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T660(E)(N23)T

NOVEMBER EXAMINATION

NATIONAL CERTIFICATE

INDUSTRIAL ELECTRONICS N5

(8080175)

23 November 2016 (X-Paper) 09:00–12:00

This question paper consists of 6 pages and 1 formula sheet of 6 pages.

(8080175) -2- T660(E)(N23)T

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DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA

NATIONAL CERTIFICATE INDUSTRIAL ELETRONICS N5

TIME: 3 HOURS MARKS: 100

INSTRUCTIONS AND INFORMATION 1. 2. 3. 4. 5. 6. 7.

Answer ALL the questions. Read ALL the questions carefully. Number the answers according to the numbering system used in this question paper. Keep questions and subsections of questions together. ALL sketches and diagrams must be large, clear and neat. Show ALL steps and calculations. Write neatly and legibly.

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QUESTION 1: ALTERNATING CURRENT THEORY 1.1 In a serial RC circuit it takes 1 ms to charge up to 63,2% of the

applied voltage VT. If R = 100 kW and VT = 6 V, calculate:

1.1.1 Value of C 1.1.1 VC after 1 ms

(2 × 2)

(4) 1.2 Three components are connected in parallel across a 10 V, 50 Hz supply.

The values of the components are as follows: R = 5 W; L = 20 mH and C = 100 µf Calculate:

1.2.1 Total impedance (4) 1.2.2 Total current flow (2) 1.2.3 Current through the three branches (6)

[16] QUESTION 2: POWER SUPPLIES 2.1 Calculate the value of the load resistor RL if the following values of a

capacitor filter are known: Vm = 15 V; f = 50 Hz after half-wave rectification; C = 470 µf and Idc = 10 mA

(6) 2.2 Calculate the percentage (%) voltage regulation of a power supply if the

output voltage is 12,4 V without a load and 11,9 V with a load.

(2) 2.3 Sketch a neat, labelled complete block diagram that shows the connection of

the 7812-regulator in a power supply.

(4) 2.4 What is meant by the term stabilising factor of a voltage? (1)

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2.5 Choose the correct word(s) from those given in brackets. Write only the

word(s) next to the question number (2.5.1–2.5.2) in the ANSWER BOOK.

2.5.1 Over-current protection protects the (control element/network)

against excess current.

2.5.2 The 7812-integrated voltage regulator is a (fixed/adjustable)

voltage regulator. (2 × 1)

(2) [15] QUESTION 3: TRANSISTOR AMPLIFIERS 3.1 The following values of a common-emitter are known:

RE = 400 W; CE = 10 µf and Ic(max) = 5 mA Calculate:

3.1.1 Supply voltage (4) 3.1.2 Minimum input frequency for the amplifier (3) 3.2 In a common-emitter with a voltage divider bias, the following information is

known: hie = 1,2 kW; hre = 2 × 10-4; hfe = 60; hoe = 20 µA/V; RB1 = 56 kW; RB2 = 5,6 kW and RC = 5 kW Determine the following with the aid of the precision method:

3.2.1 Ai (according to the current divider rule) (7) 3.2.2 Zi (3)

[17]

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QUESTION 4: OPERATIONAL AMPLIFIERS 4.1 Draw a simplified circuit diagram of an operational amplifier. (4) 4.2 Briefly describe the THREE most important characteristics of operational

amplifiers.

(3) 4.3 Calculate the value of the equal input resistors that a summing amplifier

should have for the following values: Three input voltages of 1 V, 1,5 V and 3,5 V respectively, an output voltage of 12 V and a feedback resistor of 20 kW.

(3) [10] QUESTION 5: INTEGRATED CIRCUITS 5.1 What should be done with the unused inputs of a CMOS-integrated circuit

when it is connected to a circuit?

(2) 5.2 Indicate whether the following statements are TRUE or FALSE. Choose the

answer and write only 'true' or 'false' next to the question number (5.2.1–5.2.3) in the ANSWER BOOK.

5.2.1 CMOS-integrated circuits have higher noise immunity. 5.2.2 CMOS-integrated circuits are susceptible to static charges because

of their low reactive input.

