Practical circuits — Paper 4 · A Level Physics 9702

4 An ideal transformer has 5000 turns on its primary coil. It is to be used to convert a mains supply of 230 V r.m.s. to an alternating voltage having a peak value of 9.0 V. (a) Calculate the number of turns on the secondary coil. number = ……………………………… [3] (b) The output from the transformer is to be full-wave rectified. Fig. 4.1 shows part of the rectifier circuit. A R B Fig. 4.1 On Fig. 4.1, draw (i) diode symbols to complete the diagram of the rectifier such that terminal A of the resistor R is positive with respect to terminal B, [2] (ii) the symbol for a capacitor connected to provide smoothing of the potential difference across the resistor R. [1] Examiner’s Use (c) Fig. 4.2 shows the variation with time t of the smoothed potential difference V across the resistor R. V 0 0 t1 t2 t3 t4 t Fig. 4.2 (i) State the interval of time during which the capacitor is being charged from the transformer. from time ………… to time …………… [1] (ii) The resistance of the resistor R is doubled. On Fig. 4.2, sketch the variation with time t of the potential difference V across the resistor. [2]

9 marks

Mark scheme: 4 (a) r.m.s. output = 9/√2 or peak input = 230√2 C1 NS/NP = VS/VP C1 NS = 138 → 140 turns A1 [3] (b) (i) four diodes correctly positioned regardless of output polarity M1 giving correct output polarity (all ‘point to left’) A1 [2] (ii) capacitor shown in parallel with R B1 [1] (c) (i) time t1 to time t2 B1 [1] (ii) sketch: same peak values M1 ripple reduced and reasonable shape A1 [2]

This question in 9702/41 May/June 2007

8 (a) State three characteristics of an ideal operational amplifier (op-amp). 1. … 2. … 3. … [3] (b) An amplifier circuit for a microphone is shown in Fig. 8.1. – + 120 kΩ R V OUT X Fig. 8.1 (i) Name the type of feedback used with this op-amp. … [1] (ii) The output potential difference VOUT is 5.8 V for a potential difference across the resistor R of 69 mV. Calculate 1. the gain of the amplifier circuit, gain = ……………………… [1] Examiner’s Use 2. the resistance of resistor X. resistance = ……………………… Ω [2] (iii) State one effect on the amplifier output of reducing the resistance of resistor X. … … [1]

8 marks

Mark scheme: 8 (a) e.g. infinite (voltage) gain infinite input impedance zero output impedance infinite bandwidth infinite slew rate (any three, 1 each) B3 [3] (b) (i) negative (feedback) B1 [1] (ii) 1 gain (= 5.8/0.069) = 84 B1 [1] (ii) 2 gain = 1 + 120/X C1 84 = 1 + 120/X X = 1.45 kΩ A1 [2] (iii) gain increases OR bandwidth reduced OR output increases B1 [1] GCE A/AS LEVEL – May/June 2007 9702 04

This question in 9702/41 May/June 2007

10 An analogue signal is sampled at a frequency of 5.0 kHz. Each sample is converted into a four-bit number and transmitted as a digital signal. Fig. 10.1 shows part of the digital signal. START 0010 0101 1010 1111 0100 0010 0101 1010 1111 0100 most significant bit Fig. 10.1 The digital signal is transmitted and is finally converted into an analogue signal. (a) On the axes of Fig. 10.2, sketch a graph to show the variation with time t of this final analogue signal. 18 16 signal 14 12 10 8 6 4 2 0 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 t /ms [4] Fig. 10.2 (b) Suggest two ways in which the reproduction of the original analogue signal could be improved. 1. … … 2. … … [2]

6 marks

Mark scheme: 10 (a) correct values of 2, 5, 10, 15 and 4 (–1 each error) B2 graph drawn as a series of steps M1 steps occurring at correct times A1 [4] (b) sample more frequently B1 greater number of bits B1 [2]

This question in 9702/41 May/June 2007

8 (a) Fig. 8.1 shows a circuit incorporating an ideal operational amplifier (op-amp). + 9V – + – 9V V1 VOUT V2 Fig. 8.1 The voltages applied to the inverting and the non-inverting inputs are V1 and V2 respectively. State the value of the output voltage VOUT when (i) V1 > V2, VOUT = … V (ii) V1 < V2. VOUT = … V [1] Examiner’s Use (b) The circuit of Fig. 8.2 is used to monitor the input voltage VIN. +V A +5.0V – + B +3.0V – + VIN red green Fig. 8.2 At point A, a potential of 5.0 V is maintained. At point B, a potential of 3.0 V is maintained. Complete Fig. 8.3 by indicating with a tick (✓) the light-emitting diodes (LEDs) that are conducting for the input voltages VIN shown. Also, mark with a cross ( ) those LEDs that are not conducting. VIN / V red LED green LED +2.0 +4.0 +6.0 [3] Fig. 8.3 Examiner’s Use (c) The input voltage VIN in (b) is provided by a sensor circuit. (i) Complete Fig. 8.4 to show a sensor circuit that will provide a voltage output that increases as the temperature of the sensor decreases. Show clearly the output connections from the circuit. [2] Fig. 8.4 (ii) Explain the operation of the sensor circuit. … … … … … … [3]

9 marks

Mark scheme: 8 (a) (i) - 9 V (ii) + 9 V (both (i) and (ii) correct for the mark) … B1 [1] (b) … B1 … B1 … B1 [3] (no e.c.f. from (a)) (c) (i) cct: thermistor and resistor in series ………………………………………………...M1 output connections across thermistor … A1 [2] (ii) as temperature decreases, thermistor resistance increases … B1 p.d. across thermistor = RT / (R + RT) × V … M1 as RT increases, output increases … A1 [3]

This question in 9702/41 Oct/Nov 2007

11 In a cellular phone network, a country is divided into a number of cells, each with its own base station. Fig. 11.1 shows a number of these base stations and their connection to a cellular exchange. cell cellular exchange base station Fig. 11.1 (a) Suggest and explain why the country is divided into a number of cells. … … … … [2] (b) Outline what happens at the base station and the cellular exchange when a mobile phone handset is switched on, before a call is made. … … … … … … … [4]

6 marks

Mark scheme: 11 (a) carrier frequencies can be re-used (simultaneously without interference) … B1 so that number of handsets possible is increased … B1 OR anything sensible e.g. UHF used (B1) so ‘line of sight’ (B1) [2] (b) handset sends out an (identifying) signal … M1 communicated by base stations to (computer at) exchange … A1 computer selects base station with strongest signal … B1 and allocates a (carrier) frequency … B1 [4]

This question in 9702/41 Oct/Nov 2007

9 A block diagram for an electronic sensor is shown in Fig. 9.1. output device Fig. 9.1 (a) Complete Fig. 9.1 by labelling the remaining boxes. [2] (b) A device is to be built that will emit a red light when its input is at +2 V. When the input is at –2 V, the light emitted is to be green. (i) On Fig. 9.2, draw a circuit diagram of the device. input either + 2V or – 2V [2] Fig. 9.2 (ii) Explain briefly the action of this device. … … [1]

5 marks

Mark scheme: 9 (a) blocks labelled sensing device / sensor / transducer B1 processor / processing unit / signal conditioning B1 [2] (b) (i) two LEDs with opposite polarities (ignore any series resistors) M1 correctly identified as red and green A1 [2] (ii) correct polarity for diode to conduct identified M1 hence red LED conducts when input (+)ve or vice versa A0 [1]

This question in 9702/41 May/June 2008

10 (a) The circuit for an amplifier incorporating an ideal operational amplifier (op-amp) is shown For in Fig. 10.1. Examiner’s Use R2 R1 +9 V – P + VIN –9 V VOUT Fig. 10.1 (i) State 1. the name of this type of amplifier circuit, … [1] 2. why the point P is referred to as a virtual earth. … … … … [3] (ii) Show that the gain G of this amplifier circuit is given by the expression R2 G = – ––– . R1 Explain your working. [4] (b) The circuit of Fig. 10.1 is modified by connecting a light-dependent resistor (LDR) as For shown in Fig. 10.2. Examiner’s Use R2 R1 +9 V – + VIN = +1.2 V –9 V VOUT V Fig. 10.2 The resistances R1 and R2 are 5.0 kΩ and 50 kΩ respectively. The input voltage VIN is +1.2 V. A high-resistance voltmeter measures the output VOUT. The circuit is used to monitor low light intensities. (i) Determine the voltmeter reading for light intensities such that the LDR has a resistance of 1. 100 kΩ, reading = … V [3] 2. 10 kΩ. reading = … V [2] (ii) The light incident on the LDR is provided by a single lamp. Use your answers in (i) For to describe and explain qualitatively the variation of the voltmeter reading as the Examiner’s lamp is moved away from the LDR. Use … … … … … [3]

16 marks

Mark scheme: 10 (a) (i) 1. inverting (amplifier) B1 [1] 2. gain of op-amp is very large / infinite B1 non-inverting input is at earth / 0 V B1 for amplifier not to saturate, P must be at about earth / 0 V B1 [3] (ii) input resistance is very large B1 (so) current in R1 = current in R2 B1 I = VIN / R1 B1 I = – VOUT / R2 (minus sign can be in either of the equations) B1 hence gain = VOUT / VIN = –R2 / R1 A0 [4] (b) (i) 1. feedback resistance = 33.3 kΩ C1 gain (= 33.3 / 5) = 6.66 C1 VOUT (= 6.66 × 1.2) = 8.0 V (+ or – acceptable, allow 1 s.f.) A1 [3] 2. feedback resistance = 8.33 kΩ C1 VOUT (= {6.66 × 1.2} / 5) = 2.0 V (+ or – acceptable, allow 1 s.f.) A1 [2] (ii) (Increase in lamp-LDR distance gives) decrease in intensity M1 Feedback / LDR resistance increases M1 voltmeter reading increases / becomes more negative A1 [3]

This question in 9702/41 Oct/Nov 2008

10 (a) By reference to an amplifier, explain what is meant by negative feedback. … … … [2] (b) An amplifier circuit incorporating an ideal operational amplifier (op-amp) is shown in Fig. 10.1. – + V IN 120kΩ V OUT R Fig. 10.1 The supply for the op-amp is ± 9.0 V. The amplifier circuit is to have a gain of 25. Calculate the resistance of resistor R. resistance = … Ω [2] (c) State the value of the output voltage VOUT of the amplifier in (b) for input voltages VIN of (i) – 0.08 V, VOUT = … V [1] (ii) +0.4 V. VOUT = … V [1]

6 marks

Mark scheme: 10 (a) (part of) the output is added to /returned to / mixed with the input B1 and is out of phase with the input / fed to inverting input B1 [2] (b) 25 = 1 + (120 / R) C1 R = 5 kΩ A1 [2] (c) (i) –2 V A1 [1] (ii) 9 V A1 [1]

This question in 9702/41 May/June 2009

9 An amplifier incorporating an operational amplifier (op-amp) has three inputs A, B and C, as shown in Fig. 9.1. 8.0 kΩ A 2.0 kΩ RF +9 V B 4.0 kΩ – C 8.0 kΩ + VA VB –9 V VOUT VC Fig. 9.1 Negative feedback is provided by the resistor RF of resistance 8.0 k . For each of the inputs A, B and C, the amplifier may be considered as a single input amplifier. That is, each input is independent of the other two. When the amplifier is not saturated, the output potential VOUT is given by the expression VOUT = –(4VA + GVB + VC), where VA, VB and VC are the input potentials of the inputs A, B and C respectively and G is a constant. (a) State two effects of negative feedback on an amplifier. 1. … … 2. … … [2] (b) In the expression for the output potential VOUT , the constant G is the gain associated For with input B. Show that the numerical value of G is 2. Examiner’s Use [1] (c) The input potentials VA, VB and VC are either zero or 1.0 V. The magnitudes of some output potentials for different combinations of VA, VB and VC are shown in Fig. 9.2. VA / V VB / V VC / V VOUT / V 0 0 1 1 0 1 0 …… 1 0 0 4 1 0 1 5 1 1 0 …… 1 1 1 …… Fig. 9.2 (i) Complete Fig. 9.2 for the three remaining values of VOUT . [1] (ii) Suggest a use for this circuit. … [1]

5 marks

Mark scheme: 9 (a) e.g. reduces gain increases bandwidth less distortion greater stability ………(1 each, max 2) … B2 [2] (b) gain = – RF / RI = – 8.0 / 4.0 … M1 numerical value is 2 … A0 [1] (c) (i) 2, 6 and 7 … A1 [1] (ii) e.g. digital-to-analogue converter (allow DAC) adding / mixing signals with ‘weighting’ … B1 [1] [Total: 5]

This question in 9702/42 Oct/Nov 2009

12 A block diagram representing part of a mobile phone network is shown in Fig. 12.1. For Examiner’s Use X X Public Switched X Y Telephone Network (PSTN) mobile phone X handset X Fig. 12.1 (a) State what is represented by (i) the blocks labelled X, … [1] (ii) the block labelled Y. … [1] (b) A user of a mobile phone is making a call. Explain the role of the components in the boxes labelled X and Y during the call. … … … … … … … … [5]

7 marks

Mark scheme: 12 (a) (i) base stations … B1 [1] (ii) cellular exchange … B1 [1] (b) base station / X sends / receives signal to / from handset … … B1 call relayed to cellular exchange / Y (and on to PSTN) … B1 computer at cellular exchange monitors signal from base stations … B1 selects base station with strongest signal … B1 allocates a (carrier) frequency / time slot for the call … B1 [5] [Total: 7]

This question in 9702/42 Oct/Nov 2009

6 A student is asked to design a circuit by which a direct voltage of peak value 9.0 V is obtained For from a 240 V alternating supply. Examiner’s The student uses a transformer that may be considered to be ideal and a bridge rectifier Use incorporating four ideal diodes. The partially completed circuit diagram is shown in Fig. 6.1. 240 V + load – Fig. 6.1 (a) On Fig. 6.1, draw symbols for the four diodes so as to produce the polarity across the load as shown on the diagram. [2] (b) Calculate the ratio number of turns on the secondary coil . number of turns on the primary coil ratio = … [3]

5 marks

Mark scheme: 6 (a) all four diodes correct to give output, regardless of polarity M1 connected for correct polarity A1 [2] (b) NS / NP = VS / VP C1 V0 = √2 × Vrms C1 ratio = 9.0 / (√2 × 240) = 1/38 or 1/37 or 0.027 A1 [3] GCE AS/A LEVEL – May/June 2010 9702 41

This question in 9702/41 May/June 2010

9 The circuit diagram of Fig. 9.1 is an amplifier circuit incorporating an operational amplifier (op-amp). 4.2 kΩ +9 V 1.0 kΩ – + + 1.5 V –9 V V – Fig. 9.1 (a) (i) On Fig. 9.1, mark, with the letter X, the virtual earth. [1] (ii) Explain what is meant by a virtual earth. … … … … [3] (b) In bright sunlight, the light-dependent resistor (LDR) has resistance 200 Ω. (i) Calculate, for the LDR in bright sunlight, the voltmeter reading. reading = … V [3] (ii) The sunlight incident on the LDR becomes less bright. For State and explain the effect on the voltmeter reading of this decrease in Examiner’s brightness. Use … … … … [3]

10 marks

Mark scheme: 9 (a) (i) point X shown correctly B1 [1] (ii) op-amp has very large / infinite gain M1 non-inverting input is at earth (potential) / earthed / at 0 V M1 if amplifier is not to saturate, inverting input must be (almost) at earth potential / 0 (V) same potential as inverting input A1 [3] (b) (i) total input resistance = 1.2 kΩ C1 (amplifier) gain (= –4.2 / 1.2) = –3.5 C1 (voltmeter) reading = –3.5 × –1.5 = 5.25 V A1 [3] (total disregard of signs or incorrect sign in answer, max 2 marks) (ii) (less bright so) resistance of LDR increases M1 (amplifier) gain decreases M1 (voltmeter) reading decreases A1 [3] GCE AS/A LEVEL – May/June 2010 9702 41

This question in 9702/41 May/June 2010

9 (a) Negative feedback may be used in amplifier circuits. State (i) what is meant by negative feedback, … … … [2] (ii) two effects of negative feedback on an amplifier incorporating an operational amplifier (op-amp). 1. … … 2. … … [2] (b) Fig. 9.1 is a circuit for an amplifier that is used with a microphone. P microphone 120 kΩ V OUT R Fig. 9.1 The output potential difference VOUT is 4.4 V when the potential at point P is 62 mV. Determine (i) the gain of the amplifier, gain = … [1] (ii) the resistance of the resistor R. For Examiner’s Use resistance = … Ω [2] (c) The maximum potential produced by the microphone at point P on Fig. 9.1 is 95 mV. The power supply for the operational amplifier may be either +/– 5 V or +/– 9 V. State which power supply should be used. Justify your answer quantitatively. … … … … [3]

10 marks

Mark scheme: 9 (a) (i) fraction of the output (signal) is added to the input (signal) M1 out of phase by 180° / π rad / to inverting input A1 [2] (ii) e.g. reduces gain increases bandwidth greater stability reduces distortion (any two, 1 mark each) B2 [2] (b) (i) gain = 4.4 / 0.062 = 71 A1 [1] (ii) 71 = 1 + 120/R C1 R = 1.7 × 103 Ω A1 [2] (c) for the amplifier not to saturate B1 maximum output is (71 × 95 × 10–3 =) approximately 6.7 V M1 supply should be +/– 9 V A1 [3]

This question in 9702/42 May/June 2010

9 (a) Negative feedback may be used in amplifier circuits. State (i) what is meant by negative feedback, … … … [2] (ii) two effects of negative feedback on an amplifier incorporating an operational amplifier (op-amp). 1. … … 2. … … [2] (b) Fig. 9.1 is a circuit for an amplifier that is used with a microphone. P microphone 120 kΩ V OUT R Fig. 9.1 The output potential difference VOUT is 4.4 V when the potential at point P is 62 mV. Determine (i) the gain of the amplifier, gain = … [1] (ii) the resistance of the resistor R. For Examiner’s Use resistance = … Ω [2] (c) The maximum potential produced by the microphone at point P on Fig. 9.1 is 95 mV. The power supply for the operational amplifier may be either +/– 5 V or +/– 9 V. State which power supply should be used. Justify your answer quantitatively. … … … … [3]

10 marks

Mark scheme: 9 (a) (i) fraction of the output (signal) is added to the input (signal) M1 out of phase by 180° / π rad / to inverting input A1 [2] (ii) e.g. reduces gain increases bandwidth greater stability reduces distortion (any two, 1 mark each) B2 [2] (b) (i) gain = 4.4 / 0.062 = 71 A1 [1] (ii) 71 = 1 + 120/R C1 R = 1.7 × 103 Ω A1 [2] (c) for the amplifier not to saturate B1 maximum output is (71 × 95 × 10–3 =) approximately 6.7 V M1 supply should be +/– 9 V A1 [3]

This question in 9702/43 May/June 2010

9 An amplifier circuit incorporating an operational amplifier (op-amp) is shown in Fig. 9.1. R2 +9 V – + –9 V VOUT VIN R1 Fig. 9.1 (a) State (i) the name of this type of amplifier circuit, … [1] (ii) the gain G in terms of resistances R1 and R2. … [1] (b) The value of R1 is 820 Ω. The resistor of resistance R2 is replaced with a light-dependent For resistor (LDR). Examiner’s The input potential difference VIN is 15 mV. Use Calculate the output potential difference VOUT for the LDR having a resistance of (i) 100 Ω (the LDR is in sunlight), VOUT = … V [2] (ii) 1.0 MΩ (the LDR is in darkness). VOUT = … V [1]

5 marks

Mark scheme: 9 (a) (i) non-inverting (amplifier) B1 [1] (ii) (G =) 1 + R2 / R1 B1 [1] (b) (i) gain = 1 + 100 / 820 C1 output = 17 mV A1 [2] (ii) 9 V A1 [1] (R2 / R1 scores 0 in (a)(ii) but possible 1 mark in each of (b)(i) and (b)(ii) (1 + R1 / R2) scores 0 in (a)(ii), no mark in (b)(i), possible 1 mark in (b)(ii) (1 – R2 / R1) or R1 / R2 scores 0 in (a)(ii), (b)(i) and (b)(ii))

This question in 9702/41 Oct/Nov 2010

9 An amplifier circuit incorporating an operational amplifier (op-amp) is shown in Fig. 9.1. R2 +9 V – + –9 V VOUT VIN R1 Fig. 9.1 (a) State (i) the name of this type of amplifier circuit, … [1] (ii) the gain G in terms of resistances R1 and R2. … [1] (b) The value of R1 is 820 Ω. The resistor of resistance R2 is replaced with a light-dependent For resistor (LDR). Examiner’s The input potential difference VIN is 15 mV. Use Calculate the output potential difference VOUT for the LDR having a resistance of (i) 100 Ω (the LDR is in sunlight), VOUT = … V [2] (ii) 1.0 MΩ (the LDR is in darkness). VOUT = … V [1]

5 marks

Mark scheme: 9 (a) (i) non-inverting (amplifier) B1 [1] (ii) (G =) 1 + R2 / R1 B1 [1] (b) (i) gain = 1 + 100 / 820 C1 output = 17 mV A1 [2] (ii) 9 V A1 [1] (R2 / R1 scores 0 in (a)(ii) but possible 1 mark in each of (b)(i) and (b)(ii) (1 + R1 / R2) scores 0 in (a)(ii), no mark in (b)(i), possible 1 mark in (b)(ii) (1 – R2 / R1) or R1 / R2 scores 0 in (a)(ii), (b)(i) and (b)(ii))

This question in 9702/42 Oct/Nov 2010

10 (a) State three properties of an ideal operational amplifier (op-amp). For Examiner’s 1. … Use 2. … 3. … [3] (b) A circuit incorporating an ideal op-amp is to be used to indicate whether a door is open or closed. Resistors, each of resistance R, are connected to the inputs of the op-amp, as shown in Fig. 10.1. +3 V S R R +9 V – + –9 V R R R Fig. 10.1 The switch S is attached to the door so that, when the door is open, the switch is open. The switch closes when the door is closed. (i) Explain why the polarity of the output of the op-amp changes when the switch For closes. Examiner’s Use … … … … [3] (ii) A red light-emitting diode (LED) is to be used to indicate when the door is open. A green LED is to indicate when the door is closed. On Fig. 10.1, 1. draw symbols for the LEDs to show how they are connected to the output of the op-amp, [1] 2. identify the green LED with the letter G. [1] Please turn over for Question 11.

8 marks

Mark scheme: 10 (a) e.g. infinite input impedance / resistance zero output impedance / resistance infinite gain infinite bandwidth infinite slew rate (any three, 1 each) B3 [3] (b) (i) with switch open, V – is less (positive) than V + M1 output is positive A1 with switch closed, V – is more (positive) than V + so output is negative A1 [3] (allow similar scheme if V – more positive than V + treated first) (ii) 1. diodes connected correctly between output and earth M1 2. green identified correctly A1 [2] (do not allow this mark if not argued in (i))

This question in 9702/43 Oct/Nov 2010

9 (a) An operational amplifier (op-amp) may be used as a comparator. State the function of a comparator. … … … [2] (b) The variation with temperature θ of the resistance R of a thermistor is shown in Fig. 9.1. 4.0 3.0 R / kΩ 2.0 1.0 0 0 5 10 15 20 25 30 θ / °C Fig. 9.1 The thermistor is connected into the circuit of Fig. 9.2. For Examiner’s +5V Use X 2.0kΩ +9V – + –9V 2.0kΩ V OUT Fig. 9.2 The op-amp may be considered to be ideal. (i) The temperature of the thermistor is 10 °C. Determine the resistance of the variable resistor X such that the output potential VOUT is zero. resistance = … Ω [2] (ii) The resistance of the resistor X is now held constant at the value calculated in (i). Describe the change in the output potential VOUT as the temperature of the thermistor is changed from 5 °C to 20 °C. … … … … … [4]

8 marks

Mark scheme: 9 (a) to compare two potentials / voltages M1 output depends upon which is greater A1 [2] (b) (i) resistance of thermistor = 2.5 kΩ C1 resistance of X = 2.5 kΩ A1 [2] (ii) at 5 ˚C / at < 10 ˚C, V – > V + M1 so VOUT is –9 V A1 at 20 ˚C / at > 10 ˚C, V – < V + and VOUT is +9 V B1 VOUT switches between negative and positive at 10 ˚C B1 [4] (allow similar scheme if 20 ˚C treated first)

This question in 9702/41 May/June 2011

9 (a) State two effects of negative feedback on the gain of an amplifier incorporating an operational amplifier (op-amp). 1. … … 2. … … [2] (b) An incomplete circuit diagram of a non-inverting amplifier using an ideal op-amp is shown in Fig. 9.1. +9 V – + 12 kΩ –9 V R Fig. 9.1 (i) Complete the circuit diagram of Fig. 9.1. Label the input and the output. [2] (ii) Calculate the resistance of resistor R so that the non-inverting amplifier has a voltage gain of 15. resistance = … Ω [2] (c) On Fig. 9.2, draw a graph to show the variation with input potential VIN of the output For potential VOUT . Examiner’s You should consider input potentials in the range 0 to +1.0 V. Use 16 VOUT / V 12 8 4 0 0 0.2 0.4 0.6 0.8 1.0 VIN / V Fig. 9.2 [2] (d) The output of the amplifier circuit of Fig. 9.1 may be connected to a relay. State and explain one purpose of a relay. … … … [2]

10 marks

Mark scheme: 9 (a) e.g. reduced gain increased stability greater bandwidth or less distortion (allow any two sensible suggestions, 1 each, max 2) B2 [2] (b) (i) V – connected to midpoint between resistors B1 VOUT clear and input to V+ clear B1 [2] (ii) gain = 1 + RF/R 15 = 1 + 12000/R C1 R = 860 Ω A1 [2] (c) graph: straight line from (0,0) to (0.6,9.0) B1 straight line from (0.6,9.0) to (1.0,9.0) B1 [2] (d) either relay can be used to switch a large current/voltage M1 output current of op-amp is a few mA/very small A1 [2] or relay can be used as a remote switch (M1) for inhospitable region/avoids using long heavy cables (A1)

This question in 9702/41 Oct/Nov 2011

9 (a) State two effects of negative feedback on the gain of an amplifier incorporating an operational amplifier (op-amp). 1. … … 2. … … [2] (b) An incomplete circuit diagram of a non-inverting amplifier using an ideal op-amp is shown in Fig. 9.1. +9 V – + 12 kΩ –9 V R Fig. 9.1 (i) Complete the circuit diagram of Fig. 9.1. Label the input and the output. [2] (ii) Calculate the resistance of resistor R so that the non-inverting amplifier has a voltage gain of 15. resistance = … Ω [2] (c) On Fig. 9.2, draw a graph to show the variation with input potential VIN of the output For potential VOUT . Examiner’s You should consider input potentials in the range 0 to +1.0 V. Use 16 VOUT / V 12 8 4 0 0 0.2 0.4 0.6 0.8 1.0 VIN / V Fig. 9.2 [2] (d) The output of the amplifier circuit of Fig. 9.1 may be connected to a relay. State and explain one purpose of a relay. … … … [2]

