CHAPTER 4
DC Meter.
School of Computer and Communication Engineering,
UniMAP
Prepared By:
Amir Razif b. Jamil Abdullah
EMT 113: V-2008
1
4.0 DC Meters.
4.1 Introduction to Meters.
4.2 Analogue Meter
4.3 Introduction to DC Meters.
4.4 D’Arsonval Meter Movement in DC Meters.
4.5 Ayrton Shunt.
4.6 Voltmeter Loading Effect.
4.7 Ammeter Insertion Effect.
4.8 Ohmmeter.
2
4.1 Introduction to Meters.
 A meter is any device built to accurately detect and display an
electrical quantity in a form readable by a human being.
(i) Pointer (analogue).
(ii) Series of lights (analogue).
(iii) Numeric display (digital).
 In this chapter students will familiarized with the d’Arsonval meter
movement, its limitations and some of its applications.
 Electrical meters;
(i) DC, AC average quantities:
-Voltmeter
-Ammeter
-Ohmmeter
(ii) AC measurements:
-Oscilloscope
3
Cont’d…
 A meter is any device built to accurately
detect and display an electrical quantity in
a form readable by a human being.
 In the analysis and testing of circuits,
there are meters designed to measure the
basic quantities of voltage, current, and
resistance.
 Most modern meters are "digital" in
design, meaning that their readable
display is in the form of numerical digits.
 Older designs of meters are
mechanical in nature, using some kind
of pointer device to show quantity of
measurement.
 The first meter movements built were
known as galvanometers, and were
usually designed with maximum
sensitivity in mind.
Figure 4.1: Galvanometer.
Figure 4.2: Voltmeter
4
Cont’d…
Galvanometer
 A very simple galvanometer may be made from a magnetized
needle (such as the needle from a magnetic compass) suspended
from a string, and positioned within a coil of wire.
 Current through the wire coil will produce a magnetic field which
will deflect the needle from pointing in the direction of earth's
magnetic field. An antique string galvanometer is shown in Figure
4.1.
 The term "galvanometer" usually refers to any design of
electromagnetic meter movement built for exceptional sensitivity,
and not necessarily a crude device such as that shown in Figure 4.1.
 Practical electromagnetic meter movements can be made now where
a pivoting wire coil is suspended in a strong magnetic field, shielded
from the majority of outside influences. Such an instrument design
is generally known as a permanent-magnet, moving coil, or
PMMC movement.
5
4.2 Analogue Meters.
 The analogue meters are mostly based on moving coil meters.
The typical structure consists of a wire wound coil placed between
two permanent magnets, Figure 4.3.
 When current flows through the coil in the presence of a
magnetic field, a force is exerted on the coil;
F = Bil
 This force is directly proportional
to current flowing in the coil.
If the coil is free to rotate, the
force causes a deflection of the
coil that is proportional to the
current.
 By adding an indicator (e.g. needle)
and a display, the level of current
can be measured.
Figure 4.3: Analogue Meter
6
Cont’d…
 For a given meter, there is a maximum rated current that produces
full-scale deflection of the indicator; FSD rating.
 By adding external circuit components, the same basic moving
coil meter can be used to measure different ranges of voltage or
current.
 Most meters are very sensitive. That is, they give full-scale
deflection for a small fraction of an amp (A) for example a typical
FSD current rating for a moving coil meters is 50 μA, with internal
wire resistance of 1 kΩ.
 With no additional circuitry, the maximum voltage that can be
measured using this meter is
50 x 10-6x 1000W = 0.05V.
 Additional circuitry is needed for most practical measurements.
7
4.3 Introduction DC Meters.
 The meter movement will have a pair of metal connection
terminals on the back for current to enter and exit.
 Meter;
(i) Polarity sensitive.
(ii) Polarity-insensitive.
 Most meter movements are polarity-sensitive, one direction of
current driving the needle to the right and the other driving it to
the left.
 Some meter movements are polarity-insensitive, relying on the
attraction of an un-magnetized, movable iron vane toward a
stationary, current-carrying wire to deflect the needle. Such meters
are ideally suited for the measurement of alternating current
(AC).
 A polarity-sensitive movement would just vibrate back and forth
uselessly if connected to a source of AC.
8
Cont’d…
