CHAPTER 2
DC and AC Meters
1
PART 1 (DC)
•
•
•
•
•
•
Introduction to DC
meters
D’ Arsonval meter
movement
DC ammeter
DC voltmeter
DC ohmmeter
Loading effects of DC
meter
PART 2 (AC)
•
•
•
•
Introduction to AC
meters
D’ Arsonval meter
movement (half-wave
rectification)
D’ Arsonval meter
movement (full-wave
rectification)
Loading effects of AC
meter
STRUCTURE FOR CHAPTER 2
2
PART 1 – DC METERS
Meters
Digital Meters
Measure the continuous voltage/ current signal at discrete points in time.
The signal converted from analog signal (continuous in time) to a digital signal
(discrete instants in time)
Analog Meters
Based on the d’Arsonval meter movement which implements the readout
mechanism.
3
INTRODUCTION
HOW CAN WE MEASURED CURRENT AND
VOLTAGE?
Im
Rm
Torque (T)
PMMC instrument
T = BxAxNxI
[N.m]
4
ADVANTAGES AND DISADVANTAGES
OF MOVING COIL INSTRUMENT
AMMETER, VOLTMETER AND OHMMETER?

DC Ammeter : The shunting resistor Rsh and d’Arsonval
movement form a current divider

DC Voltmeter : Series resistor Rs and d’Arsonval movement
form a voltage divider.

Ohmmeter
: Measures the current to find the resistance
Rs
Rsh
Rs
6
DC AMMETER

D’Arsonval movement, Rm || (shunt resistor), Rsh

To limit the amount of the current in the movement’s
coil by shunting some of it through Rsh
I
Rsh
Im
Rm
d'Arsonval movement
Ish
Rsh =
resistance of the shunt
Rm =
internal resistance of the meter
movements (movable coil)
Ish
=
shunt current
Im
=
full scale deflection current of the
meter movement
I
=
full-scale deflection current for
the
ammeter
| | = Parallel symbol
7
DC AMMETER
Vm = ImRm
Vsh = IshRsh
Vsh = Vm
IshRsh = ImRm
Ish
Rsh
Im
Rm
d'Arsonval movement
Ammeter Terminal
I
Rsh = ImRm / Ish (Ω) ----(a)
I = Ish + Im
Ish = I – Im
Therefore, Rsh = ImRm/(I – Im)
Purpose I >>n Im , n = multiplying factor
n=I/Im
I = nIm ---(b)
Substitute b to a
Rsh = ImRm/(nIm – Im)
Rsh= Rm/(n-1) -----(c)
8
EXAMPLE
1 (DC AMMETER)
Example
1: DC Ammeter
A 100uA meter movement with an internal
resistance of 800Ω is used in a 0 - 100 mA
ammeter . Find the value of the required shunt
resistance.
Solution:
n = I/Im = 100 mA / 100 µA = 1000
Thus,
Rsh = Rm / (n – 1) = 800 / 999 = 0.8 Ω
9
THE ARYTON SHUNT
Rsh = Ra + Rb + Rc
Most sensitive
range
1A
S
Ra
Im
1 mA
5A
Rb
Rshunt
+
Rm
Rsh 
Rm
n 1
----(c)
50
10A
-
Rc
•Used in multiple range ammeter
•Eliminates the possibility of the moving coil to
be in the circuit without any shunt resistance
10
THE ARYTON SHUNT
At point B, (Rb+Rc)||(Ra+Rm)
Ra
I1
Middle
sensitive
range
+
I2
S
Im
B
Rshunt
Rb
VRb  Rc  VRa  Rm
Rm
(Rb + Rc )(I2 -Im) = Im(Ra +Rm)
I3
-
Since,
Rc
Ra = Rsh – (Rb + Rc),
yield,
I2 (Rb + Rc ) – Im(Rb+Rc) = Im [Rsh – (Rb + Rc ) + Rm]
Rb  Rc 
I m ( Rsh  Rm )
I2
----(d)
11
THE ARYTON SHUNT
At point C,
I1
Ra
Im
I2
Rb
S
I3
-
C
Rc
Rshunt
+
Rm
Rc||(Ra+Rb+Rm)
VRc  VRa  Rb  Rm
(I3-Im)Rc = Im(Ra+Rb+Rm)
I3Rc = Im(Ra+Rb+Rc+Rm)
I3Rc = (Rsh+Rm)
I m ( Rsh  Rm )
Rc 
I3
----(e)
12
THE ARYTON SHUNT

Substitute eqn (d) into eqn (e), yields
1 1
Rb  I m ( Rsh  Rm )   
 I 2 I3 
----(f)
Ra = Rsh – (Rb+Rc)
----(g)
13
EXAMPLE 2: THE ARYTON SHUNT
Calculate the value for Ra, Rb and Rc as shown, given the
value of internal resistance, Rm=1kΩ and full scale current of
the moving coil = 100 µA. The required range of current are:
I1 = 10 mA, I2 = 100 mA and I3 = 1A.
I1
S
Ra
I2
B
Rb
Rshunt
+
Im
Rm
I3
-
Rc
14
AMMETER INSERTION EFFECT
R1
R1
X
Connect
Ammeter
Ie
X
Im
E
E
Ie 
E
R1
Y
InsertionError 
I e  I m  100%
Ie
Rm
Im 
E
R1  Rm
Y
Im
R1