10 marks

Mark scheme: 9 (a) e.g. reduced gain increased stability greater bandwidth or less distortion (allow any two sensible suggestions, 1 each, max 2) B2 [2] (b) (i) V – connected to midpoint between resistors B1 VOUT clear and input to V+ clear B1 [2] (ii) gain = 1 + RF/R 15 = 1 + 12000/R C1 R = 860 Ω A1 [2] (c) graph: straight line from (0,0) to (0.6,9.0) B1 straight line from (0.6,9.0) to (1.0,9.0) B1 [2] (d) either relay can be used to switch a large current/voltage M1 output current of op-amp is a few mA/very small A1 [2] or relay can be used as a remote switch (M1) for inhospitable region/avoids using long heavy cables (A1)

This question in 9702/42 Oct/Nov 2011

5 The components for a bridge rectifier are shown in Fig. 5.1. For Examiner’s Use supply load Fig. 5.1 (a) Complete the circuit of Fig. 5.1 by showing the connections of the supply and of the load to the diodes. [2] (b) Suggest one advantage of the use of a bridge rectifier, rather than a single diode, for the rectification of alternating current. … … [1] (c) State (i) what is meant by smoothing, … … [1] (ii) the effect of the value of the capacitance of the smoothing capacitor in relation to smoothing. … … … [2]

6 marks

Mark scheme: 5 (a) supply connected correctly (to left & right) B1 load connected correctly (to top & bottom) B1 [2] (b) e.g. power supplied on every half-cycle greater average/mean power (any sensible suggestion, 1 mark) B1 [1] (c) (i) reduction in the variation of the output voltage/current B1 [1] (ii) larger capacitance produces more smoothing M1 either product RC larger or for the same load A1 [2]

This question in 9702/43 Oct/Nov 2011

9 (a) The resistance of a light-dependent resistor (LDR) is approximately 500 Ω in daylight. Suggest an approximate value for the resistance of the LDR in darkness. resistance = … Ω [1] (b) An electronic light-meter is used to warn when light intensity becomes low. A light-dependent resistor is connected into the circuit of Fig. 9.1. +4.5 V 1.7 kΩ +9 V – P +2.5 V + R R –9 V red green Fig. 9.1 The operational amplifier (op-amp) is ideal. The resistors R are to ensure that the light-emitting diodes (LEDs) do not over-heat. (i) On Fig. 9.1, mark the polarity of the point P for the red LED to be emitting light. [1] (ii) The LDR is in daylight and has a resistance of 500 Ω. State and explain which diode, red or green, will be emitting light. … … … … [3] (iii) The intensity of the light decreases and the LDR is in darkness. State and explain the effect on the LEDs of this change in intensity. … … … [2]

7 marks

Mark scheme: 9 (a) any value greater than, or equal to, 5 kΩ B1 [1] (b) (i) ‘positive’ shown in correct position B1 [1] (ii) V + = (500/2200) × 4.5 ≈ 1 V B1 V – > V + so output is negative M1 green LED on, (red LED off) A1 [3] (allow full ecf of incorrect value of V +) (iii) either V + increases or V + > V – M1 green LED off, red LED on A1 [2]

This question in 9702/43 Oct/Nov 2011

10 A student designs an electronic sensor that is to be used to switch on a lamp when the light intensity is low. Part of the circuit is shown in Fig. 10.1. +5 V +5 V X – + –5 V 240 V sensing device processing unit output device Fig. 10.1 (a) State the name of the component labelled X on Fig. 10.1. … [1] (b) On Fig. 10.1, draw the symbols for (i) two resistors to complete the circuit for the sensing device, [2] (ii) a relay to complete the circuit for the processing unit. [2] (c) (i) State the purpose of the relay. … … [1] (ii) Suggest why the diode is connected to the output of the operational amplifier (op-amp) in the direction shown. … … … [2]

8 marks

Mark scheme: 10 (a) light-dependent resistor (allow LDR) B1 [1] (b) (i) two resistors in series between +5 V line and earth M1 midpoint connected to inverting input of op-amp A1 [2] (ii) relay coil between diode and earth M1 switch between lamp and earth A1 [2] (c) (i) switch on/off mains supply using a low voltage/current output B1 [1] (allow ‘isolates circuit from mains supply’) (ii) relay will switch on for one polarity of output (voltage) C1 switches on when output (voltage) is negative A1 [2]

This question in 9702/41 May/June 2012

9 An operational amplifier (op-amp) may be used as part of the processing unit in an electronic sensor. (a) State four properties of an ideal operational amplifier. 1. … 2. … 3. … 4. … [4] (b) A comparator circuit incorporating an ideal op-amp is shown in Fig. 9.1. +5 V – + VIN –5 V VOUT Fig. 9.1 The variation with time t of the input potential VIN is shown in Fig. 9.2. 6 potential / V 4 2 VINVIN 0 t –2 –4 –6 Fig. 9.2 On the axes of Fig. 9.2, draw a graph to show the variation with time t of the output potential VOUT . [3] (c) The output potential VOUT is to be displayed using two light-emitting diodes (LEDs). A For Examiner’s diode emitting red light is to indicate when VOUT is positive and a diode emitting green light is to be used to indicate when VOUT is negative. Use Complete Fig. 9.3 to show the connections of the two LEDs to the output of the op-amp. Label each LED with the colour of light that it emits. VOUT Fig. 9.3 [3]

10 marks

Mark scheme: 9 (a) e.g. infinite input impedance/resistance zero output impedance/resistance infinite (open loop) gain infinite bandwidth infinite slew rate (any four, one mark each) B4 [4] (b) graph: square wave M1 180° phase change A1 amplitude 5.0 V A1 [3] (c) correct symbol for LED M1 diodes connected correctly between VOUT and earth A1 diodes identified correctly A1 [3] (special case: if diode symbol, not LED symbol, allow 2nd and 3rd marks to be scored)

This question in 9702/42 May/June 2012

12 An incomplete simplified block diagram of the circuitry for a mobile-phone handset is shown For in Fig. 12.1. Examiner’s Use aerial switch tuning circuit amplifier amplifier X oscillator demodulator parallel- Y to-serial converter ADC DAC amplifier amplifier microphone loudspeaker Fig. 12.1 (a) State the name of the block labelled (i) X, … [1] (ii) Y. … [1] Question 12 continues on page 24. (b) Explain the purpose of For Examiner’s (i) the switch, Use … … [1] (ii) the parallel-to-serial converter. … … … [2]

5 marks

Mark scheme: 12 (a) (i) modulator B1 [1] (ii) serial-to-parallel converter (accept series-to-parallel converter) B1 [1] (b) (i) enables one aerial to be used for transmission and receipt of signals A1 [1] (ii) all bits for one number arrive at one time B1 bits are sent out one after another B1 [2]

This question in 9702/42 May/June 2012

10 A student designs an electronic sensor that is to be used to switch on a lamp when the light intensity is low. Part of the circuit is shown in Fig. 10.1. +5 V +5 V X – + –5 V 240 V sensing device processing unit output device Fig. 10.1 (a) State the name of the component labelled X on Fig. 10.1. … [1] (b) On Fig. 10.1, draw the symbols for (i) two resistors to complete the circuit for the sensing device, [2] (ii) a relay to complete the circuit for the processing unit. [2] (c) (i) State the purpose of the relay. … … [1] (ii) Suggest why the diode is connected to the output of the operational amplifier (op-amp) in the direction shown. … … … [2]

8 marks

Mark scheme: 10 (a) light-dependent resistor (allow LDR) B1 [1] (b) (i) two resistors in series between +5 V line and earth M1 midpoint connected to inverting input of op-amp A1 [2] (ii) relay coil between diode and earth M1 switch between lamp and earth A1 [2] (c) (i) switch on/off mains supply using a low voltage/current output B1 [1] (allow ‘isolates circuit from mains supply’) (ii) relay will switch on for one polarity of output (voltage) C1 switches on when output (voltage) is negative A1 [2]

This question in 9702/43 May/June 2012

9 An operational amplifier (op-amp) may be used as part of the processing unit in an electronic sensor. (a) State three properties of an ideal op-amp. 1. … 2. … 3. … [3] (b) A comparator circuit incorporating an ideal op-amp is shown in Fig. 9.1. +5 V – + V1 VOUT –5 V V2 Fig. 9.1 (i) In one application of the comparator, V2 is kept constant at +1.5 V. The variation with time t of the potential V1 is shown in Fig. 9.2. The potential V2 is also shown. 10 For Examiner’s Use 8 potential / V 6 4 2 V2 0 t –2 V1 –4 –6 –8 –10 Fig. 9.2 On Fig. 9.2, show the variation with time t of the output potential VOUT . [4] (ii) Two light-emitting diodes (LEDs) R and G are connected to the output of the op-amp in Fig. 9.1 such that R emits light for a longer time than G. On Fig. 9.1, draw the symbols for the two diodes connected to the output of the op-amp and label the diodes R and G. [3]

10 marks

Mark scheme: 9 (a) e.g. zero output impedance / resistance infinite input impedance / resistance infinite (open loop) gain infinite bandwidth infinite slew rate 1 each, max. 3 B3 [3] (b) (i) graph: square wave M1 correct cross-over points where V2 = V1 A1 amplitude 5 V A1 correct polarity (positive at t = 0) A1 [4] (ii) correct symbol for LED M1 diodes connected correctly between VOUT and earth A1 correct polarity consistent with graph in (i) A1 [3] (R points ‘down’ if (i) correct)

This question in 9702/41 Oct/Nov 2012

9 An operational amplifier (op-amp) may be used as part of the processing unit in an electronic sensor. (a) State three properties of an ideal op-amp. 1. … 2. … 3. … [3] (b) A comparator circuit incorporating an ideal op-amp is shown in Fig. 9.1. +5 V – + V1 VOUT –5 V V2 Fig. 9.1 (i) In one application of the comparator, V2 is kept constant at +1.5 V. The variation with time t of the potential V1 is shown in Fig. 9.2. The potential V2 is also shown. 10 For Examiner’s Use 8 potential / V 6 4 2 V2 0 t –2 V1 –4 –6 –8 –10 Fig. 9.2 On Fig. 9.2, show the variation with time t of the output potential VOUT . [4] (ii) Two light-emitting diodes (LEDs) R and G are connected to the output of the op-amp in Fig. 9.1 such that R emits light for a longer time than G. On Fig. 9.1, draw the symbols for the two diodes connected to the output of the op-amp and label the diodes R and G. [3]

10 marks

Mark scheme: 9 (a) e.g. zero output impedance / resistance infinite input impedance / resistance infinite (open loop) gain infinite bandwidth infinite slew rate 1 each, max. 3 B3 [3] (b) (i) graph: square wave M1 correct cross-over points where V2 = V1 A1 amplitude 5 V A1 correct polarity (positive at t = 0) A1 [4] (ii) correct symbol for LED M1 diodes connected correctly between VOUT and earth A1 correct polarity consistent with graph in (i) A1 [3] (R points ‘down’ if (i) correct)

This question in 9702/42 Oct/Nov 2012

6 A bridge rectifier consists of four ideal diodes A, B, C and D, connected as shown in Fig. 6.1. For Examiner’s Use A B R X D C Y Fig. 6.1 An alternating supply is applied between the terminals X and Y. (a) (i) On Fig. 6.1, label the positive (+) connection to the load resistor R. [1] (ii) State which diodes are conducting when terminal Y of the supply is positive. diode … and diode … [1] (b) The variation with time t of the potential difference V across the load resistor R is shown in Fig. 6.2. +8 +6 V / V +4 +2 0 t −2 −4 −6 −8 Fig. 6.2 The load resistor R has resistance 2700 Ω. For Examiner’s (i) Use Fig. 6.2 to determine the mean power dissipated in the resistor R. Use power = … W [3] (ii) On Fig. 6.1, draw the symbol for a capacitor, connected so as to increase the mean power dissipated in the resistor R. [1] (c) The capacitor in (b)(ii) is now removed from the circuit. The diode A in Fig. 6.1 stops functioning, so that it now has infinite resistance. On Fig. 6.2, draw the variation with time t of the new potential difference across the resistor R. [2]

8 marks

Mark scheme: 6 (a) (i) connection to ‘top’ of resistor labelled as positive B1 [1] (ii) diode B and diode D B1 [1] (b) (i) VP = 4.0 V C1 mean power = VP2/2R C1 = 42 / (2 × 2700) = 2.96 × 10–3 W A1 [3] (ii) capacitor, correct symbol, connected in parallel with R B1 [1] (c) graph: half-wave rectification M1 same period and same peak value A1 [2]

This question in 9702/43 Oct/Nov 2012

9 A student designs an electronic sensor to monitor whether the temperature in a refrigerator is above or below a particular value. The circuit is shown in Fig. 9.1. A D +5 V – + –5 V B C sensing processing output device unit device Fig. 9.1 (a) Name the components used in the output device. … [1] (b) An operational amplifier (op-amp) is used as the processing unit. Describe the function of this processing unit. … … … [2] (c) State the function of (i) the resistors C and D, … … [1] (ii) the resistor B. … … [1] (d) The output device of the circuit in Fig. 9.1 is changed so that the new output device may For be used to switch on a high-voltage circuit. Examiner’s Use (i) State the component that is used in the new output device. … … [1] (ii) Draw on Fig. 9.2 to show how the component in (i), together with a diode, are connected so that the high voltage may be switched on when the output of the op-amp is negative. +5 V – + –5 V connections to high-voltage circuit output device Fig. 9.2 [2]

8 marks

Mark scheme: 9 (a) light-emitting diode (allow LED) B1 [1] (b) gives a high or a low output / +5 V or –5 V output M1 dependent on which of the inputs is at a higher potential A1 [2] (c) (i) provides a reference/constant potential B1 [1] (ii) determines temperature of ‘switch-over’ B1 [1] (d) (i) relay A1 [1] (ii) relay connected correctly for op-amp output and high-voltage circuit B1 diode with correct polarity in output from op-amp B1 [2]

