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EKT212/4 - PRINCIPLE OF MEASUREMENTAND INSTRUMENTATION

Voltage Measurement

Semester IIAcademic session 2016/2017

School of Computer and Communication EngineeringUniversiti Malaysia Perlis (UniMAP)

Email: said@unimap.edu.my

Outline

Introduction

Voltmeter design

DC VoltmeterMulti-range VoltmeterLoading Effects

AC VoltmeterIntroductionHalf-wave RectificationFull-wave Rectification

Introduction

I A voltmeter is an instrument used for measuring thepotential difference between two points in an electric circuit.

I A voltmeter is placed in parallel with a circuit element tomeasure the voltage drop across it.

I Voltmeter must be designed to draw very little current fromthe circuit so that it does not appreciably change the circuit itis measuring.

Voltmeter design

Voltmeter design...continued

I A galvanometer full-scale current is very small: on the orderof mA.

I To accomplish this, a large resistor is placed in series with thegalvanometer.

I Its value is chosen so that the design voltage placed across themeter will cause the meter to deflect to its full-scale reading.

I To allow meter movement to measure a greater voltage, weneed a voltage divider circuit (series resistances) toproportion the total measured voltage into a lesser fractionacross the meter movements connection points.

I A resistor is connected in series with the meter movement andit is called a multiplier resistor because it multiplies theworking range of the meter movement as it proportionatelydivides the measured voltage across it.

DC Voltmeter

I The basic D’Arsonval meter movement can be converted to aDC voltmeter by connecting a multiplier Rs in series with themeter movement.

I The purpose of the multiplier:I to extend the voltage range of the meterI to limit current through the D’Arsonval meter movement to a

maximum full-scale deflection

V = Im(Rs +Rm)

Rs =V

Im−Rm

DC Voltmeter...continued

I Since multiplier, Rs limits the current through the movementto not exceed the value of full scale deflection, Ifsd, this gives;

Im = Ifsd

Rs =V

Ifsd−Rm

I Hence, the sensitivity, S of the meter movement can be givenby

S =1

Ifsd(Ω/V )

DC VoltmeterEXAMPLE

Calculate the value of the multiplier resistance on the 50V rangeof a DC voltmeter that used a 500µA meter movement with aninternal resistance of 1kΩ.

Given that, V = 50V, Rm = 1kΩ and Im = Ifsd = 500µA,

Rs =V

Ifsd−Rm =

50

500× 10−6− 1× 10−3 = 99000

The multiplier, Rs is equal to 99kΩ.

Multi-range Voltmeter

I In DC voltmeter, to obtain multi-range voltmeter, aconnection of a number of multipliers along with a rangeswitch can be implemented.

I Figure shows a multi-range voltmeter in individual multipliersformation.

Multi-range VoltmeterEXAMPLE

A D’Arsonval movement with a full scale deflection of 50µAand having an internal resistance of 500Ω is to be convertedinto a multi-range voltmeter. Determine the value of multiplierrequired for 0 – 20V, 0 – 50 V and 0 – 100V using individualmultipliers for each range. Calculate the values of the individualresistor.

Given that Im = 50µA and Rm = 500ΩFor range 0 – 20V;

Rs =V

Im−Rm =

20

50× 10−6− 500 = 399.5kΩ

For range 0 – 50V; Rs = 999.5kΩFor range 0 – 100V; Rs = 1999.5kΩ

Multi-range Voltmeter

I Another formation of multi-range voltmeter is the multipliersare connected in series.

I In this arrangement, all multipliers except the first one canhave a standard resistance value.

Multi-range VoltmeterEXERCISE

Convert a basic D’Arsonval movement with a full scaledeflection of 10mA and having an internal resistance of 100Ωinto a multi-range voltmeter with ranges from 0 – 5V, 0 – 50 Vand 0 – 100V.

Loading Effects

I An ideal voltmeter has infinite resistance, hence, it draws nocurrent from the circuit under test.

I However, practically, this is not the case.

I When a voltmeter is used to measure the voltage across acircuit component, the voltmeter circuit itself is in parallelwith the circuit component.

Loading Effects

I Since the parallel combination of two resistors is less thaneither resistor alone, the resistance seen by the source is lesswith the voltmeter connected. Therefore, the voltage acrossthe component is less whenever the voltmeter is connected.

I The decrease in voltage may be negligible or it may beappreciable, depending on the sensitivity of the voltmeterbeing used.

I This effect is called voltmeter loading. The resulting error iscalled a loading error.

Loading EffectsEXAMPLE

Two different voltmeters are used to measure the voltage acrossresistor RB in the circuit. The meters are as follows.Meter A: S = 1kΩ/V, Rm = 0.2kΩ, range = 10VMeter B: S = 20kΩ/V, Rm = 1.5kΩ, range =10VCalculate:a. Voltage across RB without any meter connected across it.b. Voltage across RB when meter A is used.c. Voltage across RB when meter B is usedd. Error in voltmeter readings.

