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NATIONAL
SENIOR CERTIFICATE
GRADE 12
ELECTRICAL TECHNOLOGY: ELECTRONICS
NOVEMBER 2019
MARKING GUIDELINES
MARKS: 200
These marking guidelines consist of 17 pages.
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Electrical Technology Grade 12 NSC November 2019 Memo (Eng)
Grade 12 Electrical Technology memorandum from the NSC November Exam 2019 (English), 17 pages. Read online or download the free PDF.
Electrical Technology · Grade 12 · NSC November Exam · 2019. Memorandum, 17 pages. Read online or download the PDF.
- Subject
- Electrical Technology
- Grade
- Grade 12
- Document type
- Memorandum
- Year
- 2019
- Exam period
- NSC November Exam
- Curriculum
- CAPS
- Pages
- 17
- File size
- 438.7 KB
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Electrical Technology: Electronics 2 DBE/November 2019
NSC – Marking Guidelines
INSTRUCTIONS TO THE MARKERS
1. All questions with multiple answers imply that any relevant, acceptable
answer should be considered.
2. Calculations:
2.1 All calculations must show the formulae.
2.2 Substitution of values must be done correctly.
2.3 All answers MUST contain the correct unit to be considered.
2.4 Alternative methods must be considered, provided that the correct
answer is obtained.
2.5 Where an incorrect answer could be carried over to the next step,
the first answer will be deemed incorrect. However, should the
incorrect answer be carried over correctly, the marker has to re-
calculate the values, using the incorrect answer from the first
calculation. If correctly used, the candidate should receive the full
marks for subsequent calculations.
2.6 Markers should consider when and where a candidate has rounded
off in a calculation, as well as the subsequent effect it has on the
final answer obtained. The calculation should therefore be awarded
marks on merit.
3. This marking guideline is only a guide with model answers. Alternative
interpretations must be considered and marked on merit. However, this
principle should be applied consistently throughout the marking session at
ALL marking centres.
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Electrical Technology: Electronics 3 DBE/November 2019
NSC – Marking Guidelines
QUESTION 1: OCCUPATIONAL HEALTH AND SAFETY
1.1 • Teamwork helps a team to meet its goals. ü
• Teamwork helps a team deliver quality work ü
• Teamwork can win the respect of your co-workers.
• Working together as a team can improve discipline in the workshop.
• Teamwork will lead to improved productivity. (2)
1.2 1.2.1 The purpose of the Act is to:
• Provide for health and safety of persons at work. ü
• Protect against hazards arising from the activities of other
people at work. ü
• Establish an advisory council for occupational health and safety
and related matters.
• Provide for health and safety for persons in connection with the
use of plant and machinery (2)
1.2.2 Workplace can be defined as any premises or place where a
person performs work ü during the time of his employment. ü (2)
1.3 1.3.1 Employee. ü Learner, Supervisor, Team leader, Safety
representative (1)
1.3.2 Health and safety representative. ü Inspector. (1)
1.4 • Remain calm ü
• Stop whatever you are doing and switch off machinery. ü
• Notify a responsible adult or teacher.
• If there is an emergency stop nearby activate it if necessary.
• Move in an orderly manner to the nearest assembly point.
NOTE: Procedure can be listed in any order. (2)
[10]
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Electrical Technology: Electronics 4 DBE/November 2019
NSC – Marking Guidelines
QUESTION 2: RLC CIRCUITS
2.1 2.1.1 Power factor is the ratio ü of the true power to the apparent
power. ü
NOTE: If the formula is written 1 mark must be awarded (2)
2.1.2 The quality factor of an inductor refers to the ratio of the inductor's
reactance ü to its internal resistance during resonance. ü
If the formula is written 1 mark must be awarded
The ratio of the inductive/capacitive voltage to the supply voltage
during resonance
The ratio of the inductive/capacitive reactance to the impedance
during resonance (2)
2.2 • Wattless voltage dividers ü
• Timing circuits ü
• Filter circuits
• Oscillating circuits
• Radio-tuning circuits (2)
2.3 2.3.1 VT = VR + (VC − VL )
2 2 ü
= 12 2 + (24 − 16 ) ü
2
= 14,42 V ü (3)
2.3.2 VL
XL = ü
I
16
= ü
3
= 5,33 Ω ü (3)
2.3.3 The circuit is capacitive ü because the voltage drop across the
capacitor is greater than the voltage across the inductor. ü
NOTE: If X L and X C are calculated and stated as a capacitive
circuit, 2 marks should be awarded.
