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NATIONAL
SENIOR CERTIFICATE
GRADE 12
SEPTEMBER 2020
ELECTRICAL TECHNOLOGY: ELECTRONICS
MARKING GUIDELINE
MARKS: 200
This marking guideline consists of 13 pages.
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Electrical Technology Grade 12 Eastern Cape Mock 2020 Memo
Grade 12 Electrical Technology memorandum from the Eastern Cape Mock Exam 2020 (English), 13 pages. Read online or download the free PDF.
Electrical Technology · Grade 12 · Eastern Cape Mock Exam · 2020 · English. Memorandum, 13 pages. Read online or download the PDF.
- Subject
- Electrical Technology
- Grade
- Grade 12
- Language
- English
- Document type
- Memorandum
- Year
- 2020
- Exam period
- Eastern Cape Mock Exam
- Curriculum
- CAPS
- Pages
- 13
- File size
- 537.1 KB
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2 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2020)
INSTRUCTIONS TO 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 that candidates' answers may deviate slightly
from the marking a guideline depending on how and where in the
calculation rounding off was used.
3. These marking guidelines are only a guide with model answers.
4. 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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(EC/SEPTEMBER 2020) ELECTRICAL TECHNOLOGY: ELECTRONICS 3
QUESTION 1: OCCUPATIONAL HEALTH AND SAFETY
1.1 Accident means an incident arising out of and in the course of an
employee’s employment and resulting in a personal injury, illness or the
death of the employee. (2)
1.2 Take reasonable care for the health and safety of him/herself and of
other persons who may be affected by his or her act.
Cooperate with the employer or persons to enable that any duty given
by the requirements and procedures.
Carry out any lawful order given to him, and obey the health and safety
rules and procedures laid down by his employer or by anyone
authorised there by his employer, in the interest of health or safety.
If any situation which is unsafe or unhealthy comes to his/her attention,
as soon as practicable report this to his/her employer, or a healthy and
safety representative who should report it to the employer. (2)
1.3 Horseplay
Running in the workshop.
Throwing things.
Leaving bags, stools or material in a walkway.
Spilling a liquid or oil without cleaning.
Any relevant answer (2)
1.4 Third degree burn: where all layers of skin have been burned, causing
permanent skin damage affecting fat, muscle and even bone. (2)
1.5 To review the effectiveness of health and safety measures.
To identify hazards and potential major incidents at the workplace. (2)
[10]
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4 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2020)
QUESTION 2: RLC CIRCUIT
2.1 2.1.1 The current and the voltage are out of phase with each other and
the current is leading the voltage by 90°. (2)
2.1.2 The current and the voltage are out of phase with each other and
the current is lagging the voltage by 90°. (2)
2.1.3 The current and voltage are in phase and their phase angle is
zero. (2)
2.2 2.2.1 𝑋𝐿 = 2𝜋𝐹𝐿
= 2𝜋 × 60 × 44 × 10−3
= 16,587 Ω (3)
2.2.2 𝑋𝑐 = 1
2𝜋𝐹𝐶
1
𝑋𝑐 =
