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LNGR SPKRHIFI. TWO WAYS. HANDCRAFTED.
Knowledge / Understand components
LNGR · Learning workshop

The LED

An LED makes current visible. For a long working life, that current must suit the LED.

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Stylised package · verify terminal identification

Start without prior knowledgeUnderstand the model limitsDE / EN
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Circuit 05 · 5 small steps

An LED with a series resistor

An LED with a series resistorThe LED is forward biased: anode above, cathode below. We use a simplified constant 2 V forward drop. This is not a universal LED value.+−5 VAB · 0 VR1 = 330 ΩLED: U_F = 2 VMA_DK
The LED is forward biased: anode above, cathode below. We use a simplified constant 2 V forward drop. This is not a universal LED value.

  1. Which component limits current?

    Hint

    The LED itself is not a current limiter.

    Answer and working

    R1 limits current. Without it, a voltage source can overload the LED.

  2. What voltage appears across R1?

    Hint

    Source voltage minus LED forward drop.

    Answer and working
    Calculation notes
    U_R=U-U_F
    {U_R=5\,\mathrm{V}-2\,\mathrm{V}=3\,\mathrm{V}}

  3. What is the LED current?

    Hint

    I = UR / R1.

    Answer and working
    Calculation notes
    I=\frac{U_R}{R_1}
    I=\frac{3\,\mathrm{V}}{330\,\Omega}
    {I\approx0.00909\,\mathrm{A}=9.09\,\mathrm{mA}}

  4. How much power is dissipated in the resistor?

    Hint

    PR = UR × I.

    Answer and working
    Calculation notes
    P_R=U_R\cdot I
    P_R\approx3\,\mathrm{V}\cdot0.00909\,\mathrm{A}
    {P_R\approx0.0273\,\mathrm{W}=27.3\,\mathrm{mW}}

  5. Increase R1 to 680 Ω. Calculate and check the LED model.

    Hint

    The simplified model still gives 3 V across R1.

    Answer and working
    Calculation notes
    I=\frac{U-U_F}{R_1}
    I=\frac{5\,\mathrm{V}-2\,\mathrm{V}}{680\,\Omega}
    {I\approx4.41\,\mathrm{mA}}

    The LED usually becomes dimmer. Brightness is not exactly proportional to current.

Think it through before opening the answer. Always write the formula, substituted values and units.

01

The mental picture

An LED is not a tiny incandescent bulb with fixed resistance. A small voltage change can cause a large current change. The resistor takes the remaining voltage and limits current.

02

The principle

Forward voltage depends on colour, technology, temperature and current. On an untrimmed standard LED the longer lead is often the anode, but the datasheet is more reliable. Brightness, current and perceived brightness are not identical linear quantities.

03

Try it yourself

Interactive experiment

The LED

Set 5 V, a 2 V forward drop and 330 Ω. Then change the forward drop to 3 V. The same supply and resistor produce a different current.

Model limits: Fixed assumed LED forward voltage and resistive current limiter. No thermal or optical model.

04

A worked example

With 5 V and an assumed 2 V LED drop, the resistor sees 3 V. A 10 mA target gives 300 Ω. A nearby standard value of 330 Ω produces slightly less current, provided the particular LED is suitable.

A closer look · equations and assumptions
Formula notes
R=\frac{U_V-U_F}{I}
P_R=I^2R
P_{\mathrm{LED}}\approx U_F I

For robust sizing, check the highest supply voltage and lowest plausible forward voltage. That combination gives maximum current. Higher-power LEDs or widely varying supplies often use constant-current drivers. Parallel LEDs sharing one resistor do not automatically share current evenly.

A common pitfall. A microcontroller pin is not an unlimited voltage source. Check pin, port and total current limits as well as output voltage drop.

05

Workshop takeaway

An indicator often needs less current than expected. Calculate, start low and judge brightness in its actual mounting position.

Sources & transparency

Original illustrations and educational calculations. No measurements of a specific product. Edited 11 October 2026.

Work it out · answers unfold

Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.

04 · LED with assumed 2 V forward drop: 5 V source, target 10 mA. Find series resistor.

Show worked answer
  1. Resistor must drop 3 V.
  2. Current in A: 0.01 A.
  3. R = 300 Ω.
Calculation
R=\frac{5-2}{0.010}=300\,\Omega

05 · Use 330 Ω instead of 300 Ω. Find LED current and resistor dissipation in the 2 V model.

Show worked answer
  1. I = 3/330 = 9.09 mA.
  2. P = 3²/330.
  3. P = 27.27 mW.
Calculation
P_R=\frac{3^2}{330}=27.27\,mW

06 · LED forward voltage varies from 1.8 to 2.2 V. Source 5 V, R = 330 Ω. Find current range.

Show worked answer
  1. Lower forward voltage gives maximum current.
  2. Maximum I = 3.2/330 = 9.70 mA.
  3. Minimum I = 2.8/330 = 8.48 mA.
Calculation
8.48\,mA\leq I\leq9.70\,mA

07 · Why is 0.7 V not a universal hard diode turn-on threshold?

Show worked answer
  1. Current changes continuously with voltage.
  2. Temperature and diode type affect the curve.
  3. 0.7 V is a simplified calculation assumption.

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