EC-305 : Electronic Measurement and

Instrumentation

Instructor- Prof. F A Talukdar

1
Introduction

Instrumentation : Instrumentation is the use

of measuring instruments to monitor and
control a process. It is the art and science of
measurement and control of process
variables within a production, laboratory, or
manufacturing area.
Significance of Measurement
“When you can measure, what you are speaking and
express it in numbers, you know something about and
can express it in numbers, you know something about it;
when you cannot measure it, when you can not express
in it numbers, your knowledge is of a meagre and
unsatisfactory kind” – Lord Kelvin

The measurement confirms the validity of a hypothesis

and also add to it the understanding. This eventually
leads to new discoveries that require new and
sophisticated measuring techniques.

Through measurement a product can be designed or a

process be operated with max. efficiency , minimum
cost and with desired degree of reliability and
maintainability
Few Definitions
Measurement: It is the act, or the result of quantitative comparison between a
predetermined std. and or an unknown magnitude. Since two quantities are
compared and the result are expressed in numerical value.

Measurand: The physical quantity or the characteristic conditions which is the

object of measurement in an instrumentation system is termed as measurand or
measurement variable or process variable, e.g.,
Fundamental Quantity: length, mass, time etc.
Derived Quantity: Speed, Velocity, Pressure etc.

Measurand (Qty. to be measured)

Std. Unknown Quantity Result (Read out)

Process of Comparison
Contd..

Measured Value: Any value or any reading calculated from measurement

system or measuring instrument.

True value: Any value calculated from rated value known as True value of
Actual Value, e.g. Motor Actual Speed

True Value Measured Value

Measuring Instrument

Error : Any deviation of measured

value from true value
Measured Value-True Value
Methods of Measurement

Method of Measurement

Direct Method Indirect Method

The unknown quantity (measurand) In this method the comparison
is directly compared against a standard. is done with a standard
The result is expressed as a numerical through the use of a calibration
number and a unit. Direct methods are s/m. These methods are used
common for the measurement of those cases where the desired
physical quantities like length, mass and parameter to be measured.
time e.g. acceleration, power
Direct Methods are Classified as:

Deflection methods
Deflection method” includes the deflection of pointer on a scale
due to the quantity to be measured. Example: Wattmeter,
ammeter voltmeter

Comparison methods
“Comparison method” include the comparison of the quantity
under measurement with a pre-defined standard quantity which
gives measurement. Example: potentiometer
Sensor VS transducer
Functional Elements of an Instruments

Any instrument or measuring can be represented by

block diagram, that indicates necessary elements and
its functions.
The entire operation of the measuring system
can be understood from the bock diagram
Data storage element

Primary Variable Variable Data Data

sensing conversion manipulation transmission presentation
Qty. to be element element element element element
measured

Data conditioning element Observer

Take an example:

 Just take an example of an Analog meter (Ammeter)

which measures current.

Current Moving Magnets and other Force Mechanical Pointers and

Coil components Linkages scale

Primary Data Data Observer

Sensing conditioning Transmission

BASIC SCHEMATIC OF AMMETER

Classification of Instruments

Measurement involve the use of instruments as a physical

means of determining quantities or variables.

 Absolute/ Secondary Instruments

 Analog/ Digital Instruments
 Mechanical/Electrical or Electronic Instruments
 Active/Passive Instruments
 Manual/Automatic Instruments
 Self contained /Remote Indicating Instruments
 Deflection/null o/p instruments
Active/Passive Instruments
What Are Active Instruments? What Are Passive Instruments?

Active instruments are the Passive instruments are

instruments in which the quantity to instruments where the output is
be measured activates the magnitude produced completely by the
of external power input source that quantity that is measured.
produces the measurement.
An example of a passive
An example of an active instrument is instrument is the pressure
a float-type petrol-tank level measuring device. The pressure
indicator. In this instrument, the of the fluid is translated into
change in petrol level moves a movement of a pointer against
potentiometer arm and the output a scale. The energy expended
signal consists of a proportion of the in moving the pointer is derived
external voltage source applied entirely from the change in
across the two ends of the pressure measured; there are
potentiometer. The energy in the no other energy inputs to the
output signal comes from the system.
external power source; the primary
transducer float system is merely
modulating the value of the voltage
from this external power source.
Absolute or Primary/Secondary
Instruments
Absolute Instruments
 It gives the magnitude of quantity under measurement in
terms of physical constants of the instrument e.g. Tangent
Galvanometer
 In this type of instruments no calibration
or comparison with other instruments is necessary.
 They are generally not used in laboratories and
are seldom used in practice by electricians and engineers.

Secondary Instruments
 These instruments are so constructed that the quantity being measured
can only be determined by the output indicated by the instrument.
 These instruments are calibrated by comparison with an absolute instrument or
another secondary instrument, which has already been calibrated against an
absolute instrument.
e.g. Ammeter, Voltmeter etc.
Classification of Secondary Instruments
(a) Classification based on the various effects of electric current (or voltage)
upon which their operation depend.

Magnetic effect: Used in ammeters, voltmeters, watt-meters, integrating meters etc

Heating/thermal effect: Used in ammeters and voltmeters
Electromagnetic field of attraction/repulsion
Electrostatic effect: Used in voltmeters
Electromagnetic induction effect: Used in ac ammeters, voltmeters, watt meters and
integrating.

(b) Classification based on the Nature of their Operations

•Indicating instruments: Indicating instruments indicate, generally the quantity to be

measured by means of a pointer which moves on a scale. Examples are ammeter,
voltmeter, wattmeter etc.
•Recording instruments: These instruments record continuously the variation of
any electrical quantity with respect to time. In principle, these are indicating
instruments but so arranged that a permanent continuous record of the indication
is made on a chart or dial
Classification of Secondary Instruments

Integrating instruments: These instruments record the consumption of the total

quantity of electricity, energy etc., during a particular period of time. : Ampere-hour
meter: kilowatt thour (kWh) meter, kilovolt-ampere-hour (kVARh) meter.
(c) Classification based on the Kind of Current that can be Measurand.
• Direct current (dc) instruments
• Alternating current (ac) instruments

(d) Classification based on the method used

Direct measuring instruments: These instruments converts the energy of the
measured quantity directly into energy that actuates the instrument and the
value of the unknown quantity is measured or displayed or recorded directly
Examples are Ammeter, Voltmeter, Watt meter etc.
•Comparison instruments: These instruments measure the unknown quantity by
comparison with a standard. Examples are dc and ac bridges and potentiometers.
They are used when a higher accuracy of measurements is desired
Analog /Digital Instruments

Analogue Instruments: The signal of an analog unit vary in a

continuous fashion and can take an infinite no. of values in a
given range. E.g. ammeters, voltmeter, wrist watch,
speedometer etc.

Digital instruments: Signals varying in discrete steps and

taking on a finite no. of different values in a given range are
digital signals e.gs timer on a score board, odometer of an
automobile
Characteristics of Instruments
The performance of an instrument is described by
means of a quantitative qualities termed as
characteristics. These are broken down into:
1. Static Characteristics: These characteristics
pertain to a system where the quantities to be
measured are constant or vary slowly with time
2. Dynamic Characteristics: Performance criteria
based on dynamic relations (involving rapidly
varying quantities)
Static Characteristics

 Accuracy: It is the closeness with which an instrument

reading approaches the true value of the quantity
measured.
 Precision : The degree to which repeated measurements
show the same results.

