The standard practice for calculating the ripple voltage of a simple rectifier circuit with a reservoir capacitor (*Fig. 1*) is to linearize the problem. A linear approximation of the voltage across the reservoir capacitor (*Fig. 2*) assumes that the discharge time is equal to the period of the input sine wave, which is only approximately true for small ripple voltages. (This linearization also might misleadingly suggest the voltage across the reservoir capacitor can change instantaneously, which of course is physically impossible.)

If the amplitude of the input sine wave is U_{I,P, t} is the time constant RC, and T is the period of the input sine wave, then the output ripple voltage’s peak-to-peak value given by linearizing is:

where R1 = 1k, C = 100 µF, t = 0.1 s, T = 20 ms, and U_{I,AC,P} = 20 V, which gives U_{R,AC,PTP} = 4 V.

The approximation obtained by linearization improves as the ripple voltage shrinks. Unfortunately, ripple voltages aren’t always small. Too small a ripple voltage results in very high peak currents in the diode(s) of the rectifier and the preceding circuit, which may create unnecessary and difficult-to-cure EMC problems.

For larger ripple voltages, you could use a circuit simulator such as Spice to check the approximation. But if you don’t have a Spice version at your desk, a simple handheld calculator with an “Ans” key, such as the Casio FX-82MS, is all you need. An “Ans” key inserts the result of the previous calculation into your current calculation, which makes it very convenient for performing iterations.

The calculation uses three formulas derived from *Figure 3*, where U_{I,P} is the amplitude of the input sine wave, t is the time constant RC, T is the period of the input sine wave, and UR,AC,PTP is the peak-to-peak ripple voltage.

Rearranging Equations 2, 3, and 4 yields the time t_{1} during which the reservoir capacitor is discharging:

The calculator can iteratively solve this equation. Be sure the calculator is in “rad” mode and enter an initial guess for t_{1}. Any numerical value between 15 ms and 20 ms will do. For instance, using the same parameters as the linearization example and an initial guess of 18 ms, the keying sequence would be:

18 EXP (-) 3 = 15 EXP (-) 3 + sin,^{-1} ( e^{x} ( ( -) Ans ÷ 0.1) ) ÷ ( 2 × p × 50)

Press the “=” key a few times until the display no longer changes digits.

I performed this calculation on my Casio and got 18.14012782 ms. Using this value in Equation 3 gives U_{R,AC,PTP} = 3.32 V.

Just to get an idea of this method’s accuracy, I “Spiced” a halfwave rectifier for several values of t (*see the table*). The results indeed show that the method is better than using the linear approximation. Even for high values of U_{R,AC,PTP}, this method gives an acceptable approximation despite not taking into account the time constant for charging the reservoir capacitor.

If you want to use this method for a full-wave rectifier with a reservoir capacitor, the only change is that in Equations 4 and 5, 3T/4 simply becomes T/4.

One final note: To determine C for a given R and a desired value of U_{R,AC,PTP}, one shouldn’t linearize the problem nor use the iterative method mentioned above. Instead, determine t2 directly from the equations. The value of C is then easily obtained from: