_______________________

Name:
_
Devices
_______________________
Class:
_

_______________________
Date:
_

Time: 675 min.

Marks: 532 marks

Comments:

Page 1 of 47
Q1.
(a) Two logic inputs, A and B, feed into the logic circuit shown in Figure 1. The logic
output from the circuit is Q.

Figure 1

Deduce the Boolean expression for the output of this logic circuit in terms of inputs
A and B.
Include all the logic operations that take place between the inputs and the output.

Q = ____________________
(2)

(b) The truth table shows some of the logic states for the logic gates in Figure 1.

Complete the truth table.

B A C D E F Q

0 0 1 0 0

0 1 1 1 1

1 0 0 0 1

1 1 0 0 0
(2)

(c) Figure 2 shows a different logic circuit that produces the same logic output as that
of Figure 1.

Figure 2

Page 2 of 47
 A manufacturer wants to produce a system that uses this logic function, but is
undecided as to which circuit to use.

Suggest, giving reasons, two benefits of using the logic circuit in Figure 2 compared
to the logic circuit in Figure 1.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)

(d) State the single logic gate that would perform the same logic function as the circuits
shown in Figure 1 and Figure 2.

___________________________________________________________________
(1)
(Total 7 marks)

Q2.
(a) An ultrasound sensor produces an output that needs to be amplified to 3.0 V
The amplifier used has a voltage gain of 40

Calculate the input voltage Vin to the amplifier from the sensor.

Vin = ____________________ V
(1)

(b) An operational amplifier in non-inverting mode is used to amplify the output of the
sensor. The partially completed circuit diagram is shown below.

Page 3 of 47
 Complete the circuit diagram above by adding and labelling two resistors, Rin and
Rf, so that the operational amplifier is correctly configured in its non-inverting
mode.

The power lines should not be shown in the completed diagram.

(2)

(c) Determine, using resistors selected from the list below, how the voltage gain of 40
can be achieved by the non-inverting amplifier of the diagram.

1 kΩ 3.6 kΩ 10 kΩ 39 kΩ 150 kΩ

Rin = ____________________ kΩ

Rf = ____________________ kΩ
(2)

(d) The ultrasound frequency detected by the sensor is 50 kHz

For this operational amplifier

gain × bandwidth = 1.0 MHz

Discuss whether this operational amplifier is suitable for amplifying the sensor’s
output voltage.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Page 4 of 47
 ___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)
(Total 7 marks)

Q3.
Figure 1 shows an operational amplifier used as an inverting amplifier.

Figure 1

(a) Label Figure 1 with an X to show the point which is a virtual earth.
(1)

(b) Name the input pin shown by a (+) on the operational amplifier.

___________________________________________________________________
(1)

(c) Derive the expression for the inverting amplifier gain

(2)

(d) Figure 2 shows the inverting amplifier modified to make a summing amplifier that is
to form part of a two-channel audio mixer.

Figure 2

Page 5 of 47
 Calculate the voltage gain produced by channel 1.

voltage gain (channel 1) = ____________________

(1)

(e) The mixer is tested using the input signals to channels 1 and 2 with the amplitudes
shown in Figure 2.

Calculate the amplitude of the output voltage Vout produced in the test.

Vout = ____________________ V
(2)

(f) Describe how the function of the audio mixer could be improved by changing the
two input resistors from fixed values to variable values.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(1)
(Total 8 marks)

Q4.
A die, where dots on the faces of a cube indicate the numbers 1 to 6, is shown in Figure 1
and is used in many games.

Figure 1

Page 6 of 47
A student makes an electronic version of this by feeding pulses from a pulse generator
into a 4-bit binary counter.

The circuit uses the first three outputs of the counter A (least significant bit), B and C.

By feeding the outputs from the counter through logic gates, the seven LEDs shown in
Figure 2 can be made to display the numbers 1 to 6 in sequence.

Figure 2

Figure 3 shows the sequence of numbers.

Figure 3

The black dots show which LEDs are lit for

each of the numbers 1 to 6.

The partially completed truth table below shows which of the LEDs
(L1 to L6) are ON (logic 1) and which are OFF (logic 0) during the counting sequence.

Number Logic inputs Logic outputs

shown on
C B A L1 L2 L3 L4 L5 L6 L7
die
1 0 0 0 0 0 0 0 1
2 0 0 1 0 0 0 0 0
3 0 1 0 0 0 0 0 1
4 0 1 1 0 1 1 0 0
5 1 0 0 0 1 1 0 1
6 1 0 1 1 1 1 1 0
Reset
6→1

(a) Complete the table to show the logic outputs for the lamps L1 and L6.
(2)

(b) Deduce the simplest Boolean expression that can be used to show how output L7
can be controlled by the logic inputs.

Page 7 of 47
 ___________________________________________________________________
(1)

(c) Figure 4 shows some of the input and output pins of the 4-bit binary counter.

Figure 4

The data sheet for the counter indicates that the counter resets when the reset pin R
is taken from logic 0 to logic 1.

Draw on Figure 4 the logic gate needed and the connections required from the
outputs to the reset pin R on the counter so that the counter cycles as required.
(2)

(d) The output of both L3 and L4 can be written as (A.B.C̅ ) + (B̅ .C)

Figure 5 shows part of a logic circuit needed to represent this Boolean expression.