LowAccuracy Low Accuracy

Low Precision High Precision

HighAccuracy High Accuracy

Low Precision High Precision
Accuracy and Precision

Accuracy may be specified in terms of inaccuracy or limit of errors

and can be expressed in the following ways:
1. Point Accuracy
2. Accuracy as “Percentage of Scale Range”
3. Accuracy as “Percentage of True value”

Indication of Precision
Significant Figures: It is an indication of precision of
measurement. It conveys the actual information regarding the
magnitude and the measurement precision of a quantity. The
more the significant figures, the greater the precision.
e.g. 302 A = 3 S.F.
302.10 V = 5 S.F.
0.00030 = 5 S.F.
or ?
Static Characteristics

Resolution or Discrimination: The smallest detectable

incremental change of the input parameter that can be
detected in the output signal. Eg; Scale, Multi range meters.
Sensitivity: For an instrument or sensor with input x and
output y. Sensitivity = dy/dx

output output

Input Input

Static sensitivity = Infinitesimal change in output /infinitesimal change in

input
Static Characteristics

 Repeatability: Closeness of output reading when the same input is

applied repeatedly over a short period of time with the same
measurement conditions, same instrument and observer, same
location and same conditions of use maintained throughout.
 Reproducibility: Closeness of output readings for the same input
when there are changes in method of measurement, observer,
location, conditions of use, and time of measurement.
 Span & Range:
Range : Range of a Instrument is the capacity to measure the
value more accurately between calibrated scale of its minimum
and maximum value. High measurement possible
Span : Difference between max. and min measurement possible e.g.,
Thermocouple (700 0C to 1200 0C), Ammeter (0 to 10 A)
 Dead zone : The largest of a measured variable for which the
instrument does not respond Cause: friction in mechanical
measurement system
Static Characteristics

 Dead Time :The time before the instrument begins to respond after the
measured quantity has been changed. e.g: Camera, Data acquisition
card,Ammeter
Static Characteristics

 Drift : It is an undesired gradual departure of the instrument o/p over a

period of time that is unrelated to changes in i/p , operating conditions or
load.
The drift may be caused by the following factors:
1) Mechanical vibrations
2) Temp. changes
3) Wear and Tear etc.
Classification:
1) Zero drift : If the whole of instrument calibration/ characterstics gradually
shifts by same amount. It may be due to undue warming up of tube of
electronic tube circuits or slippage and can be corrected by shifting
pointer position
.
Characteristics with zero drift

output
Normal characteristics

Zero Drift
Static Characteristics

2) Span or senstivtity Drift : If the calibration from zero upwards changes

proportionally

output
Normal characteristics

Span drift

3) Zonal Drift : When the drift occurs only over a portion of span of an
instrument.

output
Normal characteristics

zonal drift
Static Characteristics

Linearity: If the calibration from zero upwards changes

proportionally. If input-output relationship is a straight line passing
through origin
• Nonlinearity causes lot of problem during signal conditioning even though it is
more accurate in some cases e.g. LVDT (linear) , Thermistor (Non-linear)

Actual calibration curve

Output
Idealised St. Line

Input

Any departure from straight line relationship is non-linearity

Static Characteristics

 Error: Error is the degree to which a measurement conforms to the

expected or true value .Errors are due to measuring instruments (causing
the change in the value of the parameter being measured) or due to
persons carrying out the measurements (human errors).Errors may be
expressed as absolute or percentage.

Types of Errors
 Gross errors
- Human errors
 Systematic Error
Instrument errors
Environmental errors
Observational errors
 Random errors
Static Characteristics

 Error: Error is the degree to which a measurement conforms to the

expected or true value .Errors are due to measuring instruments (causing
the change in the value of the parameter being measured) or due to
persons carrying out the measurements (human errors).Errors may be
expressed as absolute or percentage.

Types of Errors
 Gross errors [generally, human errors]
 Systematic errors
- Instrument errors
- Environmental errors
- Observational errors
 Random errors
 TYPES OF STATIC ERROR
1) Gross Error
 causes by human mistakes in reading/using instruments
 may also occur due to incorrect adjustment of the
instrument and the computational mistakes
 cannot be treated mathematically
 cannot be eliminated but can be minimized e.g., Improper use
of an instrument.
 This error can be minimized by taking proper care in
reading and recording measurement parameter.
 In general, indicating instruments are affected by ambient
conditions to some extent when connected into a complete
circuit.
 Therefore, several readings (at three readings) must be taken
to minimize the effect of ambient condition changes.
 TYPES OF STATIC ERROR (contd)

2) Systematic
Error
- due to shortcomings of the instrument (such as
defective or worn parts, ageing or effects of the
environment on the instrument)
• In general, systematic errors can be subdivided into static and
dynamic errors.
• Static – caused by limitations of the measuring device or
the physical laws governing its behavior.
• Dynamic – caused by the instrument not responding very
fast enough to follow the changes in a measured variable.
 TYPES OF STATIC ERROR (contd)

Three types of systematic errors are :-

1. Instrumental error
2. Environmental error
3. Observational error
 TYPES OF STATIC ERROR (cont)

(i) Instrumental error

Inherent while measuring instrument because of their
mechanical structure (eg: in a D‘Arsonval meter, friction
in the bearings of various moving component, irregular
spring tension, stretching of spring, etc)
Error can be avoided by:
a) selecting a suitable instrument for the particular
measurement application
b) apply correction factor by determining instrumental
error
c) calibrate the instrument against standard instrument
 TYPES OF STATIC ERROR (contd)
(ii) Environmental error
- due to external condition effecting the measurement
including surrounding area condition such as change in
temperature, humidity, barometer pressure, etc
- to avoid the error :-
a) use air conditioner
b) sealing certain component in the instruments
c) use magnetic shields
(iii) Observational error
- introduces by the observer most common: parallax
error and estimation error (while reading the scale)
e.g: an observer who tend to hold his head too far to the
left while reading the position of the needle on the
scale.
 TYPES OF STATIC ERROR (contd)
3) Random error

- due to unknown causes, occur when all systematic error has

been accounted
-accumulation of small effect, require at high degree of accuracy
- can be avoid by
(a)increasing number of reading
(b) use statistical means to obtain best approximation of true
value
Errors in Measurement

Static Error/Absolute Error- It is defined as the difference between the measured

value and the true value of the quantity. Then:
∆ A= Am-At (1)
Where ∆ A= error
Am = measured value of
quantity At = True value of
quantity
∆ A is also absolute static error of
(2)
quantity A we have ɛ0 = ∆ A
Where ɛ0 = absolute static error of
quantityA (3)
Relative Static Error
ɛr = absolute error/ true value
= ∆ A/ At
= ɛ0 /At
Percentage static error % ɛr = ɛ x 100
r (4)
We have At = Am - ∆ A

= Am - ɛ 0= Am - ɛr At = Am/(1+ ɛr ) (5)
Errors in Measurement

Equation (5) can also be written as

At = Am (1- ɛr) (6)

Static Correction
∆ C= At -Am (7)
Question

1. Which of the following instrument is more quality instrument?

InstrumentA Instrument B
∆ A= 1A ∆ A= 10A
At = 2 amp At= 1000 amp

a) OnlyA
b) Only B
c) Both A andB
d) None of above
Problem Set-I (Due date 15 Sept 21)
1. Find the resolution of a 10-bit ADC, if it is excited by a 10V source.

2. The accuracy specified for a pressure gauge of range 0-10kPa is 2%. Find the
maximum error in measurement in Pa if it gives a reading of 4.0 kPa.

3. Justify the following statements:

(a) A potentiometer is a zero-th order device.
(b) A bare thermocouple is a first order device.
(c) An accelerometer is a second order device.