 An increase in measured current will drive the needle to point
further to the right.
 A decrease will cause the needle to drop back down toward its
resting point on the left.
 Most of the mechanical meter movements are based on
electromagnetism ; electron flow through a conductor creating a
perpendicular magnetic field,
 A few are based on electrostatics; the attractive or repulsive force
generated by electric charges across space.
9
Cont’d…
(a) Permanent Magnet Moving Coil (PMMC).
Figure 4.4: Permanent Magnet Moving Coil (PMMC) Meter Movement.
 In the PMMC-type instruments, Figure 4.4. Current in one
direction through the wire will produce a clockwise torque on
the needle mechanism, while current the other direction will
produce a counter-clockwise torque.
10
Cont’d…
(b) Electrostatic Meter Movement.
 In the electrostatics, the attractive or repulsive force generated by
electric charges across space, Figure 4.5.
 This is the same phenomenon exhibited by certain materials; such
as wax and wool, when rubbed together.
 If a voltage is applied between two conductive surfaces across
an air gap, there will be a physical force attracting the two
surfaces together capable of moving some kind of indicating
mechanism.
 That physical force is directly proportional to the voltage applied
between the plates, and inversely proportional to the square of the
distance between the plates.
Figure 4.5: Electrostatic Meter Movement.
11
Cont’d…
 The force is also irrespective of polarity, making this a polarityinsensitive type of meter movement.
 Unfortunately, the force generated by the electrostatic attraction is
very small for common voltages.
 It is so small that such meter movement designs are impractical
for use in general test instruments.
 Typically, electrostatic meter movements are used for measuring
very high voltages; many thousands of volts.
 One great advantage of the electrostatic meter movement,
however, is the fact that it has extremely high resistance, whereas
electromagnetic movements (which depend on the flow of
electrons through wire to generate a magnetic field) are much lower
in resistance.
12
Cont’d…
 Some D'Arsonval movements have full-scale deflection current
ratings as little as 50 µA, with an (internal) wire resistance of less
than 1000 Ω.
 This makes for a voltmeter with a full-scale rating of only 50
millivolts (50 µA X 1000 Ω).
Figure 4.6: Voltmeter.
13
4.4 D’Arsonval Meter Movement in
DC Meter.
 The basic d’Arsonval meter movement has only limited usefulness
without modification.
 By modification on the circuit it will increase the range of
current that can be measured with the basic meter movement.
 This is done by placing the low resistance in parallel with the
meter movement resistance Rm.
 The low resistance shunt (Rsh) will provide an alternate path for
the total meter current I around the meter movement.
 The Ish is much greater than Im.
Where
Rsh = resistance of the shunt
Rm = internal resistance of the meter movement
(resistance of the moving coil)
Ish = current through the shunt
Im = full-scale deflection current of the
meter movement
14
Figure 4.7: D’Ársonval Meter Movement
I = full-scale deflection current for the ammeter
Used in Ammeter Circuit
Cont’d…
 The voltage drop across the meter movement
is
Vm = ImRm
 Since the shunt resistor is in parallel with the meter movement, the
voltage drop across the shunt is equal to the voltage drop across
the meter movement. That is,
Vsh = Vm
 The current through the shunt is equal to the total current minus
the current through the meter movement:,
Ish = I – Im
 Knowing the voltage across, and the current through, the shunt
allows us to determine the shunt resistance as
Rsh = Vsh/Ish
= ImRm/Ish = (Im/Ish)(Rm)
= Im/(I – Im)*Rm W
15
Example 4.1: D’Arsonval Movement.
A D'Arsonval meter movement having a full-scale deflection rating of
1 mA and a coil resistance of 500 Ω:
Ohm's Law (E=IR), determine how much voltage will drive this
meter movement directly to full scale,
Solution:
E  I *R
E  (1mA) * (500W)
E  0.5V
.
16
Example 4.2: D’Arsonval Meter.
Calculate the value of the shunt resistance required to convert a 1-mA
meter movement, with a 100 W internal resistance, into a 0 - to 10 mA
ammeter.
Solution:
Calculate Vm.
Vm  I m Rm
 1mA * 100W   0.1V
Vm is in parallel with Vsh.
Vsh  Vm  0.1V
KCL
I sh  I  I m
 10mA  1mA  9mA
.
Rsh 
Vsh 0.1V

