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DC & AC METERS(PART-II)
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CHAPTER OUTLINE
3.5 DC Ohmmeter
3.6 Introduction to AC meter
3.7 dArsonval meter movement (half-waverectification)
3.8 dArsonval meter movement (full-wave
rectification)
3.9 Loading effects of AC meter
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OBJECTIVES
At the end of this chapter, students should be able to:
Explain the purpose of OhmmeterDescribe the construction and operation of a
basic OhmmeterDescribe the operation of half-wave rectifier
circuit
Trace the current path in a full-wave bridge
rectifier circuitCalculate ac sensitivity and the value of
multiplier resistors for half-wave and full-waverectification
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3.5 DC OHMETER
The purpose of an Ohmmeter is to measure resistance
Resistance reading is indicated trough a mechanicalmeter movement which operates on electric current.
Thus, Ohmmeter must have an internal source ofvoltage to create current necessary to operate themovement.
It also must have an appropriate ranging resistors to
allow just the right amount of current. A simple Ohmmeter comprises battery and meter
movement as in figure below:
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3.5 DC OHMETER
When there is infinite resistance (no continuity between testleads), there is zero current through the meter movement, andthe needle points toward the far left of the scale.
In this regard, the ohmmeter indication is "backwards" becausemaximum indication (infinity) is on the left of the scale, whilevoltage and current meters have zero at the left of their scales.
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3.5 DC OHMETER
So, to avoid this, add series resistance to the meters circuit sothat the movement just registers full-scale when the test leads are
shorted together
To determine the proper value for R, calculate the Rtotalneeded tolimit current to only 1mA (full-scale) with 9V of potential from the
battery, then subtract the movement's internal resistance:
99
1total
V VR k
I mA
500 8.5total
R R k
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3.5 DC OHMETER
Ohmmeters only function correctly when measuringresistance that is not being powered by a voltage or
current source.
In other words, you cannot measure resistance with anohmmeter on a "live" circuit!
The reason for this is simple: the ohmmeter's accurateindication depends on the only source of voltage being
its internal battery. The presence of any voltage across
the component to be measured will interfere with theohmmeter's operation.
If the voltage is large enough, it may even damage theohmmeter.
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EVALUATION
Find the value of R, scale, scale and
scale of this Ohmmeter?
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SUMMARY
In this sub-topic, we have learned about:
Ohmmeters contain internal sources of voltage tosupply power in taking resistance measurements.
An analog ohmmeter scale is "backwards" from that ofa voltmeter or ammeter, the movement needle readingzero resistance at full-scale and infinite resistance at
rest.
Ohmmeters should neverbe connected to anenergized circuit (that is, a circuit with its own source of
voltage). Any voltage applied to the test leads of an
ohmmeter will invalidate its reading.
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3.6 INTRODUCTION TO AC
METERReview
Most meters use a moving coil.
A coil of wire to which the pointer is attached ispivoted between the poles of a permanent magnet.
When a current is present in the coil, it sets up amagnetic field that interacts with the field of the magnet
to cause the coil to turn.
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3.6 INTRODUCTION TO AC
METERAn ac meter measures ac voltages; a dc metermeasures dc voltages.
The actual scale calibration of commercially made
(ac) voltmeters is almost always in terms of RMS (rootmean square) values useful when variates are positive and
negative, e.g., sinusoids
For sine waves this is satisfactory, and useful sinceRMS is the standard measure at power-line frequency.
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3.6 INTRODUCTION TO AC
METER
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3.6 INTRODUCTION TO AC
METER
Half Waveform
Full WaveformAC Waveform
There are two types of ac meters that will bediscussed.
i) Half-wave rectifier voltmeter
ii) Full-wave rectifier voltmeter
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3.6 INTRODUCTION TO AC
METERAC Waveforms
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3.6DARSONVAL METER MOVEMENT(HALF-WAVE RECTIFICATION)
Rs
Rm Im
Im
+
-
Still remember our DC Voltmeter, using dArsonval MM?
To measure ac with dArsonval MM, rectify the ac current byusing a diode rectifier
This process will produce a uni-directional current flowSeveral types of diode rectifiers are available: -
copper oxide, vacuum diode, semiconductor diode etc.
The dArsonval meter movement
used in a DC voltmeter
1
dc
fsd
SI
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3.6 DARSONVAL METER MOVEMENT
(HALF-WAVE RECTIFICATION)
If we add a diode to a DC Voltmeter, then we have ameter circuit capable of measuring ac voltage.
RS
Rm Im
+
_
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3.6 DARSONVAL METER MOVEMENT
(HALF-WAVE RECTIFICATION)
In half wave rectification, either the positive or negative half ofthe AC wave is passed, while the other half is blocked
Because only one half of the input waveform reaches theoutput, it is very inefficient if used for power transfer
Half-wave rectification can be achieved with a single diode in aone phase supply, or with three diodes in a three-phase supply.
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3.6 DARSONVAL METER MOVEMENT
(HALF-WAVE RECTIFICATION)
Now, suppose we replace the 10-Vdc with 10Vrms,what will happen?
The voltage across the MM is just the positive cycle
of the sine wave because of rectifying action of thediode.
The peak value of the ac sine wave is :
1.414p rmsV V x
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3.6 DARSONVAL METER MOVEMENT
(HALF-WAVE RECTIFICATION)
The MM will respond to the average value of sinewave where the average, or DC value equal to 0.318
times the peak value.
