1 mV.
Assuming the precision of 10 mV or 0.01 V, the display would be 8.13 V or 8.14 V; i.e.,
there would be only three significant figures.
Which of the measurements expresses a better precision the one with four or three SFs?
8 / 14
Measurement Error Combinations
(a) Sum of quantities - (V1 + V2 ) ± (∆V1 + ∆V2 )
Illustration: Three resistors have a nominal resistance of 330 Ω each and are connected in
series. One has a ±5% tolerance, and the other two are ±10%. Calculate the maximum and
minimum values of the total resistance.
9 / 14
Solution: Let
R1 = 330 Ω ± 5%
= 330 Ω ± 16.5
R2 = R3 = 330 Ω ± 10%
= 330 Ω ± 33
RTotal = (330 Ω ± 16.5) + 2(330 Ω ± 33)
= 990 ± 82.5
(8)
(9)
(10)
min = 907.5 and R max = 1072.5
Thus, we have RTotal
Total
(b) Difference of quantities ⇒ (V1 − V2 ) ± (∆V1 + ∆V2 )
(c) Product of quantities ⇒ Given P = EI , % error in P = (% error in I ) + (% error in E )
Quotient of quantities ⇒ %error in EI = (% error in I ) + (% error in E )
Quantity raised to power ⇒ % error in AB = B(% error in A)
10 / 14
Illustration: An 820 Ω resistor with an accuracy of ±10% carries a current of 10 mA. The
current was measured by an analog ammeter on a 25 mA range with an accuracy of ±2% of
full scale. Calculate the power dissipated in the resistor, and determine the accuracy of the
result.
Solution:
P = I 2R
= (10 mA)2 × 820 Ω
= 82 mW
(11)
Error in R = ±10%
The absolute error of the instrument can be computed as
εa = εr ρt
= ±2% × 25 = ±0.5 mA
(12)
The magnitude of the current being measured is 10 mA, and the corresponding relative error
at this current can be computed as, that is,
11 / 14
±0.5 mA
% = ±5%
10 mA
(13)
error in I 2 = 2(±5%) = ±10%
(14)
error in I =
error in P = (% error in I 2 ) + (% error in R)
= ±(10% + 10%) = ±20%
(15)
12 / 14
References I
[1] David A. Bell. Electronic Instrumentation and Measurements, Second Edition.
Prentice-Hall, 2003.
13 / 14
Electromechanical Instruments
1/9
Permanent Magnet Moving Coil (PMMC)
A typical PMMC instrument comprises a
lightweight coil of copper wire suspended
in the field of a permanent magnet.
Current passing through wire causes the
coil to generate a magnetic field that
interacts with the field from the
permanent magnet.
The interaction causes a partial rotation of
the coil. A pointer connected to the coil
deflects over a calibrated scale, indicating
the level of current flowing in the wire.
Basically, the PMMC is a low-level DC
ammeter. However, it can also be made
to measure a wider range of direct current
and function as a voltmeter or ohmmeter.
Figure 1: DC/AC moving coil ammeter
2/9
Fundamentals of Deflection Instruments
In PMMC instruments, three operating forces are involved for a pointer deflection to occur
over a calibrated scale, namely deflection force, controlling force, and damping force.
Deflection force
The deflection force is required for moving
the pointer from its zero position when a
current flows.
It occurs due to the interaction between
the magnetic field produced by the coil
pivoted between the poles of the
permanent magnet.
As a result, a force is exerted on the coil
turns, causing the coil to rotate on its
pivots .
Figure 2: The deflecting force in a PMMC
instrument.
3/9
Fundamentals of Deflection Instruments Contd.
Controlling force
In the absence of a controlling force, the pointer will
swing beyond the final position for any magnitude
of current, causing the deflection to be indefinite.
The controlling force provided by spiral springs
produces a force equal and opposite to the
deflecting force at the final steady position of the
pointer to make the pointer deflection definite for a
particular magnitude of current.
Also, the controlling force brings back the coil and
pointer to their zero position when the deflecting
force is removed.
Figure 3: Controlling force from spiral spring
4/9
Fundamentals of Deflection Instruments Contd.
Damping force
The deflection and controlling forces are produced by
systems that have inertia; hence, the pointer and the
coil tend to oscillate for some time before settling down
at their final position. - Fig. 4(a)
A damping force is required to damp out the oscillations.
