Level 6 — Electronic Skills — Jul/Aug 2025

Level 6 • July/August 2025 • Written assessment

Work through each question before checking the answer. These revision answers are prepared for learners; they are not an official marking scheme.

Question 1

Explain valence and conduction bands using silicon covalent bonding.

Answer

Silicon atoms share four valence electrons in covalent bonds. In a large crystal, interacting atomic energy levels split into closely spaced allowed levels, forming bands. The valence band contains bonding-electron states; the conduction band contains higher-energy states in which electrons can move through the crystal. A forbidden energy gap separates the bands.

Question 2

Explain silicon’s behaviour at absolute zero and room temperature.

Answer

At 0 K, ideal intrinsic silicon has a full valence band and an empty conduction band, so there are no thermally generated mobile carriers. At room temperature, some electrons gain sufficient thermal energy to cross the band gap. Each promoted electron leaves a hole in the valence band; both the electrons and holes contribute to conduction. The relatively small semiconductor band gap permits this limited carrier generation.

Question 3

Find the uncompressed memory size of a 1920 × 1080 image at 24 bits per pixel.

Answer

Step-by-step calculation

  1. Pixels = 1920 × 1080 = 2,073,600.
  2. Bits = 2,073,600 × 24 = 49,766,400.
  3. Bytes = 49,766,400 / 8 = 6,220,800.
  4. Using decimal megabytes: 6,220,800 / 1,000,000 = 6.2208 MB. Using 1,048,576 bytes per MiB gives 5.9326 MiB. This excludes file headers and row padding.

Question 4

Three resistors 27 Ω, 33 Ω and 56 Ω are in series across 240 V. Find the voltage across 33 Ω.

Answer

Step-by-step calculation

  1. Total series resistance = 27 + 33 + 56 = 116 Ω.
  2. Current = V/R = 240/116 = 2.06897 A.
  3. Voltage across 33 Ω = IR = 2.06897 × 33 = 68.28 V.

Question 5

Explain how the two intrinsic semiconductor charge carriers are formed.

Answer

  • An electron becomes a mobile conduction electron after gaining enough energy to leave a covalent bond and enter the conduction band.
  • The missing bonding electron leaves a hole. Nearby valence electrons can fill the hole, making the hole appear to move as a positive charge carrier. Intrinsic generation creates electron–hole pairs.

Question 6

Name two dopants for n-type silicon.

Answer

  • Phosphorus.
  • Arsenic. Both are pentavalent donor dopants; antimony is another valid example.

Question 7

Explain resistance versus conductor cross-sectional area.

Answer

R = ρL/A for a uniform conductor. With material, length and temperature held constant, resistance is inversely proportional to area: doubling A halves R.

Question 8

Give two cache-memory advantages.

Answer

  • Lower access latency for frequently used instructions/data than main RAM.
  • Exploits locality to reduce repeated main-memory accesses and processor waiting time.

Question 9

Convert 1354B₁₆ to binary and 126.02₁₀ to hexadecimal.

Answer

Step-by-step calculation

  1. The hexadecimal digit B is valid and equals decimal 11.
  2. Map each hex digit to four bits: 1→0001, 3→0011, 5→0101, 4→0100, B→1011.
  3. 1354B₁₆ = 0001 0011 0101 0100 1011₂ (leading zeros may be omitted).
  4. Integer part: 126 ÷ 16 = 7 remainder 14 (E), so 126 = 7E₁₆.
  5. Fraction: 0.02×16 = 0.32 (digit 0); 0.32×16 = 5.12 (5); 0.12×16 = 1.92 (1); 0.92×16 = 14.72 (E); 0.72×16 = 11.52 (B); 0.52×16 = 8.32 (8).
  6. The fraction then repeats from 0.32: 126.02₁₀ = 7E.051EB851EB8…₁₆, a non-terminating hexadecimal fraction.

Question 10

Given A = 18₈ and B = 10₂ as printed, encode their sum in Excess-3.

Answer

The printed A is invalid: octal digits can only be 0–7, so 18₈ has no defined value. B = 10₂ = 2. A correction is needed before an exact sum can be calculated.

Step-by-step calculation

  1. After a valid correction, convert both operands to decimal and add.
  2. Encode each decimal digit separately by adding 3 and writing its four-bit binary value.
  3. Illustration only: if the intended A were 18₁₀, the sum would be 20; 2+3=5→0101 and 0+3=3→0011, giving 0101 0011. This is conditional, not a claimed correction to the paper.

Question 11

An 8 GB system reserves 1.5 GB for the OS. Each application instance needs 750 MB. Use 1 GB = 1024 MB to find the maximum number.

Answer

Step-by-step calculation

  1. Available RAM = (8 − 1.5) × 1024 = 6656 MB.
  2. 6656 / 750 = 8.8747.
  3. Round down to whole instances: 8 instances. They use 6000 MB; 9 would use 6750 MB, exceeding 6656 MB.

Question 12

For the shown 230 V circuit find supply current. Then analyse a 5 V silicon diode circuit with 1 kΩ and 0.7 V forward drop, evaluate the binary expressions, and explain three ROM types.