The average value of the AC sine wave is :
2 2 1.4140.9
p rmsave rms
V xV xV xV
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3.6 DARSONVAL METER MOVEMENT
(HALF-WAVE RECTIFICATION)
0.9 9ave rms
V xV V
Let say Vin=10Vrms
10 1.414 14.14p rmsV V x V
Since the diode conducts only during positive half cycle, theaverage value over the entire cycle is one half the average value of
9V 4.5V
Therefore, the pointer will deflect for a full scale if 10V dc isapplied and 4.5V when a 10 Vrmssinusoidal signal is applied
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3.6 DARSONVAL METER MOVEMENT
(HALF-WAVE RECTIFICATION)
0.45dc rms
s m m
dc dc
V VR R R
I I
In order to have a full scale deflection meter when a10-Vrmsis applied, we have to design the meter withRshaving 45% of Rsof DC Voltmeter.
Since the equivalent DC voltage is 45% of RMSvalue, we can write like this:
0.45dc rmsV xV
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EXAMPLE 3.10
Compute the value of Rsfor a 10-VrmsAC range onthe voltmeter shown in figure below.
Given: Ein= 10-Vrms, Ifsd= 1mA, Rm=200.
RS
Rm Im
+
_
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EXAMPLE 3.11 (solution)
0.45rms
s m
dc
xVR R
I
0.45 10200 4.3
1
x Vk
mA
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3.6 DARSONVAL METER MOVEMENT
(HALF-WAVE RECTIFICATION)
Commercial AC voltmeter
Rsh increase current flow through D1 during the +ve cycle
diode will be operating in linear region
improve linearity of AC meter during measurement of low
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3.6 DARSONVAL METER MOVEMENT
(HALF-WAVE RECTIFICATION)
Commercial AC voltmeter
D2 +ve cycle = no effect (Reverse-bias)
-ve cycle = provides an alternate path for reverse biased
leakage currentthat would normally flow through meter
movementand D1
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EXAMPLE 3.11
In the wave rectifier shown below, D1 and D2 havean average forward resistance of 50 and are assumedto have an infinite resistance in reverse biased.Calculate the following:
(a) Rsvalue(b) Sac(c) Sdc
Given that Ein= 10-Vrms, Rsh= 200, Ifs= Im=100A, Rm= 200
Rs
Rm
D2
D1
Rsh
Ein
IT
Ish
Im
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EXAMPLE 3.11(solution)
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SUMMARY
dArsonval MM can be used to measure both DCand AC current/ voltages.
The MM will respond to the average value of sine
wave where the average, or DC value equal to 0.318times the peak value.
Sac= 0.45Sdc
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3.8 DARSONVAL METER MOVEMENT
(FULL-WAVE RECTIFICATION)
It is more desirable to use a full-wave rectifier in ACvoltmeters because it shows higher sensitivity rating
compared to wave rectifier.
The most frequently used circuit for full-waverectification is the bridge-type rectifier, as shown below
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3.8 DARSONVAL METER MOVEMENT
(FULL-WAVE RECTIFICATION)
D1 D2
D3
D4
Rs
Vin
Rm
IfsVout
Figure 2: Full-Wave Bridge Rectifier used in an AC voltmeter
Positive cycle
Current flows from Vintrough D2, trough the MM frompositive to negative, then trough D3.
Negative cycle Current flows from Vintrough D4, trough the MM from
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3.8 DARSONVAL METER MOVEMENT
(FULL-WAVE RECTIFICATION)
Even though the polarity of Vin changes in every cycle, the direction of current that flows through
MM does not change.
Since current flows through MM on both cycles,then the deflection of MM will be greater compared
to wave rectification.
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EXAMPLE 3.12
For the circuit below, given Vin = 10Vrms, Rm= 250
and Ifs= 1mA. Calculate:
a) Vp
b) Vavec) Sdc
d) Sac
e) Rs
D1D2
D3
D4
Rs
Vin
Rm
IfsVout
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EXAMPLE 3.12 (solution)
a) Vp = 1.414 x Vrms = 1.414 x 10V = 14.14V
b) Vave = 0.9 x Vrms = 0.9 x 10V = 9V
c) Sdc = 1 / Ifsd = 1 / 1mA = 1k/V
d) Sac = 0.9 x Sdc = 0.9 x 1k/V 0.9k/V
e) Rs = Sac x rangeRm = 0.9k/V x 10V -250= 8.75
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EXAMPLE 3.13
For the circuit below, given that Vin= 10Vrms,
Rm=500, Ifs= 1mA, Rsh=500, average forward
resistance of 50 and infinite reverse resistance of
each diode, calculate the following:
a) the multiplier, Rsb) the ac sensitivity, Sacc) the dc sensitivity Sdc.
D1D2
D3
D4
Rs
Vin
Rm
Ifs
Rsh
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3.9 LOADING EFFECTS OF AC
METER
As already being discussed, the sensitivity ofAC Voltmeters, using either wave or Full
wave rectification, is always less than the
sensitivity of the DC Voltmeters.
Therefore, the loading effect of an ACVoltmeter is always greater than that
of a DC Voltmeter.
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3.9 LOADING EFFECTS OF AC
METER
a) Determine the reading obtained with DC voltmeter at RBwhen
the switch is set at point A.
b) Determine the reading at the same RBusing wave and
Full wave rectifier AC meter respectively when the switch
is set at point B.Given that Ifs = 100-mA and set at 10-V dc or rms range.
RB
E = 20V
RA 10k
5.5k
AC20Vrms
A B
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THANK YOU!!
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