This type of force must be present only when the coil is
in motion; thus, it must be generated by the motion of
the coil.
In PMMC instruments, the damping force is normally
provided by eddy currents that is induced from the coil
frame made of aluminum. - Fig. 4(b)
The eddy currents induced in the coil frame set up a
magnetic flux that opposes the coil motion, thus
damping the oscillations of the coil.
Figure 4
5/9
Assignment
Figure 5: PMMC construction.
Explain in detail the construction of a PMMC instrument shown in Figure 5.
6/9
Torque Equation and Scale.
As illustrated in Fig. 6, a force F is
exerted on each side of the coil when a
current flows through the coil. This force
can be described by (1)
F = BIL newtons,
(1)
B is the magnetic flux density in tesla, I
is the current in amperes, and L is the
length of the coil in meters.
Given N turns of coil; the force acts on
the two sides of the coil, thus, the total
force can be given as
F = 2BILN newtons.
(2)
Figure 6
The force on each side acts at radius r,
producing a deflecting torque.
TD = 2BILN r, Newton meters
= BILN (2r),
TD = BLIN D,
(3)
(4)
(5)
where D is the coil diameter.
7/9
The controlling torque TC is
proportional to the actual angle
of the deflection of the pointer.
TC = Kθ,
(6)
Note, the expression in (8) indicates that the
pointer deflection is always proportional to the coil
current and the scale is linear.
Illustration
where K is a constant. For a
A PMMC instrument has a 0.12 T magnetic flux density
given deflection, the controlling in its air gaps. The coil dimensions are D = 1.5 cm and
and deflecting torques are equal: L = 2.25 cm. Determine the number of coil turns
required to give a torque of 4.5 µN.m when the coil
Kθ = BLIN D.
(7) current is 100 µA.
From (7), all quantities are
constant except I and θ,
θ = CI,
N=
(8)
4.5 × 10−6
TD
=
BLID
0.12 × 1.5 × 10−2 × 100 × 10−6 × 2.25
≈ 1111
where C is a constant.
8/9
Galvanometer
A galvanometer is essentially a PMMC instrument
of a center-zero scale type, designed to be sensitive
to extremely low current levels (microamperes).
The pointer can deflect either right or left of zero,
depending on the direction of the current through
the moving coil.
Galvanometers are often employed to detect zero
current or voltage in a circuit rather than to
measure the actual level of current or voltage.
Galvanometers find their principal application in
bridge and potentiometer measurements, where
their function is to indicate zero current.
A galvanometer must be protected from excessive
current flow that might occur when the voltage
across the instrument terminal is not close to zero.
Figure 7
Protect is provided by an
adjustable resistance connected
in shunt with the instrument.
9/9
PMMC Instruments Contd
1/8
DC Ammeter
Ammeter circuit
Ammeters are connected in series with a circuit
in which current is to be measured. They
should have resistance much lower than the
circuit resistance in so that they cause a small
voltage drop and consequently absorb small
power.
However, maximum pointer deflection is
produced by a very small current, and the coil
is usually wound of thin wire that would be
quickly destroyed by large currents.
When heavy currents are to be measured, the
major part of the current is bypassed through a
low resistance called shunt.
Figure 1: Ammeter circuit
The shunt Rs is connected in parallel
with the instrument coil as depicted
in Figure 1. Im , Is and Rm are the
meter current, shunt current, and
coil/meter resistance, respectively.
2/8
Illustration
0.05 mA × 99 Ω
Vm
An ammeter depicted in Figure 1, has a
=
= 4.95 mA
Is =
R
1
Ω
s
PMMC instrument with a coil resistance of
Rm = 99 Ω and FSD current of 0.1 mA, shunt Thus, I = (0.05 + 4.95) mA = 5 mA
resistance Rs = 1 Ω. Determine the total
What is the value of I at 0.25 FSD?
current passing through the ammeter at (a)
✓ Going by the above illustration, the meter
FSD and (b) 0.5 FSD and (b) 0.25 FSD.
can be calibrated to indicate 10 mA.
