GCE A level 1145/01 ELECTRONICS ET5. P.M. THURSDAY, 31 May hours. Centre Number. Candidate Number. Surname. Other Names

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Surname Other Names Centre Number 0 Candidate Number GCE A level 1145/01 ELECTRONICS ET5 P.M. THURSDAY, 31 May 2012 1 1 2 hours For s use Question Maximum Mark Mark Awarded 1. 6 2. 9 3. 8 4. 6 1145 010001 ADDITIONAL MATERIALS In addition to this examination paper, you will need a calculator. INSTRUCTIONS TO CANDIDATES Use black ink or black ball-point pen. Write your name, centre number and candidate number in the spaces at the top of this page. Answer all questions. Write your answers in the spaces provided in this booklet. 5. 8 6. 4 7. 8 8. 8 9. 7 10. 6 Total 70 INFORMATION FOR CANDIDATES The total number of marks available for this paper is 70. The number of marks is given in brackets at the end of each question or part-question. You are reminded of the necessity for good English and orderly presentation in your answers. You are reminded to show all working. Credit is given for correct working even when the final answer given is incorrect. CJ*(S12-1145-01)

2 Preferred Values for resistors INFORMATION FOR THE USE OF CANDIDATES The figures shown below and their decade multiples and sub-multiples are the E24 series of preferred values. 10, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 43, 47, 51, 56, 62, 68, 75, 82, 91. Standard Multipliers Prefix Multiplier T 10 12 G 10 9 M 10 6 k 10 3 Prefix Multiplier m 10 3 μ 10 6 n 10 9 p 10 12 Alternating Voltages V o = V rms 2 Silicon Diode V F 0 7V R F R IN RF Operational amplifier G = Inverting amplifier G = 1 + R 1 Non-inverting amplifier V OUT = V DIFF V V OUT = R F R R V L V Z 1 + Emitter follower V OUT = V IN 0 7V 1 1 R F 1 V2 V3 + + R R Difference amplifier Summing amplifier Stabilised power supply 1 Filters f b = Break frequency for high pass 2 RC and low pass filters 1 X C = 2 fc Thyristor phase control φ = tan 1 R tan φ R = X C i/p voltage range Signal conversion resolution = ADC X C R R F 1 Capacitive reactance V Power amplifier P MAX = where V S is the rail-to-rail R S 2 8 L voltage 2 2 n 3

3 PIC Information The PIC programs include equate statements that define the following labels: Label Description PORTA input / output port A PORTB input / output port B TRISA the control register for port A TRISB the control register for port B STATUS the status register INTCON the interrupt control register W the working register (= h 0 ) F the file register (= h 1 ) RP0 the register page selection bit 0 Z the zero flag status bit GIE the global interrupt controller bit INTE the external interrupt enable bit Pin out for 16F84 PIC IC: List of commands: Mnemonic bcf bsf btfss call clrf goto movf movlw movwf retfie Operands f, b f, b f, b k f k f, d k f RA2 RA3 RA4 MCLR VSS RB0/INT RB1 RB2 RB3 Comparison of TASM and MPASM languages: 16F84 RA1 RA0 CLK IN CLK OUT VDD RB7 RB6 RB5 RB4 Description Clear bit b of file f Set bit b of file f Test bit b of file f, skip next instruction if bit is set Call subroutine k Clear file f Branch to label k Move file f (to itself if d = 1, or to working register if d = 0) Move literal k to working register Move working register to file f Return from interrupt service routine and set global interrupt enable bit GIE 1145 010003 Number system notation Version Opcode Notation Structure of the INTCON register Decimal Hex Binary TASM 153 $2B %10010110.equ.org.end label: MPASM d'153' h'2b' or 0x2B b'10010110' equ org end label Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0 GIE EEIE TOIE INTE RBIE TOIF INTF RBIF Structure of the STATUS register Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0 IRP RP1 RP0 TO PD Z DC C Turn over.