 11.11W
I sh 9mA
17
4.5 Ayrton Shunt.
 The purpose of designing the shunt circuit is to allow to measure a
current I that is some number n times larger than Im, Figure 4.8.
 The number n is called a multiplying factor and relates total
current and meter current as the Ayrton Shunt.
I  nI m
Substituting for I in previous equation,
yields
Rm I m
Rm
Rsh 
(nI  I m )

(n  1)
Advantage:
(i) it eliminates the possibility
of the meter movement being in the
circuit without any shunt resistance.
(ii) May be used with a wide range of meter
movements.
Figure 4.8: Aryton Shunt.
18
Cont’d…
 The individual resistance values of the shunts are calculated by
starting with the most sensitive range and working toward the least
sensitive range.
 The shunt resistance is,
Rsh  Ra  Rb  Rc
 On this range the shunt resistance is equal to Rsh and can be
computed by the equation,
R
Rsh 
m
(n  1)
 The equation needed to compute the value of each shunt, Ra, Rb,
and Rc, can be developed from the circuit in Figure 4.8.
 Since the resistance Rb + Rc is in parallel with Rm + Ra, the voltage
across each parallel branch should be equal and can be written as
VRb  VRc  VRa  VRm
19
Cont’d…
 In current and resistance terms we can write
or
( Rb  Rc )( I 2  I m )  I m ( Ra  Rm )
I2(Rb + Rc) - Im(Rb + Rc) = Im[Rsh-(Rb + Rc)+Rm]
Multiplying through by Im on the right yields
I2(Rb + Rc) - Im(Rb + Rc) = ImRsh- Im(Rb + Rc)+ImRm
 This can be rewritten as
Rb+ Rc = Im (Rsh+ Rm)/I2
Having already found the total shunt resistance Rsh, we can now
determine Ra as
Ra = Rsh – (Rb + Rc)
The current I is the maximum current for the range on which
the ammeter is set. The resistor Rc can be determined from
Rc = Im(Rsh+ Rm)/I3
 The resistor Rb can now be computed as, Rb = (Rb + Rc) – Rc
20
Example 4.3: Aryton Shunt.
Compute the value of the shunt resistors for the circuit below. I3 =
1A, I2 = 100 mA, I1 = 10 mA, Im = 100 uA and Rm = 1K Ohm.
Solution:
The total shunt resistance is found from
Rsh 
Rm
1KW

 10.1W
n  1 100  1
This is the shunt for the 10 mA range. When the meter is set on the
100-mA range, the resistor Rb and Rc provide the shunt . The total
shunt resistance is found by the equation.
Rb  Rc 

I m ( Rsh  Rm )
I2
(100uA) * (10.1W  1KW)
 1.01W
100mA
21
Cont’d…
The resistor Rc , which provides the shunt resistance on the 1-A range
can be found by the same equation, however the current I will now be
1A.
I (R  R )
Rc 
m
sh
m
I3

(100uA) * (10.1W  1KW)
 0.101W
1A
The resistor Rb is found from the equation below;
Rb  ( Rb  Rc )  Rc
 1.01W  0.101W  0.909W
The resistor Ra is found from;
Ra  Rsh  ( Rb  Rc )
 10.1W  (0.909W  0.101W)  0.909W
Verify the above result.
Rsh  Ra  Rb  Rc
 9.09W  0.909W  0.101W  10.1W
.
22
Example 4.4 (T2-2005): Aryton Shunt.
Figure below is an Aryton Shunt circuit. Given that R1 = 0.5 W, R2 = 6.5
W, R3 = 55.5 W, Rm = 1 kW, Im=100 mA and n = 16. Calculate the value
of, I1, I2, I3 and I4.
Solution:
Rsh 
Rm
1KW

 66.67W
n  1 16  1
 Find Rsh
R1  R2  R3 
 I3
I3 
I m ( Rsh  Rm )
I3
I m ( Rsh  Rm )
R1  R2  R3
100 mA(66.67W  1KW)
0.5W  6.5W  55.5W
I 3  1.707 mA

Can be easily derived
23
R1  R2 
 I2
I2 
I m ( Rsh  Rm )
I2
I m ( Rsh  Rm )
R1  R2
100 mA(66.67W  1KW)
0.5W  6.5W
I 2  15.238mA
I2 
 I3
R1 
I m ( Rsh  Rm )
I1
I1 
I m ( Rsh  Rm )
R1
100 mA(66.67W  1KW)
0.5W
I 1  313.33mA
I1 
 I4
R1  R2  R3  R4 
I4 
I m ( Rsh  Rm )
I4
I m ( Rsh  Rm )
R1  R2  R3  R4
100 mA(66.67W  1KW)
0.5W  6.5W  55W  4.17W
I 4  1.60mA