This question in 9702/43 Oct/Nov 2012

12 (a) Suggest applications, one in each case, for the transmission of signals using For Examiner’s (i) a wire pair, Use … [1] (ii) a coaxial cable, … [1] (iii) a microwave link. … [1] (b) A cable used for the transmission of a signal has an attenuation per unit length of 2.1 dB km–1. There are no amplifiers along the cable. The input power of the signal is 450 mW. (i) Calculate the output power of the signal for the cable of length 40 km. output power = … W [3] (ii) The minimum acceptable signal power in the cable is 7.2 × 10–11 W. Calculate the maximum uninterrupted length of the cable. length = … km [2]

8 marks

Mark scheme: 12 (a) (i) e.g. linking a (land) telephone to the (local) exchange B1 [1] (ii) e.g. connecting an aerial to a television B1 [1] (iii) e.g. linking a ground station to a satellite B1 [1] (b) (i) attenuation = 10 lg (P2 / P1) C1 total attenuation = 2.1 × 40 (= 84 dB) C1 84 = 10 lg ({450 × 10–3} / P) P = 1.8 × 10–9 W A1 [3] (answer 1.1 ×108 W scores 1 mark only) (ii) maximum attenuation = 10 lg ({450 × 10–3} / {7.2 × 10–11}) = 98 dB C1 maximum length = 98/2.1 = 47 km A1 [2]

This question in 9702/43 Oct/Nov 2012

6 A simple transformer is illustrated in Fig. 6.1. For Examiner’s Use load input resistor primary secondary coil coil laminated iron core Fig. 6.1 (a) State (i) why the iron core is laminated, … … … [2] (ii) what is meant by an ideal transformer. … … [1] (b) An ideal transformer has 300 turns on the primary coil and 8100 turns on the secondary coil. The root-mean-square input voltage to the primary coil is 9.0 V. Calculate the peak voltage across the load resistor connected to the secondary coil. peak voltage = … V [2]

5 marks

Mark scheme: 6 (a) (i) to reduce power loss in the core B1 due to eddy currents/induced currents B1 [2] (ii) either no power loss in transformer or input power = output power B1 [1] (b) either r.m.s. voltage across load = 9.0 × (8100 / 300) C1 peak voltage across load = √2 × 243 = 340 V A1 [2] or peak voltage across primary coil = 9.0 × √2 (C1) peak voltage across load = 12.7 × (8100/300) = 340 V (A1)

This question in 9702/43 May/June 2013

9 (a) State three properties of an ideal operational amplifier (op-amp). 1. … 2. … 3. … [3] (b) An amplifier circuit is shown in Fig. 9.1. 10.8 k1 +9 V – + –9 V V OUT V IN 1.2 k1 Fig. 9.1 (i) Calculate the gain of the amplifier circuit. gain = … [2] (ii) The variation with time t of the input potential VIN is shown in Fig. 9.2. For Examiner’s 10 Use potential / V 5 0 t V IN – 5 –10 Fig. 9.2 On the axes of Fig. 9.2, show the variation with time t of the output potential VOUT . [3]

8 marks

Mark scheme: 9 (a) e.g. zero output impedance / resistance infinite input impedance / resistance infinite (open loop) gain infinite bandwidth infinite slew rate (1 each, max. 3 ) B3 [3] (b) (i) gain = 1 + (10.8 / 1.2) C1 = 10 A1 [2] (ii) graph: straight line from (0,0) towards VIN = 1.0 V, VOUT = 10 V B1 horizontal line at VOUT = 9.0 V to VIN = 2.0 V B1 correct +9.0 V → –9.0 V (and correct shape to VIN = 0) B1 [3]

This question in 9702/43 Oct/Nov 2013

11 (a) A circuit incorporating an ideal operational amplifier (op-amp) is shown in Fig. 11.1. P – + V IN V OUT Fig. 11.1 (i) State the name of this circuit. … [1] (ii) Explain why the point P is referred to as a virtual earth. … … … … [3] (b) The circuit of Fig. 11.1 is modified, as shown in Fig. 11.2. RC C RB B RA A 1.0 k1 – + V IN V Fig. 11.2 The voltmeter has infinite resistance and its full-scale deflection is 1.0 V. The input potential to the circuit is VIN. The switch position may be changed in order to have different values of resistance in the circuit. (i) The input potential VIN and the switch position are varied. For each switch position, the reading of the voltmeter is 1.0 V. Complete Fig. 11.3 for the switch positions shown. switch position VIN / mV resistance A 10 RA = … B 100 RB = … C … RC = 1.0 kΩ Fig. 11.3 [3] (ii) By reference to your answers in (i), suggest a use for the circuit of Fig. 11.2. … … [1]

8 marks

Mark scheme: 11 (a) (i) inverting amplifier B1 [1] (ii) gain is very large/infinite B1 V+ is earthed/zero B1 for amplifier not to saturate, P must be (almost) earth/zero B1 [3] (b) (i) RA = 100 kΩ A1 RB = 10 kΩ A1 VIN = 1000 mV A1 [3] (ii) variable range meter B1 [1] GCE AS/A LEVEL – May/June 2014 9702 41

This question in 9702/41 May/June 2014

10 (a) State the function of a comparator circuit incorporating an operational amplifier (op-amp). … … … … [3] (b) An ideal op-amp is incorporated into the circuit of Fig. 10.1. +1.5 V 1.2 k1 +5 V – + G –5 V V IN 2.4 k1 R Fig. 10.1 (i) On Fig. 10.1, draw a circle around the part of the circuit that is being used as an output device. [1] (ii) Show that the potential at the non-inverting input of the op-amp is 1.0 V. [1] (iii) The variation with time t of the potential VIN at the inverting input of the op-amp is shown in Fig. 10.2. 6 potential 4 / V VIN 2 +1.0 0 t 1 t 2 time t –2 –4 –6 Fig. 10.2 1. On the axes of Fig. 10.2, draw the variation with time t of the output potential of the op-amp. [3] 2. State whether each diode is emitting light or is not emitting light at time t1 and at time t2. At time t1, diode R will … and diode G will … . At time t2, diode R will … and diode G will … . [2]

10 marks

Mark scheme: 10 (a) compares the potentials/voltages at the (inverting and non-inverting) inputs B1 either output (potential) dependent on which input is the larger or V+ > V–, then VOUT is positive B1 states the other condition B1 [3] (b) (i) ring drawn around both the LEDs (and series resistors) B1 [1] (ii) V– R (1.5 × 2.4) / (1.2 N 2.4) R 1.0 V B1 [1] (allow 1.5 × 2.4 / 3.6 R 1.0 V) (iii) 1. VOUT switches at N1.0 V B1 maximum VOUT is 5.0 V B1 when curve is above N1.0 V, VOUT is negative (or v.v.) B1 [3] 2. at time t1, diode R is emitting light, diode G is not emitting B1 at time t2, diode R is not emitting, diode G is emitting B1 [2] (must be consistent with graph line. If no graph line then 0 / 2)

This question in 9702/42 May/June 2014

11 (a) A circuit incorporating an ideal operational amplifier (op-amp) is shown in Fig. 11.1. P – + V IN V OUT Fig. 11.1 (i) State the name of this circuit. … [1] (ii) Explain why the point P is referred to as a virtual earth. … … … … [3] (b) The circuit of Fig. 11.1 is modified, as shown in Fig. 11.2. RC C RB B RA A 1.0 k1 – + V IN V Fig. 11.2 The voltmeter has infinite resistance and its full-scale deflection is 1.0 V. The input potential to the circuit is VIN. The switch position may be changed in order to have different values of resistance in the circuit. (i) The input potential VIN and the switch position are varied. For each switch position, the reading of the voltmeter is 1.0 V. Complete Fig. 11.3 for the switch positions shown. switch position VIN / mV resistance A 10 RA = … B 100 RB = … C … RC = 1.0 kΩ Fig. 11.3 [3] (ii) By reference to your answers in (i), suggest a use for the circuit of Fig. 11.2. … … [1]

8 marks

Mark scheme: 11 (a) (i) inverting amplifier B1 [1] (ii) gain is very large/infinite B1 V+ is earthed/zero B1 for amplifier not to saturate, P must be (almost) earth/zero B1 [3] (b) (i) RA = 100 kΩ A1 RB = 10 kΩ A1 VIN = 1000 mV A1 [3] (ii) variable range meter B1 [1] GCE AS/A LEVEL – May/June 2014 9702 43

This question in 9702/43 May/June 2014

10 (a) An ideal operational amplifier (op-amp) has infinite open-loop gain and infinite input resistance (impedance). State three further properties of an ideal op-amp. 1. … … 2. … … 3. … … [3] (b) The circuit of Fig. 10.1 is used to detect changes in temperature. 740 1 +9 V A – + –9 V V Fig. 10.1 The voltmeter has infinite resistance. The variation with temperature θ of the resistance R of the thermistor is shown in Fig. 10.2. 4.0 R / k1 3.0 2.0 1.0 0 5 10 15 20 25 30 e / °C Fig. 10.2 (i) When the thermistor is at a temperature of 1.0 °C, the voltmeter reads +1.0 V. Show that, for the thermistor at 1.0 °C, the potential at A is −0.20 V. [4] (ii) The potential at A remains at −0.20 V. Determine the voltmeter reading for a thermistor temperature of 15 °C. voltmeter reading = … V [2] (c) The voltmeter reading for a thermistor temperature of 29 °C is 0.35 V. (i) Assuming a linear change of voltmeter reading with change of temperature over the range 1 °C to 29 °C, calculate the voltmeter reading at 15 °C. voltmeter reading = … V [1] (ii) Suggest why your answers in (b)(ii) and (c)(i) are not the same. … … [1]

11 marks

Mark scheme: 10 (a) e.g. zero output resistance / impedance infinite bandwidth infinite slew rate 1 mark each, max. 3 B3 [3] (b) (i) at 1.0 °C, thermistor resistance is 3.7 kΩ B1 amplifier gain = –R / 740 = –3700 / 740 (negative sign essential) C1 = –5.0 C1 potential = 1.0 / –5.0 = –0.20 V A1 [4] (ii) at 15 °C, R = 2.15 kΩ (allow ±0.05 kΩ) C1 reading = (2150 / 740) × 0.2 = 0.58 V (0.59 V → 0.57 V) A1 [2] (c) (i) 0.68 V A1 [1] (ii) resistance (of thermistor) does not change linearly with temperature B1 [1]

This question in 9702/41 Oct/Nov 2014

13 A simplified block diagram of a mobile phone handset is shown in Fig. 13.1. aerial switch tuning circuit r.f. amplifier r.f. amplifier modulator oscillator demodulator parallel-to- serial-to- serial parallel converter converter ADC DAC a.f. amplifier a.f. amplifier loudspeaker microphone Fig. 13.1 State the purpose of (a) the switch, … … … … [2] (b) the tuning circuit. … … …

2 marks

Mark scheme: 13 (a) either for transmission and reception of signal or switching between transmitted and received signals M1 either so that one aerial may be used or so that transmission and reception can occur in quick succession A1 [2] (b) gives large signal for one (input) frequency M1 (and) rejects / very small signal for all other frequencies A1 [2]

This question in 9702/41 Oct/Nov 2014

10 (a) An ideal operational amplifier (op-amp) has infinite open-loop gain and infinite input resistance (impedance). State three further properties of an ideal op-amp. 1. … … 2. … … 3. … … [3] (b) The circuit of Fig. 10.1 is used to detect changes in temperature. 740 1 +9 V A – + –9 V V Fig. 10.1 The voltmeter has infinite resistance. The variation with temperature θ of the resistance R of the thermistor is shown in Fig. 10.2. 4.0 R / k1 3.0 2.0 1.0 0 5 10 15 20 25 30 e / °C Fig. 10.2 (i) When the thermistor is at a temperature of 1.0 °C, the voltmeter reads +1.0 V. Show that, for the thermistor at 1.0 °C, the potential at A is −0.20 V. [4] (ii) The potential at A remains at −0.20 V. Determine the voltmeter reading for a thermistor temperature of 15 °C. voltmeter reading = … V [2] (c) The voltmeter reading for a thermistor temperature of 29 °C is 0.35 V. (i) Assuming a linear change of voltmeter reading with change of temperature over the range 1 °C to 29 °C, calculate the voltmeter reading at 15 °C. voltmeter reading = … V [1] (ii) Suggest why your answers in (b)(ii) and (c)(i) are not the same. … … [1]

11 marks

Mark scheme: 10 (a) e.g. zero output resistance / impedance infinite bandwidth infinite slew rate 1 mark each, max. 3 B3 [3] (b) (i) at 1.0 °C, thermistor resistance is 3.7 kΩ B1 amplifier gain = –R / 740 = –3700 / 740 (negative sign essential) C1 = –5.0 C1 potential = 1.0 / –5.0 = –0.20 V A1 [4] (ii) at 15 °C, R = 2.15 kΩ (allow ±0.05 kΩ) C1 reading = (2150 / 740) × 0.2 = 0.58 V (0.59 V → 0.57 V) A1 [2] (c) (i) 0.68 V A1 [1] (ii) resistance (of thermistor) does not change linearly with temperature B1 [1]

This question in 9702/42 Oct/Nov 2014

13 A simplified block diagram of a mobile phone handset is shown in Fig. 13.1. aerial switch tuning circuit r.f. amplifier r.f. amplifier modulator oscillator demodulator parallel-to- serial-to- serial parallel converter converter ADC DAC a.f. amplifier a.f. amplifier loudspeaker microphone Fig. 13.1 State the purpose of (a) the switch, … … … … [2] (b) the tuning circuit. … … …