If the learner wrote, the circuit is inductive, but give the same
reason, that motivation is deemed incorrect. (2)
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Electrical Technology: Electronics 5 DBE/November 2019
NSC – Marking Guidelines
2.3.4
VL = 16 V
ü
VR = 12 V
θü IT = 3 A
ü
(VC-VL) ü
=8V VT = 14,42 V
VC = 24 V
NOTE: No label, no marks (4)
2.3.5 When the impedance of the circuit is increased, the power factor
will decrease ü therefore increasing the value of the phase
angle. ü (2)
2.4 2.4.1 2 ü
IT = IR + (IL − IC ) 2
ü
= 6 2 + (4 − 3 )
2
ü
= 6,08 A (3)
2.4.2 I ü
Cos θ = R
IT
I ü
θ = Cos −1 R
IT
6
= Cos −1
6,08 ü
= 9,30 (3)
2.4.3 The phase angle is lagging ü because the inductive current is
larger than the capacitive current. ü
NOTE: If only the second part of the answer is given = 1 mark (2)
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Electrical Technology: Electronics 6 DBE/November 2019
NSC – Marking Guidelines
2.5 2.5.1 At resonance X L = X C = 50 Ω ü
1
C= ü
2 π f XC
1 ü
=
2 × π × 2000 × 50
= 1,59 μF ü (4)
NOTE: If the learner only substituted 50 Ω in the place of X C , the
correct substitution will be awarded 2 marks
If the learner calculated the inductance first, and used the resonant
frequency’s formula to work out the capacitance full marks must be
awarded.
XL
L=
2 × πf
50
L=
2 × π × (2000)
L = 3,98 × 10−3 H
1
fr =
2π LC
1
2000 =
2π 3,97 × 10 − 3 × C
( )
2000 2π 3,97 × 10 − 3 × C = 1
C = 1,6 μF
2.5.2 At resonance R = Z = 12 Ω ü
V
I= T ü
Z
120
= ü
12
= 10 A ü (4)
If R is used in the place of Z, full marks are awarded
2.5.3 If the resistance is doubled, the current will be halved. ü (1)
2.5.4 Z=R ü
VL = VC ü
ü
VR = VT
XL = X C
Cos θ = 1
θ = 0°
I = maximum
Z = minimum
(3)
[40]
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Electrical Technology: Electronics 7 DBE/November 2019
NSC – Marking Guidelines
QUESTION 3: SEMICONDUCTOR DEVICES
3.1 N-type ü semiconductor material. (1)
3.2 N-channel JFET.
ü
DRAIN (D)
ü
GATE (G)
ü
SOURCE (S)
NOTE: The symbol must be drawn correctly to receive marks for labelling. (3)
3.3 3.3.1 P-channel. ü (1)
3.3.2 The drain must be connected to positive. ü
The source must be connected to negative. ü
The gate voltage must be positive. ü (3)
3.4 3.4.1 Switch ü
Oscillators
Phase Control circuits
Timing circuits (1)
3.4.2 The UJT triggers when the voltage across the capacitor / emitter
rises ü to 0,7 V above the intrinsic standoff ratio (V X or V BB ). ü
The UJT triggers on when the voltage across the capacitor / emitter
rises to the UJT's peak voltage V P . (2)
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Electrical Technology: Electronics 8 DBE/November 2019
NSC – Marking Guidelines
3.4.3 VC
Input
ü
Output ü
ü
(3)
3.5 3.5.1 Pin 1 is identified by being the first pin to the left and below the
indentation. ü
OR
Being closest to the dot in the body of the IC. (1)
3.5.2 Input stageü or differential amplifier
Intermediate stage üor high gain differential amplifier
Output stage ü or common collector (3)
3.6 3.6.1 R
VOUT = VIN × − F ü
R IN
100 × 10 3 ü
= 2 × −
3
12 × 10 ü
= −16,67 V (3)
3.6.2 The amplifier is driven into saturation ü which causes the tops and
bottoms to be clipped. ü
If the learner identified that the input is fed into the inverting input,
there will be a 180 degree phase shift on the output = 2 marks (2)
3.6.3 +15 voltsü –15 voltsü (2)
3.7 3.7.1 Dual in-line ü package (1)
3.7.2 16 V ü (1)
3.7.3 5Vü (1)
3.8 When the trigger voltage rises above the threshold voltage, the output of a
555 timer will change state ü from high to low. ü
NOTE: If the answer is given as “Switch off” only = 1 mark (2)
[30]
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Electrical Technology: Electronics 9 DBE/November 2019
NSC – Marking Guidelines
QUESTION 4: SWITCHING CIRCUITS
4.1 4.1.1 A – Bistable multivibrator ü
B – Monostable multivibrator ü (2)
4.1.2 ü
+V
0
t
ü
-V
2 marks = 1 mark for each correct half cycle
If waveform is drawn correctly, but inverted, 2 marks will be
awarded
NOTE: Due to the +V instead of -V typo on the answer sheet, the
following responses are also accepted. (2)
+V +V +V
ü ü ü ü
ü ü
0 0
0
4.2 4.2.1 Resistors R 1 and R 2 are pull-up resistors. üü
R 1 and R 2 hold both trigger pin 2 and reset pin 4 high. (2)
4.2.2 When set switch S 1 is pressed, trigger pin 2 will be pulled low ü
and cause the IC output to 'flip' and rise high ü turning the LED
on. ü (3)
4.2.3 Threshold pin 6 is purposefully held at 0 V ü causing the IC not to
reset, ü keeping the output high ü when S 1 is pressed, (3)
4.3 4.3.1 The output signal represents an inverting Schmitt trigger ü
because the output signal is inverted ü with reference to the input
signal. ü
OR
The output signal is at -V CC when triggered by the upper trigger
voltage and swings to +V CC when triggered by the lower trigger
voltage. (3)
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Electrical Technology: Electronics 10 DBE/November 2019
NSC – Marking Guidelines
4.3.2
ü
ü + VCC
VIN ü
ü VOUT
ü RF ü
-VCC
R1 ü
0V
NOTE: Where a portion of the circuit is incorrect, he/she will lose
those marks and will be awarded marks for the correct section /
labels.