Complete the logic circuit in Figure 5 by adding AND, OR and NOT gates.

Figure 5

Page 8 of 47
 (3)
(Total 8 marks)

Q5.
Figure 1 shows the first-stage filter circuit for a simple AM receiver. The circuit can be
adjusted to resonate at 910 kHz so that it can receive a particular radio station.

Figure 1

(a) Calculate the value of the capacitance when the circuit resonates at a frequency of
910 kHz.

Page 9 of 47
 capacitance = ____________________ pF
(2)

(b) Draw on Figure 2 an ideal response curve for the resonant circuit, labelling all
relevant frequency values based upon a 10 kHz bandwidth.

Figure 2

(3)

(c) The Q-factor for the practical tuning circuit has a smaller value than the ideal one
assumed in question (b).

Discuss the changes the listener might notice when tuning to this station due to the
practical Q-factor being smaller.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)
(Total 7 marks)

Q6.
A photodiode forms part of a light meter used for checking light levels in an office. Figure
1 shows the circuit diagram for the light meter.

Page 10 of 47
 Figure 1

(a) State the mode in which the photodiode is being used in Figure 1.

___________________________________________________________________
(1)

(b) In which mode is the operational amplifier being used in Figure 1?

Tick (✔) the correct box.

Non-inverting amplifier

Comparator

Summing amplifier

Difference amplifier

(1)

(c) Figure 2 shows an extract from a data sheet of the characteristics for a photodiode
under different light levels measured in lux.

Figure 2

Page 11 of 47
 For a particular lighting condition, the current through the photodiode in Figure 1
was 0.10 mA.

Estimate, using the information in Figure 2, the light level needed to cause this
reverse current through the photodiode.

light level = ____________________ lux

(1)

(d) Calculate the voltage at point X in the circuit shown in Figure 1 for the light level in
question (c).

voltage = ____________________ V
(1)

(e) The 10kΩ linear potential divider shown in Figure 1 is set to give 1.75 V at point Y.

Assume that the operational amplifier has ideal characteristics.

Deduce whether the output LED would be switched ON or OFF when the current
through the photodiode is 0.10 mA.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)
(Total 6 marks)

Q7.
The diagram shows part of the motor from a computer disk drive.

Page 12 of 47
On each rotation a small magnet passes a Hall-effect sensor H which detects the change
in magnetic field and produces an output potential difference (pd) that varies with time as
shown in Figure 1.

Figure 1

(a) Determine the speed of the motor in revolutions per second.

speed of motor = ____________ rev s−1

(2)

(b) Explain why the output from the magnetic field sensor is unsuitable to be applied
directly to the logic circuit of the motor controller.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(1)

Page 13 of 47
 (c) The signal from the magnetic field sensor is applied to the circuit shown in Figure 2.

Figure 2

Show that the potential of point A is about +3 V.

(1)

(d) Draw on Figure 1 the waveform showing the variation with time of the output
voltage Vout of the circuit shown in Figure 2.
(2)

Q8.
The diagram shows a constant-current generator connected to a 22 MΩ resistor.

(a) The output from the current generator is 5.0 × 10−10 A.

Show that the magnitude of the pd across the 22 MΩ resistor is about 10 mV.

Page 14 of 47
 (1)

(b) A student tries to measure this voltage with a digital voltmeter set on the 200 m V
range. The resistance of the meter on this range is 1.0 MΩ.

Explain why the meter does not read 10 mV.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)

(c) The student then decides to use a non-inverting operational amplifier to interface to
the digital voltmeter.

Discuss the important property that makes this configuration a suitable choice.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)

(d) Figure 1 shows a partly-drawn circuit diagram for the non-inverting amplifier.

Figure 1

Complete Figure 1 to form a non-inverting amplifier by adding two labelled resistors

Page 15 of 47
 and any connections that are needed.
(2)

(e) The values of the two resistors added to Figure 1 are to give the non-inverting
amplifier a voltage gain of 20. The resistance of the smaller of these two resistors is
500 kΩ.

Which of the following is the correct resistance for the other resistor?
Tick (✔) the correct answer in the right-hand column.

✔ if correct

1.0 MΩ

9.5 MΩ

10.0 MΩ

10.5 MΩ
(1)

(f) Justify your choice of resistance value by calculation.

(1)

Q9.
A student needs to monitor the temperature of a pond using a remote link. To provide the
link the student decides to use radio transmitter and receiver modules.

The radio modules available to the student are available with either amplitude modulation
(AM) or with frequency modulation (FM).

(a) Describe what is meant by amplitude-modulated and by frequency-modulated radio

waves.

In your answer you should:

• indicate the principal features of each of these two types of modulation

• explain the differences between them
• refer to the advantages and disadvantages of using each type of modulation.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Page 16 of 47
 ___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(6)

(b) The radio modules chosen by the student transmit a signal that is amplitude
modulated having a carrier frequency of 434 MHz. The data sheet for the modules
specifies quarter-wave aerials for both the transmitter and receiver.