2𝜋 × 60 × 120
𝑋𝑐 = 22,105 Ω (3)
2.2.3 𝑍 = √𝑅2 + (𝑋𝐿 − 𝑋𝑐)2
= √252 + (22,105 − 16,58)2
𝑍 = 25,6 Ω (3)
2.2.4 𝐼𝑇 = 𝑉𝑇
𝑍𝑇
120
=
25,6
= 4,69 𝐴 (3)
2.3 The capacitive reactance will decrease because the capacitive reactance is
inversely proportional to frequency. (3)
2.4 2.4.1 𝐼𝑇 = √𝐼𝑅2 + (𝐼𝐿 − 𝐼𝐶)2
= √62 + (5 − 4)2
= 6,08 𝐴 (3)
2.4.2 𝐶𝑂𝑆𝜃 = 𝐼𝑅
𝐼𝑧
6
𝜃 = 𝐶𝑜𝑠 −1
6,08
= 9,305° (3)
2.4.3 𝑋𝐿 = 𝑉𝑇
𝐼𝐿
240
𝑋𝐿 =
5
𝑋𝐿 = 48 Ω (3)
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(EC/SEPTEMBER 2020) ELECTRICAL TECHNOLOGY: ELECTRONICS 5
2.4.4 𝑋𝑐 = 𝑉𝑇
𝐼𝐶
240
𝑋𝐶 =
4
= 60 Ω (3)
2.5 2.5.1 𝐹𝑅 = 1
2𝜋√𝐿𝐶
1
𝐹𝑅 =
2𝜋√0.2×0,05−6
𝐹𝑅 = 1 591,5 𝐻𝑧 (3)
2.5.2 𝑋𝐿 = 2𝜋𝐹𝐿
𝑋𝐿 = 2𝜋 × 1 591,5 × 0,2
𝑋𝐿 = 1 999,938 Ω
𝑋𝐿
𝑄=
𝑅
1 999,938
𝑄=
10
𝑄 = 200 (4)
2.5.3 𝐵𝑊 = 𝐹𝑟
𝑄
1 591,5
𝐵𝑊 =
200
= 7,958 Hz (3)
2.6 2.6.1 𝑄 = 𝐹𝑟
𝐵𝑊
95×106
𝑄=
200×103
Q = 475 Hz (3)
2.6.2 𝑋𝑐 = 1
2𝜋𝐹𝐶
1
𝑋𝑐 =
2𝜋×95×106 ×2,5×106
𝑋𝑐 = 22,105 Ω (2)
2.6.3 𝑋𝐿
𝑅=
𝑄
670
𝑅=
475
= 1,4466 Ω (2)
[50]
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6 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2020)
QUESTION 3: SEMICONDUCTOR DEVICES
3.1 The field effect transistor is a device that relies on the creation of an
electrical field within its body to control current flow. (2)
3.2 The main advantage of a FET over a BJT is that it has significantly lower
levels of power making it more efficient.
FET’s also have very high input resistances thus drawing very little
current from the previous stage. (4)
3.3
D
√
√
A
C √
(3)
3.4 UJT as switch
UJT as sawtooth (2)
3.5 Very low output resistance.
Very wide bandwidth.
Infinite gain
Extremely high input resistance (3)
3.6 Signal applied to the non-inverting terminal will appear at the output
terminal in the same direction as when it entered the op-amp.
A signal applied to the inverting terminal will appear at the output terminal in
an inverted direction. (2)
3.7 C
√
√
B
√
√
E
(4)
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(EC/SEPTEMBER 2020) ELECTRICAL TECHNOLOGY: ELECTRONICS 7
3.8 When negative feedback is used the overall gain of an op-amp circuit is
reduced and controlled so that the op amp is able to operate as a normal
linear amplifier. Negative feedback also provides control of the circuit input
and output impedance as well as its operating frequency range. (3)
3.9 𝑅𝐹
𝐴𝑣 = 1 +
𝑅𝑖𝑛
120 × 108
=1+
12 × 108
= 11 (3)
3.10 3.10.1 𝐹𝑅
𝑉𝑔𝑎𝑖𝑛 = 𝑉𝑖𝑛 × −
𝐹𝑖𝑛
200 × 1 000
= 0,02 × −
200
= −20 𝑉 (3)
3.10.2 𝑅𝑓
𝐴𝑣 = −
𝑅𝑖𝑛
200 000
𝐴𝑣 = −
2 000
𝐴𝑣 = −100 (3)
3.11 Comparator compares the voltage at its input terminal, depending on the
difference between them the out voltage goes either high or low. (2)
3.12 Astable mode
Monostable mode
Bistable mode (3)
3.13 Three series connected resistors of identical value (5 kΩ) use voltage-
division to provide two stepped down voltages of 2/3 and 1/3 of the
supply voltage to the rest of the internal circuit. (3)
[40]
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8 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2020)
QUESTION 4: SWITCHING CIRCUITS
4.1 An astable multivibrator generates a continuous series of pulses without
the need for a trigger pulse. (2)
4.2
Rf
+ 15 V
V OUT
R1
C1 - 15 V
R2
0V (5)
4.3 4.3.1 Resistors R1 and R2 are pull-up resistors.