4. Followings are the excerpts from the specifications of a laser displacement sensor:
(a) Measurement range: ± 10mm (b) Measurement point: 40mm (c) Resolution: 3 μ m
(d) Linearity: 1% Full Scale, (e) Response time: 0.15ms, (f) Linear output: 4-20mA
Answer the following questions
a) Explain the meaning of each term
b) Suppose, the distance between the sensor and the object is 35mm. Then what
would be output in mA?
c) What is the error due to nonlinearity under the above condition?
d) Find out the sensitivity of the sensor in mA/mm
Errors in sum and Difference of Quantities

Error in the sum of quantities Error in the difference of quantities

equal the sum of absolute errors equal the sum of absolute errors

Do it for multiplication and division

Errors in product and Quotient of Quantities
Dynamic Characteristics

1) Speed of Response: It is defined as the rapidity with which a measurement system

responds to changes in the measurement quantity.

2) Measurement Lag: It refers to retardation or delay in the response of

measurement system to changes in measured quantity. The lag is caused by
conditions such as capacitance, inertia or resistance.
Measuring lag are of two types:
a) Retardation type lag
b) Time delay type lag

3) Fidelity: Fidelity of a system is defined as the ability of the system to

reproduce the output in the same form as the input. It is defined as the degree
to which a measurement system indicates changes in the measured quantity
without any dynamic error.
4) Dynamic error or measurement error :The dynamic error is the difference
between the true value of the quantity changing with time and the value
indicated by the instrument if no static error is assumed. However, the total
dynamic error of the instrument is the combination of its fidelity and the time
lag or phase difference between input and output of the system.
Order of the Instruments
Order of the Instruments
 First-Order Systems: Step Response
Steady-state
response
Error Fraction
Suppose we rewrite the equation as
y(t) KA y(t)  y()
em(t)    e t /
y0  KA y(0)  y()

63.2% response to
input response
y(t)  KA t /
e
y(0) KA

90%
99%
Order of the Instruments
Numericals

 The output voltage of a 5 V DC supply is measured as 4.9 V. Find (1) Absolute error (2)
Percent error (3) Relative accuracy and (4) Percent accuracy
 The three resistors R1 , R2 and R3 have the following ratings:
R1= 25Ω± 4 %
R2= 65Ω± 4%
R3= 45Ω± 4%
Determine the following
a) Limiting value of resultant resistance
b) % Limiting error of series combinations of resistance.
Deflection /Null o/p Instruments

Deflection
Null
 Only one source of input reqd.
Require two input- measurand
 Output reading is based on the
and balance input
deflection from the initial condition of
Must have feedback operation that
the instrument
compares the measurand with std.
• The measurand value of the qty.
value Most accurate and sensitive
depends on the calibration of the
instrument
Essential Requirements of Indicating Instruments

1. Deflecting torque (Td) : Deflecting torque causes the moving

system and pointer of the instrument to move from its zero
position. Production of deflecting torque depends upon the type of
indicating instrument and its principle of operation
2. Controlling torque (Tc) : Controlling torque limits the movement of
pointer and ensures that the magnitude of deflection is unique and is
always same for the given value of electrical quantity to be measured.

Two methods of Controlling Torque

a) Spring Control method

b) Gravity control method

2
6
Spring Control Method

 Two phosphor bronze hair springs of

spiral shapes are attached to the
spindle of the moving system of the
instrument.
 They are wound in opposite direction
 Pointer is attached to the spindle of
the moving system

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Gravity Control Method

 In gravity control method, a

small weight is attached to
the spindle of the moving
system
 Due to the gravitational
pull, a control torque
(acting in opposite
direction to the deflecting
torque) is produced
whenever the pointer tends
to move away from its
initial position.

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Essential Requirements of Indicating Instruments
Gravity Control Spring Control
1 Adjustable small weight is used which Two hair springs are used which exert controlling
produces the controlling torque torque
2 Controlling torque can be varied Controlling torque is fixed
3 Performance is not temperature dependent Performance is temperature dependent

4 Scale is non-uniform Uniform scale

5 Controlling torque is proportional to Sin Ɵ Controlling torque is proportional to Ɵ

6 Readings can not be taken accurately Readings can be taken very accurately

7 System must be used in vertical position System need not be necessarily in vertical position
only
8 Proper levelling is required in gravity No levelling is required
control
9 Simple, cheap but delicate Simple, rigid but costlier compared to gravity
control
10 Rarely used for indicating and portable Very popularly used in most of the instruments
instruments
Essential Requirements of Indicating Instruments

3. Damping Torque: Damping torque minimizes the oscillations of the pointer about
the final steady state deflection and makes it steady.. In the absence of this torque,
pointer continues oscillating to its final position after reaching to its final position.
Depending on the magnitude of damping, it can be classified as underdamped,
over damped and critically damped
Damping Methods

Air friction Damping

Fluid Friction Damping
 Electromagnetic/ Eddy
current Damping
Air Friction Damping
Fluid Friction Damping
Eddy Current Damping
Types of Instruments

1. Permanent Magnet Moving Coil (PMMC)

type Instrument.
2. Moving Iron type Instrument
3. Electro Dynamometer type Instrument
4. Hot wire type Instrument
5. Thermocouple type Instrument
6. Induction type Instrument
7. Electrostatic type Instrument
8. Rectifier type Instrument
D’ARSORVAL METER MOVEMENT

• Also called Permanent-Magnet Moving Coil (PMMC).

• Based on the moving-coil galvanometer constructedby
Jacques d’ Arsonval in 1881.

• Can be used to indicate the value of DC and ACquantity.

• Basic construction of modernPMMC can be seen in
Figure .

61
Operation of D'Arsonval Meter

• When current flows through the coil, the core will rotate.
• Amount of rotation is proportional to the amount of current
flows through the coil.

• The meter requires low current (~50uA) for a full scale

deflection, thus consumes very low power (25-200uw).

• Its accuracy is about 2% -5% of full scale deflection

62
Permanent Magnet Coil Instrument (PMMC)
Permanent Magnet Coil Instrument (PMMC)-
Torque Equation
Torque Equation for PMMC Instrument:
Let,
N = Number of turns of the coil
l = Vertical length of the coil (m)
d = Horizontal length of the coil (m)
i = Current flowing through the meter
B = Flux density between the poles (wb/m3)
α = Angle made by the conductor with the magnetic field
In galvanometer the field present is radial. Therefore, α = 90°. From Lorentz's
force equation, we have, the force exerted on each coil side is,
F = N B Ii sinα
F = N B Ii sin90°
F = N B Ii ...(1)
Therefore, deflecting torque is given by, Td = F × Distance = F × d ...(2)
Substituting equation 1 in 2, we get,Td = N B Il × d
But the area of the coil is given by, A = l × d
Td = N B Ai
∴ Td = GI (G = galvanometer constant = NBA)
The controlling torque is obtained by the spring control action.∴ Tc = Kθ
Assuming θ as deflection and K as spring constant.
At steady-state deflection, we know that Tc = Td i.e., Kθ = GI.
∴ Deflection, θ = GI/K
Permanent Magnet Coil Instrument (PMMC)-
Torque Equation
Controlling Torque: The value of
control torque depends on the
mechanical design of the control
device. For spiral springs and strip
suspensions, the controlling torque is
directly proportional to the angle of
deflection of the coil.
Permanent Magnet Coil Instrument (PMMC)-
Torque Equation
It is provided by the induced currents in a metal former or core on which the coil is
wound or in the circuit of the coil itself.
Errors

 Frictional Error
 Temperature Error
 Errors due weakening of permanent magnet
 Error due to ageing of spring
 Stray magnetic field error
Advantages of PMMC

 Low Power consumption

 Scales are uniform
 No hysteresis loss (iron loss)
 High Torque/wt. ratio
They have a very effective and efficient eddy
current damping
 Range can be extended with shunts or
multipliers
Disadvantages of PMMC

 Use only for dc

 The cost of these instruments is higher than that of
moving iron instrument
Numericals