5.2.3 As soon as you work on the circuit of a CMOS-integrated circuit,

the power supply to the circuit must be switched off. (3 × 1)

(3) [5]

QUESTION 6: ELECTRONIC PHASE CONTROL Draw a neat, labelled block diagram of a typical closed-loop system. [5]

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QUESTION 7: TEST EQUIPMENT 7.1 Illustrate, using a 4-bit binary code, how a successive approximation

A/D converter would measure an initial unknown of 11 V.

(5) 7.2 Show the output waveform of the D/A converter in the circuit of a staircase

voltmeter.

(2) [7]

QUESTION 8: OSCILLATORS 8.1 8.1.1 Draw a neat, labelled circuit diagram of an RC-phase shift

oscillator.

(4) 8.1.2 Calculate the values of the resistors for the RC-phase shift

oscillator if the oscillating frequency is 50 kHz and the capacitor values are 10 nF.

(3)

8.2 8.2.1 Draw a neat, labelled circuit diagram of a Schmitt trigger. (4) 8.2.2 Name an application of the circuit in QUESTION 8.2.1. (1)

[12] QUESTION 9: TRANSDUCERS 9.1 Transducers can be classified into TWO groups.

Name the groups and briefly describe the difference between them.

(4) 9.2 Briefly describe the operating principle of an infrared detector. (4) 9.3 Draw the circuit diagram to show how a strain gauge must be connected in a

wheatstone bridge, as well as the differential amplifier that observes the displacement equivalent signal.

(5) [13]

TOTAL: 100

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INDUSTRIAL ELECTRONICS N5 FORMULA SHEET

T = RC T =

RVI =

RVRIIVP2

2 ===

...... 332211321 +++=+++= RIRIRIVVVVT

......3

3

2

2

1

1321 +++=+++=

RV

RV

RVIIIIT

RL

dtdvRCVR = ò= dtv

RCV iC

1

fLXL p2=fC

XC p21

=

LjXRZ += CjXRZ -=

)( CL XXjRZ -+=T

TT Z

VI =

RIV TR = )( LTL jXIV =

)( CTC jXIV -=LC

fr p21

=

12

1ff

fCL

RRX

RX

VV

VVQ rCL

T

C

T

L-

======

12 ffBW -=21

111ZZZT

+=

21

22ZZZZZT +

=21

21 ZV

ZVIIIT +=+=

L

LT jXR

jXRZ+

=)(

LT Xj

RZ-=

11

LRT jIII -=L

T XVj

RVI -=

C

CT jXR

jXRZ--

=)(

CT Xj

RZ+=

11

CRT jIII +=C

T XVj

RVI +=

(8080175) -2- T660(E)(N23)T

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a + jb = r /q

Q =

÷÷ø

öççè

æ--=

CLT XXj

RZ1111

)( CLRT IIjII --= ÷÷ø

öççè

æ--=

CLT X

VXVj

RVI

=+ -

abba 122 tan/

)sin(cos/ qqq jrr += qtan

2

2121

LR

LCf -=

p 1CRL

dZ =

mmrms VVV 707,021

== mmdc VVV 637,02==

p

P

S

S

P

S

PII

NN

VV

== mmdc VVV 318,01==

p

mVPIV = mVPIV 2=

mrmsr VR 385,0)( = mrmsr VV 305,0)( =

dc

rmsr

VV

r )(=32)(

)(ppr

rmsrV

V -=

2)( ppr

mdcV

VV --=

fCIVV dc

mdc 2-=

fCIVV dc

mdc 4-=

L

dcdcrmsr fCR

VfC

IV3232)( ==

L

dcdcrmsr fCR

VfC

IV3434)( ==

Ldc

dcfCRfCV

Ir321

32==

Ldc

dcfCRfCV

Ir341

34==

dcL

Ldc V

RRR

V .'+

= )(22)( .' rmsrC

Crmsr V

XR

XV+

=

'

')('

41

21

dc

rmsrCC

V

Vr

fCX

fCX ===

pp

)()( .' rmsrC

rmsr VRXV = ÷÷

ø

öççè

æ +=

L

LC RR

RRrXr.

'

(8080175) -3- T660(E)(N23)T

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1' RIVV dcdcdc -= dcL

Ldc V

RRRV .'

