AC Meters Chapter 03 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.

2 Outline Introduction: What is AC? dArsonval with Half-wave Rectification dArsonval with Full-wave Rectification 3 Outline Electrodynamometer movement.

Loading effects of AC Voltmeters Summary 4 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: 5

Introduction No Meter Movement DC Use AC Use Applications 1

Electrodynamometer 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 DArsonval YES YES-w/ Voltage, currents, rectifiers resistance, etc 6 Introduction AC Waveforms

7 dArsonval MM with Wave Rectification In the previous chapter, we have discussed in detail dArsonval MM (PMMC) and its applications in Ammeter, Voltmeter and Ohmmeters. Now, well learn about using the same MM to measure ac current or voltages. 8 dArsonval MM with

Wave Rectification In order to measure ac with dArsonval 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. 9 dArsonval MM with Wave Rectification Still remember our DC Voltmeter, using

dArsonval meter movement? Im + Sensitivity= 1/Ifs Rs Rm Im Figure 1: The dArsonval meter movement used in a DC voltmeter 10 dArsonval MM with

Wave Rectification 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. 11 dArsonval MM with Wave Rectification

12 dArsonval MM with Wave Rectification If we add a diode to a DC Voltmeter, then we have a meter circuit capable of measuring ac voltage. RS + Rm Im

_ 13 dArsonval MM with Wave Rectification 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? 14 dArsonval MM with

Wave Rectification The voltage across the MM is just the positive 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. 15 dArsonval MM with Wave Rectification 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/ =0.45x Ex Erms 16 dArsonval MM with Wave Rectification 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. 17 dArsonval MM with Wave Rectification 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 wave rectification is only approximately 45% sensitive as a DC Voltmeter.

18 dArsonval MM with Wave Rectification 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 = (0.45x EErms/Idc) -Rm 19

dArsonval MM with Wave Rectification Example 1 Compute 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=300. RS + Rm Im _

20 dArsonval MM with Wave Rectification Example 2 In the wave rectifier shown below, D1 and D2 have an average forward resistance of 50 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 = 200, Ifs = 100mA, Rm = 200 Rs D1

IT Im Ish Ein D2 Rsh Rm Make it as your exercise at home 21

Conclusion dArsonval 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 = 0.45x ESdc 22