ac dc meters -2

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DC & AC METERS(PART-II)

5/28/2014 1RUSAYL INSTITUTE LLC


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CHAPTER OUTLINE

3.5 DC Ohmmeter

3.6 Introduction to AC meter

3.7 dArsonval meter movement (half-waverectification)

3.8 dArsonval meter movement (full-wave

rectification)

3.9 Loading effects of AC meter



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OBJECTIVES

At the end of this chapter, students should be able to:

Explain the purpose of OhmmeterDescribe the construction and operation of a

basic OhmmeterDescribe the operation of half-wave rectifier

circuit

Trace the current path in a full-wave bridge

rectifier circuitCalculate ac sensitivity and the value of

multiplier resistors for half-wave and full-waverectification



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3.5 DC OHMETER

The purpose of an Ohmmeter is to measure resistance

Resistance reading is indicated trough a mechanicalmeter movement which operates on electric current.

Thus, Ohmmeter must have an internal source ofvoltage to create current necessary to operate themovement.

It also must have an appropriate ranging resistors to

allow just the right amount of current. A simple Ohmmeter comprises battery and meter

movement as in figure below:



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3.5 DC OHMETER

When there is infinite resistance (no continuity between testleads), there is zero current through the meter movement, andthe needle points toward the far left of the scale.

In this regard, the ohmmeter indication is "backwards" becausemaximum indication (infinity) is on the left of the scale, whilevoltage and current meters have zero at the left of their scales.



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3.5 DC OHMETER

So, to avoid this, add series resistance to the meters circuit sothat the movement just registers full-scale when the test leads are

shorted together

To determine the proper value for R, calculate the Rtotalneeded tolimit current to only 1mA (full-scale) with 9V of potential from the

battery, then subtract the movement's internal resistance:

99

1total

V VR k

I mA

500 8.5total

R R k



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3.5 DC OHMETER

Ohmmeters only function correctly when measuringresistance that is not being powered by a voltage or

current source.

In other words, you cannot measure resistance with anohmmeter on a "live" circuit!

The reason for this is simple: the ohmmeter's accurateindication depends on the only source of voltage being

its internal battery. The presence of any voltage across

the component to be measured will interfere with theohmmeter's operation.

If the voltage is large enough, it may even damage theohmmeter.



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EVALUATION

Find the value of R, scale, scale and

scale of this Ohmmeter?



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SUMMARY

In this sub-topic, we have learned about:

Ohmmeters contain internal sources of voltage tosupply power in taking resistance measurements.

An analog ohmmeter scale is "backwards" from that ofa voltmeter or ammeter, the movement needle readingzero resistance at full-scale and infinite resistance at

rest.

Ohmmeters should neverbe connected to anenergized circuit (that is, a circuit with its own source of

voltage). Any voltage applied to the test leads of an

ohmmeter will invalidate its reading.



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3.6 INTRODUCTION TO AC

METERReview

Most meters use a moving coil.

A coil of wire to which the pointer is attached ispivoted between the poles of a permanent magnet.

When a current is present in the coil, it sets up amagnetic field that interacts with the field of the magnet

to cause the coil to turn.



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METERAn ac meter measures ac voltages; a dc metermeasures dc voltages.

The actual scale calibration of commercially made

(ac) voltmeters is almost always in terms of RMS (rootmean square) values useful when variates are positive and

negative, e.g., sinusoids

For sine waves this is satisfactory, and useful sinceRMS is the standard measure at power-line frequency.



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METER



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METER

Half Waveform

Full WaveformAC Waveform

There are two types of ac meters that will bediscussed.

i) Half-wave rectifier voltmeter

ii) Full-wave rectifier voltmeter

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METERAC Waveforms

5/28/201415RUSAYL INSTITUTE LLC


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3.6DARSONVAL METER MOVEMENT(HALF-WAVE RECTIFICATION)

Rs

Rm Im

Im

+

-

Still remember our DC Voltmeter, using dArsonval MM?

To measure ac with dArsonval MM, rectify the ac current byusing a diode rectifier

This process will produce a uni-directional current flowSeveral types of diode rectifiers are available: -

copper oxide, vacuum diode, semiconductor diode etc.

The dArsonval meter movement

used in a DC voltmeter

1

dc

fsd

SI



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3.6 DARSONVAL METER MOVEMENT

(HALF-WAVE RECTIFICATION)

If we add a diode to a DC Voltmeter, then we have ameter circuit capable of measuring ac voltage.

RS

Rm Im

+

_



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In half wave rectification, either the positive or negative half ofthe AC wave is passed, while the other half is blocked

Because only one half of the input waveform reaches theoutput, it is very inefficient if used for power transfer

Half-wave rectification can be achieved with a single diode in aone phase supply, or with three diodes in a three-phase supply.

http://www.falstad.com/circuit/e-rectify.html5/28/2014 18RUSAYL INSTITUTE LLC
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Now, suppose we replace the 10-Vdc with 10Vrms,what will happen?

The voltage across the MM is just the positive cycle

of the sine wave because of rectifying action of thediode.

The peak value of the ac sine wave is :

1.414p rmsV V x



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The MM will respond to the average value of sinewave where the average, or DC value equal to 0.318

times the peak value.

