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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.
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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 +=
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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
-=
(8080175) -4- T660(E)(N23)T
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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
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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
Copyright reserved
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 =
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