Calculate the length of a quarter-wave aerial.

length of aerial = ____________ m

(2)

(c) The output from the temperature sensor in the pond is connected to an astable
oscillator. The frequency f (in Hz) of the astable varies with temperature θ (in °C)
according to the equation

f = 7.3(6θ + 273)

During the course of the year the temperature of the pond varies between 4°C and
22°C. The highest frequency produced by the astable is 3.0 kHz.

Calculate the lowest frequency in kHz produced by the astable.

lowest frequency = ____________ kHz

(1)

(d) Deduce the maximum bandwidth required by the AM carrier wave produced by the
radio transmitter as the signal from the astable is transmitted.

Page 17 of 47
 (2)
(Total 11 marks)

Q10.
(a) Draw connections on Figure 1 to show how the 4-bit binary counter could be used
to form a binary coded decimal (BCD) counter.

Figure 1

(2)

An automatic bread maker has 10 discrete processes. These are shown in the following
table. The processes are controlled by the BCD counter.

Process
Process Action
number

0 OFF add ingredients

1 mix ingredients motor on

2 warm ingredients heater on

3 add yeast add ingredients

4 mix dough motor on

5 warm ingredients heater on

6 mix dough motor on

7 warm ingredients heater on

8 cook heater on

alert that bread is

9 switch heater OFF
cooked

(b) Identify and write down the process numbers from the table that correspond to each
of the actions when the heater is on. Convert each of these numbers into their
binary values.

___________________________________________________________________

Page 18 of 47
 ___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(1)

(c) Show, by reference to the labelled terminals on the counter in Figure 1, that the
Boolean expression for the conditions when the heater is on is

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)

Q11.
A student experiments with the circuit shown in Figure 1.

Figure 1

Y is a NAND gate for which an input of less than 2.5 V is logic state 0 and input greater
than 2.5 V is logic state 1.

Switch S is set to position A for a long time so that the capacitor is uncharged.

(a) Explain why the output of Y is logic state 1.

___________________________________________________________________

___________________________________________________________________

Page 19 of 47
 ___________________________________________________________________
(1)

(b) When S is moved to position B, the capacitor charges through the 100 kΩ resistor.

Show that the capacitor will be charged to 2.5 V after about 7 s.

(2)

(c) Suggest an application for a circuit such as that shown in Figure 1.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(1)
(Total 4 marks)

Q29.
A fridge is fitted with a temperature-sensing unit to indicate whether the temperature
inside the fridge is too high, too low, or at a safe temperature.
The system consists of a temperature sensor that produces a 2−bit binary output, a logic
circuit and a low current, common cathode 7−segment display.
Figure 1 shows a block diagram of the system.

Figure 1

Table 1 shows the operation of the system.

Table 1

Page 20 of 47
(a) Complete Table 2 to show the logic signals required on lines a to g to display the
specified characters.

Table 2

(3)

(b) Circle the single logic gate which would generate the required signal for segment a.

AND EXOR OR NAND NOR NOT

(1)

(c) The LEDs in the 7−segment display must be protected by current limiting resistors.
Figure 2 shows two methods, A and B, of connecting current limiting resistors.

Figure 2

Page 21 of 47
 (i) State one disadvantage of method A.

______________________________________________________________

______________________________________________________________
(1)

(ii) Calculate the value of the current limiting resistors required in method B to
limit the current in each segment to 20 mA.
Assume the voltage from the logic circuit is 5 V and the forward voltage drop
across each LED in the 7−segment display is 2.2 V.

______________________________________________________________

______________________________________________________________

______________________________________________________________
(2)

(iii) Circle the appropriate value for these resistors from the following list of E24
resistors.

110 Ω 150 Ω 270 Ω 1.1 kΩ 1.5 kΩ

(1)
(Total 8 marks)

Q30.
A Zener diode is used to produce a stabilized 5.1 V from an unregulated 12 V supply to
power a project that requires 80 mA.
Part of the circuit is shown in the diagram.

Page 22 of 47
(a) Draw on the diagram the Zener diode connected correctly between points X and Y.
(2)

(b) The Zener diode requires at least 5 mA to maintain its Zener voltage of 5.1 V.

(i) Calculate the minimum current flowing through R when switch S is closed.

______________________________________________________________
(1)

(ii) Calculate the voltage across resistor R under these conditions.

______________________________________________________________

______________________________________________________________
(1)

(iii) Calculate the value of resistor R.

______________________________________________________________

______________________________________________________________

______________________________________________________________
(2)

(c) The circuit in the diagram above is now constructed using a value of 75 Ω for
resistor R.

(i) Show that the power dissipated in the resistor is approximately 0.6 W.

______________________________________________________________

______________________________________________________________

______________________________________________________________
(2)

(ii) The project is disconnected by turning switch S off, but the 12 V supply

Page 23 of 47
 remains connected.

Calculate the current that now flows through the Zener diode.

______________________________________________________________

______________________________________________________________

______________________________________________________________
(2)
(Total 10 marks)

Q31.
The diagram shows a logic circuit with three inputs A, B and C.

(a) Write the Boolean expressions for the signals at the intermediate points D, E, and
G in terms of the inputs A, B and C only.

D _________________________________________________________________

E _________________________________________________________________

G _________________________________________________________________
(3)

(b) Complete the truth table below for the logic signals at the intermediate points D, E
and G.

