Entering three of the four values for the input power frequency (AC f), the peak to peak ripple voltage (Vpp), the load current (I Load), or the size of the filter capacitor (C) finds the missing value.
Unless otherwise specified, the capacitor’s tolerance is assumed to be 20%. If you wish to specify a different tolerance, enter it as a percentage (25% = 25).
The solver determines the ripple voltage by finding the amount the capacitors voltage decreases based on the load current and capacitance over a time period equal to one half of the AC input cycle time. The equation used is shown in the following figure.
Note that this solver only applies to full wave rectified AC voltages. For half wave rectified voltages, use the Power Supply Ripple (Half Wave) solver. Also note that this solver works well for the magnitude of ripple voltages typically used in power supply design. For large ripple voltages (greater than ≈10% of the peak voltage), this solver may overestimates the size of the capacitor. The solver errs on the safe side.
To reduce the power supply’s equivalent DC resistance, it is common practice to use two (or more) parallel capacitors whose total capacitance is equal to the required value. In addition, the use of smaller valued capacitors of a different type (ceramic and/or tantalum) in parallel with the typical aluminum electrolytic capacitors used in power supply filters will help the power supply remain stable under higher frequency loads as well as help attenuate high frequency noise on the AC power input.