27 POWERSUPPLYCRCUITS.3.6 Fault Fin

27 POWERSUPPLYCRCUITS.

3.6 Fault Finding Techniques and Typical Fault

Conditions

When a faulty power unit is returned for repair, the

fault has to be isolated to some particular portion of

the unit. The fault may lie in the transformer, the

rectifier, the filter section, or thegu1ator, and

measurement with a voltmeter will be necessary to

locate the fault.

However its probably best to start diagnosis with

a few rather obvious but often overlooked checks.

First measure the d.c. output voltage. If this is

zero, the next check should be on the mains input.

Is the mains supply reaching the transformer primary?

If it isn’t, there is the possibility of a faulty plug

(relatively simple to repairs), open circuit mains

wires, or a blown fuse. If the fuse i suspected always

test its continuity with an ohmmeter, never rely on

Just a visual Inspection It is also worth noting that

both the live and neutral wires may have a fuse in

circuit, so make sure bath are checked.

lithe fuse is blown it has done so because of some

fault condition and the fault must be cleared before

anew fuse is fitted. Resistance checks ‘(with’ the mains

unplugged!) must be used to locatštich a fault. Use

an ohmmeter to measure the resistance of the trans

former primary, the secondary, the rectifiers, and so

on. The winding resistance depends, of course, on

the size of the transformer. The primary resistance,

for a medium size transformer, should be low,

typically about 50 12. The secondary, usually supply.

ing a lower voltage, may have a resistance of only a

few ohms. Detecting shorted turns on a winding can

therefore be. quite’difficu1 Wherever possible com

pare the measured resistance with any available data

on the type of transformer being used. Another Use-

. ful check is to run the transformer off load and test

for overheating.

When using an ohmmeter take care to use the .

correct polarity for resistance checks where diodes,

electrolytic capacitors and transistors are presént. It

is all too easy to get misleading results, For example,

in Fig. 3.123 if the meter is used to measure the

. resistance of the ‘unstabffized line, the positive prod

(connected inside the meter to the positive plate of a’

battery) should be placed on the positive line and

:theñegative prod to earth. If the meter is reversed

there will be a low reištanee path through the

rectifiers and a leakage path through’ the’ capacitor.

Returning to the faulty power unit, Suppose how

ever that the fuse is intact, and that the mains is

‘reaching the primary. The, next Step is to measure the

secondaryaæ. voltage, the unstabijzedd.c. voltage

then the d.c. voltage in the regulator and so on, until

the fault is located. . ,

Table 3.1 lists some typical faults together with ‘

the associated symptoms, The faults are only a sample

of those which may occur. Locating a faulty com

ponent from a given set of symptoms will come with

practice and the following exercises are designed for

that purpose.

To regulator

Multmeter set

to ohms ronge

Red

Fig. 3.12 Using an ohmmeter to measure the resistance

across the unstabjljzed line

. . /‘









3.7 Exercise: Power Unit with a Simple Linòar

Regulator (Fig. 3.13) .

This unit incorporates most of the features discussed.

.earlier•and.is designed to give an outpûtof.12 Vat

1QO mA. The output resistance.is less than 0.5 c2,

the load tegulation:better than OS%, and the ripple.

. less than 5 mV peak töpeakon full load. ..

, The unstabilizedd.c. is obtained fröm abridge

rectifier circuit and a ryircaacitor of 3300 zF.

The transformer has a secondary voltage of 12 V

r.m.s. so the unstabffized.yoìtage across C1 will be

approximately l2J2 V, i.e. about 16 V.

The reference voltage is provided by a 5.6 V zener

diode; a 400 mW device such as a BZYS8 C5V6 is .

ideal. Tr1 is the d.c..error amplifier which compares

a portion of the d.c. output voltage, the voltage .

across R4, with the reference. Any difference between

the two voltages is amplified by.Tr1 and the amplifled

signal is fed to the base .of Tr2. Consider for example

the case when the d.c. output falls when more load ‘.

current is taken; the base voltage of Tr1 decreases

and Tr1 conducts less current. Therefore Tr1 collector

voltage rises, and this rise in voltage is coupled

through Tr2, which acts as an eWitter follower to

cõunteract the origijial fall in outpút. Thus the .

circuit operates to málntain the output a:nearly .

constant as possible. “ . - , .