Inputs Intermediate points

C B A D E G

0 0 0

0 0 1

0 1 0

0 1 1

Page 24 of 47
 1 0 0

1 0 1

1 1 0

1 1 1
(5)
(Total 8 marks)

Q32.
An LDR is being used as a light sensor in a system that will switch on a porch light when it
gets dark.
The characteristic for the LDR is shown in Figure 1.

Figure 1

(a) (i) Explain how the use of the logarithmic scale in Figure 1 is helpful when
displaying this characteristic.

______________________________________________________________

______________________________________________________________
(1)

(ii) The LDR has a resistance of 60 kΩ when the light level causes the system to
switch on the porch light.

Page 25 of 47
 State the value of this light level by reading from the graph in Figure 1.

light level ____________________ lux

(1)

(b) Figure 2 shows the circuit for detecting the light level.
The design makes use of an op-amp acting as a comparator.
A red LED acts as an output indicator to aid testing of the detector circuit.

Figure 2

Draw on Figure 2 the connections from points X and Y to the op-amp inputs so
that the red LED switches on when the light level falls below the required value.
(1)

(c) (i) Calculate the voltage at point X when the red LED switches on.

______________________________________________________________

______________________________________________________________

______________________________________________________________
(2)

(ii) The reference voltage at Y is produced by two fixed-value resistors.

Calculate the value for resistor R1 in order to achieve the required circuit
operation.

______________________________________________________________

______________________________________________________________

______________________________________________________________
(2)

(d) The red LED was found to stay on dimly even when the light level was well above
the value expected to switch it off.

Explain why this might happen and how the problem could be solved.

Page 26 of 47
 ___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)
(Total 10 marks)

Q33.
Figure 1 shows a simplified diagram of a road safety system for traffic travelling towards a
road tunnel. The tunnel is too narrow for two-way traffic and too low for lorries.

Figure 1

C and L are laser beam sensors placed at different heights on the road just before the
tunnel. When a beam is broken, the sensor produces a logic 1.

Cars will break the beam at sensor C only. Lorries will break the beams at both sensor L
and sensor C.

M is an electronic message display that tells lorries to take a diversion. The message
display lights up when it receives a logic 1.

T is a sensor buried in the road inside the tunnel. It produces a logic 1 when an oncoming
car is in the tunnel.

The red stop light R comes on when a lorry is detected or when there is an oncoming car
in the tunnel. R will light up when it receives a logic 1.

The green go light G comes on when a car is detected and there are no oncoming cars in
the tunnel. G will light up when it receives a logic 1.

(a) Complete the truth table.

Some of the data has already been entered for you.

Input Output

Page 27 of 47
 Message Red Green
Sensor Sensor Sensor display stop go light
T C L M light G
R

0 0 0 0 0 1

0 0 1 0 0 1

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 0 1 0

1 1 0 1

1 1 1 1
(4)

(b) Write the simplest Boolean expression for the red stop light R in terms of T, C and L.

___________________________________________________________________
(2)

(c) The expression for the green go light G could be written as G =

Draw on Figure 2 the logic diagram for this expression using only NOT, AND and
OR gates.

Figure 2

(3)
(Total 9 marks)

Q34.
Figure 1 shows an inverting op-amp amplifier subsystem.

Figure 1

Page 28 of 47
(a) (i) Write in the box the letter that corresponds to the virtual earth point in Figure
1.

(1)

(ii) Explain the meaning of the term virtual earth point.

______________________________________________________________

______________________________________________________________
(2)

(iii) State the input resistance of this amplifier subsystem.

______________________________________________________________
(1)

(b) Calculate the value of Rf needed to give the amplifier subsystem a voltage gain of −
47.

___________________________________________________________________

___________________________________________________________________
(3)

(c) The amplifier subsystem in part (b) is used to increase the signal voltage from an
electric guitar.

The voltage from the guitar to the amplifier input is shown in Figure 2.

Draw onto the lower part of Figure 2 the output signal from the amplifier subsystem.

Figure 2

Page 29 of 47
 (4)
(Total 11 marks)

Page 30 of 47
Mark schemes

Q1.
(a) Q = (A̅ .B) + (A.B̅ ) ✔✔ (allow written format)
Do not allow (A ⊕ B)
• Correct two terms ✔
• Correct operator for OR gate ✔
2

(b)
B A C D E F Q

0 0 1 1 0 0 0

0 1 0 1 0 1 1

1 0 1 0 1 0 1

1 1 0 0 0 0 0

✔ ✔
Correct column C
Correct column E
2

(c) Any two criteria ✔✔

Less complex circuit – easier to manufacture

Only uses one type of chip – more economical to buy
Uses fewer ICs so saves space
Uses fewer ICs so saves on power consumption
Accept any other valid reason
2

(d) EOR ✔
Also allow EXOR, XOR
1
[7]

Q2.
(a) Voltage in = Voltage out / Voltage gain

= 3 V / 40

= 75 × 10–3 V ✔
1

Page 31 of 47
 (b)
Two resistor chain, correctly labelled connected between
output and ground
Inverting input connected to mid-point of resistor chain
2

(c)

calculation to give resistor ratio of 39 ✔

Rin = 1 kΩ ; Rf = 39 kΩ ✔
2

(d) Desired gain × bandwidth is 40 × 50 kHz = 2 MHz ✔

The Op Amp can only provide ½ the amplification needed. Not suitable. ✔
1 mark - relevant calculation
1 mark - reference to only providing ½ require amplification /
gain so not suitable
2
[7]

Q3.

