Voltage Tripler

Charge capacitors on alternating half-cycles to obtain approximately 3 times the AC peak voltage, less diode losses.

V1 supplies a ±5 V, 1 kHz sine wave. The 3 diodes steer charge into 3 capacitors of 1 µF. RL = 1 MΩ gives a light load so the voltage-multiplication principle is easy to see.

$$V_{\mathrm{out}}\approx 3(V_{\mathrm{pk}}-V_F)$$ $$I_{\mathrm{load}}=\frac{V_{\mathrm{out}}}{R_L}$$

The upper two-diode branch produces approximately +2 peak voltages. The lower single-diode branch produces approximately −1 peak voltage. RL is connected between these outputs, so their difference is approximately three peak voltages, around 13 V after diode losses.

Read the chart. Vin is green. Vout is red and is measured directly across RL; Iload is blue. The output reference marker identifies the negative load terminal. That terminal is not ground. The simulation runs for 40 ms; the final 10 ms show the charged circuit and its ripple. The saved current axis starts at zero to show the DC level. For the heavier-load experiment, enable Current in the chart's Y-axis autoscale menu so the larger current stays visible. The capacitors start uncharged in the source's zero-voltage operating point.

Model limits. Loaded output is below the ideal multiplication ratio. Source impedance, diode resistance, capacitor leakage and ESR add further sag in hardware. These ideal capacitors do not enforce voltage ratings.

Try it. Set RL from 1 MΩ to 100 kΩ; compare the voltage sag and ripple. Restore the original values after comparing the results.