Solution
I = Im + Is
0.1 mA × 99 Ω
Vm
=
= 9.9 mA
Is =
Rs
1Ω
Thus, I = (0.1 + 9.9) mA = 10 mA
At 0.5 FSD
Illustration - Shunt Resistance
A PMMC instrument with a 750 Ω coil
resistance gives FSD with a 500 µA coil
current. Determine the required shunt
resistance to convert the instrument into a dc
ammeter with an FSD of (a) 50 mA and (b)
30 mA
Im = 0.5 × 0.1 mA = 0.05 mA
3/8
Solution
Vm = Im Rm
= 750 Ω × 500 µA = 375 mV
Is = I − Im
= 50 mA − 500 µA = 49.5 mA
Vm
375 mV
Rs =
=
Is
49.5 mA
= 7.576 Ω
Swamping Resistance:
Errors can be introduced in the ammeter
measurements due to resistance change in
the coils as a result of change in
temperature of the coil.
Figure 2: Ammeter circuit with swamp resistance
To alleviate this effect, a swamping
resistance made of manganin or
constantan is connected in series with the
coil as shown in Fig .
If the swamping resistance is nine times
the coil resistance, a 1% change in coil
resistance would result in a total
resistance change of 0.1%.
4/8
Multirange Ammeter:
The circuit development of a multirange
ammeter can be realized as depicted in
Figures 3 and 4.
A critical safety factor considered in the
two designs is that the instruments are
not left without a shunt in parallel with it
even for a brief instance.
Figure 4: Multirange ammeter - Config. 2
Figure 3: Multirange ammeter - Config. 1
In Fig. 4, the switch is at contact B, and
the resistors R1 + R2 + R2 which
constitute the ayrton shunt, are in parallel
with the instrument. The meter circuit
remains Rm
5/8
Solution
Switch at contact B
When the switch is at contact C, the
resistor R3 is in series with the meter, and Vs = Im Rm = 100 µA × 1.2 kΩ
R1 + R2 is in parallel with Rm + R3
= 0.12 V
Similarly, with the switch at contact D,
Vs
0.12 V
Is =
=
R1 is in parallel with Rm + R2 + R3 .
(R1 + R2 + R3 + R4 )
0.4 Ω
Hence, the instrument is never left
= 0.3 A
without a parallel-connected shunt.
I = Im + Is = 100 µA + 0.3 A = 300.1 mA
Illustration - Multirange ammeter
A dc ammeter consists of an Aryton shunt in
parallel with a PMMC instrument that has a
1.2 kΩ coil resistance and 100 µA FSD. The
Aryton shunt is made up of four 0.1 Ω
series-connected resistors. Calculate the
ammeter range at each setting of the shunt.
≃ 300 mA
Switch at contact E
Vs = Im (Rm + R2 + R3 + R4 ) = 120.03 mV
Vs
120.03 mV
Is =
=
= 1200.3 mA
R1
0.1 Ω
I = 100 µA + 1200.3 A = 1200.4 mA
≃ 1200 mA
6/8
DC Voltmeter
Voltmeter circuit:
The deflection of a PMMC instrument is
proportional to the current flowing
through the moving coil. The coil current
is directly proportional to the voltage
across it.
Hence, the scale of PMMC instrument
could be calibrated to indicate voltage.
The voltmeter range can be extended by
connecting a resistance in series with the
instrument as shown in Fig. 5.
Because it increases the range of the
voltmeter, the multiplier resistance, the
series resistance is termed a multiplier
resistance.
Figure 5: Multirange voltmeter
Illustration
A PMMC instrument with 900 Ω coil
resistance and FSD of 75 µA is to be used as a
dc voltmeter. Calculate the individual
multiplier resistance to give an FSD of (a)
100 V (b) 30 V, and (c) 5 V.
7/8
Solution
At 100 V
V = Im (Rm + R1 )
100 V
V
− Rm =
− 900 Ω
R1 =
Im
75 µA
= 1.33 MΩ
Rectifier Voltmeter
8/8
PMMC Instruments Contd
1 / 10
Quiz
In the circuit illustrated above, all the diodes are of silicon, each with a forward resistance of
2 Ω. Calculate the current through the 50 Ω resistor.
2 / 10
Rectifier Voltmeter
Full-Wave (FW) Rectifier Voltmeter:
The FW rectifier voltmeter as depicted in
Figure 1, uses a series multiplier resistor
to limit the current flow though the
PMMC instrument.
The meter deflection is proportional to the
average current, which is 0.637 times peak
current. → Idc = Iav = 2Iπm
However, the actual current (or voltage)
to be indicated in ac measurement is
normally the rms quantity, which is 0.707
of the peak value or 1.11 times the
Im
average value. → Irms = √
2
Since there are direct relationships
between rms, peak, and average values,
Figure 1: Full-Wave rectifier voltmeter
the meter scale can be calibrated to indicate
rms volts.
