Draw and interpret circuit diagrams containing diodes and light-emitting diodes (LEDs) and know how these components behave in the circuit

4.3.1 Circuit Diagrams and Circuit Components – Diodes & LEDs

Learning objectives (AO1‑AO3)

• Draw and label circuit diagrams that contain cells, batteries, switches, resistors, lamps, ammeters, voltmeters, diodes and light‑emitting diodes (LEDs).
• Identify the correct IEC/BSI symbols for every component required by the IGCSE syllabus.
• Explain how a diode and an LED behave when they are forward‑biased and when they are reverse‑biased.
• Calculate the value of a series resistor that will allow an LED to operate safely at a chosen current.
• Spot common mistakes that cause a circuit not to work or that damage the components.

1. IEC/BSI symbols (with ASCII equivalents)

The exam board uses the IEC (International Electrotechnical Commission) symbols shown below. The ASCII description is provided for quick typing.

Component	IEC/BSI symbol	ASCII description	Key features to note
Cell (single‑cell battery)		—|—	Longer line = positive terminal, shorter line = negative.
Battery (two or more cells)		—|——|—	Longer line = positive, shorter line = negative.
Switch (SPST)		⎯⎯⎯⏚⎯⎯⎯	Open = gap; closed = the two lines touch.
Resistor		⎯⎯⎯⎯⎯⎯⎯	Often labelled with its resistance value (e.g. 470 Ω).
Incandescent lamp		⎔	Circle with a filament line inside.
Diode		►|	Triangle (arrow) points from anode to cathode (the vertical line).
LED		►| ↗↘	Same as diode plus two small arrows indicating light emission.
Ammeter		A	Connected in series with the component whose current is to be measured.
Voltmeter		V	Connected in parallel with the component whose voltage is to be measured.

2. How a diode works

• Unidirectional conduction: current can flow only from the anode (+) to the cathode (–).
• Forward bias

  • Anode is at a higher potential than the cathode.
  • Current starts when the applied voltage reaches the forward voltage \$V_F\$ (≈ 0.7 V for a silicon diode, ≈ 0.3 V for germanium).
  • Beyond \$V_F\$ the diode behaves almost like a short circuit – its dynamic resistance is very low.

• Reverse bias

  • Cathode is at a higher potential than the anode.
  • The diode blocks current; only a tiny leakage (µA) flows.
  • If the reverse‑breakdown voltage is exceeded the diode can be damaged. This is not examined in the IGCSE, but it is useful to remember.

3. Light‑Emitting Diodes (LEDs)

LEDs are specialised diodes that emit visible light when forward‑biased. Their main differences from a plain silicon diode are shown below.

Why a series resistor is essential

LEDs are non‑linear: a voltage increase of only a few tenths of a volt above \$V_F\$ can cause a large rise in current. Without a resistor the current may exceed the LED’s rating and the device will burn out.

4. Calculating the series resistor for an LED

For a simple series circuit (battery → resistor → LED → battery) the current is:

\[

I = \frac{V{\text{supply}} - VF}{R}

\]

Re‑arranged to give the required resistor value:

\[

R = \frac{V{\text{supply}} - VF}{I_{\text{desired}}}

\]

When the resistor is chosen, also check its power rating:

\[

P = I^{2}R \quad\text{or}\quad P = I\,V_R

\]

A ¼ W (0.25 W) resistor is usually sufficient for LED currents up to 30 mA.

5. Complete example circuit (battery, switch, resistor, LED, voltmeter)

The diagram below shows a typical IGCSE‑style circuit that incorporates all the required symbols.

6. Worked calculations

6.1 LED with a 9 V supply (red LED, \$V_F=2.0\$ V, \$I=20\$ mA)

\[

R = \frac{9.0\ \text{V} - 2.0\ \text{V}}{0.020\ \text{A}} = 350\ \Omega

\]

6.2 Two LEDs in series (red \$VF=2.0\$ V, green \$VF=2.2\$ V, 12 V supply, \$I=20\$ mA)

\[

R = \frac{12.0 - (2.0+2.2)}{0.020} = 390\ \Omega

\]

6.3 Silicon diode with 1 kΩ resistor across 12 V

\[

I = \frac{12.0\ \text{V} - 0.7\ \text{V}}{1.0\ \text{k}\Omega} \approx 11.3\ \text{mA}

\]

7. Common mistakes and how to avoid them

• Reversing the LED. The cathode (flat side/short lead) must face the negative side of the supply.
• Leaving out the series resistor. The LED will draw excessive current and burn out.
• Choosing a resistor that is too large. The LED will be dim or may not light at all.
• Assuming every diode has \$V_F=0.7\$ V. LED forward voltages vary with colour (0.8–3.5 V).
• Connecting a diode directly across a power source. In reverse bias the diode can break down if the voltage exceeds its rating.
• Forgetting the voltmeter’s internal resistance. A voltmeter is always placed in parallel; it does not affect the circuit’s operation if its resistance is high (≥ 10 MΩ).

8. Summary checklist (exam‑ready)

1. Identify and draw the correct IEC symbol for every component required.
2. Mark polarity on diodes and LEDs (anode → cathode direction of the arrow).
3. State the forward voltage \$V_F\$ for the specific diode/LED you are using.
4. Use \(R = \dfrac{V{\text{supply}} - VF}{I_{\text{desired}}}\) to find the series resistor.
5. Check the resistor’s power rating with \(P = I^{2}R\) (¼ W is usually enough for LED circuits).
6. Verify the circuit layout: battery → switch → resistor → LED → battery, with any measuring devices correctly placed (ammeter in series, voltmeter in parallel).

9. Practice questions

1. Draw a complete circuit diagram that lights a green LED (\$V_F = 2.2\text{ V}\$) from a 5 V supply using a 150 Ω resistor. Include a switch and a voltmeter across the LED, and clearly label anode and cathode.
2. A silicon diode (\$V_F = 0.7\text{ V}\$) is placed in series with a 1 kΩ resistor across a 12 V battery. Calculate the current through the diode (assume it is forward‑biased).
3. Explain why an LED will not emit light when it is connected in reverse bias, even though a very small leakage current may still flow.
4. Two LEDs (red \$VF = 2.0\text{ V}\$, blue \$VF = 3.2\text{ V}\$) are to be powered from a 9 V supply. Determine the value of a single series resistor that will give each LED a current of 15 mA.
5. Identify the error(s) in the following schematic description: “A 5 V battery is connected directly to a yellow LED (forward voltage 2.1 V) with the cathode towards the positive terminal.”