GCSE Bitesize Examinations
General Certificate of Secondary Education
Design and Technology
Electronic Products
Total marks: 83
•
•
Mark allocations are shown in brackets
Explanations shown in green
Question 1 – Component Symbols
1 – 11
= 2 marks
12 – 21
= 4 marks
22 correct = 6 marks
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
(6 marks)
l)
m)
n)
o)
p)
q)
r)
s)
t)
u)
v)
Battery
Voltage Rails
Switch (single pole single throw)
Push switch
Variable Resistor
Potentiometer
LDR (light dependent resistor)
Thermistor
Capacitor (Electrolytic)
Diode
Transistor (NPN)
Thyristor
Lamp (bulb)
Motor
Relay
Solenoid
LED (Light emitting diode)
Voltmeter
AND Gate
OR Gate
Inverter
Operational Amplifier
* In the exam you will not have as many components to identify.
Question 2 - Resistors
2a)
(ii) 1k
Brown Black Red
1
0
00
(1 mark)
Gold
=1000R or 1k (ohms)
* The third colour band refers to the number of zeros. The k (kilo) means 1000s.
2b)
(iii) 2k2
(1 mark)
Red Red Red
2
2
00
Gold
=2200R or 2k2 (ohms)
* The k is used instead of a decimal point which is much more obvious on circuit diagrams
2c)
(ii) 150k
(1 mark)
Brown Green Yellow
1
5
0000
Gold
=150000R or 150k (ohms)
2d)
(iv) Tolerance
* These resistors have a ‘Gold’ tolerance which means they are accurate to 5%
Question 3 - Potential dividers
3a)
Award one mark for each correct answer.
(1 mark)
(3 marks)
Voltmeter 1 = 9 volts
Voltmeter 2 = 4.5 volts
Voltmeter 3 = 4.5 volts
* Voltmeter 1 is connected around the battery so will display the battery’s voltage which is 9
volts.
Resistor 1 and resistor 2 are connected as a potential divider, their resistances are the same as the
supply voltage is shared equally between the two resistors. Half of the supply voltage across R1
is 4.5 volts and the other half of the supply across R2 is again 4.5 volts.
Voltmeter 2 is connected around R1 and will measure 4.5 volts. Voltmeter 3 will also measure
4.5 volts because it is around resistor 2.
This also works with any two identical value resistors.
3b)
(iii) Just above zero volts.
(1 mark)
* The supply voltage 9v is unequally divided between resistors R1 and R2. R2 is much smaller
in value than R1 and therefore will drop a much smaller voltage. If the lower end of R2 is held at
zero volts (return to the battery), then the voltage on the meter will be only slightly higher than
this or just above zero.
You could calculate or simulate the exact voltage here but it is also a useful skill to be able to
approximate voltages around a circuit especially when fault tracing.
3c)
Two marks for the correct formula
Two marks for showing how the formula is used
(4 marks)
R2
V (out) = ---------------- x V (in)
R1 + R2
For this potential divider:
V (out) =
100
-------------- x 9
1000 + 100
V (out) =
100
-------------- x 9
1100
V (out) =
1
-------------- x 9
11
V (out) =
0.090 x 9
V (out) =
0.818 volts
* Always show as much of the working out as possible, if you give just the answer and it is
wrong the examiner will find it difficult to give part marks.
Question 4 - Capacitors
4a)
(i) Plot a
(1 mark)
* The 1K resistor will allow far more charging current to flow than the 10k resistor. The
capacitor will be filled far faster.
4b)
(iii) 100µF capacitor and 100k (ohm) resistor.
* The largest capacitor will take longer to fill through the highest value resistor.
(1 mark)
4c)
(iii) By using this method the capacitor is drained safely through the 10k resistor.
(1 mark)
* For (i) This is an electrolytic capacitor which can be permanently damaged if the polarity
reversed, the higher voltage types can even burst their casing.
In (ii) a short circuit will instantly discharge the capacitor and all its stored energy suddenly
released. This can also damage the capacitor internally.
Finally (iv) yes this method would work but the input resistance of most modern test equipment
is over 1M (ohms). In this case it would take over 90 seconds.
Question 5 – Diodes and LEDs
5a)
One mark for each correct answer.
(4 marks)
(i) True
(ii) True
* Most LEDs can be damaged by currents in excess of 20mA
(iii) False
(iv) True
* The lower the current being taken from the battery the longer it lasts. Reducing the current
from 20mA to say 5mA will extend the batteries life by 4 times!
5b)
(iv) D2 and D6 light
(1 mark)
5c)
Award one mark for each correct answer.
(i) True
(4 marks)
* In the diagram below the diode is disconnected and the switch is opened. The plot shows a
large downward voltage spike (negative). This spike is what damages switches, transistors and
other sensitive semiconductors.
------------------------------------------------------------------------------------------------------In this circuit the diode is connected and the negative spike is absorbed by the diode.
The negative going is now limited by the diode to about 0.7 volts, which is far less of a nuisance.
Most components will safely cope with this. Also the actual spike lasts for a very small fraction
of a second.
(ii) False.
* This is a diode not an LED.
(iii) True.
* Yes the diode absorbs high voltage spikes as shown in the graphs above.
(iv) False.
* The diode is connected in parallel around the motor and also in wrong direction for any current
to flow, so it will have no effect on the circuit or the motor.
Question 6 – The transistor
6a)
1 – 9
= 2 marks
10 – 16
= 4 marks
17 correct = 6 marks
(6 marks)
When the push switch SW1 is pressed a small amount of current flows through resistor R1 and
into the base connection of the transistor.
This current then leaves the transistor through the emitter and returns to the power supply. The
transistor is now turned on or biased.