2 marks

Mark scheme: 13 (a) either for transmission and reception of signal or switching between transmitted and received signals M1 either so that one aerial may be used or so that transmission and reception can occur in quick succession A1 [2] (b) gives large signal for one (input) frequency M1 (and) rejects / very small signal for all other frequencies A1 [2]

This question in 9702/42 Oct/Nov 2014

10 An electronic sensor may be represented by the block diagram of Fig. 10.1. sensing processing output device unit device Fig. 10.1 (a) State suitable sensing devices, one in each case, for the detection of (i) change of temperature, … [1] (ii) pressure changes in a sound wave. … [1] (b) The ideal operational amplifier (op-amp) shown in Fig. 10.2 is to be used as a processing unit. +5 V – + V IN –5 V V OUT Fig. 10.2 (i) State the value of the output potential VOUT for an input potential VIN of +0.5 V. Explain your answer. … … … … [3] (ii) A sensing device produces a variable potential VIN . The variation with time t of VIN is shown in Fig. 10.3. 10 8 potential / V 6 4 V IN 2 0 0 t –2 –4 –6 –8 –10 Fig. 10.3 On the axes of Fig. 10.3, sketch the variation with time t of the output potential VOUT . [3]

8 marks

Mark scheme: 10 (a) (i) thermistor / thermocouple B1 [1] (ii) quartz crystal / piezoelectric crystal or transducer / microphone B1 [1] (b) (i) VOUT = –5 V A1 inverting input is positive or V– is positive or V– > V+ so VOUT is negative B1 op-amp has very large / infinite gain and so saturates B1 [3] (ii) sketch: VOUT switches from (+) to (–) when VIN is zero B1 VOUT is +5 V or –5 V M1 VOUT is negative when VIN is positive (or v.v.) A1 [3]

This question in 9702/43 Oct/Nov 2014

10 An operational amplifier (op-amp) is used in the comparator circuit of Fig. 10.1. +4.5 V 4.2 k1 +5 V + – –5 V V IN V OUT 1.2 k1 R Fig. 10.1 (a) (i) Show that the potential at the inverting input of the op-amp is +1.0 V. [1] (ii) Explain why the potential difference across resistor R is + 5 V when VIN is greater than 1.0 V and is zero when VIN is less than 1.0 V. VIN > 1.0 V: … … … VIN < 1.0 V: … … … [4] (b) The variation with time t of the input voltage VIN is shown in Fig. 10.2. 6 5 4 voltage / V 3 V IN 2 1 +1 V 0 0 time t Fig. 10.2 (i) On the axes of Fig. 10.2, draw the variation with time t of the output potential VOUT. [2] (ii) Suggest a use for this type of circuit. … … [1]

8 marks

Mark scheme: 10 (a) (i) (potential =) 1.2 / (1.2 + 4.2) × 4.5 = +1.0 V A1 [1] (ii) (for VIN > 1.0 V) V+ > V– B1 output (of op-amp) is +5 V or positive M1 diode conducts giving +5 V across R or Vout is +5 V A1 (for VIN < 1.0 V) output of op-amp –5 V / negative so diode does not conduct, giving Vout = 0 or 0 V across R A1 [4] (b) (i) square wave with maximum value +5 V and minimum value 0 M1 vertical sides in correct positions and correct phase A1 [2] (ii) re-shaping (digital) signals/regenerator (amplifier) B1 [1]

This question in 9702/41 May/June 2015

10 An operational amplifier (op-amp) is used in the comparator circuit of Fig. 10.1. +4.5 V 4.2 k1 +5 V + – –5 V V IN V OUT 1.2 k1 R Fig. 10.1 (a) (i) Show that the potential at the inverting input of the op-amp is +1.0 V. [1] (ii) Explain why the potential difference across resistor R is + 5 V when VIN is greater than 1.0 V and is zero when VIN is less than 1.0 V. VIN > 1.0 V: … … … VIN < 1.0 V: … … … [4] (b) The variation with time t of the input voltage VIN is shown in Fig. 10.2. 6 5 4 voltage / V 3 V IN 2 1 +1 V 0 0 time t Fig. 10.2 (i) On the axes of Fig. 10.2, draw the variation with time t of the output potential VOUT. [2] (ii) Suggest a use for this type of circuit. … … [1]

8 marks

Mark scheme: 10 (a) (i) (potential =) 1.2 / (1.2 + 4.2) × 4.5 = +1.0 V A1 [1] (ii) (for VIN > 1.0 V) V+ > V– B1 output (of op-amp) is +5 V or positive M1 diode conducts giving +5 V across R or Vout is +5 V A1 (for VIN < 1.0 V) output of op-amp –5 V / negative so diode does not conduct, giving Vout = 0 or 0 V across R A1 [4] (b) (i) square wave with maximum value +5 V and minimum value 0 M1 vertical sides in correct positions and correct phase A1 [2] (ii) re-shaping (digital) signals/regenerator (amplifier) B1 [1]

This question in 9702/43 May/June 2015

10 The output potential VOUT from an operational amplifier is to be monitored using an output device. The output VOUT can be either +5 V or −5 V. (a) On Fig. 10.1, draw a circuit for the output device that consists of two light-emitting diodes B and G. Diode B alone is to emit light when VOUT is +5 V. Diode G alone is to emit light when VOUT is −5 V. VOUT Fig. 10.1 [3] (b) On Fig. 10.2, draw a circuit of the output device that consists of a relay and a diode such that a high-power lamp is switched on only when VOUT is −5 V. VOUT high power lamp power supply Fig. 10.2 [4]

7 marks

Mark scheme: 10 (a) correct LED symbol B1 separately connected between VOUT and earth with opposite polarities M1 diode B ‘pointing’ from VOUT to earth A1 [3] (ignore protective resistors) (b) diode in VOUT line M1 diode ‘pointing’ towards VOUT from earth A1 relay coil connected between VOUT and earth M1 switch connected across lamp A1 [4] (if a diode is placed across the relay it must point down otherwise max. 2/4; one diode but wrong direction max. 3/4)

This question in 9702/43 Oct/Nov 2015

12 A transmission system for speech may be represented by the block diagram of Fig. 12.1. P Q X Y Z analogue-to- parallel-to- digital serial converter converter Fig. 12.1 (a) Name the component labelled (i) block X, … [1] (ii) block Y, … [1] (iii) Z. … [1] (b) The variation with time of part of the signal at the input P to the analogue-to-digital converter (ADC) is shown in Fig. 12.2. 16 14 signal 12 / mV 10 8 6 4 2 0 0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 time / ms Fig. 12.2 Each number of the output from the ADC is a digital number where the smallest bit represents 1 mV. State (i) the minimum number of bits in each digital number so that the signal in Fig. 12.2 can be sampled fully, number = … [1] (ii) the digital number produced by the ADC at time 0.50 ms. number = … [1] (c) The ADC samples the signal in Fig. 12.2 at a frequency of 4.0 kHz. The first sample is taken at time zero. Using data from Fig. 12.2, draw, on the axes of Fig. 12.3, the variation with time of the output at point Q for time zero to time 1.5 ms. 16 14 12 output 10 level 8 6 4 2 0 0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 time / ms Fig. 12.3 [4] Please turn over for Question 13.

9 marks

Mark scheme: 12 (a) (i) serial-to-parallel converter B1 [1] (ii) digital-to-analogue converter or DAC B1 [1] (iii) (audio) amplifier or AF amplifier B1 [1] (b) (i) 4 A1 [1] (ii) 1011 A1 [1] (c) correct levels at 0.25 ms intervals 0, 8, 11, 10, 15 A1 and 7, 4 A1 series of steps, each of depth 0.25 ms M1 voltage levels shown in correct intervals A1 [4]

This question in 9702/43 Oct/Nov 2015

8 (a) State two effects of negative feedback on the gain of an amplifier incorporating an operational amplifier (op-amp). 1. … … 2. … … [2] (b) An incomplete diagram of an amplifier circuit incorporating an ideal operational amplifier is shown in Fig. 8.1. +8.0 V – + 7.2 k1 –8.0 V VIN VOUT Fig. 8.1 The amplifier has a voltage gain of +5.0. (i) Complete the circuit diagram of Fig. 8.1. [2] (ii) Calculate the resistance of any additional resistor you have drawn on Fig. 8.1. resistance = … kΩ [2] (iii) Fig. 8.2 shows the variation with time of the input potential VIN. 15 potential / V 10 5 VIN 0 0 1 2 3 4 5 time / ms Fig. 8.2 On Fig. 8.2, draw the variation with time of the output potential VOUT. [2] [Total: 8]

8 marks

Mark scheme: 8 (a) decreases gain B1 increases bandwidth / decreases distortion / increases (operating) stability B1 [2] (b) (i) additional resistor connected between 7.2 kΩ resistor and earth B1 V – joined to lower end of 7.2 kΩ resistor and V + joined to VIN B1 [2] (ii) either 5 = 1 + (7.2 / R) or 5 = 1 + (7200 / R) C1 R = 1.8 kΩ A1 [2] (iii) horizontal line from (0, 8.0) to (1.8, 8.0) B1 straight line from (1.8, 8.0) to (5.0, 0) B1 [2] (allow a tolerance of ± ½ small square when marking the graph)

This question in 9702/42 Feb/March 2016

8 The circuit of an inverting amplifier incorporating an ideal operational amplifier (op-amp) is shown in Fig. 8.1. R1 +V R2 P – + V IN –V V OUT Fig. 8.1 (a) Explain why point P is known as a virtual earth. … … … [3] (b) Derive an expression, in terms of the resistances R1 and R2, for the gain of the amplifier circuit. Explain your working. [3] (c) A relay and the output terminals of the amplifier circuit are shown in Fig. 8.2. V OUT Fig. 8.2 On Fig. 8.2, show how the relay may be connected to the amplifier output so that the relay operates only when VOUT is positive. [3] [Total: 9]

9 marks

Mark scheme: 8 (a) op-amp has infinite/(very) large gain B1 op-amp saturates if V + ≠ V – M1 V+ is at earth potential so P (or V–) must be at earth A1 [3] (b) input resistance to op-amp is very large or current in R2 = current in R1 B1 VIN (– 0) = IR2 and (0) – VOUT = IR1 M1 VOUT / VIN = –R1 / R2 A1 [3] (c) relay coil connected between VOUT and earth M1 correct diode symbol connected between VOUT and coil or between coil and earth M1 correct polarity for diode (‘clockwise’) A1 [3]

This question in 9702/41 May/June 2016

11 A bridge rectifier contains four ideal diodes A, B, C and D, as shown in Fig. 11.1. % input $ 9 ' & / 1 Fig. 11.1 The output of the rectifier is connected to a load L of resistance 2.4 kΩ. (a) On Fig. 11.1, mark with the letter P the positive terminal of the load. [1] (b) The variation with time t of the potential difference V across the input to the rectifier is shown in Fig. 11.2. 8 6 LQSXW 9 / V 4 2 0 W –2 –4 –6 –8 Fig. 11.2 Calculate the root-mean-square (r.m.s.) current in the load L. r.m.s. current = … A [2] (c) The potential difference across the load L is to be smoothed using a capacitor. (i) On Fig. 11.1, draw the symbol for a capacitor, connected to produce smoothing. [1] (ii) The minimum potential difference across the load L with the smoothing capacitor connected is 3.0 V. On Fig. 11.2, sketch the variation with time t of the potential difference across the load L. [3] [Total: 7]

7 marks

Mark scheme: 11 (a) point P shown at ‘lower end’ of load B1 [1] (b) Vr.m.s. = 6.0 / √2 = 4.24 V C1 Ir.m.s. = 4.24 / (2.4 × 103) = 1.8 × 10–3 A A1 [2] (c) (i) capacitor in parallel with load B1 [1] (ii) line from peak to curve at 3.0 V for either half- or full-wave rectified M1 correct curvature on line (gradient becoming more shallow) A1 line drawn as for full-wave rectified A1 [3]

This question in 9702/42 May/June 2016

8 The circuit of an inverting amplifier incorporating an ideal operational amplifier (op-amp) is shown in Fig. 8.1. R1 +V R2 P – + V IN –V V OUT Fig. 8.1 (a) Explain why point P is known as a virtual earth. … … … [3] (b) Derive an expression, in terms of the resistances R1 and R2, for the gain of the amplifier circuit. Explain your working. [3] (c) A relay and the output terminals of the amplifier circuit are shown in Fig. 8.2. V OUT Fig. 8.2 On Fig. 8.2, show how the relay may be connected to the amplifier output so that the relay operates only when VOUT is positive. [3] [Total: 9]

9 marks

Mark scheme: 8 (a) op-amp has infinite/(very) large gain B1 op-amp saturates if V + ≠ V – M1 V+ is at earth potential so P (or V–) must be at earth A1 [3] (b) input resistance to op-amp is very large or current in R2 = current in R1 B1 VIN (– 0) = IR2 and (0) – VOUT = IR1 M1 VOUT / VIN = –R1 / R2 A1 [3] (c) relay coil connected between VOUT and earth M1 correct diode symbol connected between VOUT and coil or between coil and earth M1 correct polarity for diode (‘clockwise’) A1 [3]

This question in 9702/43 May/June 2016

8 A circuit incorporating an ideal operational amplifier (op-amp) is shown in Fig. 8.1. 50 kΩ RA +9 V 100 Ω – RB + 10 kΩ VIN –9 V V Fig. 8.1 The supply to the op-amp is +9 V / –9 V. The output of the amplifier is measured using a voltmeter having a range 0 – 5.0 V. A switch enables the inverting input to the op-amp to be connected to either resistor RA or resistor RB. (a) A positive potential +VIN is applied to the input to the circuit. On Fig. 8.1, mark with the letter P the positive connection of the voltmeter such that the voltmeter shows a positive reading. [1] (b) Calculate the potential VIN such that the voltmeter has a full-scale deflection when the inverting input to the op-amp is connected to (i) resistor RA of resistance 100 Ω, VIN = … V [2] (ii) resistor RB of resistance 10 kΩ. VIN = … V [1] (c) Suggest a use for this type of circuit. … … [1] [Total: 5]