Loading EffectsEXAMPLE ...continued

a. Voltage across RB without any meter connected across it.

VRB=

RB

RB +RA× E =

5× 103

(25× 103) + (5× 103)× 30 = 5V

Loading EffectsEXAMPLE ...continued

b. Voltage across RB when meter A is used.

Given that S = 1kΩ/V, Rm = 0.2kΩ, range = 10VTotal meter resistance,

RmA = Rs +Rm =V

Ifsd= SV = 1× 10 = 10kΩ

Total resistance across RB is RB||RmA,

Req =RB +RmA

RB ×RmA= 3.33kΩ

Hence,

VRB=

Req

Req +RA× E =

3.33× 103

(3.33× 103) + (25× 103)× 30 = 3.53V

Loading EffectsEXAMPLE ...continued

c. Voltage across RB when meter B is used

VRB= 4.9V

d. Error in voltmeter readings

Voltmeter A error = 29.4%Voltmeter B error = 2%

AC VoltmeterIntroduction

I Generally, AC voltmeter make use of PMMC metermovement and rectifier.

I Rectifier arrangement usually employs silicon diodes due to itslow reverse current and high forward current ratings.

I Two types of PMMC meter used in AC measurement:I Half-wave rectificationI Full-wave rectification

I AC voltmeter using half wave and full wave rectification aresuitable for measuring only sinusoidal voltages.

AC VoltmeterIntroduction ...signal properties

I Due to inertia, the meter indicates a steady deflectionproportional to the average value of the current.

I Important relations are as follows:

Irms =Im√

2

Iav =2Imπ

I For half-wave rectification, it expected that average value willreduce to half.

Half-wave Rectification

I It can be obtained by addingdiode to the DC voltmetercircuit such in figure.

I Diode D1 conducts only inpositive half cycle of theinput AC voltage.

Half-wave RectificationAnalysis

Due to the rectification, the peak voltage is given by

Vp = Vrms ×√

2 = 10×√

2 = 14.14

Hence, the effective DC voltage is equal to average value given by

Vav =2Vpπ

=2× 14.14

π= 9.002

But, the circuit above consist of D1 where it only conducts onpositive half cycle, therefore the effective DC voltage is

Vdc =Vav2

= 4.5

Half-wave RectificationAnalysis... continued

Based on the voltmeter above, the sensitivity is given by

Sdc =1

Ifsd= 1000

With the given Rs = 10, 000Ω, the 10V DC voltage will move themeter to its full-scale.But, when AC voltage of 10V rms is applied, the effective DCvoltage is only 4.5V.

Sac = 0.45Sdc

Hence, the value of multiplier resistor required for AC voltmeterusing half-wave rectifier can be calculated as follows:

Rs =VdcIdc−Rm =

0.45× Vrms

Idc−Rm

Half-wave RectificationEXAMPLE

Compute the value of the multiplier resistor for a 15Vrms ACrange on the voltmeter with internal resistance of 300Ω andfull-scale deflection current of 1mA.

Given that Vrms=15V, Rm=300Ω and Ifsd=1mA, therefore, themultiplier resistor is given by

Rs =0.45× Vrms

Im−Rm =

0.45× 15

1× 10−3− 300 = 6.45kΩ

Half-wave RectificationGeneral rectifier type voltmeter

I Commercially produced half-wave AC voltmeter is based onthis circuit.

I Diode D2 and shunt, Rsh were added to the circuit.

I In positive half cycle, D1 conducts and the meter deflectaccording to the average value of this half cycle.

I The purpose of Rsh in the circuit is to draw more currentthrough diode D1.

I In negative half cycle, the diode D2 conducts and the currentflows through the measuring circuit.

Full-wave Rectification

Due to the rectification, the peak voltage is given by

Vp = Vrms ×√

2 = 10×√

2 = 14.14

Hence, the effective DC voltage is equal to average value given by

Vav =2Vpπ

=2× 14.14

π= 9.002

Therefore, for full-wave rectification, the 10V rms AC voltage willturn the meter to the approximately 9V mark.

Full-wave RectificationEXAMPLE

Compute the value of the multiplier resistor for a 10Vrms ACrange on the full-wave rectification voltmeter with internalresistance of 500Ω and the DC sensitivity of 1000 Ω/V.

Given the Rm=500Ω and Sdc=1000. For full-wave rectificationAC voltmeter,

Sac = 0.9× Sdc = 900Ω/V

Hence, the multiplier resistor is given by

Rs = Sac × Rangeac −Rm

Rs = 900× 10− 500 = 8500Ω