4 1. (a) Each stage in a synchronous counter is clocked at the same time. What advantage does this give synchronous counters over ripple counters? [1] (b) Synchronous counters are prone to a problem caused when unused states are also stuck states. (i) Explain what is meant by: unused states; [1] stuck states. [1] (ii) When does the problem caused by stuck states occur? [1]

5 (c) The behaviour of a synchronous counter is defined by its state diagram. Use the table below to draw the state diagram for this 3-bit synchronous counter. [2] Current Output Next Output State C B A D C D B D A 0 0 0 0 0 0 1 1 0 0 1 0 1 1 2 0 1 1 1 1 0 3 1 1 0 1 0 0 4 1 0 0 0 0 0 5 0 1 0 0 0 0 6 1 0 1 1 1 1 7 1 1 1 1 0 0 1145 010005 Turn over.

6 BLANK PAGE

7 2. A sequence generator is controlled by the following Boolean expressions. D C = C.A D B = C + A D A = A + C.B (a) Complete the circuit diagram for this sequence generator. (Credit will be given for using the minimum number of gates.) [5] C B A D C Q D B Q D A Q Q Q Q 1145 010007 Clock input (b) Complete the table to show the sequence of outputs generated by these Boolean expressions: [4] Current Output Next Output State C B A D C D B D A 0 0 0 0 1 2 3 4 5 6 7 Turn over.

8 3. (a) The following code is written to the data direction registers of a PIC microcontroller: bsf movlw movwf movlw movwf bcf STATUS,RP0 b'00111' TRISA b'00000000' TRISB STATUS,RP0 Describe the effect of this code on PORT A of the microcontroller. [1] (b) How does the microcontroller locate the Interrupt Service Routine? [1] (c) The PIC microcontroller has LEDs attached to all bits of PORT B. A three second delay subroutine called threesec is available. A Reset switch is connected to PORT A bit 0. Here is the Interrupt Service Routine. The numbers in the first column are line numbers in the program listing. 101 inter movwf Wstore 102 loop movlw b'11001100' 103 movwf PORTB 104 call threesec 105 clrf PORTB 106 call threesec 107 btfss PORTA,0 108 goto loop 109 movf Wstore,W 110 retfie (i) What is the purpose of the instructions in lines 101 and 109 in this program? [1]

9 (ii) Describe exactly what is observed at the output when lines 102 to 106 of the program are executed. [3] (iii) State the purpose and effect of the instructions in lines 107 and 108 in this program. [2] 1145 010009 Turn over.

10 4. Here is the circuit diagram for a Digital-to-Analogue Converter (DAC). This system uses a 10 V signal to represent logic 1 and a 0 V signal to represent logic 0. The output of the op-amp saturates at +10 V and 10 V. A B C 100 kω 50 kω 25 kω 10 kω V 1 0V (a) Calculate voltage V 1 when the digital number 001 is applied to the inputs. [1] V 1 =... (b) What is the value of V 1 when the input is the binary number 111? [1] V 1 =...

11 (c) Use the axes provided to draw a graph to show how voltage V 1 changes as the binary input increases, in steps, from 000 to 100. [2] 10 8 6 4 2 Voltage V 1 / V 0 000 001 010 011 100 Binary input 2 4 6 8 1145 010011 10 (d) Modify the circuit diagram by adding a second amplifier, with a gain of 1, to invert the signal V 1. Label any resistors used with suitable values. [2] Turn over.

12 5. A high quality power supply incorporates line and load regulation. (a) Explain what is meant by line regulation and load regulation. [3] (b) The diagram shows the circuit for a power supply. h FE = 50 I OUT AC mains supply + V S V R V OUT LOAD V Z Calculate the output voltage V OUT when V S = 14 V V Z = 9.2 V [1]

13 (c) The performance can be improved by adding an op-amp voltage amplifier to the circuit. Show how this is done by modifying the following circuit diagram. Your modification should make it possible to vary the output voltage, V OUT. No component values are needed. [4] AC mains supply + V OUT Turn over.

