2012 Technological Studies. Higher. Finalised Marking Instructions

Transcription

1 0 Technological Studies Higher Finalised Marking Instructions Scottish Qualifications Authority 0 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial basis. If it is to be used for any other purposes written permission must be obtained from SQA s NQ Delivery: Exam Operations. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the necessary copyright clearance. SQA s NQ Delivery: Exam Operations may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 Section A Question Mark Allocation Marks. (a) A B C Z marks for all z output marks 7 or 6 z correct mark 5 or 4 z correct 0 marks 3 or less correct 3 (b) (c) Z A.B.C A.B.C A.B. C Alternatively A B C Z (A B). C mark for each correct combination 3 mark for logic operators 4 (if operators swapped, deduct mark) (C not C, deduct mark) (missing brackets, deduct mark) Z AND gate equivalents and connections 3 NOT gate equivalents and connections 3 OR gate equivalent and connections Cancellation of redundant gates 9 (6) See Supplementary Sheet for part (c) from simplified Boolean (page 3) Page

3 Supplementary Sheet Question Mark Allocation Marks. (b) Z = (A+B) C (simplified form) (c) A Z B for NAND, for NAND inverters for correct connections ( mark only for OR function) for NAND for NAND inverter for correct connections ( mark only for AND function) C for NAND inverter, for connections ( mark only for NOT function) Page 3

4 Stress Question Mark Allocation Marks. (a) A = 00 mm answer F A = = 0 N/mm 0 N/mm 30 N/mm 40 N/mm 50 N/mm Δl ε = l any of the above (units not necessary) = = σ E = ε any of the above (must be two matching substitutions) E = 00 kn/mm answer and unit 5 (b) Soft Brass (c) C A elastic range B yield point B C ultimate load D D plastic range A 4 (d) A Brittle trace Ductile trace Correct labels Correct axis Strain B 4 (4) Page 4

5 3. (a) Close: TIME = 0 for b0 = to 0 (or COUNTER) Including next b0 below for b= to 0 high 4 pause 0 both for mark low 4 pause TIME next b TIME = TIME + if pin = then finish next b0 finish: return including label close above (b) (i) Pulse Width Modulation (PWM) (ii) Valve closes at a reducing speed either until closed or a fixed time has elapsed. (c) Mark = 0 ms Max Space 0 ( 0) 50 ms Mark: Space ratio = :5 (6) 4. (a) (i) Darlington Pair/Driver (ii) Protection diode to prevent current generated at switch-off damaging transistor. (b) V be 0.7 V by inspection V V calculation V 0.9 b = = R 3600 all substitutions b 0.5 ma answer I I I b substitutions 0.0 A answer P Pump resistancer P I substitutions 0 answer Ic 3 h FE = = substitutions I 0 0 b = 06.5 answer Overall current gain = h FE h FE = substitutions = 850 answer See Supplementary Sheet, page 6 (5) Page 5

6 Supplementary Sheet Question Mark Allocation Marks 4. Quantity Working Vin 3 V Vbe 0 7 V by inspection Ib V 3600 =.3-4 = 0 9 V Calculation Ib V R All substitutions Ib = 0 5 ma Answer Ib Ib = Substitutions = 0 ma (0 0 A) Answer I p Pump Resistance 3 A P Rp I Substitutions h FE 80 = 0 Ω Answer h FE h FE I I c b Substitutions = 06 5 (no unit) Answer Overall Current Overall gain: h FE h FE Gain = Substitutions = 850 (no unit) Answer () Page 6

7 5. (a) To provide an appropriate signal within the correct voltage range. (b) Data is held whether powered or not (ie non volatile) Data may be erased (electronically). (c) How rapidly data is changing; memory available; consequence of missing an event (any acceptable) (d) tempmonitor select page 0 select flux temp sensor adcread eewrite delay 3.8s select page select solder temp sensor adcread eewrite for both return 0 If flowchart box is incorrect, do not award a mark (to a maximum of 4) (5) Page 7

8 6. (a) Summing (b) (i) R substitutions R = 7880 Ω = 7.9 kω answer (ii) substitutions R 0.03 R g g R g = 80. Ω answer (iii) R 5 0k answer 00 R 5.5 k 40 answer (c) V V out OR V out substitutions = 0.0 V answer (0.03 ( 40)) (0.0 ( 50)) (0.0 ( 00)) substitutions V out..05 simplification 4.5V answer 7 (4) Page 8

9 7. (a) (i) NODE Y XY H xy V xy ZY V xy = 90 H xy = ZY ZY XY XY = 90 sin N 90 substitution answer H xy YZ H XY 345.7cos57 substitution H XZ 88 88N answer 88 XZ cos7 substitution 608N answer 6 (ii) NODE Z 608sin7 V = 608sin7 V = 578 N answer NODE X V Applied force 790 N Components F n = 64 N; F v = 773 N H F 773 V F V F H F = 64 (578 90) 773 = 95 F = = 90 N º 90 (iii) tanθ = F θ 64 Page 9 substitution answer (4)

