Diodes and LEDs

LED Current Limiter

resistor sets the branch current

A fixed LED drop leaves the rest of the source voltage for the current-limiting resistor. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

The resistor gets the remaining voltage

The source voltage minus the LED forward drop leaves 10 volts across the limiter.

VR=12 V−2 V=10 VV_R = 12\ \text{V} - 2\ \text{V} = 10\ \text{V}
LED current limiterThe LED drop leaves the rest for the limiter.12 V2 V5 ohm2 A+-2 V+-10 V

Ohm's law gives the branch current

The limiter is 5 ohm, so 10 volts gives 2 amperes.

I=VRR=10 V5 ohm=2 AI = \frac{V_R}{R} = \frac{10\ \text{V}}{5\ \text{ohm}} = 2\ \text{A}
LED current limiterThe current arrow equals the checked limiter current.12 V2 V5 ohm2 A+-2 V+-10 V

The resistor limits the LED current

In this ideal LED circuit, the resistor is the element that turns the remaining voltage into a checked current. These values are schematic training values, not component recommendations.

ILED=2 AI_{\text{LED}} = 2\ \text{A}
LED current limiterThe current arrow equals the checked limiter current.12 V2 V5 ohm2 A+-2 V+-10 V

More resistance means less current

Keep the source and LED drop fixed. The remaining voltage is the same, so the resistor controls the branch current.

VsVfRVRI12 V2 V10 ohm10 V1 A12 V2 V5 ohm10 V2 A12 V2 V2 ohm10 V5 A\begin{array}{c|c|c|c|c}V_s&V_f&R&V_R&I\\12\ \text{V}&2\ \text{V}&10\ \text{ohm}&10\ \text{V}&1\ \text{A}\\12\ \text{V}&2\ \text{V}&5\ \text{ohm}&10\ \text{V}&2\ \text{A}\\12\ \text{V}&2\ \text{V}&2\ \text{ohm}&10\ \text{V}&5\ \text{A}\\\end{array}
LED current limiterThe middle table row is the checked diagram.12 V2 V5 ohm2 A+-2 V+-10 V