A larger current path can now flow from the power supply through resistor R2 and the LED D1
into the collector connection of the transistor, where it joins the small base current.
This combined current leaves the transistors emitter connection and back to the power supply.
The LED D1 is now lit and its brightness or current is controlled by resistor R2.
6b)
(4 marks)
When the device was switched on the buzzer sounded all the time. The nail had caused the
positive supply from the battery to be directly connected to the base of the transistor.
The current flow into the base connection had exceeded its maximum safe
working level. The transistor suffered damage which caused it to be permanently turned on.
6c)
This is the modified circuit.
(4 marks)
The resistor R1 is to keep the base current down to a safe level even if the water probes (or the
cable) become shorted.
* Base protection resistors are an excellent and cheap method of protecting transistors.
6d)
The diagram below shows the modified circuit.
(6 marks)
The water probe is replaced by a potential divider consisting of VR1 (variable resistor) and R2
the LDR (light dependent resistor). When the LDR is being lit by the LED its resistance is quite
low (a few hundred ohms) so nearly all of the current flowing through VR1 returns to the battery
and very little arrives at the transistor to turn it ‘on’.
When the float rises and blocks off the light path the resistance of the LDR becomes quite high.
Most of the current flows to the input of the transistor (base) through the limiting resistor R2.
Providing there is sufficient current, the transistor turns on and the buzzer will sound.
Question 7 – The NE555 Integrated circuit
7a)
(i) The NE555 is working in monostable mode.
(6 marks)
(ii) The two components which control the timing period are R1 (the timing resistor) and C1 (the
timing capacitor).
(iii) The name given to R2 is a pull up resistor (keeps the trigger pin voltage normally high). If
you want to make this circuit touch sensitive try making this resistor very high 1 – 10M (ohms)
and connecting pin 2 to a conducting metal plate (on odd bit of PCB would do!)
(iv) The circuit can be modified so the LED D1 remains off for the timing period like this:
The LED and its current control resistor are wired from the + supply rail to the output pin of the
NE555.
7b)
Award one mark for each part of the question.
(2 marks)
(i) This is the correct pin order.
Pin 1
Pin 2
Pin 8
NE555
Pin 7
Pin 3
Pin 6
Pin 4
Pin 5
(ii) The IC is a dual in line type (two rows of four pins).
7c)
(i) The NE555 is working in astable mode.
(2 marks)
(ii) The two main components which control the timing period are R2 and C1.
7d)
This is the modified circuit for many LEDs.
(2 marks)
7e)
(4 marks)
The current used by one LED needs to be calculated first. There are many methods to do this.
One of the easiest is to measure the voltage drop across the LED current control resistor R3
while the LED is turned on. Using ohms law calculate the current.
Resistance (R)
voltage (V)
= --------------Current (I)
or
Voltage (V)
Current = -----------Resistance (R)
In this case the voltage measured was just over 7 volts, therefore
7
Current = --------1000 (1k)
=
Current = 0.007 A or 7mA
If the NE555 can output up to 300mA then 300 mA ÷ 7 = 42 LEDs
* Whether the NE555 could continually run this number would push the NE555 near to its limits,
over heating might be a problem but this number is possible in theory.
Question 8 – Operational Amplifiers
8a)
(i)
VR1 is a variable resistor
(ii)
R1 is a thermistor
(iii)
R2 and R3 form a potential divider
(iv)
IC1 is an operational amplifier
8b)
(i) Input a is called the non inverting input
(4 marks)
(2 marks)
(ii) Input b is called the inverting input
8c)
(8 marks)
(i) The resistance of the thermistor changes with temperature. The thermistors resistance falls as
temperature increases. (A thermistor is a negative temperature coefficient device.)
(ii) At input ‘a’ (non – inverting) the voltage changes as the temperature of the thermistor
changes.
* This is because the variable resistor VR1 and the thermistor form a potential divider circuit.
The potential dividers output voltage changes with a change of resistance. As the temperature of
the thermistor rises its resistance falls. The thermistor is in the lower part of the potential divider
so the voltage going into input ‘a’ is pulled down.
(iii) At input ‘b’ the voltage is constant at 4.5 volts.
* This is because input ‘b’ is supplied a voltage from another potential divider made up from R2
and R3. As both of the resistors are fixed and are of the same value the voltage at input ‘b’ will
always be half of the supply voltage. In this case, 9 volts ÷ 2 = 4.5 volts.
(iv)
When the output of the IC1 (operational amplifier) is high, current flows from the output through
LED D2 and resistor R5. D2 lights and its brightness is controlled by R5.
When IC1 goes low, current flows from the positive rail, through LED D1, resistor R4 into IC1
where it is returned back to the power supply. D1 will light and its brightness is controlled by
R4.
* Some operational amplifiers (741) are not capable of outputting much current, just a few milli–
amps (mA). The resistors limiting the LED current will need to be kept quite high in order to
avoid damage to the IC. With careful selection there are now op-amps that will provide much
higher output currents and have lots of advantages over the 741.
(v)
The variable resistor VR1 is provided to set the point where the output LEDs switch. In this
example the switching needs to be at 63˚ C. So with the thermistor at this temperature VR1 will
need to be adjusted so the output just switches. In other words the non-inverting input ‘a’ is just
at 4.5 volts.
8d)
(3 marks)
The Darlington pair circuit lacks ‘sensitivity’. Most thermistors have only a very small change in
resistance for a change in temperature of 1˚ C. It would take a swing of several degrees to cause
the state of the LED to change.
The ‘sensitivity’ of the operational amplifier is much higher and if set-up correctly will be
sensitive to a degree.