(a)
Correct position of X:
1

(b) The non-inverting input

(non-inverting)
1

(c) I = (Vin – Vx) / Rin = (Vx – Vout) / Rf

But Vx = 0 V (a virtual earth)
I = Vin / Rin = – Vout / Rf

Page 32 of 47
 Making use of: Iin = - IF

Making use of virtual earth concept

2

(d) Voltage gain (Channel 1) = – Rf / Rin 1

–(150 kΩ / 7.5 kΩ)

–20
Both number and sign must be correct
1

(e) Vout = – Rf (Vin Ch1 / R1 + Vin Ch2 / R2)

= – 150 kΩ (( 15mV / 7.5kΩ) + (–100 mV / 30 kΩ))

= – ((0.3) + (–0.5)) = 0.2 Volts

Evidence of correct method
Answer and correct sign
1

(f) By using variable resistors

The gain can easily be changed

or
the relative levels of the two channels can be set
or
the required balance between the two signals can be made
One relevant point made
1
[6]

Q4.
(a)
Number
shown on Logic inputs Logic outputs
die
C B A L1 L2 L3 L4 L5 L6 L7
1 0 0 0 0 0 0 0 0 0 1
2 0 0 1 1 0 0 0 0 1 0
3 0 1 0 1 0 0 0 0 1 1
4 0 1 1 1 0 1 1 0 1 0
5 1 0 0 1 0 1 1 0 1 1
6 1 0 1 1 1 1 1 1 1 0
Reset
6→1
One mark for each full pattern of L1 and L6:
2

(b) L7 = NOT A; Accept: L7 = 𝐴̅

1

Page 33 of 47
 (c)
1 mark for reset condition from B and C
1 mark for use of a single 2-input AND gate
(accept correct implementation of the full reset code A̅ .B.C
for 1 mark)
2

(d)
1 mark - NOT gate from B:
1 mark - AND gate from B̅ and C:
1 mark - OR gate connecting the two conditions:
3
[8]

Q5.
(a) f = 1 / (2π √LC)
C = 1/ f24π2L
C = 1/ (910 × 103)2 × 4 × π2 × 1.1 × 10–3
C = 27.8 pF (accept 28pF)

Page 34 of 47
 Formula with correct substitution / evidence of correct
working
Answer
1
1

(b)
General shape around f0 and to max of 1.0 on relative
voltage gain axis
1
10 kHz bandwidth
at 0.71 gain
1
Frequencies (905 – 910 – 915) kHz (identified / used)
1

(c) Smaller Q factor leads to:

(Any two from)

(i) Broader bandwidth
(ii) More noise / (hiss) detected
(iii) Less selectivity
(iv) More susceptible to crosstalk from neighbouring stations
on the frequency spectrum.
(v) Less gain due to energy loss / loss of signal detail
2
[7]

Q6.
(a) Photoconductive (accept reverse bias)
1

(b)
Tick (✔) if
correct
Non-inverting amplifier
Comparator ✔
Summing amplifier
Difference amplifier
1

(c) Light level ~ 1000 lux +/- 10%

1

Page 35 of 47
 (d) Vx = IR; Vx = 100 μA × 20 kΩ = 2 V
1

(e) Rule that if V– > V+ then Vout is 0 V (low)

1
Voltage drop across LED so LED is ON
Do not allow LED is ON if supported by incorrect reason
1
[6]

Q7.
(a) Period T ( = 11.3 − 3.2) = 8.1 ± 0.2 ms.✔

Number of revolutions per second = 120 (± 10) (rev s−1).✔

2

(b) Signal is noisy OR output is not at appropriate levels for logic circuit.✔
1

(c) ✔
1

(d) Output pd switches between 0 and +5V.✔

Output pd is inverted compared to input pd.✔
Waveform has vertical sides cutting curve at about 3V.✔
MAX 2
[6]

Q8.
(a) VB (= IR = 5 × 10−10 × 22 × 106) = 0.011 V (11 mV)✔
1

(b) Resistance of voltmeter and 22 MΩ in parallel ≈ 1MΩ ✔

Current of 5 × 10−10 A in this combination gives pd ≈ 0.5 mV✔

Qualitative answer can score both marks.
2

(c) Operational amplifier has very large input resistance.✔

This ensures that the 22 MΩ resistor is not shunted by a lower resistance
or ensures that (almost) all the current is in resistor rather than shared.✔
2

(d) Two resistors labelled e.g. R1 and RF in correct connections.✔✔

i.e. RF between –ve input and output, R1 between 0V and –ve
input.
1 mark only if resistors are labelled wrong way round or are
unlabelled.
2

(e) 9.5 MΩ✔

1

Page 36 of 47
 (f)
from which 10 = 0.5 + RF (giving RF = 9.5 M Ω)✔
1
[9]

Q9.
(a) The mark scheme gives some guidance as to what statements are expected to
be seen in a 1 or 2 mark (L1), 3 or 4 mark (L2) and 5 or 6 mark (L3) answer.
Guidance provided in section 3.10 of the ‘Mark Scheme Instructions’
document should be used to assist marking this question.