..
Rsh  R1  R2  R3  R4  66.67W
24
4.6 Voltmeter Loading Effect.
(a) D’Arsonval Meter Movement Used in a DC Voltmeter.
 The basic d’Arsonval meter movement can be converted to a dc
voltmeter by connecting a multiplier Rs in series with the meter
movement, as in Figure 4.10 below.
 The multiplier will extend the voltage range of the meter to limit
current through the d’Arsonval meter movement to a full scale
deflrction current.
 The value of the multiplier resistor can be found by determine the
sensitivity.
Sensitivity 
1
(W / V )
I fs
Figure 4.10: The d’Arsonval meter movement used in the DC voltmeter.
25
Cont’d…
Voltmeter Design.
 Consider a moving coil meter with FSD rating of 1 mA and coil
resistance, Rc, of 500 Ω.
 The maximum voltage required to produce FSD is 0.5 V.
 The voltage range is increased by adding a series resistor,
Figure 4.11: Voltmeter.
 The voltage that can be applied to the – and + terminals before
FSD current flows, is then increased to:
VFSD  I FSD ( Rc  Rm )
 Rm is called a multiplier resistor because it multiplies the working
range of the meter.
26
Cont’d…
 For a given required FSD voltage, say VFSD, the multiplier
resistance, Rm, is chosen as:
 VFSD 
  Rc
Rm  
 I FSD 
 For example, to provide a voltmeter with FSD reading of 10 V with
the given meter (IFSD = 1 mA, Rc= 500 Ω):
V 
 10V 
Rm   FSD   Rc  
  500W  9.5kW
 1mA 
 I FSD 
 With exactly 10 V applied, there will be exactly 1 mA of current
flowing, thereby producing full-scale deflection.
 There is only the maximum allowed voltage of 0.5V dropped
across the moving coil meter.
 The scale of the meter must be changed to indicate the new range
of the circuit.
27
Example 4.5: Voltmeter Sensitivity.
Calculate the sensitivity of 100 uA meter movement which is to be used
as dc voltmeter.
Solution.
The sensitivity is compute as
Sensitivity, S 
1
1

 10kW / V
I fs 100mA
 .
 The unit of sensitivity is expressed as the value of multiple
resistance for 1-V range.
 To calculate the value of the multiplier for voltage range greater
than 1V, multiply the sensitivity by the range and subtract the
internal resistance of the meter movement.
Rs  S * Range  Internal Re sis tan ce
28
Example 4.6: Sensitivity Voltmeter Range.
Calculate the value of multiplier resistance on the 50-V range of a dc
voltmeter that used a 500-uA meter movement with an internal
resistance of 1kW.
Solution.
Figure 4.12: The Volmeter.
The sensitivity of 500uA movement is,
Sensitivity, S 
1
1
kW

2
I fs 500mA
V
The value of the multiplier Rs is calculated as,
Rs  S * Range  Internal Re sis tan ce
.
Rs 
2kW
* 50V  1kW  99kW
V
29
Cont’d…
(b) Voltmeter Loading Effect.
 When the voltmeter is used to measure the voltage across a circuit
component the voltmeter circuit itself is in parallel with the circuit
component.
 The parallel combination of two resistor is less than either resistor
alone.
 The resistance seen by the source is less with the voltmeter
connected than without.
 The voltage across the source is less when the voltmeter is
connected. This effect is called voltmeter loading. The resulting
error is called loading error.
30
Example 4.7: Voltmeter Loading Effect.
In the voltmeter loading effect experiment, the two different voltmeters
are used to measure the voltage across resistor RB in the circuit Figure
4.13 below, The meters are as follow,
Meter A: S = 1 k W /V, Rm = 0.2 k W , range = 10V
Meter B: S = 20 k W /V, Rm = 1.5 k W , range = 10V
Calculate,
(i) Voltage across RB without any meter connected across it.
(ii) Voltage across RB when meter A is used.
(iii) Voltage across RB when meter B is used.
(iv) Error in voltmeter readings.
Solution.
Figure 4.13: The Voltmeter Loading
Effect.
31
(i) Voltage across RB without any meter connected across it.
VRB  E
RB
5kW
 (30V )
 5V
R A  RB
25kW  5kW
(ii) Voltage across RB when meter A is used.
RS  S * Range  Rm 
1kW
(10V )  1kW  9kW
V
The parallel combination of RB and meter A,

Re1 
( RB )( RTA ) (5kW)(10kW)

 3.33kW
RB  RTA
5kW  10kW
Therefore the voltage reading obtained with meter A, determine by
the voltage divider equation is

VRB
Re1
3.33kW
E
 (30V )
 3.53V
Re1  R A
3.33kW  25kW
32
(iii) Voltage across RB when meter B is used.
 The total resistance that meter B presents to the circuit is
RTB  S * Range 
20kW
(10V )  200kW
V
The parallel combination of RB and meter B is
Re 2 
R B RTB
(5kW)(200kW)

 4.88kW
R B  RTB
5kW  200kW
the voltage reading obtain with meter B, determine by use of the
voltage divider equation
Re 2
4.88kW

V E
 (30V )
 4.9V
RB
Re 2  R A
4.88kW  25kW
(iv) Error in voltmeter readings.

