3.7 Exercise: Power Unit with a Simple LinòarRegulator (Fig. 3.13) .Th terjemahan - 3.7 Exercise: Power Unit with a Simple LinòarRegulator (Fig. 3.13) .Th Bahasa Indonesia Bagaimana mengatakan

3.7 Exercise: Power Unit with a Sim

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

,

0/5000
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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,
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3.7 Latihan: Power Unit dengan Linòar Sederhana (. Gambar 3.13). Regulator Unit ini menggabungkan sebagian besar fitur yang dibahas. .earlier • and.is dirancang untuk memberikan outpûtof.12 Vat 1QO mA. Output resistance.is kurang dari 0,5 c2, yang tegulation beban:. Lebih baik dari OS%, dan riak. Kurang dari 5 mV puncak töpeakon beban penuh. .., Unstabilizedd.c The. diperoleh dari meringkas rangkaian penyearah dan ryircaacitor dari 3300 zF. transformator memiliki tegangan sekunder 12 V r.ms sehingga unstabffized.yoìtage di C1 akan sekitar l2J2 V, yaitu sekitar 16 V. Tegangan referensi disediakan oleh 5,6 V zener diode; perangkat 400 mW seperti BZYS8 C5V6 adalah. ideal. TR1 adalah penguat dc.error yang membandingkan sebagian dari tegangan output dc, tegangan. Di R4, dengan referensi. Perbedaan antara dua tegangan diperkuat by.Tr1 dan amplifled sinyal dimasukkan ke basis .dari Tr2. Pertimbangkan misalnya kasus ketika output dc jatuh saat beban lebih '. Saat ini diambil; tegangan basis dari TR1 menurun dan TR1 melakukan kurang saat ini. Oleh karena itu TR1 kolektor tegangan naik, dan kenaikan ini tegangan digabungkan melalui Tr2, yang bertindak sebagai pengikut eWitter untuk menangkal jatuh origijial output. . Jadi sirkuit beroperasi untuk málntain output sebuah:. Hampir konstan mungkin. ". -., Sejak arus beban output 'hanya 100 mA, maka Tr2 (BFY5I) tidak perlu dipasang pada heat sink. Rangkaian ini no.t disediakan dengan "batas saat ini, arus yang berlebihan sehingga shouid.an ditarik, mengatakan oleh korslet keluaran dia, maka .Tt2 wciùldundoubedly bum keluar Batas saat ini dapat ditambahkan jika diperlukan sebagai ditunjukkan pada Gambar 3 8, tapi untuk keperluan latihan kami tidak bertanggung batas saat Tegangan de biasa diukur dengan standar multi-range meteran yang 'sebagai foil AWS:. . '. . Uji titik 1 2 3 4 Volfige '0,16' '13, "5'. ' 12.2 Pertama mari kita mempertimbangkan kondisi followíng.fault: '. 4. TP .1.2 ..'. V '17,5 17,5': 0 '28 FAULT ELECTRONIC DIAGNOSIS TABLE 3.1. Kesalahan khas pada Power Supply Unit FAULT GEJALA transformator Mains, terbuka sirkuit utama pr sekunder DC outputzero. Kedua aryaæ. nol, resistensi yang tinggi. primer atau sekunder, Induk transformator, korsleting Dua kemungkinan: (a) malas menyala primer atau sekering ditiup atau (b) dc rendah 'sekunder output dan transformator lebih pemanasan karena berlebihan saat ditarik. transformator Mains, gulungan .























































































































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