5 marks

Mark scheme: 8 (a) P shown between earth symbol and voltmeter B1 [1] (b) (i) gain = (50 × 103) / 100 = 500 C1 VIN (= 5.0 / 500) = 0.010 V A1 [2] (ii) VIN (= 5.0 / 5.0) = 1.0 V A1 [1] (c) e.g. multi-range (volt)meter c.r.o. sensitivity control amplifier channel selector B1 [1]

This question in 9702/42 Oct/Nov 2016

11 (a) State Faraday’s law of electromagnetic induction. … … … [2] (b) An alternating current is passed through an air-cored solenoid. An iron core is inserted into the solenoid and then held stationary within the solenoid. The current in the solenoid is now smaller. Explain why the root-mean-square (r.m.s.) value of the current in the solenoid is reduced as a result of inserting the core. … … … … … [3] (c) Practical transformers are very efficient. However, there are some power losses. State two sources of power loss within a transformer. 1. … … 2. … … [2] [Total: 7]

7 marks

Mark scheme: 11 (a) (induced) e.m.f. proportional/equal to rate M1 of change of (magnetic) flux (linkage) A1 [2] (b) (for same current) iron core gives large(r) (rates of change of) flux (linkage) B1 e.m.f induced in solenoid is greater (for same current) M1 induced e.m.f. opposes applied e.m.f. so current smaller/acts to reduce current A1 [3] or same supply so same induced e.m.f. balancing it (B1) (rate of change of) flux linkage is same (M1) smaller current for same flux when core present (A1) (c) e.g. (heating due to) eddy currents in core (heating due to current in) resistance of coils hysteresis losses/losses due to changing magnetic field in core Any two of the above marking points, 1 mark each B2 [2]

This question in 9702/42 Oct/Nov 2016

6 (a) The slew rate of an ideal operational amplifier (op-amp) is said to be infinite. Explain what is meant by infinite slew rate. … … … [2] (b) The circuit of Fig. 6.1 is designed to indicate whether the temperature of the thermistor is above or below 24 °C. 2.00 kΩ +5 V – 4.5 V + VOUT R 3.00 kΩ –5 V Fig. 6.1 The operational amplifier (op-amp) is assumed to be ideal. At 24 °C, the resistance of the thermistor is 1.50 kΩ. (i) Determine the resistance of resistor R such that the output VOUT of the op-amp changes at 24 °C. resistance = … Ω [2] (ii) On Fig. 6.1, 1. draw two light-emitting diodes (LEDs) connected so as to indicate whether the output VOUT of the op-amp is either +5 V or −5 V, [2] 2. label with the letter G the LED that will be emitting light when the temperature is below 24 °C. Explain your working. … … … [3] [Total: 9]

9 marks

Mark scheme: 6 (a) reference to input (voltage) and output (voltage) B1 there is no time delay between change in input and change in output B1 [2] or reference to rate at which output voltage changes (B1) infinite rate of change (of output voltage) (B1) (b) (i) 2.00 / 3.00 = 1.50 / R C1 or V+ = (3.00 × 4.5) / (2.00 + 3.00) = 2.7 2.7 = 4.5 × R / (R + 1.50) (C1) resistance = 2.25 kΩ A1 [2] (ii) 1. correct symbol for LED M1 two LEDs connected with opposite polarities between VOUT and earth A1 [2] 2. below 24 °C, RT > 1.5 kΩ or resistance of thermistor increases/high B1 V– < V+ or V– decreases/low (must not contradict initial statement) M1 VOUT is positive/+5 (V) and LED labelled as ‘pointing’ from VOUT to earth A1 [3]

This question in 9702/43 Oct/Nov 2016

8 A student designs a circuit incorporating an operational amplifier (op-amp) as shown in Fig. 8.1. +6 V component C X R +5 V – + Y R –5 V B G RV 0 V Fig. 8.1 (a) (i) On Fig. 8.1, draw a circle around the output device. [1] (ii) State the purpose of this circuit. … … … [2] (b) The resistors X and Y each have resistance R. When conducting, the LED labelled B emits blue light and the LED labelled G emits green light. (i) State whether blue light or green light is emitted when the resistance of component C is greater than the resistance RV of the variable resistor. Explain your answer. … … … … … [3] (ii) State and explain what is observed as the resistance of component C is reduced. … … … … … [3] (c) Suggest the function of the variable resistor. … … [1] [Total: 10]

10 marks

Mark scheme: 8(a)(i) circle around both diodes B1 8(a)(ii) indicates (whether) temperature M1 (is) above or below a set value A1 8(b)(i) (when resistance of C > RV,) V– > V+ or V+ < 3 V or p.d. across RV < p.d. across R/Y/3 V or p.d. across C > p.d. across R/ X/3 V M1 op-amp output is negative M1 (only) green A1 8(b)(ii) resistance of C becomes less than RV or V– < V+ B1 green (LED) goes out A1 blue (LED) comes on A1 8(c) changes/determines temperature at which LEDs switch B1

This question in 9702/42 May/June 2017

7 (a) Feedback is used frequently in amplifier circuits. State (i) what is meant by feedback, … … … [2] (ii) two benefits of negative feedback in an amplifier circuit. 1. … … 2. … … [2] (b) An amplifier circuit incorporating an ideal operational amplifier (op-amp) is used to amplify the output of a microphone. The circuit is shown in Fig. 7.1. +5.0 V P + – 92.5 kΩ –5.0 V V OUT R Fig. 7.1 When the potential at point P is 48 mV, the output potential difference VOUT is 3.6 V. (i) Determine 1. the gain of the amplifier circuit, gain = … [2] 2. the resistance of resistor R. resistance = … Ω [2] (ii) State and explain the effect on the amplifier output when the potential at P exceeds 68 mV. … … … [2] [Total: 10]

10 marks

Mark scheme: 7(a)(i) (part of) the output is combined with the input M1 reference to potential/voltage/signal A1 7(a)(ii) • increased (operating) stability • increased bandwidth/range of frequencies over which gain is constant • less distortion (of output) Any 2 points. B2 7(b)(i) 1. gain = 3.6 / (48 × 10–3) C1 = 75 A1 2. gain = 1 + RF / R 75 = 1 + (92.5 × 103) / R C1 R = 1300 Ω A1 7(b)(ii) for 68 mV, gain × VIN = 5.1 (V) or output voltage would be greater than the supply voltage M1 amplifier would saturate (at 5.0 V) or output voltage = 5.0 (V) A1

This question in 9702/42 Oct/Nov 2017

7 The circuit of an amplifier incorporating an ideal operational amplifier (op-amp) is shown in Fig. 7.1. R2 +9.0 V R1 P – + V IN –9.0 V V OUT Fig. 7.1 (a) By reference to the properties of an ideal op-amp, (i) explain why point P is referred to as a virtual earth, … … … … … [4] (ii) derive an expression, in terms of the resistances R1 and R2, for the gain of the amplifier circuit. [4] R2(b) In the circuit of Fig. 7.1, the ratio is 4.5. R1 The variation with time t of the input potential VIN is shown in Fig. 7.2. 14 12 10 8 6 potential / V 4 2 VINVIN 0 t –2 –4 –6 –8 –10 –12 –14 Fig. 7.2 On Fig. 7.2, show the variation with time t of the output potential VOUT. [3] [Total: 11]

11 marks

Mark scheme: 7(a)(i) gain of amplifier is very large B1 V+ is at earth (potential) B1 for amplifier not to saturate M1 difference between V– and V+ must be very small or V– must be equal to V+ A1 or if V– ≠ V+ then feedback voltage (M1) acts to reduce gap until V– = V+ when stable (A1) 7(a)(ii) input impedance is infinite B1 (so) current in R1 = current in R2 B1 (VIN – 0) / R1 = (0 – VOUT) / R2 B1 (gain =) VOUT / VIN = – R2 / R1 B1 7(b) graph: correct inverted shape (straight diagonal line from (0,0) to a negative potential, then a horizontal line, then a straight diagonal line back to the t-axis at the point where VIN = 0) B1 horizontal line at correct potential of (–)9.0 V B1 both ends of horizontal line occur at correct times (coinciding with when VIN = 2.0 V) B1

This question in 9702/43 Oct/Nov 2017

8 (a) Two properties of an ideal operational amplifier (op-amp) are infinite bandwidth and infinite slew rate. Explain what is meant by (i) infinite bandwidth, … … [1] (ii) infinite slew rate. … … [1] (b) An ideal op-amp is incorporated into the circuit of Fig. 8.1. +4 V T 1.8 kΩ +9 V – + 800 Ω 1.2 kΩ –9 V VOUT Fig. 8.1 (i) Determine the resistance RT of the thermistor T at which the output potential difference VOUT is zero. RT = … Ω [1] (ii) The temperature of the thermistor is gradually increased so that its resistance decreases from 1.5RT to 0.5RT. On Fig. 8.2, draw a line to show the variation of the output potential difference VOUT with the thermistor resistance. 12 VOUT / V 9 6 3 0 0.5RT RT 1.5RT −3 thermistor resistance −6 −9 −12 Fig. 8.2 [2] (iii) On Fig. 8.1, draw the symbol for a light-emitting diode (LED), connected at the output of the circuit, such that it emits light when the resistance of the thermistor is less than RT. [2] [Total: 7]

7 marks

Mark scheme: 8(a)(i) all frequencies have the same gain B1 8(a)(ii) output changes at the same time as input changes B1 8(b)(i) RT / 800 = 1.8 / 1.2 RT = 1200 Ω A1 8(b)(ii) stepped from –9 V to +9 V or v.v. B1 Vout negative < RT and Vout positive > RT B1 8(b)(iii) correct LED symbol with connection between VOUT and earth B1 diode pointing upwards B1

This question in 9702/42 Feb/March 2018

8 (a) Negative feedback is often used in amplifiers incorporating an operational amplifier (op-amp). State (i) what is meant by negative feedback, … … … [2] (ii) two effects of negative feedback on the gain of an amplifier. 1. … … 2. … … [2] (b) An ideal op-amp is incorporated into the amplifier circuit shown in Fig. 8.1. 9600 Ω +6 V 800 Ω – + VIN –6 V VOUT Fig. 8.1 (i) Calculate the gain G of the amplifier circuit. G = … [2] (ii) Determine the output potential difference VOUT for input potential differences VIN of 1. – 0.10 V, VOUT = … V 2. +1.3 V. VOUT = … V [2] (iii) The gain of the amplifier shown in Fig. 8.1 is constant. State one change that can be made to the circuit of Fig. 8.1 so that the amplifier circuit monitors light intensity levels, with the magnitude of the gain decreasing as light intensity increases. … … [1] [Total: 9]

9 marks

Mark scheme: 8(a)(i) (fraction of) output is combined with the input M1 output (fraction) subtracted/deducted from input A1 8(a)(ii) any two valid points e.g.: • greater bandwidth/gain constant over a larger range of frequencies/greater bandwidth • smaller gain B2 8(b)(i) gain = (–)9600 / 800 C1 = –12 A1 8(b)(ii) 1. 1.2 V B1 2. –6 V B1 8(b)(iii) replace the 9600 Ω resistor with an LDR B1

This question in 9702/41 May/June 2018

7 (a) Negative feedback is often used in amplifiers. State (i) what is meant by negative feedback, … … … [2] (ii) two effects of negative feedback on the gain of an amplifier. 1. … … 2. … … [2] (b) An ideal operational amplifier (op-amp) is incorporated into the circuit shown in Fig. 7.1. 6400 Ω +9.0 V – + –9.0 V VOUT VIN 800 Ω Fig. 7.1 (i) Calculate the gain G of the amplifier circuit. G = … [1] (ii) Determine the output potential difference VOUT for an input potential difference VIN of 1. +0.60 V, VOUT = … V 2. –2.1 V. VOUT = … V [2] (iii) The gain of the amplifier shown in Fig. 7.1 is constant. State one change that may be made to the circuit of Fig. 7.1 so that the amplifier circuit monitors temperature with the gain decreasing as the temperature rises. … … [1] [Total: 8]

8 marks

Mark scheme: 7(a)(i) (fraction of) output is combined with the input M1 output (fraction) subtracted/deducted from input A1 7(a)(ii) Any two valid points e.g. • greater bandwidth/gain constant over a larger range of frequencies • smaller gain B2 7(b)(i) gain = 1 + (6400 / 800) = 9.0 A1 7(b)(ii) 1. (+)5.4 V A1 2. –9.0 V A1 7(b)(iii) replace the 6400 Ω resistor with a thermistor B1

This question in 9702/42 May/June 2018

8 (a) Negative feedback is often used in amplifiers incorporating an operational amplifier (op-amp). State (i) what is meant by negative feedback, … … … [2] (ii) two effects of negative feedback on the gain of an amplifier. 1. … … 2. … … [2] (b) An ideal op-amp is incorporated into the amplifier circuit shown in Fig. 8.1. 9600 Ω +6 V 800 Ω – + VIN –6 V VOUT Fig. 8.1 (i) Calculate the gain G of the amplifier circuit. G = … [2] (ii) Determine the output potential difference VOUT for input potential differences VIN of 1. – 0.10 V, VOUT = … V 2. +1.3 V. VOUT = … V [2] (iii) The gain of the amplifier shown in Fig. 8.1 is constant. State one change that can be made to the circuit of Fig. 8.1 so that the amplifier circuit monitors light intensity levels, with the magnitude of the gain decreasing as light intensity increases. … … [1] [Total: 9]