L3 All three bullet points of The student presents

5-6 the question are relevant information
marks addressed in good coherently, employing
detail. structure, style and
sp&g to render
meaning clear. The text
is legible.

L2 The answer includes The student presents

3-4 some discussion of two relevant information
marks or three of the three and in a way which
bullet points. assists the
communication of
meaning. The text is
legible. Sp&g are
sufficiently accurate not
to obscure meaning.

L1 The answer addresses The student presents

1-2 one bullet point in some some relevant
marks detail but the others information in a simple
may be neglected. form. The text is usually
legible. Sp&g allow
meaning to be derived
although errors are
sometimes obstructive.

0 marks Little or no discussion The student’s

of relevant content. presentation, spelling,
punctuation and
grammar seriously
obstruct understanding.
The following statements could be present:
• description of variable amplitude of AM carrier wave
• description of variable wavelength of FM carrier wave.
For AM :
• amplitude of carrier wave varies at frequency of
superimposed signal whilst frequency (or wavelength)
of carrier wave is constant
• advantages of AM:
• modulation is simpler to produce so equipment is
less complex

Page 37 of 47
 • fewer transmitters needed
• disadvantages of AM
• more subject to interference and distortion
• waveband has limited capacity.
For FM:
• frequency of carrier wave varies at frequency of
superimposed signal whilst amplitude of carrier wave is
constant
• advantages of FM:
• less prone to interference
• clearer reception
• waveband has greater capacity
• disadvantages of FM:
• signal less diffracted because of shorter
wavelength so more difficult to receive
• require more transmitters.
6

(b) ✔

Length of aerial = 0.173 (m).✔

2

(c) At 4 °C f = 7.3(6θ + 273) gives f = 7.3(6 × 4 + 273) = 2170 Hz = 2.2 (kHz).✔

1

(d) Bandwidth calculation based on higher frequency signal fH i.e 3.0 kHz ✔ required
bandwidth = 2 fH = 6.0 kHz.✔
2
[11]

Q10.
(a) Connections shown from QB and QD to AND gate inputs.✔
Connection shown from output of AND gate to re-set terminal R of counter.✔
2

(b) Process numbers identified as 2, 5, 7, 8 and converted to 10, 101, 111, 1000
respectively.✔
1

(c) QA, QB, QC and QD [or A, B, C, D] identified with 1, 2, 4 and 8 respectively [or with 1,
10, 100, 1000 or 20, 21, 22 and 23].✔

Interpretation of expression as

D̅ .C̅ .B.A̅ meaning (0+0+2+0) = 2

OR D̅ .C.B̅ .A meaning (0+4+0+1) = 5

OR D̅ .C.B.A meaning (0+4+2+1) = 7

OR D.C̅ .B̅ .A̅ meaning (8+0+0+0) = 8

(i.e. 2 OR 5 OR 7 OR 8).✔

Page 38 of 47
 Allow 1 mark for confusing whether A or D is the least
confusing bit.
Allow ECF from incorrect process numbers in part (b) but
only allow MAX 1 if 3 process numbers or fewer are
identified.
2
[5]

Q11.
(a) No pd across 100 kΩ (since no current through it) so both inputs are 0 V / logic 0
and output there for logic 1.✔
1

(b) ✔

Solution gives t = 6.93 s.✔

2

(c) Any appropriate time delay application that leaves a device switched ON or OFF for
a short time after activation or deactivation.✔
e.g. Car interior light as driver leaves vehicle, intruder alarm
exit delay.
1
[4]

Q29.
(a)
A B a b c d e f g Display

0 0 0 0 0 1 1 1 0 L

0 1 1 0 1 1 0 1 1 S

1 0 1 0 1 1 0 1 1 S

1 1 0 1 1 0 1 1 1 H

1 mark for row L

1 mark for row S (both)
1 mark for row H
3

(b) EXOR gate

1

(c) (i) Different combinations produce different brightness

1 Disadvantage
1

(ii) R=V/I; (5V − 2.2V) / 20mA; 2.8V / 20mA = 140Ω

1 mark for 2.8V drop

Page 39 of 47
 1 mark for answer
2

(iii) E24 = 150Ω

1 mark for answer
1
(8)

Q30.
(a)

1 mark for Zener symbol

1 mark for orientation
2

(b) (i) 80mA + 5mA = 85mA

Answer − 1
1

(ii) 12V − 5.1 V = 6.9V

Calculation and answer − 1
1

(iii) R = 6.9V / 85mA = 81Ω

Calculation and answer − 2
2

(c) (i) P = V2 / RP = (6.9 × 6.9) / 75P = 0.64W

Hence P is approx. 0.6W
Calculation and answer − 2
2

(ii) I=V/R I = 6.9 / 75 I = 92mA

Calculation and answer − 2
2
[10]

Q31.
(a) D=C+B
1
E=
1
G=

Page 40 of 47
 1
3

(b)
INPUTS INTERMEDIATE OUTPUTS

C B A D E G

0 0 0 0 1 1

0 0 1 0 0 0

0 1 0 1 1 0

0 1 1 1 0 0

1 0 0 1 1 1

1 0 1 1 0 0

1 1 0 1 1 0

1 1 1 1 0 0

2 marks for each of correct columns D & G

1 mark for column E
5
(Total 8 marks)

Q32.
(a) (i) Log graph enables a wide range of values to be displayed on the same axis.