1. A permanent magnet moving coil instrument has a coil of dimensions 15mm x 12

mm. The flux density in the air gap is 1.8 x 103 Wb/ 2 and the spring constant is
0.14 x 10 −6 Nm/rad. Determine the number of turns required to produce an
angular deflection of 90 degrees when a current of 5mA is flowing through the
coil.
2. The control spring of an instrument has the following dimensions:
Length of strip =370 mm , thickness of strip =0.073 mm, width of strip=
0.51mm The young modulus is 112.8 GN/ 2. Estimate the torque exerted by
spring when it is turned through 90𝑜.
3. The coil of a moving coil voltmeter is 40mm long and 30mm wide and has 100
turns on it. The control spring exerts a torque of 240 x 10 −6 N-m when the
deflection is 100 divisions on full scale. If the flux density of the magnetic field in
the air gap is 1.0 wb/ 2, estimate the resistance that must be put in series with
the coil to give one volt per division. The resistance of the voltmeter coil may be
neglected.
Ammeter Shunts

DC Ammeter
 Its is always connected in series
 low internal resistance
 maximum pointer deflection is produced by a very small current
 For a large currents, the instrument must be modified by connecting a very
low shunt resister
Extension of Ranges of Ammeter
- Single Shunt Type of Ammeter
Ammeter Shunts
V sh  V m
I s h R sh  I m R m
I m Rm
R sh 
I sh
Ish  I  I m
I m Rm
 R sh 
I  Im

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Ammeter Shunts

 Multi-range Ammeters Make-before-break switch

The instrument is not left without a shunt in parallel with it.
During switching there are actually two shunts in parallel with
the instrument.
Ayrton or Universal Shunts
Numerical

Design an Aryton shunt to provide an ammeter with a current ranges 1A, 5A and
10A. A basic meter resistance is 50 ohms and full scale deflection current is
10mA.
Numerical
Voltmeter Multipliers

A basic d’Arsonval movement can be converted into dc voltmeter by adding

in series resistor multiplier as shown in figure.
Multirange dc Voltmeter

A DC voltmeter can be converted into a multirange voltmeter by connecting a

number of resistors (multipliers) in series with the meter movement. A practical
multi-range DC voltmeter is shown in Figure

,
Ammeter/Voltmeter Sensitivity

 Ammeter sensitivity is determined by the amount of current required by the

meter coil to produce full-scale deflection of the pointer.
 The smaller the amount of current required producing this deflection, the
greater the sensitivity of the meter.

 The sensitivity of a voltmeter is given in ohms per volt. It is determined by dividing

the sum of the resistance of the meter (Rm), plus the series resistance (Rs), by the
full-scale reading in volts. In equation form, sensitivity is expressed as follows:

 This is the same as saying the sensitivity is equal to the reciprocal of the full-scale
deflection current. In equation form, this is expressed as follows:
Numericals

 Calculate the value of the shunt resistance required to convert a 1-mA meter
movement, with a 100 Ohm internal resistance, into a 0- to 10 mA ammeter

 Compute the value of the shunt resistors for the circuit below. I3 = 1A, I2 = 100 A,
I1 = 10
mA, Im = 100 uA and Rm = 1K Ohm.
Solutions

 Solution 1: Vm  I m Rm Vsh  Vm 0.1V

 1mA* 100  0.1V I sh  I  I m

 Solution 2 : This is the shunt for the 10 mA  10mA 1mA  9mA

range. When the meter is set on the 100-mA Vsh  0.1V 11.11
range, the resistor Rb and Rc provide the Rsh 
shunt. The total shunt resistance is found by I sh 9mA
the equation
Moving Iron Instruments-Torque Equation

Classification
1. Moving Iron Attraction Type Instruments
1. Moving Iron Repulsion Type Instruments.
The working principle of Similar to attraction force there will be
attraction type moving iron repulsion force when same magnetic poles are
instrument is based on magnetic placed near each other. Based upon this
attraction, which attracts an iron repulsion force between two like poles the
piece when placed near a repulsion type moving iron instruments were
magnetic field. Here, the developed.
magnetic field will be produced Repulsion type moving iron instruments are
by an electromagnet used both for ac and dc measurements. In
these instruments, when current flows through
the coil, the two vanes i.e., fixed vane and
movable vane are magnetized and same
polarities are induced in it which results in a
force of repulsion between them.
Moving Iron Instruments
Moving Iron Instruments

 Radial Vane Type

 Coaxial Vane Type
Torque Equation
Suppose the initial current is I, the instrument inductance L and the deflection ș. If the current is
increased by di then the deflection changes by d ș and the inductance by dL. In order to affect a n
increment the current there must be an increase in the applied voltage given by
Advantages

 Can be used both in D.C. as well as in A.C. circuits.

 Robust and simple in construction.
 Possess high operating torque.
 Can withstand overload momentarily.
 Since the stationary parts and the moving parts of the
instrument are simple so they are cheapest.
 Suitable for low frequency and high power circuits.
 Capable of giving an accuracy within limits of both precision
and industrial grades.
Limitations

 Scales not uniform.

 For low voltage range the power consumption is higher.
 The errors are caused due to hysteresis in the iron of the operating system
and due to stray magnetic field.
 In case of A.C. measurements, change in frequency causes serious error.
 With the increase in temperature the stiffness of the spring decreases.
Errors

Errors with both A.C and D.C

Error due to hysteresis: Because of hysteresis in the iron parts of the operating system, the readings are
higher for descending values but lower for ascending values. The hysteresis error is considerably reduced
by using mu-metal or permalloy which have negligible hysteresis loss.

Error due to stray magnetic fields: Since the operating magnetic field of the moving iron instruments in
comparatively weak, therefore, stray fields (fields other than the operating magnetic field) affect these
instruments considerably. Thus the stray fields cause serious errors. These errors can be minimized by
using an iron case or a thin iron shield over the working parts.

Error due to temperature: The effect of temperature change on moving iron instrument arises mainly
from the temperature coefficient of spring. With the change in temperature stiffness of the spring varies
which causes errors. However, for voltmeters, both the temperature coefficient of spring and temperature
coefficient of resistance of voltmeter circuit may balance each other.

Errors with A.C. only:

Error in moving iron instruments due to change in frequency: The change in frequency produces a
change in impedance of the coil and change in magnitude of eddy currents. The increase in impedance of
the coil with the increase in frequency causes serious errors in case of voltmeters only. However, this
error can be eliminated by connecting a condenser of suitable value in parallel with the swamp resistance
`r’ of the instrument. The impedance of the whole circuit of the instrument becomes independent of
frequency if C = L/r2, where C is the capacitance of the condenser.
Electrodynamometer Instruments

