Fullwave rectifiers with capacitor filters are, without question, the workhorse — and the unsung hero — of the modern electronic world and for the gadgets we have come to enjoy. That being said, it is surprising — and sadly so — that a symbolic solution set describing steadystate circuit performance has been absent.
In 1943, Schade gave a graphical solution.^{[1]} In 1972, Millman outlined a few steps, stating that “no analytic expression can be given for….; it must be found graphically”.^{[2]} In 1982, Williams presented a more analytical, albeit incomplete, attempt.^{[3]} In 1991, Kassakian gave, in Example 3.1, a description for a halfwave rectifier^{[4]} and two years later, Tarter showed several mathematical steps.^{[5]} In 2004, Shepherd offer yet another result, although it was questionable.^{[6]} There have been numerous other trials too long to list, but none have come close to providing a satisfactory result.
In this article, we will present a symbolic closedform solution in its entirety, possibly for the first time. These developments have been made possible with the recognition of several crucial facts that had been sorely missed by all previous efforts. View article Equations
First of all, the source function driving the loaded filter was mistaken as Asinωt when diodes were conducting. The correct function shall be Asin(ωt+θ) since, at the time the diode commences conduction (timezero), the driving source voltage is nonzero. Next, it is understood that, under steadystate operation, the circuit alternates periodically between two structures: Fig. 1a, which corresponds to the diode in conduction, and Fig. 1b, with the diode cutting out.
Furthermore — and this is what previous efforts failed to discern — when Fig. 1a commences, the capacitor voltage begins with a finite voltage, designated as V0a in the following analysis. This finite voltage evolves into V0b when the diodes cease conduction, and the structure in Fig. 1b takes over. The key is that V0b must also cycle back to V0a when the diodes kick in again, and then the cycle repeats.
With the property of voltages wrapped around, i.e. continuity of states, in mind and referring to both Figs. 1a and 1b, the output node produces two Kirchhoff Current Law equations:
Here, t1 represents the instant diode conduction begins, and timezero for analysis. Taking the Laplace transform of both equations produces:
(See Eq. 3 below)
The inverse Laplace transform yields the outputs corresponding to two time segments, 0 < t < t_{on} and t_{on} < t < T/2:
(See Eq. 5 below)
Here, both V0a and V0b, cyclic starting states, are embedded and yet unknown. As stated previously, V0a and V0b are linked by the following constraints; continuity of states:
(See Eq. 7)
Given two constraints in two unknowns, both unknowns can be solved with t_{1} and t_{on} as parameters:
(See Eq. 9)
(See Eq. 10)
Besides the condition of continuity of states, two critical boundary conditions are also identified:
Eq. 9 and Eq. 11 allow:
(See Eq. 12)
Regrouping and collecting the appropriate terms yields:
(See Eq. 13)
Taking the ratio V0b/V0a also gives:
(See Eq. 14)
Eventually, both Eq. 13 and Eq. 14 can be consolidated, yielding an equation with a single unknown, t_{on}:
(See Eq. 15)
This equation can be solved numerically for t_{on}. Then, diode conduction time reductions follow, t_{1}, using either Eq. 13 or Eq. 14, involving tan(ωt_{1}). However, given modern mathematical software tools, e.g. MathCAD (MathSoft, Inc.), it is preferred to use V0a (t_{1}, t_{on}) and V0b (t_{1}, t_{on}) boundary Eq. 11 and solve both t_{1} and t_{on} simultaneously. The following is an example.
EXAMPLE
Given a singlephase, 60Hz, 120V sourcefeeding 200µF filter and a 57Ω load with an assumed source series resistance of 0.5 Ω, the boundary Eq. 11 yields a conduction time reduction of t_{1} = 1.743 ms and a conduction duration of t_{on} = 3.131 ms. Both boundary voltages, Eq. 9 and Eq. 10, are also obtained: V0a = 103.637 V, while V0b = 163.719 V. The rectified source, the output voltage in Eq. 5 and Eq. 6, and the rectified source current in steady state, are given in the plot in Fig. 2.
In the course of the above derivation, the capacitor's ESR was ignored. Once it is included, both the mathematical form and the formulation of boundary conditions grow more complicated. It is also to be noted that, so far, only the firstorder circuit, the capacitor filter, is involved. In the real world, inductance, however small, is omnipresent. With the inclusion of inductance, e.g. along the input source line, the circuit becomes one of a second order with two firstorder differential equations.
Other complications arise, e.g. threephase sources and phasecontrolled SCR, that can make analytical procedures extremely tedious. However, the concept of continuity of states and the need to identify the boundary conditions still correctly apply. Refer to Wu for further discussions.^{[7]}
Given the waveforms in symbolic closedform, additional analysis of harmonic contents, for example, can be easily performed. For a halfwave rectifier, readers are cautioned to apply the second boundary appropriately: V0b(T) = V0a.
By applying the concept of continuity of states and by identifying the critical boundary conditions, symbolic solutions in closedform can be obtained for singlephase/fullwave and singlephase/halfwave rectifiers with capacitive filters.
REFERENCES

O.H. Schade; “Analysis of Rectifier Operation” Proceeding of IRE, Vol. 31, No. 7, July 1943

Jacob Millman, “Integrated Electronicsanalog and digital circuits and systems” p.111, 1972, McGraw Hill

Keith L. Williams, “Mathematical Theory of Rectifier Circuits with CapacitorInput Filters” Power Conversion International, p.42, Oct. 1982

John G. Kassakian, “Principles of Power Electronics”, p.38, 1991, AddisonWesley

Ralph E. Tarter, “Solidstate Power Conversion Handbook”, p.229, 1993.

William Shepherd, “Power Converter Circuits”, p.20, 2004, Marcel Dekker

Keng C. Wu, “Power Rectifiers, Inverters, and Converters  accelerated steadystate approaches with closedform solutions” ISBN 9781435720237, Oct. 2008, lulu.com.