Half-wave rectifier simulation: One direction changes everything

Run the experiment.

Open the half-wave rectifier example. Run Transient for 60 ms at a 10 µs output step, then compare the source and load voltage traces. The positive half cycles reach the load; the negative half cycles are blocked apart from model leakage.

The short answer.

What does one diode do to an AC signal, and how much voltage reaches the load? It passes only the positive half cycles. With a 10 V peak, 50 Hz source and a 1 kΩ load, the simulated load voltage peaks at 9.29 V, not 10 V, because the diode itself drops about 0.71 V while it carries 9.3 mA.

The circuit and settings.

Four parts, one loop. The diode sits between the source and the load, anode toward the source, so current can only flow while the source is positive.

• V1: sine voltage source, 10 V peak, 50 Hz, 0 V offset
• D1: generic diode, saturation current Is = 10 fA, emission coefficient n = 1
• RL: 1 kΩ load resistor
• Analysis: Transient, 0 to 60 ms (three cycles), 10 µs output step, default adaptive time step

Predict before you run.

A first estimate treats the diode as a fixed 0.7 V drop: the load peak is about 10 V − 0.7 V = 9.3 V, and the peak current about 9.3 V ÷ 1 kΩ = 9.3 mA.

The diode model is more precise than that. Its forward voltage follows the Shockley equation, so it depends on the current. At 27 °C the thermal voltage VT is 25.87 mV, and at 9.29 mA the equation gives 0.7128 V.

V_F = n · V_T · ln(I_D / I_S + 1) = 25.87 mV × ln(9.29 mA / 10 fA) ≈ 0.713 V

What the simulation shows.

Measured in the desktop solver over the 60 ms run:

• Load peak: 9.287 V at 5.0 ms, the top of the first positive half cycle
• Diode forward drop at that moment: 0.713 V, matching the equation above
• Peak diode and load current: 9.29 mA
• Load voltage during the negative half cycles: about −10 nV, effectively zero
• Peak reverse voltage across the diode: −10.0 V, the full source peak
• The load is above 10 mV for 48.3 % of the run

Why it behaves this way.

While the source is positive and above a few hundred millivolts, the diode is forward-biased and the same current flows through it and the load. Most of the source voltage appears across the load; the rest is the diode drop, which grows only logarithmically with current.

While the source is negative, the diode is reverse-biased. Only leakage flows: the model’s 10 fA saturation current plus the solver’s 1 pS minimum junction conductance, about 10 pA at −10 V. Across 1 kΩ that is the −10 nV you can read on the load.

Conduction covers slightly less than half of each cycle because the diode needs some forward voltage before its current becomes noticeable. In this run the load first passes 0.1 V at 0.23 ms, when the source is already at about 0.72 V.

The full −10 V appears across the diode, not the load. That peak inverse voltage is what a real diode’s reverse rating must exceed.

Common mistakes.

• Reading the source trace and calling it the output. Probe the load side of the diode.
• Expecting exactly 0.7 V. The drop depends on current: smaller at light loads, larger at heavy ones.
• Expecting DC. A half-wave rectifier outputs pulses; smoothing needs a capacitor and creates ripple.
• Forgetting the ground reference. Without it, node voltages are undefined and the run cannot start.
• Using an output step too coarse for the frequency. At 50 Hz, 10 µs gives 2,000 points per cycle.

Now change one thing.

Reverse the diode and run again. Predict which half of the waveform will reach the load. In the desktop solver the load now swings to −9.29 V on the negative half cycles and stays at about 10 nV on the positive ones: the same circuit, mirrored. Then open the bundled rectifier-filter example to see what a smoothing capacitor does to the output.

Make the prediction before pressing Run. After changing a component value, the previous result is marked out of date until you run it again.

Know the assumptions.

This is a generic diode model. Its forward voltage depends on current; it is not a universal 0.7 V drop or a prediction for every physical diode.