Instrumentation Lecture 2b

AC Meters Chapter 03

Objectives  At the end of this chapter, the

students should be able to:

Describe the operation of 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. 

Outlines • Introduction: What is AC. • D’Arsonval with Half-wave Rectification. • D’Arsonval with Full-wave Rectification.

Outlines • Electrodynamometer movement. • Loading Effects of AC Voltmeters • Summary

Introduction • Several types of meter movements maybe used to measure AC current or voltage. • The five principle meter movements used in ac instruments are listed in Table below:

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

D’Arsonval

YES

YES-w/ Voltage, currents, rectifiers resistance, etc…

d’Arsonval MM with ½ Wave Rectification. • 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 unidirectional current flow. • Several types of diode rectifiers are available: copper oxide, vacuum diode, semiconductor diode etc.

d’Arsonval MM with ½ Wave Rectification. • Still remember our DC Voltmeter, using d’Arsonval meter movement? Im +

Sensitivity= 1/Ifs

Rs

Rm Im

Figure 1: The d’Arsonval meter movement used in a DC voltmeter

d’Arsonval MM with ½ 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.

d’Arsonval MM with ½ Wave Rectification.

d’Arsonval MM with ½ Wave Rectification. • If we add a diode to a DC Voltmeter, then we have a meter circuit capable of measuring ac R voltage. S

+

Rm I m

_

d’Arsonval MM with ½ Wave Rectification. • The FW biased diode will have no effects in the operations of the circuit. (ideal diode) • Now, suppose we replace the 10Vdc with 10Vrms, what will happen?

d’Arsonval MM with ½ Wave Rectification. • 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 with ½ Wave Rectification. • 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. • The ave value of the AC sine wave is :

Eave= Ep/π =0.45x Erms

d’Arsonval MM with ½ 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.

d’Arsonval MM with ½ 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 ½ wave rectification is only approximately 45% sensitive as a DC Voltmeter.

d’Arsonval MM with ½ 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 Rs of DC Voltmeter. • Since the equivalent DC voltage is 45% of RMS value, we can write like this: Rs= (Edc/Idc)-Rm = (0.45Erms/Idc) -Rm

d’Arsonval MM with ½ 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=300Ω. RS

+

Rm I m

_

d’Arsonval MM with ½ Wave Rectification. Example 2 In 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 = 200Ω Rs

IT

D1

Im Ish

Ein

D2

Rsh Rm