Any 7 correct labels (7)
4.4 4.4.1 Inverting ü comparator ü (2)
4.4.2 VIN
VREF
0V
t
+V ü ü
ü
0
t
ü
-V
1 mark = inversion
1 mark = correct wave shape
2 marks = 1 mark for each correct trigger point. (4)
4.4.3 Change the connection of resistor R 1 ü from the positive supply to
the negative supply. ü (2)
4.5 4.5.1 The gain of the amplifier is determined by the ratio ü of the
feedback resistance ü to the input resistance of each branch. ü
NOTE: Writing the formula only = 1 mark. (3)
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Electrical Technology: Electronics 11 DBE/November 2019
NSC – Marking Guidelines
4.5.2 R R R
VOUT = − V1 F + V2 F + V3 F ü
R1 R2 R3
100 × 10 3 100 × 10 3 100 × 10 3 ü
= − 0,3 × + 0,5 × + 0,4 ×
20 × 10 3 20 × 10 3 20 × 10 3
ü
= −6 V
NOTE: If the – sign is omitted in the answer, it is incorrect
because it is an inverted summing amplifier. (3)
4.5.3 The amplifier is not saturated because the output voltage is less ü
than the supply voltage. ü (2)
4.5.4 VOUT = VIN × Gain
VOUT
Gain A V = ü
VIN
VOUT
=
(V1 + V 2 + V3 )
−6
= ü
(0,3 + 0,5 + 0,4 )
=−5 ü (3)
4.5.5 With a variable resistor in the feedback loop, the gain ü of the
amplifier can be varied / controlled. ü (2)
4.5.6 If R 2 is changed to 10 kΩ, the gain for V 2 will increase ü causing
the total output voltage to increase. ü (2)
4.6 4.6.1 RC ü passive differentiator ü
NOTE: RC circuit = 1mark (2)
4.6.2 +V
Input
0
t
+V
ü
ü
Output 0
t
ü
-V
1 mark = positive pulse
1 mark = negative pulse
1 mark = correct orientation (3)
4.6.3 If the circuit time constant is increased, the capacitor will discharge
slower ü producing a sagging square wave. ü
NOTE: If the learner draws the waveform and label correctly, mark
on merit. (2)
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Electrical Technology: Electronics 12 DBE/November 2019
NSC – Marking Guidelines
4.7 4.7.1 C F provides a feedback connection from output to input. ü (1)
4.7.2 The inputs draw zero current. ü
The two inputs will have the same voltage at all times. ü
The capacitor will charge at a constant rate when a constant
current is supplied. ü
NOTE: If the learner gives a correct account of the operation of the
integrator as applied with the 741 op-amp, the answer must be
marked on merit. (3)
4.7.3 VIN
0
t
ü
t
0
ü ü
VOUT
ü
-VSAT
2 marks = correct labelling
2 marks = 1 mark for each correct half cycle
NOTE: The wave must be drawn correct before marks can be
awarded for labelling. (4)
[60]
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Electrical Technology: Electronics 13 DBE/November 2019
NSC – Marking Guidelines
QUESTION 5: AMPLIFIERS
5.1 5.1.1 Q Point
Class A Amplifier - The Q point is in the middle of the DC load
line.ü
Class B Amplifier - The Q point is on the cut-off of the DC load
line.ü
NOTE: If a learner draws the Q point instead of a narrative answer,
mark on merit (2)
5.1.2 Efficiency
Class A Amplifier - It has low efficiency therefore it has less output
power.ü
Class B Amplifier - It has higher efficiency therefore produces more
output power.ü
Class A Amplifier - 25%
Class B Amplifier - 50% (2)
5.2 5.2.1 To limit the current flowing through the collector. ü
Protects the circuit from damage.