I e R1  Rm

Ie  Im 
InsertionError 
100%
Ie
15
EXAMPLE 3: AMMETER INSERTION
EFFECTS
A current meter that has an internal resistance of 78Ω is used
to measure the current through resistor R1. Determine the
percentage of error of the reading due to ammeter insertion.
R1
1kΩ
3V
X
Im
E
Rm
Y
16
SOLUTION EX:3
17
DC VOLTMETER
DMM become VOLTMETER – multiplier Rs
in series with the meter movement.
PURPOSE
•
To extend the
voltage range
To limit current through the DMM to a
maximum full-scale deflection current
DMM = D’Arsonval Meter Movement
18
DC VOLTMETER
Rs
Im
+
Rm
1
Sensitivit y 
I fs
(Ω/V)
Ifs= Im = full scale deflection current
Rs + Rm= (S x Vrange)
Unit derivation:
Sensitivit y 
1
1
ohms


amperes  volt  volt


ohms


It is desirable to make
R(voltmeter) >>R ( circuit)
19
EXAMPLE 4: DC VOLTMETER
Calculate the value of the multiplier resistance
on the 50 V range of a dc voltmeter that used a
500µA d’Arsonval meter with an internal
resistance of 1 kΩ.
20
MULTI-RANGE VOLTMETER

A multi-range voltmeter consists of a
deflection instrument, several multiplier
resistors and a rotary switch.
R1
30 v
10 v
R2
only one of the three multiplier resistors is
connected in series with the meter at any time.
The range of this meter is
S
+
3v
R3
Im
-
V  Im( Rm  R)
Rm
Multi-range Voltmeter
Where the multiplier resistance,
R can be R1 or R2 or R3
21
MULTI-RANGE VOLTMETER
R1
30 v
10 v
R2
S
+
3v
R3
Im
-
Rm
A commercial version of a
multi-range voltmeter
The multiplier resistors are connected in series, and
each junction is connected to one of the switch
terminals. The range of this voltmeter can be also
calculated from the equation
V  Im( Rm  R)
Where the multiplier, R, now can be
R1 or (R1 + R2) or (R1 + R2 + R3)
(Note: the largest voltage range must be
associated with the largest sum of the multiplier
resistance)
22
EXAMPLE 5: MULTI-RANGE VOLTMETER
Calculate the value of the multiplier resistance for the
multiple range dc voltmeter circuit shown in Figure (a)
and Figure (b), if Ifs = 50μA and Rm = 1kΩ
3v
R1
30 v
R2
S
R1
10 v
10 v
R2
S
+
+
30 v
3v
R3
R3
Im
-
Fig a
Rm
Im
-
Rm
Fig b
23
VOLTMETER LOADING EFFECT
RA
Rs
E
VRB
Ifs= Im
Rs= (S x Vrange) - Rm
Vrange = ( Rs + Rm) Im
Im
RB
RT = Rs +Rm
Rm
Vrange 
Req = RB //RT
Rs  Rm
S
Total voltmeter resistance, RT
RT = Rs + Rm = S x Vrange
24
VOLTMETER LOADING EFFECT
Calculation:
 1) RT = Rs + Rm = S x Vrange
 2) Req = RB // RT
RB
volt-meter VRB 
 3) Without
(expected value)
R R
A
4)
 5)

With volt-meter
(measured value)
Insertion error
VRB 
m
xE
B
Req
Req  R A
xE
VRB  VRB
x100%
VRB
m
25
EXAMPLE 6: VOLTMETER LOADING
EFFECT
RA
Rs
E
VRB
RB
Im
RT = Rs +Rm
Rm
Req = RB //RT
A volt meter (0-10V) that has an internal resistance of
78Ω is used to measure the voltage across resistor RB.
Determine the percentage of error of the reading due to
voltmeter insertion. Let E = 4V, RA=RB = 1kΩ , S =
1kΩ/V
26
DC OHMMETER
Basic Ohmmeter circuit
Fixed portion
Rz
Ifs
0.1Rz
Variable
portion
Rm
0.9Rz
E
X
Y
Rx
27
DC OHMMETER
Before measuring the Rx, the Ωmeter is set to “zero”-calibration
Definition zero = shorting the terminal x-y & adjust Rz to obtain
the full-scale deflection on the meter movement.
E
I fs 
Rz  Rm
w/o Rx
I < Ifs
E
I
Rz  Rm  Rx
with Rx
28
DC OHMMETER
Relationship between full-scale deflection to
the value of Rx is :
I
Rz  Rm
P

I fs Rz  Rm  Rx
This equation is used for marking off the scale on the
meter face of the ohmmeter to indicate the value of a
resistor being measured
29
EXAMPLE 7:DC OHMMETER
A 1 mA full-scale deflection current meter movement is to be
used in an ohmmeter circuit. The meter movement has an
internal resistance, Rm, of 100Ω, and a 3 V battery will be used
in the circuit. If the measured resistor has resistance of 1kΩ,
mark off the meter face for the reading (20%, 40%, 50%, 75%
and 100%) .
30
SOLUTION EX:7
Ohm
3k
4.5k
12k
40%
50%
75%
20%
∞
1k
0
0%
100%
Full scale
percentage
31
MULTIPLE-RANGE OHMMETER
The previous section is not capable of measuring resistance
over wide range of values.
R - fixed resistance &
We need to extend our discussion
of ohmmeters
Ifs
zeroing potentiometer to include
R
multiple-range ohmmeters
z
m
R1
Rx1
R2
R x 10
R3
R x 100
E
X
Y
32
END OF PART 1
33