Since the output load current is’ ónly 100 mA,

then Tr2 (BFY5I)need not be mounted on a heat

sink. The circuit is no.t provided with”a current limit,

so shouid.an excessive current be drawn, say by

shorting he output, then .Tt2 wciùldundoubedly

bum out A current limit can be added if required as

shown in Fig 3 8, but for the purposes of the exercise

we shall assume no current limit

The normal d e voltages measured with a standard

multirange meter àrè’ as foil àws: . . ‘ . .

Test point 1 2 3 4

Volfige ‘.16’ ‘ 13 , “ 5’.’ 12.2

First let’s consider the followíng.fault condition: ‘ .

TP .1.2..’ 4. .

V ‘ 17.5 17.5 ‘ : 0 ‘

28 ELECTRONIC FAULT DIAGNOSIS

TABLE 3.1. Typical Faults on Power Supply Units

FAULT SYMPTOMS

Mains transformer, open

circuit primary pr secondary

D.C. outputzero. Second

aryaæ. zero, high resistance

primary or secondary.

, Mains transformer, shorted

Two possibilities: (a) malas

turns on primary or

fuses blown or (b) low d.c. ‘

secondary

output and transformer over

heating because of excessive

current being drawn.

Mains transformer, windings

,

Fuses blown. Low resistance

shorting to frame or screen

between windings and earth.

One diode in bridge open

Circuit behaves as a half-

circuit

wave rectifier. Lower d.c.

output with poor regulation. ‘

Increased ripple at 50Hz not

100 Hz as should be the case.

One diode in bridge short

Mains fuse blown, since

circuit

secondary winding will be

practically shorted every other

half cycle. A resistance check

across each ann of bridge is

required, measuring the

resistance of each diode in

the forward and reverse

direction.

Reservoir capacitor open

Low d.c. output with very

circuit

high valves of a.c. ripple on

output.

Reservoir caacitor short

Fuses blown. D.C. resistance

circuit

ofunstabilized line low in

both directions. .

Error amplifier in regulator

High d.c. output that is

open circuit

unregulated. No control

.

signal for the series element.

Series transistor open

Zero d.c. output. -The un-

circuit base emitter

stabilizedd.c. will be slightly

.

higher than normal since no

current is being drawn.

Reference zener short

Low d.c. output. Possibility

circuit

of series transistor

overheating.









y

/.

‘k -

Fig. 3.13 A 12V loo mA power supply

Transîormer; primary O—240 V a,c,

secondary 12V 250 mA

R,S. type 196—303 or similar

Rectifiers JN4001

is being drawn. Also TP2 is ät’the same vbige as

TN. This further ‘shows that no current at all is

flowing through R1 into the base of Tr2 The onlý

possible fault is that T2 has an open circuit base

emitter junction. Note that if R’1 were opeñ circuit

TP2 would bé at zero volts. ‘- ..

‘ Consider a fault contjon when ešt points re

at zero volts. Further inspection showš that the fuse

has.blown. ‘Resistance çhecks give the primary resis

tance as 432, the scondary as 4&2,’ but TPI to O V

is zero ohms, The fault in this case can oi1y bè C1

short circuit. ‚

QuestiónsThe following table lists a series of ‘

‘ fault conditions; in each case state which component

or components could cause thé fault and give a

‘ SUpporting reason. ‘

1 2 3 4 Other symptoms

Fault 16 15 14’S 14.5 ‘ C’

B 11 6 4.8 5 (. -‘ L. “‘-

11667558

—- —--

29 POWER SÜPPLY CIRCUITS

Switch

500 mA

240 Vrm,s

50Hz

0V’

. :

C —‘ ‘ ( f

Increased ripple

Poor regulation

TR1

‘.Prre1atjon . ‘

Hasil (Bahasa Indonesia) 1:

27 POWERSUPPLYCRCUITS.3.6 kesalahan menemukan teknik dan khas kesalahanKondisiKetika unit daya rusak akan dikembalikan untuk diperbaiki,kesalahan harus terisolasi untuk beberapa bagian tertentu dariunit. Kesalahan dapat terletak pada transformator,rectifier, Bagian filter, atau thegu1ator, danpengukuran dengan voltmeter akan diperlukan untukmencari kesalahan.Namun mungkin terbaik untuk memulai diagnosis denganmemeriksa beberapa agak jelas tetapi sering diabaikan.Pertama mengukur tegangan output DC. Jika ini adalahnol, cek berikutnya harus pada daya masukan.Pasokan listrik mencapai primer transformator?Jika tidak, ada kemungkinan plug rusak(relatif sederhana untuk perbaikan), membuka sirkuit listrikkabel, atau sekering ditiup. Jika sekring aku curiga selalutes kesinambungan dengan ohmmeter, tidak pernah mengandalkanHanya visual inspeksi hal ini juga perlu dicatat bahwakabel hidup dan netral mungkin memiliki sekering dalamsirkuit, jadi pastikan mandi dicentang.sekering lentur ditiup itu telah melakukannya karena beberapakondisi kesalahan dan kesalahan harus dibersihkan sebelumlagi sekering dilengkapi. Cek perlawanan ' (dengan ' listrikUnplugged!) harus digunakan untuk locatštich kesalahan. Penggunaanohmmeter untuk mengukur resistansi transmantan primer, sekunder, Penyearah, dan begitupada. Perlawanan berkelok-kelok tergantung, tentu saja, diukuran transformator. Perlawanan utama,untuk ukuran menengah transformator, harus rendah,biasanya sekitar 50 12. Sekunder, biasanya pasokan.ing tegangan yang lebih rendah, mungkin memiliki ketahanan hanyabeberapa ohms. Mendeteksi korsleting ternyata pada dapat berkelok-kelokoleh karena itu akan. cukup ' difficu1 mana pun mungkin comPare resistensi yang terukur dengan data yang tersediapada jenis transformator yang sedang digunakan. Penggunaan lain-. FUL check adalah untuk menjalankan transformator beban dan pengujianoverheating.Ketika menggunakan ohmmeter mengambil peduli untuk menggunakan.polaritas yang benar untuk perlawanan memeriksa mana dioda,elektrolitik kapasitor dan transistor presént. Ituterlalu mudah untuk mendapatkan menyesatkan hasil, misalnya,di gambar 3.123 jika meter digunakan untuk mengukur. perlawanan ' unstabffized baris, prod positif(terhubung di dalam meter ke piring positif 'baterai) harus ditempatkan pada garis yang positif dan: theñegative prod ke bumi. Jika meteran terbalikakan ada jalan rendah reištanee melaluiPenyearah dan jalan kebocoran melalui ' ' kapasitor.Kembali ke unit daya rusak, misalnya bagaimanapernah sekring utuh, dan bahwa listrik' mencapai dasar. Berikutnya, langkah adalah untuk mengukursecondaryaæ. tegangan, unstabijzedd.c. teganganmaka tegangan DC di regulator dan seterusnya, sampaikesalahan ini terletak. . ,Tabel 3.1 daftar beberapa kesalahan khas bersama-sama dengan 'gejala yang terkait, kesalahan adalah hanya sebuah contohorang-orang yang mungkin terjadi. Menemukan com rusakponent from a given set of symptoms will come withpractice and the following exercises are designed forthat purpose.To regulatorMultmeter setto ohms rongeRedFig. 3.12 Using an ohmmeter to measure the resistanceacross the unstabjljzed line. . /‘3.7 Exercise: Power Unit with a Simple LinòarRegulator (Fig. 3.13) .This unit incorporates most of the features discussed..earlier•and.is designed to give an outpûtof.12 Vat1QO mA. The output resistance.is less than 0.5 c2,the load tegulation:better than OS%, and the ripple.. less than 5 mV peak töpeakon full load. .., The unstabilizedd.c. is obtained fröm abridgerectifier circuit and a ryircaacitor of 3300 zF.The transformer has a secondary voltage of 12 Vr.m.s. so the unstabffized.yoìtage across C1 will beapproximately l2J2 V, i.e. about 16 V.The reference voltage is provided by a 5.6 V zenerdiode; a 400 mW device such as a BZYS8 C5V6 is .ideal. Tr1 is the d.c..error