To protect the transistor. (1)
5.2.2 When the transistor is OFF, no collector current flows through it,ü
therefore the maximum supply voltage will appear across the
transistor.ü
OR
From the formula VCC = VCE + ICRC
If I C = 0 then
(I C R C = 0) therefore
(V CC = V CE ) (2)
5.3 In Class C amplification, a transistor is biased below the cut off region ü
therefore the collector current flows for less than 180°. ü (2)
5.4 5.4.1 Class A amplifier.
+V Phase
ü
relationship
ü
0
t
ü
-V
1 mark = phase relationship
2 marks = 1 mark for each correct half cycle (3)
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Electrical Technology: Electronics 14 DBE/November 2019
NSC – Marking Guidelines
5.4.2 Class AB amplifier
+V
ü Phase
ü relationship
0
t
ü
-V
1 mark = phase relationship
2 marks = 1 mark for each correct half cycle (3)
5.5 • It can lead to the increases in currents leading to thermal runaway,
which could destroy the transistor. ü
• Crossover point distortion may occur.
• Incorrect placement of the Q-point
• Non-linear amplification (Distortion)
• Changes in the power supply voltage can cause the transistor's currents
to change.
NOTE: Bullet 5 is not an undesirable condition, but will be accepted
because it is stated in the prescribed textbook. (1)
5.6 5.6.1 The emitter is common ü to both the input and the output circuits.ü (2)
5.6.2 Input signal will be connected between point B and C. ü
The load can be connected between point E and F. ü (2)
5.6.3 Emitter capacitor (C E ) ü
Emitter resistor (R E ) ü (2)
5.6.4 ü Inversion
DC offset ü
OR
Q-point
ü
0
t
1 mark = inversion
2 marks = 1 mark for each amplified half cycle (3)
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Electrical Technology: Electronics 15 DBE/November 2019
NSC – Marking Guidelines
5.6.5 ü Inversion
+V
ü
0
t
ü
-V
1 mark = inversion
2 marks = 1 mark for each amplified half cycle (3)
5.7 • Impedance matching.ü
• Correct frequency response.ü
• DC isolation. (2)
5.8 5.8.1 POUT ü
A P = 10 log10
PIN
ü
1200
= 10 log10
3015
= − 4 dB ü (3)
5.8.2 VOUT ü
A V = 20 log10
VIN
219 ü
= 20 log10
230
= − 0,43 dB ü (3)
5.9 The gain of an amplifier is the capability of an amplifier to increase the
magnitude of the input signal ü without distortion.ü
Gain is the ratio of how much bigger or smaller the output signal will be
bigger in comparison to the input signal. (2)
5.10 A V = A V1 × A V2 × A V3 ü
= 10 × 15 × 20 ü
= 3000 ü (3)
5.11 5.11.1 Complimentary - symmetry ü Push-pull Amplifier. ü (2)
5.11.2 Cross-over distortion is eliminated by biasing the two push pullü
transistors each into class AB mode. ü
Cross-over distortion is eliminated by biasing the two push pull
transistors just above the cut-off region. (2)
5.11.3 The resistors are used for correct biasing ü (1)
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Electrical Technology: Electronics 16 DBE/November 2019
NSC – Marking Guidelines
5.11.4
+V
ü ü Phase
relationship
0
ü t
-V
1 mark = phase relationship
2 marks = 1 mark for each amplified half cycle
NOTE: If the learner indicated cross-over distortion on the
waveform, it will be accepted because of the two transistors used
in the given circuit
If the frequency is wrong, the waveform is wrong. (3)
5.12 5.12.1 RC oscillator - Does not use a resonating tank circuit but uses
selective resistors and capacitors in the feedback network. ü
LC oscillator - Uses tank circuit in the feedback network between
the Drain and the Gate of the transistor. ü (2)
5.12.2 LC oscillators - The oscillating frequency is determined by the
values of the capacitors and inductor in the tank circuit. ü
(Colpitts oscillator)
OR
The oscillating frequency is determined by the values of the
inductors and a capacitor in the tank circuit.(Hartley oscillator)
RC oscillator - the oscillating frequency is determined by the values
of the resistors and the capacitors in the feedback circuit. ü (2)
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Electrical Technology: Electronics 17 DBE/November 2019
NSC – Marking Guidelines
5.13
C ü
ü ü
L1 L2
(3)
5.14 Provide a phase shift of 180 degrees. ü
Amplifies the signal from the tank circuit. ü (2)
5.15 Both their output waveforms are purely sinusoidal. ü
They both use a transistor for amplification. ü
They both use resistors for biasing.
They both use positive feedback to operate.
Both circuit’s feedback circuits creates a 180° phase shift.
Both circuit’s transistor configuration creates 180° phase shift. (2)
[60]
TOTAL: 200
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