Ac Meters Part 1
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Transcript of Ac Meters Part 1
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AC Meters
Chapter 03
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Objectives
At the end of this chapter, the student should be able to: Describe the operation of a half-wave rectifier
circuit. Trace the current path in a full-wave bridge
rectifier circuit. Calculate ac sensitivity and the value of multiplier
resistors for half-wave and full-wave rectification.
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Outline
• Introduction: What is AC?• d’Arsonval with Half-wave
Rectification • d’Arsonval with Full-wave
Rectification
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Outline
• Electrodynamometer movement.• Loading effects of AC Voltmeters• Summary
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Introduction
Several types of meter movements maybe used to measure AC current or voltage.
The five principal meter movements used in ac instruments are listed in the table below:
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Introduction
No Meter Movement DC Use
AC Use Applications
1 Electro-dynamometer
YES YES Standard meter,
Wattmeter, etc…
2 Iron-Vane YES YES Indicator applications, etc…
3 Electro-static YES YES High voltage measurement.
4 Thermocouple YES YES Radio freq measurement
5 D’Arsonval YES YES-w/ rectifiers
Voltage, currents, resistance, etc…
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Introduction
AC Waveforms
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d’Arsonval MM with½ Wave Rectification
• In the previous chapter, we have discussed in detail d’Arsonval MM (PMMC) and its applications in Ammeter, Voltmeter and Ohmmeters.
• Now, we’ll learn about using the same MM to measure ac current or voltages.
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• In order to measure ac with d’Arsonval MM, we must first rectify the ac current by use of a diode rectifier.
• This process will produce uni-directional current flow.
• Several types of diode rectifiers are available: -copper oxide, vacuum diode, semiconductor diode etc.
d’Arsonval MM with½ Wave Rectification
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• Still remember our DC Voltmeter, using d’Arsonval meter movement?
Rs
Rm Im
Im
+
-
Figure 1: The d’Arsonval meter movement used in a DC voltmeter
Sensitivity= 1/Ifs
d’Arsonval MM with½ Wave Rectification
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• PMMC meter movements will not work correctly if directly connected to alternating current, because the direction of needle movement will change with each half-cycle of the AC.
• Permanent-magnet meter movements, like permanent-magnet motors, are devices whose motion depends on the polarity of the applied voltage.
d’Arsonval MM with½ Wave Rectification
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d’Arsonval MM with½ Wave Rectification
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• If we add a diode to a DC Voltmeter, then we have a meter circuit capable of measuring ac voltage.
RS
Rm Im
+
_
d’Arsonval MM with½ Wave Rectification
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• The FW biased diode will have no effects in the operations of the circuit. (ideal diode)
• Now, suppose we replace the 10-Vdc with 10Vrms, what will happen?
d’Arsonval MM with½ Wave Rectification
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• The voltage across the MM is just the positive ½ cycle of the sine wave because of rectifying action of the diode.
• The peak value of the ac sine wave is :
Ep= Erms X 1.414.
d’Arsonval MM with½ Wave Rectification
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• The MM will respond to the average value of the sine wave where the average, or DC value equals 0.318 times the peak value.
• The average value of the AC sine wave is :
Eave= Ep/π =0.45x Erms
d’Arsonval MM with½ Wave Rectification
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• The diode action produces an approximately half sine wave across the load resistor.
• The average value of this voltage is referred to as the DC voltage, which a DC voltmeter connected across a load resistor will respond to.
d’Arsonval MM with½ Wave Rectification
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• Therefore, we can see that the pointer that deflected full scale when a 10-V DC signal was applied, deflects to only 4.5V when we apply a 10-Vrms sine AC waveforms.
• Thus, an AC Voltmeter using ½ wave rectification is only approximately 45% sensitive as a DC Voltmeter.
d’Arsonval MM with½ Wave Rectification
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• In order to have a full scale deflection meter when a 10-Vrms is applied, we have to design the meter with the Rs having 45% of the Rs of the DC voltmeter.
• Since the equivalent DC voltage is 45% of the RMS value, we can write like this:
Rs= (Edc/Idc)-Rm = (0.45Erms/Idc) -Rm
d’Arsonval MM with½ Wave Rectification
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Example 1Compute the value of Rs for a 10-Vrms AC range on the voltmeter shown in Figure 1.
Given that Ein= 10-Vrms, Ifs= 1mA, Rm=300Ω.
RS
Rm Im
+
_
d’Arsonval MM with½ Wave Rectification
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Example 2In the ½ wave rectifier shown below, D1 and D2 have an average forward resistance of 50Ω and are assumed to have an infinite resistance in reverse biased. Calculate the following:(a) Rs value(b) Sac
(c) Sdc
Given that Ein = 10-Vrms, Rsh = 200Ω, Ifs = 100mA, Rm = 200ΩRs
Rm
D2
D1
RshEin
IT
Ish
Im
Make it as your exercise at home
d’Arsonval MM with½ Wave Rectification
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Conclusion
• d’Arsonval MM can be used to measure both DC and AC current/voltages.
• The MM will respond to the average value of sine wave where the average, or DC value equal to 0.318 times the peak value.
• Sac = 0.45Sdc