It is a transfer type
instrument that is
calibrated with dc but
can be used for ac
measurements without
any changes. The
electrodynamic
instruments are often
used in ac voltmeters and
ammeters but they can
be used as wattmeters
with slight changes. Let
us see the construction
and working principle of
the electrodynamometer
type wattmeter.
Electrodynamometer Instruments
Torque Equation
Let us assume that the current in fixed coil be I1 and that in moving coil be I2 as shown in
figure below.
Also assume that,
L1 = Self inductance of fixed coil
L2 = Self inductance of moving coil
M = Mutual inductance between fixed and moving coils
Thus,
The flux linkage of fixed coil Ø1 = L1i1 + Mi2
The flux linkage of moving coil Ø2 = L2i2 + Mi1
The electrical energy input to the instrument,
= e1i1dt + e2i2dt
But according to Faraday’s Law,
e1 = d Ø1/dt
and e2 = d Ø2/dt
Therefore energy input to the instrument
= i1d Ø1 + i2d Ø2
= i1d (L1i1 + Mi2) + i2d(L2i2 + Mi1)
= i1L1di1 + i12dL1 + i1i2dM + i1Mdi2 + i2L2di2 + i22dL2 + i1i2dM + i2Mdi1
Since L1 and L2 are constant, therefore dL1 = 0 and dL2 = 0
= i1L1di1 + i1i2dM + i1Mdi2 + i2L2di2 + i1i2dM + i2Mdi1 …………(1)
Some of the above input energy to electrodynamometer instruments are stored in the form of
magnetic energy in the coil while rest is converted into mechanical energy of moving coil.
Thus we can write,
Energy Input = Mechanical Energy + Stored Energy
Mechanical Energy = Electrical Input – Stored Energy …………(2)
Thus to find the mechanical energy, we need to find the change in stored energy in the magnetic
field of the coil. Let us assume an infinitesimally small time dt for the sake of calculation of
change in stored energy.
Change in stored energy
= d(1/2L1i12 + 1/2L2i22 + Mi1i2)
= i1L1di1+ i2L2di2+ i1Mdi2 + i2Mdi1+ i1i2dM+(i12/2)dL1 + (i22/2)dL2
But L1 and L2 are constant, therefore dL1 = 0 and dL2 = 0
= i1L1di1+ i2L2di2+ i1Mdi2 + i2Mdi1+ i1i2dM ……(3)
From equation (1), (2) and (3),
Mechanical Energy = i1i2dM
Let Td be the dflectiong torque and dƟ be the change in deflection, then mechanical energy
= TddƟ
TddƟ = i1i2dM
⇒Td = i1i2dM/dƟ
The above equation gives the deflecting torque in electrodynamics or electrodynamometer
instruments. It can be seen that deflecting torque depends upon the multiplication of
instantaneous value of current and rate of change of mutual inductance between the fixed and
moving coil.
Case-1: When DC quantity is being measured.
Let I1 and I2 be the current in fixed and moving coil respectively. Therefore deflecting torque Td =
I1I2dM/dƟ
But this deflecting torque is controlled by the spring. Spring provides the controlling torque. The
controlling torque due to spring for a deflection of Ɵ
Tc = KƟ where K is spring constant.
At equilibrium the controlling torque and deflecting torques are equal, hence
Tc = Td
⇒KƟ = I1I2dM/dƟ
⇒Ɵ = (I1I2dM/dƟ)/K
Case-2: When AC quantity is being measured.
Let i1 and i2 are sinusoidal current having a phase displacement of Ø. Therefore we can write as
i1 = Im1Sinwt
i2 = Im2Sin(wt-Ø)
Thus the instantaneous deflecting torque is given as
Td = (Im1Sinwt)[ Im2Sin(wt-Ø)]dM/dƟ
The average torque for one time period of the currents are given by
Td = (I1I2CosØ)dM/dƟ
Where I1 = RMS Value of i1
I2 = RMS value of i2
From the above two cases, we can have following conclusions:
•For sinusoidal alternating current, the deflecting torque is determined by the product of RMS
value of coil currents and the cosine of phase angle between them.
•When the instrument is used for AC, the instantaneous torque is proportional to i2. Thus the
torque varies as the current varies but the direction of torque remains the same. Because of the
inertia of the instrument, the needle does not follow the change in torque rather it takes a
position where the average torque becomes equal to the controlling torque.
Advantages

 As the coils are air cored, these instruments are

free from hysteresis and eddy current losses
 They have a precision grade security
 These instruments can be used on both a.c. and d.c. They
are also used as a transfer instruments
 Electrodynamometer voltmeter are very useful where
accurate r.m.s. values of voltage, irrespective of
waveforms, are required
 Free from hysteresis errors
 Low power Consumption
 Light in weight
Disadvantages

 These instruments have a low sensitivity due to low torque

to weight ratio. Also it introduces increased frictional
losses. To get accurate results, these errors must be
minimized.
 They are more expensive than other type of instruments.
 These instruments are sensitive to overload and
mechanical impacts. Therefore, care must be taken
while handling them.
 They have a non-uniform scale.
 The operating current of these instruments is large due to
the fact that they have weak magnetic field.
Errors

1. Torque to weight ratio

2. Frequency errors
3. Eddy current errors currents
4. Stray magnetic field error
5. Temperature error
Numericals

1. In an electrodynamometer instrument the total resistance of the voltage coil

circuit is 8.00 Ω and mutual inductance changes uniformly from -173µH at zero
deflection to +175µH at full scale, the angle of full scale being 95 degree. If a
potential difference of 100V is applied across the voltage circuit, and a current of
3A at a power factor of 0.75 is passed through the current coil, what ill be the
deflection , if the spring control constant is 4.63 x 106 − /
Electronic Instruments

Disadvantages of PMMC voltmeter

Low input impedance: Loading
effect
Insufficient sensitivity to detect low
level signal
Approach
Utilized electronic devices such as
BJT, FET or op amp to solve the
above problems

Electronic voltmeters

Analog instrument
Digital instrument
 Important limitations
Reference : Electronic Instrumentation and Measurement, DAVID A.Bell, P.86 Chapter 4

“…cannot measure very low voltage + resistance is to low for measurements in

high--impedance circuitry”
high

Moving-coil
Moving-
Instrument

Attenuator
(signal Conditioner)

Sensor

http://www.allaboutcircuits.com/vol_1/chpt_8/3.html

PAGE : 9
 Important limitations
Reference : Modern Electronic Instrumentation and Measurement Techniques, Albert D.Helfrick and William D.Cooper, P.132-133 Chapter 6

Moving-coil
Moving-
Instrument

Amplifier
(signal conditioner)

Attenuator
(signal conditioner)

Sensor

PAGE : 10
 Important limitations

Sensitivity Input Impedance
V
The sensitivity of a VOM, given in the unit ohm/volt, has been described earlier as the reciprocal of
the meter movement’s full-
full-scale current.

Example..

A 50
50--µA movement used with a multiplier resistor to make a voltmeter has a
sensitivity of 1/5 x 10-5, or 20
20,,000 Ω/V.

On low voltage ranges, then, the impedance of such an instrument tends to

be low.

In the case of the 20 20,,000 Ω/V meter with 0-to-

to-0.5-V range, the input
impedance is only (0
(0.5 V)(
V)(20
20,,000 Ω/V), or 10 kΩ

Reference : Elements of Electronic Instrumentation and Measurement, Joseph J. Carr, P.152 Chapter 7

PAGE : 11
 Rm

Voltmeter

R1

EB

Rs Basic PMMC

Ammeter Ohmmeter
D Rm

AC voltmeter
 Electronic
voltmeter

RS
Ammeter

R1

Electronic Electronic
R2 voltmeter EB R1 voltmeter

Basic Electronic
voltmeter
Voltmeter Ohmmeter
D

Electronic
voltmeter

AC voltmeter
Loading Effect

R1 5V 100kΩ 6.7 V
100kΩ
10 V 10 V
R2 5V 100kΩ 3.3 V V 100kΩ
100kΩ

100 // 100
Vmeas = 10 V = 3.3 V
100 + 100 // 100

Circuit before measurement Circuit under measurement

100kΩ 6V 100kΩ 5.2 V

10 V 10 V
100kΩ 4.8 V V
100kΩ 4V V 200kΩ 1000kΩ

200 // 100 1000 // 100

Vmeas = 10 V = 4.0 V Vmeas = 10 V = 4.8 V
100 + 200 // 100 100 + 1000 // 100
 Loading Effect
Example Find the voltage reading and % error of each reading obtained with a voltmeter on (i) 5 V
range, (ii) 10 V range and (iii) 30 V range, if the instrument has a 20 kΩ/V sensitivity, an accuracy
1% of full scale deflection and the meter is connected across Rb

