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How are electronic systems built from input, process and output components?

The systems approach to electronics using input, process and output blocks, the function of common components (switches and sensors such as LDRs and thermistors, fixed and variable resistors, capacitors, diodes and LEDs, transistors as switches, and relays), Ohm's law and basic circuit calculations, and how components are combined to make a working product.

A focused answer to the Edexcel 9DT0 content on electronic systems, covering the input-process-output model, sensors and switches, resistors, capacitors, diodes, LEDs, transistors and relays, Ohm's law and basic circuit calculations, and combining components into a product.

Generated by Claude Opus 4.814 min answer

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What this dot point is asking

Edexcel wants you to use the input-process-output systems model, know the function of common electronic components (sensors, switches, resistors, capacitors, diodes, LEDs, transistors, relays), apply Ohm's law in basic circuit calculations, and explain how components combine into a working product.

The answer

The input-process-output model

Input components: switches and sensors

  • Switches: push-to-make and push-to-break, reed (magnetic), microswitch (limit detection) and tilt switches turn a circuit on or off.
  • Light dependent resistor (LDR): resistance is high in the dark and low in light, used to sense light level.
  • Thermistor: resistance changes with temperature (a common type falls as it warms), used to sense heat.

Sensors are usually placed in a potential divider so the changing resistance produces a changing voltage that the process block can act on.

Process components

Output components

Outputs include LEDs (with a series resistor to limit current), lamps, buzzers and motors. A relay or transistor switches loads that need more current or a separate (for example mains) supply than the control circuit can provide, isolating the low-voltage electronics.

Ohm's law and circuit calculations

The classic calculation is the LED series resistor: subtract the LED's forward voltage from the supply to get the voltage across the resistor, then divide by the desired current.

Examples in context

A garden night light uses an LDR input in a potential divider, a transistor process that switches on as it gets dark (a variable resistor sets the trigger level), and an LED or, via a relay, a mains lamp as the output. A frost alarm swaps the LDR for a thermistor and sounds a buzzer when the temperature falls. A capacitor with a resistor sets the delay in a timed light, and a diode protects a circuit from reverse-connected batteries. Sizing the LED series resistor with Ohm's law, and mapping a product onto the input-process-output model with named components, are the two skills Edexcel tests most here.

Try this

Q1. State what happens to the resistance of an LDR as the light level falls. [1 mark]

  • Cue. Its resistance increases (high resistance in the dark, low in bright light).

Q2. A resistor has 6.06.0 V across it and carries 3030 mA. Calculate its resistance. [2 marks]

  • Cue. R=VI=6.00.030=200R = \dfrac{V}{I} = \dfrac{6.0}{0.030} = 200 ohms.

Q3. Explain why a relay or transistor is used to switch a motor from a sensor circuit. [2 marks]

  • Cue. The sensor circuit provides only a small current; a relay or transistor lets that small signal switch the larger current (or separate supply) the motor needs, isolating the control electronics.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20194 marksA 9 V supply drives an LED that needs 2.0 V across it and a current of 20 mA. Calculate the value of the series resistor required.
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Award marks for the voltage across the resistor, the use of Ohm's law and the final value.

Voltage across the resistor: the supply is 99 V and the LED drops 2.02.0 V, so the resistor must drop 92.0=7.09 - 2.0 = 7.0 V.

Apply Ohm's law: R=VI=7.020×103=7.00.020=350R = \dfrac{V}{I} = \dfrac{7.0}{20 \times 10^{-3}} = \dfrac{7.0}{0.020} = 350 ohms.

The nearest higher preferred (E12) value, 390390 ohms, would be chosen so the current stays at or below the maximum. Markers reward subtracting the LED voltage first, the correct use of R=V/IR = V/I with current in amperes, and a value of about 350350 ohms.

Edexcel 20216 marksExplain, using the input-process-output model, how an automatic night light could be designed, naming suitable components.
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Extended-response item marked on levels (correct systems structure, appropriate components and how they work together).

Input (sensing): a light dependent resistor (LDR) detects the ambient light level; in a potential divider its resistance rises as it gets darker, raising the voltage at the divider output.

Process (decision/control): the rising voltage switches a transistor on once it passes the transistor's threshold (a variable resistor in the divider sets the trigger light level). The transistor acts as an electronic switch.

Output: the transistor switches the load, an LED (with a series resistor) or, via a relay, a mains lamp, so the light comes on in the dark and off in daylight.

A strong answer maps each stage to a named component (LDR input, transistor process, LED or relay output), explains the potential divider sensing the dark and the transistor switching, and may add a variable resistor to set sensitivity.

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