Allow − (enables values to be displayed as straight line)

1

(ii) 7 lux
1

(b)

1 mark for connections correct way round

1

(c) (i) 60 kΩ / (60kΩ + 30kΩ)) × 12V = 8V

Working − 1
Answer − 1

Page 41 of 47
 2

(ii) R1 = 11kΩ to give same value at Y as switching voltage at X (2:1 ratio)

(No ecf on value)
Reason / calculation − 1
Answer − 1
2

(d) The op-amp is not ideal and will saturate above 0V

Saturation − 1
Need to drop voltage
Voltage drop − 1
Acceptable method
Method − 1
3
(Total 10 marks)

Q33.
(a)
INPUTS OUTPUTS

Tunnel Car Lorry Message Red Green

Sensor Sensor Sensor Display Stop Go light
T C L M light G
R

0 0 0 0 0 1

0 0 1 0 0 1

0 1 0 0 0 1

0 1 1 1 1 0

1 0 0 0 1 0

1 0 1 0 1 0

1 1 0 0 1 0

1 1 1 1 1 0

Column M − 2 marks
Column G − 2 marks
4

(b) R = T + (C.L)
1 mark for terms
1 mark for OR

Or for full expression S = T +( .C.L)

Max 1 mark
2

Page 42 of 47
 (c)

3
(Total 9 marks)

Q34.
(a) (i) A
(at inverting input)
1

(ii) a point on the circuit where the voltage is 0v / ground

but not connected to 0v / is almost 0v / simulates 0v
assuming that the op-amp has not saturated
2 max

(iii) 10kΩ (must have units unless 10,000 which assumes standard)
oe 10,000Ω / 10K etc
1

(b) correct formula rearranged

calculation / substitution
470kΩ
3 for just correct answer with units
3

(c) inverted,
same frequency,
shape shows evidence of correct gain
maximum amplitude 3v to 5v
4
[11]

Page 43 of 47
Examiner reports

Q1.
(a) This question tested the ability of students to write a Boolean expression to
represent a given logic circuit. The question was answered well (70% of students
gained both marks) and most students opted for the algebraic version, although a
few gave a written version which received equal credit. Future questions of this
nature will probably be written asking specifically for the algebraic version.

(b) This question tested the knowledge of truth tables for a range of logic gates and the
ability to analyse a combinational logic system. Overall, the question was answered
well, with 79% of students scoring two marks.

(c) This question was a little more demanding. A few students became side-tracked by
expecting the manufacturer of the system also to make the logic IC’s. However,
most students were able to produce a viable answer based on cost and space. 86%
gained at least one mark here. Students should be shown a diagram of a typical 2-
input logic gate arrangement on an IC, even if they do not use the IC.

(d) This was a straight recall question. Many students (79%) correctly identified the Q
column in the truth table as being that of the exclusive OR gate.

Q2.
(a) This proved to be a simple calculation for most students, with 84% gaining the mark.

(b) Just under 50% of the students picked up maximum marks for correctly drawing the
non-inverting amplifier. Although many produced the arrangement shown in the
mark scheme, other arrangements of the resistor chain were accepted. However, a
significant proportion of students (40%) were unable to score on this question.

(c) The first mark here was given for identifying the correct formula and then for
showing the resistor ratio to be 39. When faced with this information, it is always
worth checking to see if using 1 kΩ for the lower resistor leads to a sensible value
for the top resistor. This produced the second mark. Most students scored well on
this question, with 68% gaining both marks.

(d) The best answers for this question came from students who appreciated that there
would be a 50 kHz signal going into an operational amplifier having a gain-
bandwidth product of 1 MHz, leading to a maximum gain of only 20. This approach
comes from the question which asked candidates to: ‘Discuss whether this
operational amplifier is suitable...’ 29% of students scored both marks. However,
one mark was given if students got as far as showing that the operational amplifier
really needed to have a gain-bandwidth product of twice that of the amplifier used in
the circuit. The main error here was produced by students thinking that the 50 kHz
signal was being transmitted by an AM system wrongly leading to a total bandwidth
of 100 kHz. It will always be made clear in the question if AM/FM transmission is in
use.

Q3.
(a) The virtual earth point was identified correctly by many students.

(b) The terminal was correctly identified by most students.

Page 44 of 47
 (c) A significant number of students did not explain their thinking behind the derivation.
The key points in the mark scheme included the same current flowing in the input
and feedback resistors, and the use of the virtual earth point when considering the
voltage across the resistors.

(d) Unfortunately, a number of students ignored or forgot the sign in this relatively
straightforward calculation. Both sign and number were needed for the single mark.