SOLUTION The voltage drop across Rb with output to the voltmeter connection

Ra
45kΩ
50 V
Rb V Rm
5kΩ
Loading Effect

Range Vb . Loading Meter Total % error

(V) (V) error (V) error (V) error (V)
5 4.78 -0.22 ± 0.05 ± 0.27 ± 5.36
10 4.88 -0.12 ± 0.1 ± 0.22 ± 4.40
30 4.95 -0.05 ± 0.3 ± 0.35 ± 6.10
 Transistor Voltmeter: Emitter Follower
+
Basic concept IB
+
VBE IE = Im
Emitter
Voltage to be Vin Rs
VCC
follower
measured
Vin Rm
Vin
Ri =
IB
reduce
increase input resistance
output resistance - -

Voltage drop across meter: Vm = Vin − VBE PMMC

where VBE is base-emitter voltage ~ 0.7 V for Si Schematic diagram of emitter follower

Vin − VBE
Meter current: Im =
Rs + Rm
IE hFE = Transistor current gain (Typical
Transistor base current: IB ≈
hFE values ~ 100-200
Transistor Voltmeter: Emitter Follower
Vin V
Circuit input resistance: Ri = ≈ hFE in ≈ hFE ( Rs + Rm )
IB IE

Example The simple emitter-follower circuit has VCC = 20 V, Rs+Rm = 9.3 kΩ, Im = 1mA at
full scale, and transistor hFE = 100
(a) Calculate the meter current when Vin = 10 V
(b) Determine the voltmeter input resistance with and without the transistor.
+ SOLUTION
IB
+
VBE IE = Im

Rs
VCC
Vin Rm
Vin
Ri =
IB

- -
 Transistor Voltmeter: Emitter Follower

*The base-emitter voltage drop (VBE) introduces some limitations in using emitter
follower as a voltmeter:
•The circuit cannot measure the input voltage less than 0.6 V
•a non-proportional deflection: error
From the above experiment, if we apply Vin with 5 V, the meter should read half of full
scale I.e. Im = 0.5 mA. But, the simple calculation shows that Im = 0.46 mA

+VCC Bridge configuration

R4
Vm = VE1 − VE 2
Q1 V Q2 R5

Vin
Rs Rm
VP
where VE1 = Vin − VBE1 VE 2 = VP − VBE 2
R2 VE1 VE2 R3 R6
I2 I3 Zero adjust
-VEE
Use negative supply also to
PMMC measure Vin < 0.6 V
Practical emitter-follower voltmeter using second transistor Q2 and voltage divider R4, R5
and R6 to eliminate VBE error in Q1
Transistor Voltmeter: Emitter Follower
At the condition of Vin = 0, Vp should be set to give zero meter reading, Vm = 0.
Therefore, the potentiometer R5 is for the zero adjust.
If transistors Q1 and Q2 are identical, VBE1 = VBE2
Vm = VE1 − VE 2 = Vin − VBE1 − (V p − VBE 2 ) = Vin − V p

At Vin = 0 -> Vm = 0, give Vp = 0

Consequently, if Vp is set properly, Vm will be the same as Vin

Example An emitter-follower voltmeter circuit as shown in the previous picture has R2

= R3 = 3.9 kΩ and supply with ±12 V. Calculate the meter circuit voltage when Vin = 1 V
and when Vin = 0.5 V. Assume, both transistors have VBE = 0.7 V

SOLUTION when Vin = 1 V

when Vin = 0.5 V

 Voltage Range Changing: Input Attenuator

Input
The input attenuator accurately divides the voltage to
Range Switch
be measured before it is applied to the input transistor.
800k Ra 1V
Calculation shows that the input voltage Vin is always 1
5V
V when the maximum input is applied on any range
100k Rb
Voltage to
10V Vin To meter
be measured E
60k Rc
25V Example On the 5 V range:
40k Rd
Rb + Rc + Rd
Vin = 5 V ×
Ra + Rb + Rc + Rd
The measurement point always sees a 100 kΩ + 60 kΩ + 40 kΩ
= 5 V×
constant input resistance of 1 MΩ 800 kΩ + 100 kΩ + 60 kΩ + 40 kΩ
=1 V
FET Input Voltmeter

The addition of FET at the input gives higher input resistance than can be achieved
with a bipolar transistor
Input FET Emitter
attenuator input stage follower

+VCC
800k Ra 1V
R4
5V

Rb
100k Q1 V Q2
10V EG VG S R5
E
VS VP
60k Rc
Rs+Rm
25V I2 R2 R3
R6
I3
40k Rd
-VEE

PMMC
A FET Input Voltmeter
Vm = VE1 − VE 2 where VE1 = EG − VGS − VBE1 VE 2 = VP − VBE 2

In general, it is not simple to calculate VGS, for simplicity, we assume that VGS will be given.
 FET Input Voltmeter

Example Determine the meter reading for the FET input voltmeter in the previous figure,
when E = 7.5 V and the meter is set to its 10 V range. The FET gate-source voltage is –5 V,
VP = 5 V, Rs+Rm = 1 kΩ and Im = 1 mA at full scale

SOLUTION On the 10 V range:

Input FET Emitter

attenuator input stage follower

+VCC
800k Ra 1V
R4
5V

Rb
100k Q1 V Q2
10V EG VG S R5
E
VS VP
60k Rc
Rs+Rm
25V I2 R2 R3
R6
I3
40k Rd
-VEE
Operational Amplifier Voltmeter

Op-Amp Amplifier Voltmeter

R4
Non-inverting
amplifier
meter
circuit
Vout = (1 + )E
R3
+VCC
The voltage gain
+
R4
E Av = (1 + )
- I4 R4 Vout Rs+Rm R3
-VEE

IB
Selection of R3 and R4
I3
R3
E Vout − E
R3 = and R4 =
I3 I3

The non-inverting amplifier gives a very high input impedance and very low output
impedance. Therefore, the loading effect can be neglected. Furthermore, it can
provide gain with enabling to measure low level input voltage.
 Operational Amplifier Voltmeter

Example Design an op-amp Voltmeter circuit which can measure a maximum input of
20 mV. The op-amp input current is 0.2 µA, and the meter circuit has Im = 100 µA FSD
and Rm = 10 kΩ. Determine suitable resistance values for R3 and R4

SOLUTION To neglect the effect of IB, the condition of I4 >> IB must be satisfied.
The rule of thumb suggested I4 should be at least 100 times greater
than IB
Non-inverting meter Select I4 = 1000 x IB = 1000 x 0.2 µA = 0.2 mA
amplifier circuit

At full scale: Im = 100 µA

+VCC

E
- I4 R4 Vout Rs+Rm
-VEE

IB
I3
R3
Operational Amplifier Voltmeter

Op-Amp Amplifier Voltmeter: voltage to current converter

+VCC Since I3 >> IB, therefore Im= I3
+ E
Meter current I m = I3 =
EB
-
R3
Im
Rs+Rm
-VEE Rm
Meter voltage Vm = E
R3
IB
if Rm > R3, voltage E is amplified by the ratio of Rm/R3
I3
VR3 R3
Current Measurement with Electronic Voltmeter
Electronic
voltmeter
+VC
C

-
Rs+Rm
-VEE
E

R3

+ + RS - -
I
Ammeter
terminals

An electronic voltmeter can be used for current measurement by measuring the voltage
drop across a shunt (Rs). The instrument scale is calibrated to indicate current.
 Electronic Ohmmeter: Series Connection

standard range
resistor switch

llu
al r f
1MΩ

s c ete
M
e
100kΩ
R1
R1 1kΩ A
100Ω Rx = 0 Rx = ∞
EB +
10Ω Electronic
1.5V Rx E voltmeter
(1.5 V range)
-
Ohmmeter scale for electronic instrument
B

