1 Electrical Indicating Instruments 1 EIE 240 Electrical and Electronic Measurement Class 4, February 6, 2015 Analogue Meter’s Concept • Han Oersted, in 1820, noted his finding without any explanation. • Lord Kelvin made more sensitivity to a current. I + – N S N S B B Compass 2 Right-hand Rule “Magnetic Field for Line of Current by Biot Savart's Law”
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Electrical Indicating Instruments
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EIE 240 Electrical and Electronic MeasurementClass 4, February 6, 2015
Analogue Meter’s Concept• Han Oersted, in 1820, noted his finding
• Permanent-Magnet Moving-Coil (PMMC)developed in 1881
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CoilPointer
Permanent Magnet
• A wire coil is attached to a shaft that pivots on two jewel bearing.
• The coil can rotate in a space between a cylindrical soft-iron core and two permanent magnetic pole pieces.
• The rotation is opposed by two fine hairsprings.
Deflecting Force= Controlling Force
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• Jacques D’Arsonval’s Movement
When a current is passed through the coil it rotates, the angle through which it rotates being proportional to the current (0.0000001 – 1 A).
The magnetic field is designed (magneticThe magnetic field is designed (magnetic pole piece’s shape) that it is always at the right angles to the coil sides no matter what angle the coil has rotated through.
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• Electric Current
B
B
FF
I
IL
b
• Fleming’s Left-Hand Rule
F = L I B
Force(Newton) Magnetic Field
Moving Coil
“Magnetic Force from a Straight Current-carrying Conductor”
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(Newton) g(Tesla)
Current(Ampere)
Coil Side Length(Metre)
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• Torque (moment) is an angular force defined by linear force multiplied by a radius.
Damping Torque in Coil = F (b/2)
= B I L b / 2
Torquetotal = 2 (B I L b / 2) = B I L b
= B I A , Area A = Lb
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for N coils, Torquetotal = N B I A= Kcoil I , Kcoil = NBA
• Controlling torque in springsTorquespring = Ks
• Critical damping or balancing forces p g g(Newton’s 3rd Law)
Action = ReactionTorquetotal = Torquespring
Kcoil I = Ks= (K / K ) I
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= (Kcoil / Ks) I
• If the magnetic field is not uniform throughout the entire region, the scales are nonlinear!
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Galvanometer• Galvanometer with a zero at the center of
the scale used in DC instruments that can detect current flow in either directiondetect current flow in either direction
• Galvanometer with a zero at the left end of the scale indicates an upscale reading only for the proper way of connecting the meter into the circuit
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Equivalent Ammeter Circuit
• The resistance of the meter coil and leads introduces a departure from the idealintroduces a departure from the ideal ammeter behavior. The model usually used to describe an ammeter in equivalent circuit is a resistance Rg in series with an ideal ammeter (no resistance)
10Ammeter
Rg 50 Ideal
Movement
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Full-Scale-Deflection Currents
• Current range is 10 A – 20 mA• Shunt resistor connected in parallel
Vs = Vg
(I - Ig)Rs = IgRg
I = (R +R )/R I
Rgalvanometer
IIg
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I = (Rs+Rg)/Rs Ig
RshuntIsScaling factor
• Multi-range shunt can be made by switching into a circuit
Rs1
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Rs2
Rs3
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• Universal shunt or “Ayrton shunt”
IgI
Ayrton shunt named after its inventor William E. Ayrton is a high-resistance shunt
Rs1 Rs2 Rs3
AB
C
Is
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used in galvanometers to increase their range.
It eliminates the possibility of having a meter without a shunt which is a serious concern.
• Universal shunt or “Ayrton shunt” (Cont’d)
IgI
RangeA > B > C
A: (I - Ig)Rs1 = Ig (Rg+Rs2+Rs3)
I = (Rg+Rs1+Rs2+Rs3)/Rs1 Ig
Rs1 Rs2 Rs3
AB
C
IsA > B > C
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g g
B: (I - Ig)(Rs1+Rs2) = Ig (Rg+Rs3)
I = (Rg+Rs1+Rs2+Rs3)/(Rs1+Rs2) IgC: I = (Rg+Rs1+Rs2+Rs3)/(Rs1+Rs2+Rs3) Ig
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Full-Scale-Deflection Voltages• Since V = IR, the response of a moving coil
meter responding to a current is also proportional to the potential difference p p pacross the meter.
• However, because Rg and Ig are low, it can only be used for low voltages ( 0.05 V).