(e) This calculation was a little more complex, particularly involving the signs. The
allocation of two marks enabled students who showed proof of a correct method to
gain some credit, even if an error was made in the final step.

(f) A range of sensible answers was accepted. It was a pity that some students lost
sight of the application, but on this occasion the mark was awarded just for a simple
statement of fact.

Q4.
(a) Most students did not appear to be distracted by the context of this question and the
truth table was done well.

(b) A small number of students committed to analysing this using a complex piece of
Boolean algebra, which did not lead to the simplest solution. Most, however, spotted
the short-cut.

(c) A number of students were thrown by two aspects of this circuit which has been
designed so that ‘0’ or a blank is never shown on the display.

(i) The counter starts at 0000, but the logic process allows the display to be ‘1’ (1
dot). Effectively the display is always one ahead of the decimal equivalent of
the binary code on the counter.

(ii) The reset binary code is 0110 and only lasts for a split second, so no output is
shown for this code. However, when the reset has been completed then the
counter code will be 0000 and the display will show decimal ‘1’ (1 dot).

As a result of the above, students were still able to gain credit for using a single AND
gate, even if they did not use the correct tap off points on the counter.

(d) Most students were able to draw/complete the logic diagram form the Boolean
expression provided. This is a vital skill in the digital section of the specification.

Q5.
(a) Some students struggled with extracting C from the formula, and one or two lost
some powers of ten. Again, it is recommended that students always show all stages
of their working, so that some credit can be gained even if the final answer is wrong.

(b) Most students answered this question well, although a significant number failed to
show on the graph the appropriate frequencies matched to the ~ 0.7 voltage gain
point.

(c) Some students were rather ‘clinical’ in their response to this question and forgot to
relate their answer to the context of the ‘listener’.

Q6.

Page 45 of 47
 (a) This proved to be a straightforward question.

(b) This also proved to be straightforward.

(c) This was a complex graph, but most students produced a result that was within the
acceptable range of values given in the mark scheme.

(d) It was disappointing to see a number of students not being able to follow through an
Ohm’s Law calculation when given all necessary values. Otherwise the question
was answered well.

(e) This question required students to correctly apply the comparator rule to give 0 Volts
output, then to appreciate that the voltage drop across the LED would turn it on.
A number of students successfully gained the first mark but then wrongly assumed
that a 0 Volt output would turn the LED off, forgetting that the comparator would sink
current.

Q29.
(a) A high percentage of candidates correctly completed the truth table.

(b) Only a few candidates who achieved success in section (a) did not go on to gain the
mark here.

(c) Answers relating to the dimming of segments as more segments were lit were less
evident, but the calculation and selection of the E24 resistor proved to be straight
forward.

Q30.
(a) An easier start than last year as the position for the component was already given. A
large number of candidates were successful here although one or two protective
resistors crept in and some symbols had the circle with a free-floating centre
character which detracted from the quality of the answer.

(b) This section of the question was straight forward for the majority of candidates.

(c) The power calculations proved to be more selective in terms of response. Good
candidates spotted the need for a V2 / R approach in (i), whilst the easier calculation
in (ii) proved to be a more demanding concept as the calculation needed to be
centred on the data for the series resistor R.

Q31.
(a) This part was answered well with a high proportion of candidates gaining full marks.

(b) The truth table proved to be slightly more demanding, but still allowed a large
percentage of the candidature access to some marks.

Q32.
(a) The mark in part (i) was most often lost by the explanation being too general − e.g.
‘it’s better’. However, the reading required from the graph in part (ii) was accurately
achieved by most candidates.

(b) A significant number of candidates were able to connect the inputs to the correct
pins of the op-amp to make the circuit function as described.

Page 46 of 47
 (c) The question appeared to expose the lack of understanding of how a transistor-
based control circuit can be used to interact with an independent power circuit via an
interface (relay).

(d) This part follows the established question of looking at how saturation affects the
output voltage characteristics of real op-amps and how this differs to the
characteristic for an ideal one. Candidates are expected to formulate their answer
based on the limitations of the op-amp in comparator mode as given in the question.

The latter part of the answer should have focused on how to make the low output
voltage from the op-amp smaller than the forward voltage of the LED rather than just
reducing the current. Solutions that would have appreciably affected circuit
operation when the op-amp gave a high output or potentially cause damage to the
LED did not gain credit.

Q33.
(a) A high percentage of candidates correctly completed the truth table. The loss of just
a single mark was mainly due to the green light being left on when a lorry was
detected.

(b) Just over half of the candidates were able to construct the correct Boolean
expression for the red light from the given completed column in the truth table.

(c) Part (c) was done very well, with most candidates opting for the solution given on
the mark scheme. However, a few candidates also produced an equally valid design
based on an expansion of the given expression for G.

Q34.
(a) (i) Most candidates selected the correct option.

(ii) A large number of answers tended to repeat the word ‘earth’ used in the
question, scoring only one point.

(iii) The vast majority gave the correct answer.

(b) Many gave the correct answer, but workings showing clearly how it was arrived at
were less common.

(c) Many were able to sketch an inverted wave of the same frequency, several gave
evidence of using the gain to calculate a theoretical peak voltage, but few noticed
the power supply limitation of the circuit.

Page 47 of 47