Answer

82 Ω and 47 Ω in series with the parallel pair 27 Ω and 56 Ω across 230 V.
82 Ω and 47 Ω in series with the parallel pair 27 Ω and 56 Ω across 230 V.
Step-by-step calculation

  1. Parallel resistance = (27×56)/(27+56) = 1512/83 = 18.2169 Ω.
  2. Total = 82 + 47 + 18.2169 = 147.2169 Ω.
  3. Supply current = 230/147.2169 = 1.562 A.
  4. Forward diode current = (5 − 0.7)/1000 = 0.0043 A = 4.3 mA.
  5. With the diode reversed at 5 V, current is approximately zero in the ideal model; a real diode has small leakage, assuming reverse breakdown is not reached.
  6. 001110₂ × 011100₂ = 14×28 = 392 = 110001000₂.
  7. 0011011.011₂ + 001100.111₂ = 27.375 + 12.875 = 40.25 = 101000.010₂.
  • PROM: programmed once to hold permanent instructions/data.
  • EPROM: can be erased with ultraviolet light and reprogrammed.
  • EEPROM/flash: non-volatile memory that can be electrically erased and rewritten; flash normally erases blocks.

Question 13

Discuss silver, silicon and glass conductivity; describe NPN/PNP transistors; find power at 440 V and 4 A; add decimal 6 and 5 in Excess-3.

Answer

Material Conductivity
Silver Excellent metal conductor with many mobile electrons
Silicon Semiconductor; conductivity depends strongly on doping and temperature
Glass Electrical insulator under ordinary conditions; very few mobile carriers
NPN emitter arrow points out; PNP emitter arrow points in. Base, collector and emitter are labelled.
NPN emitter arrow points out; PNP emitter arrow points in. Base, collector and emitter are labelled.

An NPN transistor has n-type emitter and collector separated by a thin p-type base; in forward-active operation, conventional collector current enters the collector and leaves through the emitter. A PNP reverses the conductivity types and current/voltage polarities; the emitter arrow points toward the base. For either type, IE = IC + IB.

Step-by-step calculation

  1. Power = VI = 440 × 4 = 1760 W.
  2. Excess-3: 6→1001 and 5→1000. Add: 1001 + 1000 = 1 0001.
  3. There is a decimal carry; correct the remaining nibble by adding 0011: 0001 + 0011 = 0100 (Excess-3 for decimal 1).
  4. The tens digit is also decimal 1, encoded 0100. Result for 11 = 0100 0100.

Question 14

Interpret the PNP observations, give four ROM firmware-storage reasons, and calculate current, 10-minute energy and 45-minute cost for a 1500 W, 220 V blower at KSh 29/kWh.

Answer

Observation Interpretation
No base current and no collector current Cutoff: the transistor is off (apart from leakage)
Maximum circuit-limited collector current Saturation: the transistor acts as an on switch; both junctions are forward biased
  • Non-volatile: keeps firmware without power.
  • Provides stable startup instructions before general applications load.
  • Protects against accidental changes; update protection depends on the ROM technology.
  • Provides reliable repeated reading with little routine write activity.

The 1500 W value is legible in the original copy; it was obscured in the edited scan.

Step-by-step calculation

  1. Current = P/V = 1500/220 = 6.82 A.
  2. Power = 1.5 kW; 10 minutes = 10/60 hour. Energy = 1.5×10/60 = 0.25 kWh.
  3. For 45 minutes: energy = 1.5×45/60 = 1.125 kWh.
  4. Cost = 1.125×29 = KSh 32.625 ≈ KSh 32.63.

Question 15

Give five product-improvement measures; calculate capacitor C for A=0.5 m², d=2 mm, ε₀=8.85×10⁻¹² F/m; give three removable-HDD disadvantages; and add 454+234 in BCD.

Answer

  • Use integrated components to reduce size and assembly cost.
  • Design efficient power electronics and low-power operating modes.
  • Use lightweight durable materials for portability.
  • Improve useful features and ergonomics based on customer feedback.
  • Offer reliable quality, competitive pricing and repair/support options.
Step-by-step calculation

  1. Convert spacing: d = 2 mm = 0.002 m.
  2. For a vacuum parallel-plate capacitor, C = ε₀A/d.
  3. C = (8.85×10⁻¹²×0.5)/0.002 = 2.2125×10⁻⁹ F = 2.2125 nF.
  • Moving parts are vulnerable to shock and mechanical failure.
  • Random access is slower than typical solid-state storage.
  • A removable drive can be lost, stolen or disconnected; protect data with encryption and backups.
Step-by-step calculation

  1. Write each digit separately: 454 = 0100 0101 0100; 234 = 0010 0011 0100.
  2. Add units: 0100 + 0100 = 1000 (8), a valid BCD digit.
  3. Add tens: 0101 + 0011 = 1000 (8), a valid BCD digit.
  4. Add hundreds: 0100 + 0010 = 0110 (6), a valid BCD digit.
  5. Result = 0110 1000 1000 (BCD) = 688₁₀. No digit needs the +0110 BCD correction in this addition.
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