14 6. A machine tool control system uses an encoded disc to monitor how far a shaft has rotated. The disc is encoded using Gray code, and is read by three reflective optoswitches X, Y and Z. The arrangement is shown in the diagram. Z YX optoswitches (a) What is the advantage of using Gray code instead of pure binary code in this application? [1] (b) A logic system is needed to convert the Gray code output from the optoswitches into binary numbers. Z C Gray code inputs Y Logic System B Binary number outputs X A

15 The conversion table is shown below: Gray code Binary number Z Y X C B A 0 0 0 0 0 0 0 0 1 0 0 1 0 1 1 0 1 0 0 1 0 0 1 1 1 1 0 1 0 0 1 1 1 1 0 1 1 0 1 1 1 0 1 0 0 1 1 1 Obtain Boolean expressions for the binary number outputs, C and B, in terms of the Gray code inputs Z, Y and X. Output A has been done for you. Credit will be given for simplification. [3]............... C =... B =... A = Z.(Y + X) + Z.(Y +X) Turn over.

16 7. The following system is used to detect small temperature differences inside an enclosure. The circuit contains two thermistors, T 1 and T 2, a 2 kω precision resistor and a variable resistor. The circuit diagram is given below. 12 V XT 1 T 2 Y 2 kω 10 kω V OUT 0 V Initially, the resistances of the two thermistors are exactly 1.5 kω. (a) What is the value of V OUT when the variable resistor has a value of 2.00 kω? [1] V OUT =... (b) Calculate the output voltage V OUT when: thermistor T 1 has a resistance of 1.43 kω thermistor T 2 has a resistance of 1.20 kω the variable resistor has a resistance of 2.00 kω [2] V OUT =...

17 (c) Modify the circuit diagram by adding a difference amplifier connected to amplify V OUT. [2] 12 V XT 1 T 2 Y 2 kω 10 kω V OUT 0 V (d) Calculate suitable resistor values so that the amplifier has a voltage gain of 40. Label all resistors used in the amplifier with their ideal values. [3] Turn over.

18 8. A thyristor is used to control the brightness of a lamp, using phase control. The circuit diagram is shown below: V S X (a) (i) Identify component X. [1]... (ii) What is its function in this circuit? [1] (b) The graphs show two signals obtained from parts of the circuit: Voltage V S Voltage V S P Q R Time Time Signal Y Signal Z (i) Signal Y appears across which component? [1]...

19 (ii) State what is happening to the lamp: between P and Q [1]... between Q and R [1]... (iii) Signal Z appears across which component? [1]... (iv) Name the effect shown in signal Z and state how it arises. [2] Turn over.

20 9. (a) The graph shows the frequency response of an active filter. Voltage gain 100 10 1 1 10 100 1000 Frequency / Hz (i) What kind of filter is this? [1] (ii) What is the break frequency of this filter? [1] (iii) Complete the circuit diagram for the type of filter which produces the frequency response shown in the graph. Component values are not needed. [3] Input Output 0 V (iv) Why is this filter called an active filter? [1]

21 (b) The circuit for a different filter includes the RC combination shown below. 8.2 kω 20 nf Calculate the break frequency of this filter. [1] Break frequency =... Hz Turn over.

22 10. An audio system uses a power amplifier to drive the loudspeaker output stage. Two possible choices for a power amplifier are the emitter follower and the push-pull amplifier. (a) An emitter follower circuit is shown below. 12 V 100 kω V IN 0 V 100 kω V OUT The signal shown in the graph is applied to the input. Draw the output signal on the same axes. [2] Voltage / V 10 8 6 4 2 Input signal 0 Time 2 4 6 8 10

23 (b) Complete the circuit diagram for a push-pull power amplifier. [3] 12 V 0 V V IN 0 V 12 V (c) Give one advantage of a push-pull follower over an emitter follower power amplifier. [1] THERE ARE NO MORE QUESTIONS IN THE EXAMINATION

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