10 8. (a) LIGHT SENSOR for set light level for error detector for error amplifier Driver Spotlight Light Sensor for Light Sensor connections 7 (b) V ref 4.8 substitution 3.53V answer (c) At 400 Lux R LDR = 30 Ω (accept ) from Data Book 0 Difference amplifier gain gain calculation 0 V A (V V ) So 0 V ref V LDR in 8 Vref Vin 0.667V error calculation substitution.86v answer R 9 4 substitution 97.0 answer 7 (6) Page 0

11 9. (a) About R : ΣM = 0 37.R ( ) (3.3 3) substitutions R = 5 07 kn answer (b) NODE A F AB 5 07 kn F AC 5 07 = ABsin39º F AB = 8.06 kn STRUT AC = 8.06cos39 F AC = 6.6 kn TIE substitutions force including unit nature substitutions force, unit nature (c) NODE B 39º º F BE F BC = 3 kn ΣF V = 0 (stated or implicit) F BD FAB sin 39 FBC FBD sin sin 39 3 F AB F BD F BC sin (8 06 kn) (3 kn) F BD 5.53kNTIE force, unit 4 nature Δl ε = l 5 = 3700 substitutions 6 80 answer E = N/mm value from Data Book σ = E = substitutions =.77 N/mm answer A F 8000 = 77 substitution = 66.5 mm answer D d Area 4 4 d 4A D d 75 substitutions d = 69 5 answer substitutions Min wall thickness =.75 mm answer Page

12 (d) display arrow: for b0 = to 50 pins = % pause 5 pins = % pause 5 pins = % pause 5 pins = % 0000 pause 5 next b0 return given (3 pause 5) 6 (e) V R =.8 ( 0 ) = 9.8 V answer I R = 5 0 = 00 ma (units not required) 9.8 R 0. R 98 answer 3 (f) V 0k I b V answer (unit not required) V 5.3 R 00 3 = 0 53 ma answer (unit not required) 3 Ic 000 h FE 6 I b 5300 substitutions = 87 answer (must be no unit) 4 (40) Page

13 0. (a) (holding) torque accurate (angular) positioning; easily controlled by microcontroller (any acceptable answers) (b) (i) locate: if pin = 0 then locate inc label loop: if pin 0 = 0 then loop inc label gosub pulsedata b5 = DATA pause 500 gosub pulsedata if DATA = b5 then onward (OR b5 = DATA) goto error onward: gosub pulsedata b6 = DATA pause 500 Repeated block of code gosub pulsedata if DATA = b6 then onagain (OR b6 = DATA) goto error onagain: high 4 inc label gosub x-move high 6 gosub y-move ending: return inc label error: high inc label goto ending 8 (ii) xmove: for count = to b5 (OR DATA) correct variable high 5 pause 0 low 5 high & low pause 0 both pauses next count counting loop return 5 (c) The variable-resistor-voltage-divider sets light level required to switch comparator output. LDR produces increasing voltage signal as light level decreases. When picker not breaking light beam comparator output is 0 V. When light beam broken comparator output goes high. Voltage to microcontroller conditioned (by resistors) 4 Page 3

14 (d) 85% of V = 0 V saturation value R 5 = 5 expression and substitution R = 9 kω answer 3 (e) (i) R 4 = substitutions 6 5 R = 4 8 kω answer R 3 = substitutions R = 3 6 kω answer 4 (ii) Dark signal changes state when V in = V ref = 9 V R 9 = 5 3 substitutions 9 R LDR 5 5k 3 answer Light level for 5 kω = 5 lux (accept 5 6 lux) answer 4 (40) Page 4

15 . (a) Thermocouple generates a voltage signal dependent on temperature. Thermistor is resistive so requires power supply and voltage divider. (b) thermocouple : 400 ºC = 6 mv thermocouple : 00 ºC = 4 mv values R f T : R mV 00 i V todiff.amp.44 ( ) =.44 mv answer R f 70 T : 3.7 substitutions R 00 V i todiff.amp = 4.8 mv answer difference at inputs mV V out = 3.84 V Vout 3.84 AV substitutions -3 Vin = 87.5 answer no unit R f R f A V = R i = R A i 3 v 800 substitutions 87.5 R i =.85 kω answer 0 (c) (i) Threshold voltage = 3.8 V (3 80 V T 3 8 V) within range I sat =40.8 ma value (ii) offset from origin correct curve shape saturation line axis labels 4 (iii) g m ΔI = ΔV D GS 0. = substitutions 0.04 = 55 answer (no unit) Page 5

16 (d) +V +/-power rails npn correct pnp correct motor with connections 4 (e) -V ΣM pivot = 0 (+ve) ( 900cos0.6) (70sin 8.) (Fsin 770.3) 0 3 terms 3 (900cos 0.6) (70sin8.) F 0.3sin 77 manipulation F = 48 N = 4 kn answer 5 (f) ΣF V = 0( +ve) 900cos 0 70sin8 48sin 77 R V 0 3 correct terms 3 R V = R V 4855 N answer ΣF h = 0(+ve) 900sin 0 70cos 8 48cos 77 R h 0 3 correct terms 3 R h R h 544 N answer R p 4855 R P 544 R p 4885 N answer and unit 4855 tan ( ) (40) [END OF MARKING INSTRUCTIONS] Page 6