Series Ohmmeter for electronic instrument

At Rx = ∞ or open circuit, the voltmeter indicate full scale defection (E = 1.5 V) and Rx =
0 or shorted circuit, since E = 0, no defection is observed. At other values of resistance,
the battery voltage EB is potentially divided across R1 and Rx, given by
Suppose that R1 is set to 1 kΩ
Rx 1 kΩ
E = EB E = 1.5 V × = 0.75 V (50% defection)
R1 + Rx 1 kΩ + 1 kΩ
Thus if Rx = R1, half scale will be indicated
 Electronic Ohmmeter: Series Connection

Example For the electronic ohmmeter in the Figure, determine the resistance scale
marking at 1/3 and 2/3 of full scale

Rx
standard range SOLUTION From E = EB
resistor switch R1 + Rx
1MΩ
R1
100kΩ Rearrange, give us Rx =
EB −1
R1 1kΩ A E
100Ω
EB +
1.5V
10Ω
Rx E
Electronic
voltmeter At 1/3 FSD; E = EB/3
(1.5 V range)
-
R1 R
Rx = = 1
B
EB × 3
−1 2
ll

EB
u
al r f
s c ete

At 2/3 FSD; E = 2EB/3

M
e

R1/2 R1 2R1
R1
Rx = ∞ Rx = = 2 R1
Rx = 0 EB × 3
−1
2 EB
 Electronic Ohmmeter: Parallel Connection

+ At Rx = ∞ or open circuit,
R1 R2
4kΩ E = EB
R1 + R2
A
6V
1.33 kΩ
+
= 6 V× = 1.5 V
R2
Rx E
Electronic
voltmeter
4 kΩ + 1.33 kΩ
1.33kΩ (1.5 V range)
-
Therefore, this circuit give FSD, when Rx = ∞
-
B
When, Rx = 0 Ω, E = 0 V, therefore, the meter
Shunt Ohmmeter for electronic instrument gives no defection.

R2 || Rx
At any value of Rx E = EB
R1 + R2 || Rx

So, the meter indicates half-scale when Rx = R1|| R2

AC Electronic Voltmeter

Principle
Most ac measurements are made with ac-to-dc converter, which
produce a dc current/voltage proportional to the ac input being measured

Vin ac to dc converter dc meter

Classification:
Average responding
periodic signal only
Peak responding
RMS responding (True rms meter) any signal
AC Electronic Voltmeter
The scale on ac voltmeters are ordinarily calibrated in rms volts

Average responding meter

Form factor is the ratio of the rms value to the average value of the wave form

Vrms
Vin ac to dc converter dc meter Form Factor =
Vaverage

It should be noted that the rms value is calculated from Vin, while the average value is
calculated from the output of ac-dc converter.
Peak responding meter
Form factor is the ratio of the peak value to the rms value of the wave form

V peak
Crest Factor =
Vrms
Average-Responding Meter
In this type of instrument, the ac signal is rectified and then fed to a dc millimeter.
In the meter instrument, the rectified current is averaged either by a filter or by the ballistic
characteristics of the meter to produce a steady deflection of the meter pointer.
+VD- +VD- output
+ + waveform

- + - +
D1 D1
E Input E Input
waveform
output Vm waveform Vm
waveform
Vout Vout
- -

Conventional half-wave rectifier precision rectifier

For the positive cycle, Vout = E
For the positive cycle, Vout = Vm = E
Vm = E − VD
where VD = cut-in voltage ~0.6-0.7 for Si For the negative cycle, Vout = 0
For the negative cycle, Vout = E
Therefore, the voltage drop in the forward
Vm = 0 bias can be compensated by this
Since Diode D1 is revered bias, no configuration
current flow through meter
Average-Responding Meter

V2
Vin

V1

V2
Vin

V1
 Average-Responding Voltmeter

Voltage to current converter

precision
rectifier
precision
rectifier
+VCC
C1 D1 D3
+VCC
+ + VF -
C1
Im Rs+Rm
R1 +
E - D1
Rs+Rm R1
meter
E current
-VEE -
meter
current -VEE
D2 D4

R3
R3

Half-wave rectifier Full-wave rectifier

Ep Ep
Ip = Meter peak current Ip =
Meter peak current R3
R3
2
Average meter current I = 1 I = 0.318I Average meter current I av = I p = 0.637I p
av
π
p p π
 Average-Responding Voltmeter
Example The half-wave rectifier electronic voltmeter circuit uses a meter with a FSD
current of 1 mA. The meter a coil resistance is 1.2 kΩ. Calculate the value of R3 that will
give meter full-scale pointer deflection when the ac input voltage is 100 mV (rms). Also
determine the meter deflection when the input is 50 mV.
SOLUTION at FSD, the average meter current is 1 mA

precision
rectifier

+VCC
C1
+ + VF -
R1
E - D1
Rs+Rm
-VEE
meter
current

R3
Peak-Responding Voltmeter
The primary difference between the peak-responding voltmeter and the average-
responding voltmeter is the use of a storage capacitor with the rectifying diode.

dc
amplifier

VD~0.7V

+
Vin C R
Vin C VC R
- the input impedance
Discharge cycle of the dc amp
Charge cycle

In the first positive cycle: VC tracks Vin with the difference of VD, until Vin reaches
its peak value. After this point, diode is reversed bias and the circuit keeps VC at
Vp – VD. The effect of discharging through R will be minimized if its value is large
enough to yield that RC >> T.
Peak-Responding Voltmeter

VC tracks Vin

VC

Vin
RMS-Responding Voltmeter
Suitable for: low duty-cycle pulse trains
voltages of undetermined waveform
T
1 2
T ∫0
RMS value definition: Mathematic Vrms = v (t )dt

Vin Vout
x
2
∫
RMS value definition: Physical
rms voltage is equivalent to a dc voltage which generates the same amount
of heat power in a resistive load that the ac voltage does.
Millivoltmeter

TC output (mV)
Temp. rise ∝ Vrms Non-linear
Thermocouple Difficult to calibrate scale

I
heating wire
Temp(oC)
RMS-Responding Voltmeter
Null-balance technique: non-linear cancellation

Compare the heating power generated by input voltage to the heating

power generated the dc amplifier
Measuring thermocouple

ac input ac dc
voltage Amplifier Amplifier
-
- +
Balancing Feedback
thermocouple current

Heater +
Vin A Vout
& TC
-

Heater
& TC
 Negative Feedback

VT1
+ Ve
Vin Heater A Vout
& TC
-

VT2 Heater
& TC

Vout = Ve = A (VT 1 − VT 2 )

Let, VT1 = k Vin and VT2 = k Vout where k is proportional constant of the heater and TC in
the system. Note that k may depend on the level of the input signal

Vout = A ( kVin − kVout )

Vout Ak
= If A is large Vout ≈ Vin
Vin 1 + Ak
If the amplifier gain is very large, Vout is equal to Vin, this means that the dc voltage
output is therefore equal to the effective, or rms value of the input voltage
 Digital Multimeter (DMM)

Hand-held DMM

Portable Analog
Multimeter

Bench-top DMM
 Digital Voltmeter (DVM)

DVM is essentially an Analog to digital converter (A/D) with a

digital display

Digital voltmeter
Digital display
Attenuator Analog to Digital
Amplifier Converter

Digital MultiMeter = electronic Volt Ohm Millimeter with

(DMM) digital display
 Comparison of Digital and Analog Meter

Digital meter Analog meter

Leaves no doubt about the measured quantity. Wrong scale might be used or might be
read incorrectly.
Superior resolution and accuracy. Inferior resolution and accuracy.
(±0.5% or better) (±3% in common)
Indicates a negative quantity when the Pointer attempts to deflect to the left
terminal polarity is reversed when the polarity is reversed
No usually damaged by rough treatment Can be damaged when dropped from
bench level
 Analog to Digital Conversion
A/D converts an analog signal into the digital code which is
proportional to the magnitude of the coming signal.