1+=

LCf

VV rmsr

rmsr 2)(

)( )2('

p=

LCf

VV rmr

rmsr 2)(

)( )4('

p=

FL

FLNLVVVVR -

= 100% ´-

=FL

FLNLVVVVR

122 cmcm VVVV +== mmccm VVVVV 23 31 +=+=

i

oVVSDD

= ziR VVV -=

(max)

(max)(min)

z

zis I

VVR

-=

z

zz V

PI =

SZi

ZL R

VVV

R .(max)

(min) -= bero VVV -=

c

ceccc I

VVR

-=

b

beccb I

VVR

-=

b

cII

=be

e fRC

p210

³

10cc

eVV =

c

e

e

ee I

VIV

R -= ~

c

ececcc I

VVVR

--=

b

ebeccb I

VVVR

--=

b

bccbb V

VVRR

)(21

-= eb RR b

101

2 =

beeb VVV +=

cerebiebe VhihV += ceoebfec Vhiii +=

Loe

fei Zh

hA

+=1 fei hA =

÷÷ø

öççè

æ+÷÷

ø

öççè

æ+÷÷

ø

öççè

æ+

=Lc

c

bT

b

Loe

fei RR

RZRTR

Zhh

A11

Lrefeoeieie

Lfev Zhhhhh

ZhA

)( -+

-=

ie

Lfev h

ZhA

-=

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For common base, substitute all the 'e' subscripts with 'b' in the h-parameters.

CMRR = CMRR (dB) = 20log

Loe

Lrefeie Zh

ZhhhZ

+-=11 iehZ =1

sie

refeoe Rh

hhh

Z

+-

=1

2oeh

Z 12 =

ivLi

p AARRA

A -==1

2

ie

Lfep h

RhA

2

=

220 ////// ZZZRRZ LLC ==

220 //// ZZZRZ LC == 11 // ZRZ b=

121 //// ZRRZ bbi =1

1 ZRIRI

bT

ibT=

=

÷÷ø

öççè

æ+

==ieb

ibfebfe hR

IRhIhI2

20

)(

1

0IIAi =

LcL RRZ //=1

1 ZRIRI

e

ie+

=

cm

dmAA

cm

dmAA

e

ee R

VI =

2e

cII =

C

RL I

VR L= L

ie

feLm R

hh

Rg .=

if VRR

V .1

0 ÷÷ø

öççè

æ-= i

f VRR

V .11

0 ÷÷ø

öççè

æ+=

÷÷ø

öççè

æ++-= 3

32

21

10 ... V

RR

VRR

VRR

V ffffRR

VRV

RV

V ÷÷ø

öççè

æ++-=

3

3

2

2

1

10

)( 3210 VVVV ++-= fRIIIV )( 3210 ++-=

ò-= )(1)(0 tVRC

tV i )()(1)(0 act

t bib tVtVRC

tV b

a+-= ò

(8080175) -5- T660(E)(N23)T

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ft 1=

1RR

A sv -=

s

sRRRR

R+

=1

12 CR

fs

c p21

=

dttdVRCtV i )()(0 -=

c

f

XR

A =

s

f

R

RA -= tv

dtdCRtV if wsin.)(0 -=

CRt f= )(0 ir VVAV -=

)( 120 VVRR

Vs

f -= 120 VVV -=

21210 2

1RRCC

fp

=RC

fp21

0 =

10 2

1CL

fTp

= MLLLT 221 ++=

TLCf

p21

0 =21

21CCCC

CT +=

20 2

1LC

fp

=62

1RC

fp

=

RCf 5,10 =

210

11ttt

f+

==

121 7,0 CRt = 212 7,0 CRt =

RCf

4,11

0 = )(2 ONbeeci VRIV +=

RCt 1,1=CRR

fBA )2(

443,10 +=

CRt Blow )(693,0= CRRt BAhigh )(693,0 +=

(8080175) -6- T660(E)(N23)T

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Resolution =

Resolution =

T = 273 + °C

highlowT ttt +=llRRK//

DD

=

ll /D=s4/

12d

Rp

r=

ES

=sturnsofamount

1

dropvoltagetotalturnsadjacentacrossdropvoltage

TBt AeR /=

TA VRR

RV .

21

2+

= Tt

tB V

RRR

V .3+

=

BAAB VVV -=i

v VVA 0=

kIHVHall =