Illustration: A PMMC instrument with
F SD = 100 µ and Rm = 1 kΩ is to be
employed as an ac voltmeter with
F SD = 100 V (rms). Silicon diodes are used
in the bridge rectifier circuit of Figure 1.
3 / 10
Calculate the multiplier resistance value
required.
Solution: Applying KCL in Fig. 2, we have
− Vm + Im Rs + VD1 + Im Rm + VD4 = 0
Im (Rs + Rm ) = Vm − (VD1 + VD4 )
Vm − (VD1 + VD4 )
Rs =
− Rm
Im
Vm = Vrms ×
√
Figure 2
2 = 100 V × 1.414 = 141.4 V
VD1 = VD4 = 0.7
Iav
100 µA
Im =
=
= 157 µA
0.637
0.637
141.4 V − 1.4 V
Rs =
− 1 kΩ = 890.7 kΩ
157 µA
✓ Rectifier Ammeter:
✓ Series Ohmmeter:
Assignment:
Problems 3.23 and 3.24, page 94 of [1].
Submission date: 22 January 2025, 9:00 AM.
4 / 10
References I
[1] David A. Bell. Electronic Instrumentation and Measurements, Second Edition.
Prentice-Hall, 2003.
5 / 10
Analog Electronic Instruments
6 / 10
Introduction
Voltmeters constructed of moving-coil
instrument and multiplier resistor have
some limitations. They cannot measure
very low voltages due to loading effect,
and their resistance is too low for
measurements in high-impedance circuitry.
The analog electronic voltmeter
configuration include the
✓ Emitter-Follower Voltmeter
✓ FET-Input Voltmeter
These limitations are overcome by the use
of electronic circuit that offers high input
resistance, and which amplify low voltages
to measurable levels.
Electronic voltmeter can be analog
instruments, in which the measurement is
indicated by a pointer over a calibrated
scale, or digital instruments which displays
the measurement in numerical form.
Figure 3: Emitter-Follower Voltmeter
7 / 10
Introduction
The basic emitter-follower voltmeter
circuit depicted in Fig. 3, shows a PMMC
instrument and a multiplier resistance Rs
connected in series with the transistor
emitter.
The DC supply is connected - positive to
the transistor collector and negative to the
deflection meter.
The positive terminal of voltage E to be
measured is supplied to the transistor
base, and its negative is connected to the
same terminal as the power supply
negative.
The transistor base current is substantially
lower than the meter current.
IB ≈
Im
,
hF E
where hF E is the transistor current gain.
Thus the circuit input resistance is
Ri ≈
E
,
IB
which is much larger than the meter circuit
resistance (Rs + Rm )
Illustration
A simple emitter-follower circuit voltmeter
shown in Fig. 3 has VCC = 12 V, Rm = 1 kΩ,
a 2 mA meter, and a transistor of with
hF E = 80. Calculate a suitable resistance for
Rs to give full-scale deflection when E = 5 V.
Also, determine the voltmeter input resistance.
8 / 10
Solution: Applying KCL in to the left loop of
Fig. 2, we have
−E + VBE + Im (Rm + Rs ) = 0
E − VBE − Im (Rm )
Rs =
Im
5 V − 0.7 V − (2 mA)(1 kΩ)
=
= 1.15 kΩ
2 mA
Im
2 mA
=
= 25 µA
hF E
80
E
5V
= 200 kΩ
Ri =
=
IB
25 µA
IB =
The voltage drop VBE introduces an error
in the simple emitter-follower voltmeter.
Figure 4: Modified Emitter-Follower Voltmeter
The error can be eliminated using a
potential divider and an additional emitter
follower as depicted in Fig. 4.
When E = 0 V, the base voltage of Q2 is
adjusted to give zero meter current.
9 / 10
This makes VP = 0 V,
VE1 = VE2 = −0.7 V, and V = 0 V.
Hence the voltage drop is removed
Assignment:
Sketch the complete circuit of an emitter
follower voltmeter using a FET stage. Carefully
explain the circuit operation
Submission date: To be determined later.
10 / 10
Quiz
In the circuit above, R1 = 600 Ω, R2 = 1000 Ω, R3 = 400 Ω, R4 = 1000 Ω, and R5 = 500 Ω.