Vin ≈ k ×Digital output

Where k is step size or resolution
Ex. Signal from 800-1500 mV may be converted to 8-bit binary codes starting from
010100002 (8010) to 100101102 (15010). In this case, the step size k is equal to
10 mV.

Quantization error or Conversion error of a A/D

step size 1
Quantization error = × 100 = N × 100%
full scale 2 −1
Where N is the number of bit
 Analog to Digital Conversion
Conversion time, Tc time requires to convert an analog signal to the
corresponding digital code.

111

110
Digital output

101
step size = 1 V
100
Quantization error = 1/7 × 100 =14.3%
011

010 A/D will give 010 digital code. Ex. An analog inputs 1.5-2.5 V will be
represented by digital output 010
1.5 2.5
001

000
0 1 2 3 4 5 6 7 (Full scale)
Analog input (V)
An example of 3 bit ADC
 Ramp-type Digital Voltmeter
(also called single slope)
Operation principle: The measurement of the time it takes for a linear ramp voltage to
rise from O V to the level of the input voltage, or to decrease from the
level of the input voltage to zero. This time interval is measured with
an electronic time-interval counter.
Start measurement
Ramp signal Coincindence Vramp(t) = Vo – m t
Vin
Where m is the ramp rate

Vramp(t1) = Vin = Vo –m t1
Time
Vramp(t2) = 0 = Vo – m t2
∆t
Gating ∆t = t2- t1 = Vin/m
time interval
If the period of the clock is T, then during the
time interval ∆t1, the number of pulses is

Clock pulses n pulse ∆t ≈ nT or Vin ≈ nmT

to counter
•Accuracy depends on both the ramp rate and
clock period.
Voltage-to-time conversion using gated clock pulses.
 V+>V-; Vo = V(1) Logic high
Comparator V+<V-; Vo = V(0) Logic low
Vo
V(1)
Vin +
Vo
Vref
V(0) Vin
Vref

Vo
V(1)
Vref +
Vo
Vin
V(0)
Vin
Vref
 Ramp-type Digital Voltmeter

Voltage-to-time conversion
Time measurement unit

input
Ranging vin comparator Digital
DC input
voltage
and - display
Attenuator
+ start
count
clk
Oscillator Gate Counter

ground
vramp stop
Ramp comparator
- count
Generator
0V +
start ramp
Sample
reset
Rate
MV

Block diagram of a ramp-type digital voltmeter.

 Staircase Ramp Digital Voltmeter
(also called digital ramp) Compare the input voltage to the internally generated stair
case ramp.
V in + •The most simple A/D
Comp.
-
•Slow conversion and conversion time
depends on the magnitude of input signal.
V AX
TC,max = (2N − 1) × Clock period
D/A

Digital output
Vin

D/A output
VAX
Clock Counter

clock period
Control circuit
time
Block diagram
Successive Approximation Digital Voltmeter
Ex. To determine a number between 0 – 511 (9 bit binary),
given, the number to be determined is 301

No. Estimate Results

1 256 1 0000 0000 Vin > VAX
2 256+128 = 384 1 1000 0000 <
3 256+64 = 320 1 0100 0000 <
4 256+32 = 288 1 0010 0000 >
5 288+16 = 304 1 0011 0000 <
6 288+8 = 296 1 0010 1000 >
7 296+4 = 300 1 0010 1100 >
8 300+2 = 302 1 0010 1110 <
9 300+1 = 301 1 0010 1101 Finished
 Successive Approximation Digital Voltmeter
Compare the input voltage to the internally generated voltage
•The most common A/D for general
V in +
applications
Comp.
- •Conversion time is fixed (not depend on the
V AX signal magnitude) and relatively fast
TC = N × Clock period
D/A
where N is the number of bits
Digital output Full
scale

Succesive Vin
3 Full
Clock Approximation
D/A output 4 scale
Register
1 Full
2 scale VAX

Control circuit 1 Full

4 scale
1 2 3 4 5 6 7 8

Block diagram Clock period

Digital Ramp VS Successive approximation

Digital Ramp method Successive approximation method

 Dual-slope Digital Voltmeter
C C

V in R V in R
- -
V out V out
+ +
V ref V ref

Phase 1: charging C with the unknown input for Phase 2: discharging C with the reference voltage
a given time. until the output voltage goes to zero.
Assume Vc(0) = 0
Vin T Vref Tx
Vout1 = − Vout = + Vout1
RC RC
where T is the charging time find Tx at which Vout becomes zero
Charge Discharge
time
Vin T
Tx =
Vref

Vout
Phase 1 Phase 2
 Dual-slope Digital Voltmeter
Conversion time
Charge Discharge
0
Sm
time
all
inp
ut
v olt
La

ag
e •Accuracy does not depend on R C and Clock
rg
e

(high accuracy)
in
pu
tv

•Relatively slow
ol
ta

•Capable to reject noise

ge

Vout TC = Tconst + T variable

C
Zero
crossing Display
Vin R detector
- Vout
+ -
count
Vref + Counter
reset

Clock
generator

Control logic
Ex A dual slope A/D has R= 100 kΩ and C = 0.01 µF . The reference voltage is 10 volts
and the fixed integration time is 10 ms. Find the conversion time for a 6.8 volt input.

Vin T (6.8 V)(10 ms)

Tx = = = 6.8 ms
Vref (10 V)

The total conversion time is then 10 ms + 6.8 ms = 16.8 ms Ans

Ex Find the successive approximation A/D output for a 4-bit converter to a 3.217 volt
input if the reference is 5 volts.

(1) Set D3 = 1 VAX = 5/2 = 2.5 Volts

Vin > VAX leave D3 = 1
(2) Set D2 = 1 VAX = 5/2 + 5/4 = 3.75 Volts
Vin < VAX reset D2 = 0
(3) Set D1 = 1 VAX = 5/2 +5/8= 3.125 Volts
Vin > VAX leave D1 = 1
(4) Set D0 = 1 VAX = 5/2+5/8+5/16 = 3.4375 Volts
Vin < VAX reset D0 = 0
By this procedure, we find the output is a binary word of 10102 Ans
 Typical specification of DMM
General
Maximum voltage between
:600 V
terminals
Fuse protection :200mA/250V
Power :9V battery
Display :LCD 31/2 digits, updates 2-3/ sec.
Input impedance :10 MΩ
Frequency range :40-400 Hz
Measuring method Dual-slope integration
Over range indication Only figure “1” on the display
Polarity indication “-” displayed for negative polarity
Accuracy of DMM
Indicate as ± (% of reading + No. of digits)
Ex. ± (0.5% of rdg + 1 digits) sometimes simplify as ± (0.5 + 1)
Ex. For an accuracy of ± (0.5 + 1) , calculate the maximum error of in the 1.800 V reading
error = ± (0.5% of 1.800 + 0.001 V)
= ± (0.009 + 0.001 V) = ± 0.01 V or ± 0.56% of reading
Ex A 20 V dc voltage is measured by analog and digital multimeters. The analog instrument
is on its 25 V range , and its specified accuracy is ± 2%. The digital meter has 3 ½ digit
display and an accuracy of ±(0.6+1). Determine the measurement accuracy in each case.

Analog instrument:
Voltage error = ± 2% of 25 V ½ digit
= ± 0.5 V
error = ± 0.5 V × 100%
20 V
= ± 2.5%
Digital instrument:
For 20 V displayed on a 3 ½ digit display 3½ digit display
1 Digit = 0.1 V
Voltage error = ± (0.6% of reading + 1 Digit)
= ± (1.2 V + 0.1 V)
= ± 0.22 V
error = ± 0.22 V × 100%
20 V
= ± 1.1%