Resistors R1 and R2 will hold both trigger pin 2 and reset pin 4
high. (2)
4.3.2 Trigger pin 2 will be pulled low and cause the IC output to
‘flip’ and rise high turning the LED on when set switch S1 is
pressed. (3)
4.3.3 Threshold pin 6 is purposely held at O V causing the IC not to
reset, keeping the output high when S1 is pressed. (3)
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(EC/SEPTEMBER 2020) ELECTRICAL TECHNOLOGY: ELECTRONICS 9
4.4 4.4.1
(4)
4.4.2 Trigger pulse B looks like a series of ‘on’ and ‘off’ pulses. This
is known as switch bounce. (2)
4.4.3 The duration of time of trigger pulse B will not influence the
condition of the circuit and for that reason the charge tempo of
the capacitor will not be influenced. (3)
4.5 4.5.1 The output signal represents an inverting Schmitt trigger
because the output signal is inverted with reference to the
input signal.
OR
The output signal is at –Vee when triggered by the upper
trigger voltage and swings to +Vee when triggered by the
lower trigger voltage. (3)
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10 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2020)
4.5.2
(7)
4.5.3 Thermocouple
Piezo-sensor
Microphone (2)
4.5.4
(4)
4.6 4.6.1 Inverting comparator (2)
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(EC/SEPTEMBER 2020) ELECTRICAL TECHNOLOGY: ELECTRONICS 11
4.6.2
(4)
4.6.3 Change the connection of resistor R1 from the positive supply
to the negative supply. (2)
4.7 4.7.1 The gain of the amplifier is determined by the ratio of the feed-
back resistance to the input resistance of each branch. (3)
4.7.2 RF RF RF
VOUT = -[V1 + V2 + V3 ]
R1 R2 R3
100 ×103 100 ×103 100 ×103
= -[0,3 × 3 + 0,5 × 3 + 0,4 × ]
20 ×10 20 ×10 20 × 103
= -6 V (3)
4.7.3 The amplifier is not saturated because the output voltage is
less than the supply voltage. (2)
4.7.4 VOUT = VIN ˣ Gain
VOUT
Gain Av =
VIN
VOUT
=
(V1 + V2 + V3 )
-6
=
(0,3 + 0,5 + 0,4)
= -5 (3)
4.7.6 If R2 is changed to 10 kΩ, the gain for V2 will increase causing the
total output voltage to increase. (1)
[60]
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12 ELECTRICAL TECHNOLOGY: ELECTRONICS (EC/SEPTEMBER 2020)
QUESTION 5: AMPLIFIERS
5.1 5.1.1 𝑉𝐶𝐸 = 𝑉𝑐𝑐
𝑉𝐶𝐸 = 12 𝑉 (2)
5.1.2 𝑉𝑐𝑐
𝐼𝑐 =
𝑅𝑐
12
𝐼𝑐 =
1 000
𝐼𝑐 = 12 𝑚𝐴 (3)
5.2 Each half circuit is able to amplify the input signal to double its previous
greatest size.
The circuit introduces little or no distortion
The circuit is almost as efficient as a pure Class B stage amplifier (2)
5.3 5.3.1
Vin
Vs
Vout
Av
β
(5)
5.3.2 Improving the amplifier stability.
Increase the amplifier bandwidth.
Enhancing the amplifier input and output impedance.
Reducing the noise produced within the amplifier. (3)
𝐴
5.3.3 𝐴′ =
1+𝛽𝐴
100
=
1+0,09×100
= 10 (3)
5.4 Excellent impedance matching can be achieved.
Total DC isolation between stages. (2)
5.5 RF amplifiers are used to amplify a single high frequency and to suppress
all others. (2)
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(EC/SEPTEMBER 2020) ELECTRICAL TECHNOLOGY: ELECTRONICS 13
5.6 It produces a much larger output signal than a single Class A biased
transistor amplifier.
It uses Class B biasing, so efficiencies of between 70% and 75% can be
reached. (2)
5.7 5.7.1 1
𝐹𝑅 = (2)
2𝜋√6 𝑅𝐶
5.7.2 +Vcc √
RB Rc √
Output
C1 √
√
C3
C2
R3 Re
R1 R2 √ Ce
√
0v
(6)
5.7.3 Cheap, economical and compact as it uses only resistors and
capacitors.
Offers a constant gain over a wide frequency band. (2)
5.7.4 Positive feedback occurs when a portion of the output signal is
fed back in phase with the input. (3)
5.7.5 The low frequency cut-off region
The mid-frequency region
The high frequency cut off region (3)
[40]
TOTAL: 200
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