Determine the measurement error caused by the loading effect in the voltmeter when the
voltage across AB is measured. Take the voltmeter internal resistance as 9500 Ω.
1/8
Digital Instruments
2/8
Introduction
While the analog instrument display the
quantity to be measured in terms of
deflection of a pointer, digital
instruments the measurement value in
form of a decimal number.
The digital meter works based on the
principle of quantization.
For illustration purpose, the operating
principles of digital voltmeter (DVM)
(Ramp-type and dual-slope-integrator
DVMs) will be considered.
A DVM essentially consists of
analog-to-digital converter (ADC) with a
set of seven-segment numerical displays to
indicate the measured voltage. See Fig 1.
Ramp-Type DVMs
The ramp-type VDM measures unknown
voltage using a ramp signal, based on the
time it takes for the ramp to rise from a
reference voltage to the input voltage.
The input voltage Vi and the ramp signal
Vr are compared in the comparator.
When Vi ≥ Vr , V1 = 1, the gate is in an
open state and the decade counter counts
the pulses generated from the clock
circuit.
When Vi < Vr , V1 = 0, the gate is closed.
The output of the decade counter
corresponds to the number of clock
pulses, which is proportional to the
duration of the ramp transition.
3/8
Figure 1: A ramp-type DVM. (a) DVM system. (b) DVM waveforms
4/8
Dual-slope-integrator VDM
The latch locks the counter’s output at
the end of measurement, storing the count
representing the voltage for a stable and
accurate display.
A BCD to seven-segment driver converts a
BCD input into signals that control a
seven-segment display. Its function is to
translate the 4-bit BCD code
(representing decimal digits 0–9) into
appropriate outputs
Limitation of ramp-type VDM
The ramp-type VDMs require precise ramp
voltages and precise time periods, both of
which can be difficult to maintain.
The dual-slope-integrator (DSI) VDM,
illustrated in Fig 2 eliminates the
limitations of ramp-type VDM by using a
special type of ramp generator circuit (or
integrator)
The integrator capacitor is first charged
from the analog input voltage and then
discharged at a constant rate to give a
time period that is measured digitally.
The control waveform for the integrator is
derived from the clock generator by use of
a frequency divider
During time t1 the integrator capacitor is
charged to negatively from Vi , giving a
negative-going ramp.
5/8
Figure 2: A dual-slope-integrator DVM. (a) System block diagram. (b) System waveforms
6/8
The voltage V0 produced during this
phase is directly proportional to Vi
The constant current source is then
switched into the circuit to discharge the
capacitor, thus producing a positive going
ramp voltage.
The zero voltage comparator is a voltage
comparator that gives a high output when
the integrator output waveform is
negative, and low output at the end of the
positive-going ramp.
The AND gate has high inputs from both
the zero-crossing detector and the control
waveform only during the positive-going
ramp time, i.e., during t2
Pulses from the clock generator pass
through the AND gate to the counting
circuits during this time.
The counting circuit is reset to zero by
the positive-going edge of the control
wave at the commencement of t2 , so the
output of the counting circuit is a digital
measurement of time t2
Since t2 is directly proportional to V0 , and
V0 is directly proportional to Vi , the
output is a digital measurement of the
analog input.
7/8
References I
[1] David A. Bell. Electronic Instrumentation and Measurements, Second Edition.
Prentice-Hall, 2003.
8/8
Digital Frequency Meter
Accurate frequency measurement is
essential for precise control and
monitoring of systems and devices,
ensuring they operate correctly by
maintaining the proper oscillation rate or
cycles per second.
The number of counted cycles per unit
time indicates the signal frequency.
Assignment:
Explain in detail the operation of a digital
frequency meter depicted in Figure 1.
This is particularly crucial in applications
Submission Due Date:
such as communication systems, power
grids, and electronic circuits, where stable 14 February 2025 at 11:55 PM
frequencies are necessary for signal
DVM Illustration: Calculate the maximum time
integrity and efficient performance.
t1 for the digital voltmeter in Figure 1 of
To measure the frequency of a certain
Lecture 5 presentation. If the clock generator
periodic signal, the waveform of that
frequency is 1.5 MHz. Also suggest a suitable
signal is used to toggle a counter for a
frequency for the ramp generator.
certain fixed time.
1/2
Figure 1: Digital frequency meter. (a) Frequency meter system (b) System waveforms
2/2