9 marks

Mark scheme: 8(a)(i) (fraction of) output is combined with the input M1 output (fraction) subtracted/deducted from input A1 8(a)(ii) any two valid points e.g.: • greater bandwidth/gain constant over a larger range of frequencies/greater bandwidth • smaller gain B2 8(b)(i) gain = (–)9600 / 800 C1 = –12 A1 8(b)(ii) 1. 1.2 V B1 2. –6 V B1 8(b)(iii) replace the 9600 Ω resistor with an LDR B1

This question in 9702/43 May/June 2018

7 (a) An ideal operational amplifier (op-amp) has infinite bandwidth and infinite slew rate. State what is meant by (i) infinite bandwidth, … … … [2] (ii) infinite slew rate. … … … [2] (b) An incomplete circuit for a non-inverting amplifier incorporating an ideal operational amplifier is shown in Fig. 7.1. +5.0 V – + R1 –5.0 V V IN V OUT R2 Fig. 7.1 On Fig. 7.1, draw lines to show the connections between the components to complete the circuit. [2] (c) The completed amplifier of Fig. 7.1 has a voltage gain of 10. State the output voltage VOUT for an input voltage VIN of (i) –0.36 V, VOUT = … V [1] (ii) 0.56 V. VOUT = … V [1] [Total: 8]

8 marks

Mark scheme: 7(a)(i) gain is constant M1 for all frequencies A1 7(a)(ii) no time delay between input (voltage) and output (voltage) B1 clear reference to change(s) in input and/or output (voltages) B1 7(b) diagram: VIN connected to V+ only B1 midpoint between resistors R1 and R2 connected to V– only B1 7(c)(i) –3.6 V A1 7(c)(ii) (+)5.0 V A1

This question in 9702/41 Oct/Nov 2018

7 A circuit incorporating an ideal operational amplifier (op-amp) is shown in Fig. 7.1. +5.0 V +5.0 V RT 1.8 kΩ – + –5.0 V VOUT R 2.4 kΩ Fig. 7.1 The variation with temperature θ of the resistance RT of the thermistor is shown in Fig. 7.2. 3.4 3.3 Ω RT / k 3.2 3.1 3.0 2 3 4 5 6 7 θ / °C Fig. 7.2 (a) The output potential VOUT of the op-amp circuit changes sign when the temperature of the thermistor is 4.0 °C. Calculate the resistance R. R = … kΩ [2] (b) State and explain whether the output potential VOUT is +5.0 V or −5.0 V for a thermistor temperature of 2.5 °C. … … … … [3] (c) The output of the op-amp is to be displayed using two light-emitting diodes (LEDs) labelled G and B. When the temperature of the thermistor is below 4.0 °C, only the LED labelled G emits light. The LED labelled B emits light only when the temperature of the thermistor is above 4.0 °C. On Fig. 7.1, draw and label the symbols for the two LEDs. [3] [Total: 8]

8 marks

Mark scheme: 7(a) R / RT = 2.4 / 1.8 or at 4.0 °C, RT = 3.2 kΩ C1 hence R / 3.2 = 2.4 / 1.8 R = 4.3 kΩ A1 7(b) RT = 3.37 kΩ or RT is greater (than 3.2 kΩ) B1 V+ > V– M1 hence output is +5.0 V A1 Question Answer Marks 7(c) correct LED symbol B1 two diodes shown connected, in parallel and with opposite polarities, between VOUT and earth M1 diodes labelled to show correct polarities consistent with (b) (G pointing from VOUT to earth and B pointing from earth to VOUT if (b) correct) A1

This question in 9702/42 Oct/Nov 2018

7 (a) An ideal operational amplifier (op-amp) has infinite bandwidth and infinite slew rate. State what is meant by (i) infinite bandwidth, … … … [2] (ii) infinite slew rate. … … … [2] (b) An incomplete circuit for a non-inverting amplifier incorporating an ideal operational amplifier is shown in Fig. 7.1. +5.0 V – + R1 –5.0 V V IN V OUT R2 Fig. 7.1 On Fig. 7.1, draw lines to show the connections between the components to complete the circuit. [2] (c) The completed amplifier of Fig. 7.1 has a voltage gain of 10. State the output voltage VOUT for an input voltage VIN of (i) –0.36 V, VOUT = … V [1] (ii) 0.56 V. VOUT = … V [1] [Total: 8]

8 marks

Mark scheme: 7(a)(i) gain is constant M1 for all frequencies A1 7(a)(ii) no time delay between input (voltage) and output (voltage) B1 clear reference to change(s) in input and/or output (voltages) B1 7(b) diagram: VIN connected to V+ only B1 midpoint between resistors R1 and R2 connected to V– only B1 7(c)(i) –3.6 V A1 7(c)(ii) (+)5.0 V A1

This question in 9702/43 Oct/Nov 2018

7 (a) Two properties that an ideal operational amplifier (op-amp) would have are constant voltage gain and infinite slew rate. State what is meant by: (i) gain of an amplifier … … [1] (ii) infinite slew rate. … … … [2] (b) The partially completed circuit of a non-inverting amplifier, incorporating an ideal op-amp, is shown in Fig. 7.1. R1 +9 V + – VIN –9 V VOUT R2 Fig. 7.1 (i) On Fig. 7.1, complete the circuit for the non-inverting amplifier. [2] (ii) For the completed circuit of Fig. 7.1, the gain of the amplifier is 25. The resistance of resistor R1 is 12 kΩ. Calculate the resistance of resistor R2. resistance = … Ω [2] (iii) Calculate, for the amplifier gain of 25, the range of values of VIN for which the amplifier does not saturate. range from … V to … V [2] [Total: 9]

9 marks

Mark scheme: 7(a)(i) output voltage / input voltage B1 7(a)(ii) no time delay between input and output B1 clear reference to change(s) in input and / or output B1 7(b)(i) VIN only connected to non-inverting input B1 midpoint between R1 and R2 only connected to inverting input B1 Question Answer Marks 7(b)(ii) gain = 1 + (R1 / R2) 25 = 1 + (12 × 103) / R2 C1 R2 = 500 Ω A1 7(b)(iii) VMAX = 9/25 = 0.36 V C1 range is –0.36 V to + 0.36 V A1

This question in 9702/42 Feb/March 2019

7 The circuit for an inverting amplifier incorporating an ideal operational amplifier (op-amp) is shown in Fig. 7.1. 5.2 k 0.80 k +5 V – P + –5 V R VIN VOUT D Fig. 7.1 (a) For the circuit of Fig. 7.1: (i) explain why point P is known as a virtual earth … … … … [3] (ii) calculate the gain of the amplifier. gain = … [2] (b) When the op-amp is saturated, the potential difference across the LED is 2.3 V. Calculate the minimum resistance of resistor R so that the current in the LED is limited to 30 mA. resistance = … Ω [3] [Total: 8]

8 marks

Mark scheme: 7(a)(i) (amplifier) gain is very large/infinite B1 for amplifier not to saturate, V+ = V– or feedback (loop) ensures V+ = V– B1 V+ is at earth/0 V so V– is (almost) at earth/0 V B1 7(a)(ii) gain = (–)5200 / 800 or (–)5.2 / 0.80 C1 = –6.5 A1 Question Answer Marks 7(b) (at saturation,) VOUT = 5 V C1 p.d. across R = 5 – 2.3 = 2.7 (V) C1 resistance = 2.7 / (30 × 10–3) = 90 Ω A1 or Rdiode = 2.3 / 0.030 = 77 Ω Rtotal = 5.0 / 0.030 = 167 Ω (C1) Rresistor (= 167 – 77) = 90 Ω (A1) or Rdiode = 2.3 / 0.030 = 77 Ω 77 / (Rresistor + 77) × 5 = 2.3 (C1) Rresistor = 90 Ω (A1) or Rdiode = 2.3 / 0.030 = 77 Ω Rresistor = 77 × (2.7 / 2.3) (C1) Rresistor = 90 Ω (A1)

This question in 9702/41 May/June 2019

9 Part of a circuit incorporating an operational amplifier (op-amp) is shown in Fig. 9.1. +5 V device – 4.5 V + –5 V Fig. 9.1 (a) A relay is connected to the output of the op-amp circuit so that a lamp may be switched on or off. (i) Complete Fig. 9.1 to show the relay connected into the circuit. [2] (ii) State and explain whether the output of the op-amp is positive or negative for the lamp to be switched on. … … … [2] (b) State the device in Fig. 9.1 that could be used so that the circuit indicates a change in: (i) the bending of a rod … [1] (ii) the level of daylight to switch on a street light. … [1] [Total: 6]

6 marks

Mark scheme: 9(a)(i) relay coil shown connected between diode and earth B1 switch shown connected across lamp B1 9(a)(ii) Any one from: • (for diode to conduct) current flow is into output of op-amp • when earth is at higher potential diode is forward biased • diode blocks current when output positive • diode must conduct M1 so VOUT is negative A1 9(b)(i) strain gauge B1 9(b)(ii) light-dependent resistor B1

This question in 9702/42 May/June 2019

7 The circuit for an inverting amplifier incorporating an ideal operational amplifier (op-amp) is shown in Fig. 7.1. 5.2 k 0.80 k +5 V – P + –5 V R VIN VOUT D Fig. 7.1 (a) For the circuit of Fig. 7.1: (i) explain why point P is known as a virtual earth … … … … [3] (ii) calculate the gain of the amplifier. gain = … [2] (b) When the op-amp is saturated, the potential difference across the LED is 2.3 V. Calculate the minimum resistance of resistor R so that the current in the LED is limited to 30 mA. resistance = … Ω [3] [Total: 8]

8 marks

Mark scheme: 7(a)(i) (amplifier) gain is very large/infinite B1 for amplifier not to saturate, V+ = V– or feedback (loop) ensures V+ = V– B1 V+ is at earth/0 V so V– is (almost) at earth/0 V B1 7(a)(ii) gain = (–)5200 / 800 or (–)5.2 / 0.80 C1 = –6.5 A1 Question Answer Marks 7(b) (at saturation,) VOUT = 5 V C1 p.d. across R = 5 – 2.3 = 2.7 (V) C1 resistance = 2.7 / (30 × 10–3) = 90 Ω A1 or Rdiode = 2.3 / 0.030 = 77 Ω Rtotal = 5.0 / 0.030 = 167 Ω (C1) Rresistor (= 167 – 77) = 90 Ω (A1) or Rdiode = 2.3 / 0.030 = 77 Ω 77 / (Rresistor + 77) × 5 = 2.3 (C1) Rresistor = 90 Ω (A1) or Rdiode = 2.3 / 0.030 = 77 Ω Rresistor = 77 × (2.7 / 2.3) (C1) Rresistor = 90 Ω (A1)

This question in 9702/43 May/June 2019

5 (a) A section of a coaxial cable is shown in Fig. 5.1. copper braid insulation copper wire plastic covering Fig. 5.1 (i) Suggest two functions of the copper braid. 1. … … 2. … … [2] (ii) Suggest one application of a coaxial cable for the transmission of electrical signals. … … [1] (b) (i) The constant noise power in a transmission cable is 7.6 μW. The minimum acceptable signal-to-noise ratio is 32 dB. Calculate the minimum acceptable signal power PMIN in the cable. PMIN = … W [2] (ii) The input power of the signal to the transmission cable is 2.6 W. The attenuation per unit length of the cable is 6.3 dB km–1. Use your answer in (i) to determine the maximum uninterrupted length L of cable along which the signal may be transmitted. L = … km [2] [Total: 7]

7 marks

Mark scheme: 5(a)(i) provides return for the signal B1 shields signal from noise B1 5(a)(ii) e.g. connection between aerial and TV set B1 5(b)(i) gain / dB = 10 lg (P1 / P2) C1 32 = 10 lg {PMIN / (7.6 × 10–6)} PMIN = 0.012 W A1 5(b)(ii) attenuation per unit length = (1 / L) × 10 lg (P1 / P2) 6.3 = (1 / L) × 10 lg (2.6 / 0.012) C1 L = 3.7 km A1

This question in 9702/41 Oct/Nov 2019

7 (a) An ideal operational amplifier (op-amp) has infinite bandwidth and zero output impedance. State what is meant by: (i) infinite bandwidth … … [1] (ii) zero output impedance. … … [1] (b) The circuit for a non-inverting amplifier incorporating an ideal op-amp is shown in Fig. 7.1. 4.0 kΩ +5.0 V – + –5.0 V V OUT V IN 800 Ω R Fig. 7.1 The light-emitting diode (LED) emits light when the potential difference across it is at least 2.0 V. The current in the LED must not be greater than 20 mA. (i) Calculate the gain of the amplifier circuit. gain = … [2] (ii) Determine the value of VIN for which the value of VOUT is +2.0 V. VIN = … V [1] (iii) State the maximum value of the output potential VOUT. maximum potential = … V [1] (iv) When the op-amp is saturated, the potential difference across the LED is 2.2 V. Calculate the minimum resistance of resistor R so that the current in the LED is limited to 20 mA. resistance = … Ω [2] [Total: 8]

8 marks

Mark scheme: 7(a)(i) all frequencies are amplified equally B1 7(a)(ii) no drop in output voltage (when there is a current) B1 7(b)(i) gain = 1 + RF / RIN gain = 1 + (4000 / 800) C1 gain = 6.0 A1 7(b)(ii) 2.0 / VIN = 6.0 VIN = (+)0.33 V A1 7(b)(iii) 5.0 V A1 7(b)(iv) V = 5.0 – 2.2 (= 2.8 V) C1 R = V / I = 2.8 / 0.020 = 140 Ω A1

