Diode One-Way Model - Electronic Circuits Step by Step

An ideal diode conducts only when its forward direction matches the source polarity in this circuit model. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

A diode has a forward direction

With this source polarity, the shown diode direction gives a conducting path in the ideal fixed-drop model. Real diodes have curved current-voltage behavior; this model keeps one fixed forward drop.

\text{forward direction} \Rightarrow \text{conducting path}

The forward budget closes

The source is 12 volts. The LED drop is 2 volts, so the resistor gets 10 volts.

V_{\text{source}} = V_{\text{LED}} + V_R = 2\ \text{V} + 10\ \text{V} = 12\ \text{V}

Reverse orientation blocks

Reverse the same diode with the same source and limiter. The ideal model draws no branch current arrow.

I_{\text{reverse}} = 0\ \text{A}

Forward current comes from the leftover voltage

For a forward-biased ideal LED, subtract the fixed drop first. The diagram shows the middle row.

\begin{array}{c|c|c|c|c}V_s&V_f&R&V_R&I\\7\ \text{V}&2\ \text{V}&5\ \text{ohm}&5\ \text{V}&1\ \text{A}\\12\ \text{V}&2\ \text{V}&5\ \text{ohm}&10\ \text{V}&2\ \text{A}\\17\ \text{V}&2\ \text{V}&5\ \text{ohm}&15\ \text{V}&3\ \text{A}\\\end{array}

Reverse orientation keeps current at zero

Reverse-biased rows do not claim a KVL loop through the diode. Changing source voltage does not create a conducting branch in this ideal model.

\begin{array}{c|c|c|c|c}V_s&V_f&R&V_R&I\\6\ \text{V}&2\ \text{V}&5\ \text{ohm}&0\ \text{V}&0\ \text{A}\\12\ \text{V}&2\ \text{V}&5\ \text{ohm}&0\ \text{V}&0\ \text{A}\\18\ \text{V}&2\ \text{V}&5\ \text{ohm}&0\ \text{V}&0\ \text{A}\\\end{array}