Voltmeter A error ` 5V  3.53V

*100%  29.4%

5V

Voltmeter B error
5V  4.9V

*100%  2%
.
5V
33
4.6 Ammeter Insertion Effect.
 We frequently overlook the error caused by inserting an
ammeter in a circuit to obtain a current reading.
 All ammeters contain some internal resistance.
 By inserting the ammeter in the circuit means increase the
resistance of the circuit and result in reducing current in the
circuit.
 Refer to the circuit in Figure 4.14, Ie is the current without the
ammeter.
 Suppose that we connect the ammeter in the circuit (b), the current
now becomes Im due to the additional resistance introduced by the
ammeter.
Figure 4.14: (a) Expected Current Value in a Series Circuit
(b) Series Circuit with Ammeter.
34
Cont’d…
 From the circuit;
 Placing the meter in series result in;
E
Ie 
R1
E
Im 
R1  Rm
 Divide the above equations yields;
Im
R1

Ie
R1  Rm
 Insertion error,
 Im 
1   *100%
Ie 

35
Example 4.8: Ammeter Insertion Effects.
A current meter that has an internal resistance 78 Ohm is used to
measure the current through resistor Rc in Figure 4.14. Determine
the percentage of error of the reading due to ammeter insertion.
Solution.
 Look back into the circuit from terminal X and Y.
36
Solution.
The Thevenin equivalent resistance.
Rth  Rc 
Ra Rb
Ra  Rb
 1KW  0.5 KW  1.5 KW
The ratio of meter current to the expected current is,
Im
R1
1.5KW


 0.95
Ie
R1  rm 1.5KW  78W
Solving for Im yields,
Insertion
I m  0.95I e
 Im 
error  1   *100%  5.0%
Ie 

.
37
4.7 Ohmmeter.
 The d’Arsonval meter movement can be used with the battery
and resistor to construct a simple ohmmeter.
 Figure 4.15 is the basic ohmmeter circuit,
I fs
E

R z  Rm
 Introduce Rx between point X and Y so that we can calculate the
value of resistance.
E
I
R z  Rm  R x
Figure 4.15: Basic Ohmmeter Circuit.
38
Cont’d…
E /( Rz  Rm  Rx )
( R z  Rm )
I


I fs
E /( Rz  Rm )
( R z  Rm  R x )
 P represent the ratio of the current I to the full scale deflection
P
( R z  Rm )
I

I fs ( Rz  Rm  Rx )
Figure 4.16: Basic Ohmmeter Circuit with Unknown
Resistor Rx Connected Between.
39
Example 4.9: Ohmmeter.
A 1mA full-scale deflection current meter movement is to used in an
ohmmeter circuit. The meter movement has an internal resistance, Rm,
of 100 Ohm, and a 3-V battery will be used in the circuit. Mark off
the meter face for reading resistance.
Solution.
Value of Rz, which will limit current to full-scale deflection is,
Rz 
E
 Rm
I fs
3V
Rz 
 100Ohm  2.9 KOhm
1mA
Value of Rz, with 20% full-scale deflection is,
Rx 
R z  Rm
 ( R z  Rm )
P
2.9 KW  1.0 KW

 (2.9 KW  1.0 KW)
0.2
 12 KW
40
Cont’d…
 Value of Rz, with 40% full-scale deflection
is,
 Value of Rz, with 50% full-scale deflection
is,
 Value of Rz, with 75% full-scale deflection
is,
 The ohmmeter is nonlinear due to the
high internal resistance of the
ohmmeter.
Rx 
Rx 
Rx 
R z  Rm
 ( R z  Rm )
P
3 KW

 (3KW)
0.4
 4.5 KW
R z  Rm
 ( R z  Rm )
P
3 KW

 (3KW)
0.5
 3 KW
R z  Rm
 ( R z  Rm )
P
3 KW

 (3KW)
0.75
 1KW
41