This question in 9702/41 Oct/Nov 2019

5 (a) A section of a coaxial cable is shown in Fig. 5.1. copper braid insulation copper wire plastic covering Fig. 5.1 (i) Suggest two functions of the copper braid. 1. … … 2. … … [2] (ii) Suggest one application of a coaxial cable for the transmission of electrical signals. … … [1] (b) (i) The constant noise power in a transmission cable is 7.6 μW. The minimum acceptable signal-to-noise ratio is 32 dB. Calculate the minimum acceptable signal power PMIN in the cable. PMIN = … W [2] (ii) The input power of the signal to the transmission cable is 2.6 W. The attenuation per unit length of the cable is 6.3 dB km–1. Use your answer in (i) to determine the maximum uninterrupted length L of cable along which the signal may be transmitted. L = … km [2] [Total: 7]

7 marks

Mark scheme: 5(a)(i) provides return for the signal B1 shields signal from noise B1 5(a)(ii) e.g. connection between aerial and TV set B1 5(b)(i) gain / dB = 10 lg (P1 / P2) C1 32 = 10 lg {PMIN / (7.6 × 10–6)} PMIN = 0.012 W A1 5(b)(ii) attenuation per unit length = (1 / L) × 10 lg (P1 / P2) 6.3 = (1 / L) × 10 lg (2.6 / 0.012) C1 L = 3.7 km A1

This question in 9702/43 Oct/Nov 2019

7 (a) An ideal operational amplifier (op-amp) has infinite bandwidth and zero output impedance. State what is meant by: (i) infinite bandwidth … … [1] (ii) zero output impedance. … … [1] (b) The circuit for a non-inverting amplifier incorporating an ideal op-amp is shown in Fig. 7.1. 4.0 kΩ +5.0 V – + –5.0 V V OUT V IN 800 Ω R Fig. 7.1 The light-emitting diode (LED) emits light when the potential difference across it is at least 2.0 V. The current in the LED must not be greater than 20 mA. (i) Calculate the gain of the amplifier circuit. gain = … [2] (ii) Determine the value of VIN for which the value of VOUT is +2.0 V. VIN = … V [1] (iii) State the maximum value of the output potential VOUT. maximum potential = … V [1] (iv) When the op-amp is saturated, the potential difference across the LED is 2.2 V. Calculate the minimum resistance of resistor R so that the current in the LED is limited to 20 mA. resistance = … Ω [2] [Total: 8]

8 marks

Mark scheme: 7(a)(i) all frequencies are amplified equally B1 7(a)(ii) no drop in output voltage (when there is a current) B1 7(b)(i) gain = 1 + RF / RIN gain = 1 + (4000 / 800) C1 gain = 6.0 A1 7(b)(ii) 2.0 / VIN = 6.0 VIN = (+)0.33 V A1 7(b)(iii) 5.0 V A1 7(b)(iv) V = 5.0 – 2.2 (= 2.8 V) C1 R = V / I = 2.8 / 0.020 = 140 Ω A1

This question in 9702/43 Oct/Nov 2019

8 (a) An ideal operational amplifier (op‑amp) is connected to a load resistor. The op‑amp is assumed to have infinite bandwidth and zero output resistance. State: (i) what is meant by infinite bandwidth … … [1] (ii) the effect, if any, on the output voltage of increasing the load resistance. … … [1] (b) A student designs the circuit shown in Fig. 8.1 in order to indicate changes in temperature of the thermistor T. 100 kΩ T 110 Ω 150 kΩ + Q P 1.5 V – + – 40 Ω V Fig. 8.1 (i) Explain why point P is known as a virtual earth. … … … … [3] (ii) Calculate the potential at point Q. potential = … V [2] (iii) At a temperature of 13 °C, the resistance of the thermistor T is 230 kΩ. Show that the potential difference measured with the voltmeter is 0.88 V. [2] (c) The resistance of the thermistor T in (b) decreases as its temperature rises. Explain the effect of this change in temperature on the potential difference measured with the voltmeter. … … … [2] [Total: 11]

11 marks

Mark scheme: 8(a)(i) constant gain for all frequencies B1 8(a)(ii) unchanged B1 8(b)(i) (open loop) gain of op-amp is infinite B1 feedback loop ensures V+ ≈ V– or any difference between V+ and V– results in saturated output B1 non-inverting input is 0 V so inverting input also at 0 V B1 8(b)(ii) input = (40 × 1.5) / (40 + 110) C1 = 0.40 V A1 8(b)(iii) gain = (–) (100 + 230) / 150 or feedback current = 0.40 / (150 × 103) (A) C1 p.d. = [(100 + 230) / 150] × 0.40 = 0.88 V A1 8(c) (magnitude of) gain decreases M1 voltmeter reading decreases A1

This question in 9702/42 May/June 2020

8 (a) An ideal operational amplifier (op-amp) is said to have infinite bandwidth and infinite slew rate. State what is meant by: (i) infinite bandwidth … … [1] (ii) infinite slew rate. … … [1] (b) An amplifier circuit incorporating an op-amp is shown in Fig. 8.1. R +5.0 V – + –5.0 V VIN 800 Ω VOUT Fig. 8.1 The resistance of resistor R is to be fixed so that, for an input potential difference VIN of 0.40 V, the amplifier is on the point of saturation. Determine: (i) the gain of the amplifier circuit gain = … [2] (ii) the resistance of resistor R. resistance = … Ω [2] [Total: 6]

6 marks

Mark scheme: 8(a)(i) gain is the same for all frequencies B1 8(a)(ii) no (time) delay in change in output when input is changed B1 8(b)(i) (at saturation,) VOUT = 5.0 V C1 gain = 5.0 / 0.40 = 12.5 or 13 A1 8(b)(ii) 12.5 = 1 + (R / 800) C1 R = 9200 Ω A1

This question in 9702/42 Oct/Nov 2020

7 (a) Fig. 7.1 shows the circuit diagram containing an operational amplifier (op-amp). 3.6 kΩ + 3.0 V – + – 3.0 V VOUT VIN 0.72 kΩ Fig. 7.1 (i) State the name of this type of amplifier. … [1] (ii) Show that the gain of the amplifier is 6.0. [1] (iii) At time t = 0 the input potential VIN is zero. VIN then gradually increases with time t as shown in Fig. 7.2. 6 5 potential / V 4 3 2 1 0 T t Fig. 7.2 On Fig. 7.2 sketch a line to show the variation with time t of the output potential VOUT from time t = 0 to time t = T. [2] (iv) State how the circuit of Fig. 7.1 may be changed so that the gain of the amplifier is dependent on light intensity. … … [1] (b) An op-amp is to be used to switch on a high-voltage heater. (i) State the name of the component used as the output device of the op-amp. … [1] (ii) Complete Fig. 7.3 using the device named in (i) and a diode so that the heater may be switched on when the output of the op-amp is positive. + – + – connections to high-voltage heater Fig. 7.3 [3] [Total: 9]

9 marks

Mark scheme: 7(a)(i) non-inverting (amplifier) B1 7(a)(ii) gain 1 f R R = + 3.6 gain 1 6.0 0.72 = + = B1 7(a)(iii) straight line from (0,0) to (T / 2, 3) B1 line from origin to 3.0 V then horizontal line at 3.0 V to T B1 7(a)(iv) ldr / light dependent resistor replaces one of the two resistors B1 7(b)(i) relay coil B1 7(b)(ii) relay coil between op-amp and earth B1 diode with correct polarity (pointing away from output) connected between output and device and no other connections or diode with correct polarity (pointing towards earth) between device and earth and no other connections B1 switch connected to high voltage circuit B1

This question in 9702/42 Feb/March 2021

7 (a) Two properties of an ideal operational amplifier (op-amp) are infinite input impedance and infinite bandwidth. State what is meant by: (i) infinite input impedance … … [1] (ii) infinite bandwidth. … … [1] (b) A student uses a negative temperature coefficient thermistor in the circuit of Fig. 7.1 to indicate changes in temperature. 100 kΩ 1100 Ω +5.0 V 96 kΩ 1.5 V X – + 400 Ω –5.0 V V Fig. 7.1 (i) Show that the potential at point X is 0.40 V. [1] (ii) The thermistor has a resistance of 360 kΩ at a particular temperature. For this temperature of the thermistor, calculate the magnitude of the reading on the voltmeter. voltmeter reading = … V [3] (iii) The temperature of the thermistor increases. State and explain the effect of this change on the magnitude of the reading on the voltmeter. … … … [2] (iv) Explain why the amplifier circuit will no longer indicate temperature changes when the magnitude of the gain of the circuit is greater than 12.5. … … [1] [Total: 9]

9 marks

Mark scheme: 7(a)(i) no current enters/leaves the input B1 7(a)(ii) gain is the same for all frequencies B1 7(b)(i) VIN = 1.5 × 400 / (400 + 1100) = 0.40 V or VIN = 1.5 – (1.5 × 1100 / 1500) = 0.40 V or (1.5 – VIN) / 1100 = VIN / 400 so VIN = 0.40 V A1 7(b)(ii) gain = (–) Rf/Ri C1 VOUT/0.40 = (360 + 100) / 96 C1 VOUT = 1.9 V A1 7(b)(iii) resistance of thermistor decreases B1 (magnitude of) gain decreases so reading decreases B1 7(b)(iv) (at gain 12.5) VOUT is 5.0 V, so (above gain 12.5) output becomes saturated B1 Question Answer Marks

This question in 9702/42 May/June 2021

7 (a) State two properties of an ideal operational amplifier (op-amp). 1. . … … 2. . … … [2] (b) Fig. 7.1 shows a circuit that includes an ideal op-amp and two identical resistors R. +5 V R V L – + Y R X –5 V Fig. 7.1 State the names of components X and Y. X: … Y: … [1] (c) (i) Explain why the op-amp in Fig. 7.1 has only two possible output states. … … … … [2] (ii) State the name of the type of op-amp circuit in which the op-amp behaves as in (c)(i). … [1] (iii) Describe the environmental condition under which the lamp L in Fig. 7.1 will light. … … … [2] (iv) Suggest the purpose of the variable resistor V in the circuit. … … [1] [Total: 9]

9 marks

Mark scheme: 7(a) • infinite (open-loop) gain • infinite slew rate • infinite input impedance • zero output impedance • infinite bandwidth Any two points, 1 mark each B2 7(b) X: thermistor and Y: relay B1 7(c)(i) (any) difference in voltage at the inputs causes output to saturate (because gain is very large) B1 saturates positively if V+ > V– and saturates negatively if V+ < V– B1 7(c)(ii) comparator B1 7(c)(iii) temperature M1 above a particular value A1 7(c)(iv) to adjust the temperature (at which the lamp illuminates/extinguishes) B1

This question in 9702/41 Oct/Nov 2021

7 (a) An operational amplifier (op-amp) has two input terminals and one output terminal. State what is meant by the gain of an op-amp. … … … [2] (b) State two effects of negative feedback on the gain of an amplifier circuit that uses an op-amp. 1. … … 2. … … [2] (c) Fig. 7.1 shows an op-amp circuit that uses negative feedback. 1.2 kΩ +8.0 V 480 Ω VIN – VOUT + –8.0 V 0 V 0 V Fig. 7.1 (i) State the name of the type of circuit shown in Fig. 7.1. … [1] (ii) On Fig. 7.1, label with the letter X a point in the circuit that is considered to be a virtual earth. [1] (iii) Calculate the gain of the circuit in Fig. 7.1. gain = … [2] (iv) Determine the value of VIN when VOUT is +6.5 V. VIN = … V [1] (v) Determine the value of VOUT when VIN is –5.4 V. VOUT = … V [1] [Total: 10]

10 marks

Mark scheme: 7(a) output voltage / input voltage M1 input (voltage) is difference between (inverting and non-inverting) inputs A1 7(b) • reduces the gain • greater bandwidth • more stable Any two points, 1 mark each B2 7(c)(i) inverting amplifier B1 7(c)(ii) X marked anywhere between right-hand edge of 480 Ω resistor, left-hand edge of 1.2 kΩ resistor and the inverting input B1 7(c)(iii) gain = (–)Rf / Ri C1 = (–)1200 / 480 = –2.5 A1 7(c)(iv) VIN = 6.5 / (–2.5) = –2.6 V A1 7(c)(v) (–2.5) × (–5.4) = +13.5 V, and so output saturates VOUT = (+)8.0 V A1

This question in 9702/42 Oct/Nov 2021

7 (a) State two properties of an ideal operational amplifier (op-amp). 1. . … … 2. . … … [2] (b) Fig. 7.1 shows a circuit that includes an ideal op-amp and two identical resistors R. +5 V R V L – + Y R X –5 V Fig. 7.1 State the names of components X and Y. X: … Y: … [1] (c) (i) Explain why the op-amp in Fig. 7.1 has only two possible output states. … … … … [2] (ii) State the name of the type of op-amp circuit in which the op-amp behaves as in (c)(i). … [1] (iii) Describe the environmental condition under which the lamp L in Fig. 7.1 will light. … … … [2] (iv) Suggest the purpose of the variable resistor V in the circuit. … … [1] [Total: 9]

9 marks

Mark scheme: 7(a) • infinite (open-loop) gain • infinite slew rate • infinite input impedance • zero output impedance • infinite bandwidth Any two points, 1 mark each B2 7(b) X: thermistor and Y: relay B1 7(c)(i) (any) difference in voltage at the inputs causes output to saturate (because gain is very large) B1 saturates positively if V+ > V– and saturates negatively if V+ < V– B1 7(c)(ii) comparator B1 7(c)(iii) temperature M1 above a particular value A1 7(c)(iv) to adjust the temperature (at which the lamp illuminates/extinguishes) B1

This question in 9702/43 Oct/Nov 2021