The average value of the AC sine wave is :

2 2 1.4140.9

p rmsave rms

V xV xV xV



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0.9 9ave rms

V xV V

Let say Vin=10Vrms

10 1.414 14.14p rmsV V x V

Since the diode conducts only during positive half cycle, theaverage value over the entire cycle is one half the average value of

9V 4.5V

Therefore, the pointer will deflect for a full scale if 10V dc isapplied and 4.5V when a 10 Vrmssinusoidal signal is applied

Thus, ac voltmeter is not as sensitive as dc voltmeter5/28/2014 21RUSAYL INSTITUTE LLC


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0.45dc rms

s m m

dc dc

V VR R R

I I

In order to have a full scale deflection meter when a10-Vrmsis applied, we have to design the meter withRshaving 45% of Rsof DC Voltmeter.

Since the equivalent DC voltage is 45% of RMSvalue, we can write like this:

0.45dc rmsV xV



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EXAMPLE 3.10

Compute the value of Rsfor a 10-VrmsAC range onthe voltmeter shown in figure below.

Given: Ein= 10-Vrms, Ifsd= 1mA, Rm=200.

RS

Rm Im

+

_



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EXAMPLE 3.11 (solution)

0.45rms

s m

dc

xVR R

I

0.45 10200 4.3

1

x Vk

mA



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Commercial AC voltmeter

Rsh increase current flow through D1 during the +ve cycle

diode will be operating in linear region

improve linearity of AC meter during measurement of low

voltage, but further reduces the ac sensitivity5/28/2014 25RUSAYL INSTITUTE LLC


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Commercial AC voltmeter

D2 +ve cycle = no effect (Reverse-bias)

-ve cycle = provides an alternate path for reverse biased

leakage currentthat would normally flow through meter

movementand D1

.5/28/2014 26RUSAYL INSTITUTE LLC


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EXAMPLE 3.11

In the wave rectifier shown below, D1 and D2 havean average forward resistance of 50 and are assumedto have an infinite resistance in reverse biased.Calculate the following:

(a) Rsvalue(b) Sac(c) Sdc

Given that Ein= 10-Vrms, Rsh= 200, Ifs= Im=100A, Rm= 200

Rs

Rm

D2

D1

Rsh

Ein

IT

Ish

Im



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EXAMPLE 3.11(solution)



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SUMMARY

dArsonval MM can be used to measure both DCand AC current/ voltages.

The MM will respond to the average value of sine

wave where the average, or DC value equal to 0.318times the peak value.

Sac= 0.45Sdc



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(FULL-WAVE RECTIFICATION)

It is more desirable to use a full-wave rectifier in ACvoltmeters because it shows higher sensitivity rating

compared to wave rectifier.

The most frequently used circuit for full-waverectification is the bridge-type rectifier, as shown below



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D1 D2

D3

D4

Rs

Vin

Rm

IfsVout

Figure 2: Full-Wave Bridge Rectifier used in an AC voltmeter

Positive cycle

Current flows from Vintrough D2, trough the MM frompositive to negative, then trough D3.

Negative cycle Current flows from Vintrough D4, trough the MM from

positive to negative, then trough D1.5/28/2014 31RUSAYL INSTITUTE LLC


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Even though the polarity of Vin changes in every cycle, the direction of current that flows through

MM does not change.

Since current flows through MM on both cycles,then the deflection of MM will be greater compared

to wave rectification.



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EXAMPLE 3.12

For the circuit below, given Vin = 10Vrms, Rm= 250

and Ifs= 1mA. Calculate:

a) Vp

b) Vavec) Sdc

d) Sac

e) Rs

D1D2

D3

D4

Rs

Vin

Rm

IfsVout



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EXAMPLE 3.12 (solution)

a) Vp = 1.414 x Vrms = 1.414 x 10V = 14.14V

b) Vave = 0.9 x Vrms = 0.9 x 10V = 9V

c) Sdc = 1 / Ifsd = 1 / 1mA = 1k/V

d) Sac = 0.9 x Sdc = 0.9 x 1k/V 0.9k/V

e) Rs = Sac x rangeRm = 0.9k/V x 10V -250= 8.75



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EXAMPLE 3.13

For the circuit below, given that Vin= 10Vrms,

Rm=500, Ifs= 1mA, Rsh=500, average forward

resistance of 50 and infinite reverse resistance of

each diode, calculate the following:

a) the multiplier, Rsb) the ac sensitivity, Sacc) the dc sensitivity Sdc.

D1D2

D3

D4

Rs

Vin

Rm

Ifs

Rsh



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3.9 LOADING EFFECTS OF AC

METER

As already being discussed, the sensitivity ofAC Voltmeters, using either wave or Full

wave rectification, is always less than the

sensitivity of the DC Voltmeters.

Therefore, the loading effect of an ACVoltmeter is always greater than that

of a DC Voltmeter.



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3.9 LOADING EFFECTS OF AC

METER

a) Determine the reading obtained with DC voltmeter at RBwhen

the switch is set at point A.

b) Determine the reading at the same RBusing wave and

Full wave rectifier AC meter respectively when the switch

is set at point B.Given that Ifs = 100-mA and set at 10-V dc or rms range.

RB

E = 20V

RA 10k

5.5k

AC20Vrms

A B



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ac dc meters -2