amplifier which comparesa portion of the d.c. output voltage, the voltage .across R4, with the reference. Any difference betweenthe two voltages is amplified by.Tr1 and the amplifledsignal is fed to the base .of Tr2. Consider for examplethe case when the d.c. output falls when more load ‘.current is taken; the base voltage of Tr1 decreasesand Tr1 conducts less current. Therefore Tr1 collectorvoltage rises, and this rise in voltage is coupledthrough Tr2, which acts as an eWitter follower tocõunteract the origijial fall in outpút. Thus the .circuit operates to málntain the output a:nearly .constant as possible. “ . - , .Since the output load current is’ ónly 100 mA,then Tr2 (BFY5I)need not be mounted on a heatsink. The circuit is no.t provided with”a current limit,so shouid.an excessive current be drawn, say byshorting he output, then .Tt2 wciùldundoubedlybum out A current limit can be added if required asshown in Fig 3 8, but for the purposes of the exercisewe shall assume no current limitThe normal d e voltages measured with a standardmultirange meter àrè’ as foil àws: . . ‘ . .Test point 1 2 3 4Volfige ‘.16’ ‘ 13 , “ 5’.’ 12.2First let’s consider the followíng.fault condition: ‘ .TP .1.2..’ 4. .V ‘ 17.5 17.5 ‘ : 0 ‘28 ELECTRONIC FAULT DIAGNOSISTABLE 3.1. Typical Faults on Power Supply UnitsFAULT SYMPTOMSMains transformer, opencircuit primary pr secondaryD.C. outputzero. Secondaryaæ. zero, high resistanceprimary or secondary., Mains transformer, shortedTwo possibilities: (a) malasturns on primary orfuses blown or (b) low d.c. ‘secondaryoutput and transformer overheating because of excessivecurrent being drawn.Mains transformer, windings,Fuses blown. Low resistanceshorting to frame or screenbetween windings and earth.One diode in bridge openCircuit behaves as a half-circuitwave rectifier. Lower d.c.output with poor regulation. ‘Increased ripple at 50Hz not100 Hz as should be the case.One diode in bridge shortMains fuse blown, sincecircuitsecondary winding will bepractically shorted every otherhalf cycle. A resistance checkacross each ann of bridge isrequired, measuring theresistance of each diode inthe forward and reversedirection.Reservoir capacitor openLow d.c. output with verycircuithigh valves of a.c. ripple onoutput.Reservoir caacitor shortFuses blown. D.C. resistancecircuitofunstabilized line low inboth directions. .Error amplifier in regulatorHigh d.c. output that isopen circuitunregulated. No control.signal for the series element.Series transistor openZero d.c. output. -The un-circuit base emitterstabilizedd.c. will be slightly.higher than normal since nocurrent is being drawn.Reference zener shortLow d.c. output. Possibilitycircuitof series transistoroverheating.y/.‘k -Fig. 3.13 A 12V loo mA power supplyTransîormer; primary O—240 V a,c,secondary 12V 250 mAR,S. type 196—303 or similarRectifiers JN4001is being drawn. Also TP2 is ät’the same vbige asTN. This further ‘shows that no current at all isflowing through R1 into the base of Tr2 The onlýpossible fault is that T2 has an open circuit baseemitter junction. Note that if R’1 were opeñ circuitTP2 would bé at zero volts. ‘- ..‘ Consider a fault contjon when ešt points reat zero volts. Further inspection showš that the fusehas.blown. ‘Resistance çhecks give the primary resis
tance as 432, the scondary as 4&2,’ but TPI to O V

is zero ohms, The fault in this case can oi1y bè C1

short circuit. ‚

QuestiónsThe following table lists a series of ‘

‘ fault conditions; in each case state which component

or components could cause thé fault and give a

‘ SUpporting reason. ‘

1 2 3 4 Other symptoms

Fault 16 15 14’S 14.5 ‘ C’

B 11 6 4.8 5 (. -‘ L. “‘-

11667558

—- —--

29 POWER SÜPPLY CIRCUITS

Switch

500 mA

240 Vrm,s

50Hz

0V’

. :

C —‘ ‘ ( f

Increased ripple

Poor regulation

TR1

‘.Prre1atjon . ‘