• Multiplier resistor can be connected in series
15V = IgRm + IgRg = (Rm+Rg) Ig
Rmultiplier Rgalvanometer
Ig
VmVg
• Multi-range voltmeter
Rm1
Rm2
Rm3
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Rm1 Rm2 Rm3
• Multi-range voltmeter with a chain arrangement
m2 m3
AB
C
A: V = IgRg + IgRm1 = (Rm+Rg) IgB: V = I R + I R + I R = (R +R +R ) I
RangeA < B < C
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B: V = IgRg + IgRm1 + IgRm2 = (Rm1+Rm2+Rg) IgC: V = IgRg + IgRm1 + IgRm2 + IgRm3
= (Rm1+Rm2+Rm3+Rg) Ig
Temperature on Moving-Coil Meter• Higher temperature of coil (tin copper wire), higher
resistance, lower reading decrease spring tension
• Measured value is decreased 0.2% for increasing of g1C temperature.
• Swamp resistor (manganin wire), whose resistance changes slowly with temperature, connected in series
• However, the sensitivity is decreased.
RgS R i t
RSwamp > 3 Rg
18Rs
Swamp Resistors
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Sensitivity on Moving-Coil Meter
• Sensitivity = Pointer Change / Input Change
S = / I
or S = 1 / Ifsd
= R / Vfsd
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Ammeter & Voltmeter Loading
• Systematic loading error C l l ti b i Thé i ’ th i Calculation by using Thévenin’s theorem in
Lecture 3
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Sources and Loads• An electrical circuit consists of nothing
more than “sources” and “loads”.
• Source is to produces an electrical energySource is to produces an electrical energy, e.g. a chemical battery, an electronic power supply, or a mechanical generator.
• Load is to be powered by that electrical energy, e.g. a light bulb, an electronic clock, an electric fan, or just a resistor.
• What sources are AC? And what kinds of loads do not care which way current flows through them? 21
• A current flows one way, then the other way, continually reversing direction.
• The usual waveform of an
Alternating Current
The usual waveform of an AC power is a sine wave that the change is so regular. The average value is zero (integration in a period).
• Main electricity in Thailand has a frequency of 50 Hz y q y(0.02 sec/cycle), and 60 Hz in US.
• Voltage is continually changing between positive (+) and negative (–). The effective voltage in Thailand is 220 V, and 120 V in US.
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Why Use Alternating Current?• In transmission of power, Pload = IV, the
overhead wires are not a perfect conductor and exhibit some resistance R Theand exhibit some resistance, R. The absorbed energy is dissipated as waste heat. The lost power is Plost = I2R.
• At the same power, if the current is doubled (voltage reduced by half), Pload = (2I)(V/2), ( g y ), load ( )( ),the lost power is four times greater, Plost = (2I)2R = 4I2R. To minimize that loss, we have to use much larger wires, and pay a high price for all that extra copper. 23
Why Use Alternating Current? (Cont’d)• The solution is to use a transformer that can convert AC
power at a higher/lower voltage with very slight losses. In practice, AC generators create electricity at a reasonable voltage then we use transformers to step it up to very highvoltage, then we use transformers to step it up to very high levels ( 100,000 V) for long-distance transmission with a low current, and then use additional transformers to step it back down for local distribution to individual homes in safe value. Vp / Np = Vs / Ns
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Averaging Power for AC
• P(t) = I2(t) R is just an instantaneous power at a time t.
• Mean power or the total energy converted in one cycle is
P = 1/T t=0T P dt , T = period= 1/T I2R dt
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= 1/T t=0T I2R dt= R [ 1/T t=0T I2 dt ]= R [ sqrt(1/T t=0T I2 dt) ]2
= R (Irms)2
Sine Wave
• y = sin() = sin(t)
i (2 t/T) 2 /T f i= sin(2t/T) , = 2/T for sinewave= sin(2ft) , f = 1/T
• I(t) = Ioffset + Ipeaksin(t – )
I(t)
Phase
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t
Ioffset
PeriodT
0
DC BiasIpeak
Phase
Amplitude
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RMS or Effective ValueFor only sinusoidal wave, = 2/TIroot-mean-square = sqrt(I2average) = sqrt(1/Tt=0T I2(t)dt)
= sqrt(1/T t=0T [ Ipeaksin(t) ]2 dt )
= sqrt((Ipeak)2/T t=0T sin2(t) dt )
= sqrt((Ipeak)2/2T t=0T 1-cos(2t) dt )
= sqrt( (Ipeak)2/2T { T – [sin(4)-sin(0)] } )I( )
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p
= sqrt( (Ipeak)2/2 )
= Ipeak / sqrt(2)= 0.707 Ipeak (70.7% of peak current)
t
I(t)Ipeak
T0 T/23T/2
2T
Positive Half-Cycle AveragingFor only sinusoidal wave, T=2 and =2/T=1I+ half cycle = Iaverage = 1/(T/2) t=0T/2 I(t) dt