Half-Wave Rectifier

Pass the positive half-cycles of an AC voltage through a diode to make a pulsating, unfiltered DC output.

V1 supplies a 5 V peak, 100 Hz sine wave. D1 passes current toward R1 = 1 kΩ during positive half-cycles. Negative half-cycles reverse-bias the diode. Vin is measured before D1; Vout is measured across R1.

$$v_{\mathrm{out}}\approx\max(0,v_{\mathrm{in}}-V_F)$$ $$T=\frac1f=10\,\mathrm{ms}$$

The output contains one positive pulse every 10 ms. Its peak is about 4.31 V for this generic diode model and load. The forward drop varies with current, so the first equation is an approximation. With no filter capacitor, the output falls back to nearly zero between pulses.

Read the chart. Vin is green and Vout red. Iin is blue and equals the load current. The three-cycle view makes the blocked negative half-cycles clear; reverse leakage may produce tiny negative values near zero.

Try it. Change R1’s Resistance to 2 kΩ. Current approximately halves and the lower current slightly reduces diode forward drop, raising the output peak. Restore 1 kΩ.

Model limits. D1 uses saturation current 10 fA and emission coefficient 1 at 27 °C. Its junction capacitance, transit time and series resistance are omitted. This is a resistively loaded, unfiltered rectifier; it does not supply a smooth DC rail.