Rectifier: Appendix

Ripple of Resistive Loads

If the load behaves approximately like a pure resistance, the ripple equation can be derived from time-domain analysis of a discharging RC circuit. We start with Kirchoff’s Voltage Law and time-domain differential equations [1, p. 4]. Suppose that $t = 0$ is at the peak of the wave.

$$\begin{align*} v_R(t) + v_C(t) &= 0 \\ R i(t) + v_C(0) + \frac{1}{C} \int_0^t i(\tau) \; d\tau &= 0 \end{align*}$$

Next, take derivative with respect to time:

$$\begin{align*} R \frac{di(t)}{dt} + \frac{i(t)}{C} &= 0 \\ RC \frac{di(t)}{dt} + i(t) &= 0 \end{align*}$$

Solve by separation of variables to give the general solution (where $A$ is some constant):

$$\begin{align*} \frac{t}{RC} &= -\ln \left| i(t) \right| + A \\ i(t) &= A e^{-t/RC} \end{align*}$$

To find the specific solution, solve for $A$ when $t = 0$ and the voltage is at its maximum.

$$\begin{align*} i(0) &= A = \frac{V}{R} \\ \\ i(t) &= \frac{V}{R} e^{-t/RC} \\ v_R(t) &= V e^{-t/RC} \end{align*}$$

We need to figure out at which point the lowest voltage occurs before the capacitor begins recharging. Effectively, we need to find the point where the mains wave voltage and capacitor voltage are equal. The wave is at peak voltage at $t = 0$, hence the cosine function is used.

$$\begin{align*} V e^{-t/RC} - V \cos(2\pi f t) &= 0 \tag{$2 \pi f t \in [\frac{3\pi}{2}, 2\pi]$} \\ e^{-t/RC} - \cos(2\pi f t) &= 0 \tag{$t \in [\frac{3}{4} f^{-1}, f^{-1}]$} \end{align*}$$

The above expression cannot be evaluated analytically. Let $x(RC, f)$ be the solution. This gives us everything to write out ripple voltage is both volts and percent.

$$\begin{align*} \Delta V &= V - V e^{-x/RC} \\ &= V (1 - e^{-x/RC}) \\ \end{align*}$$

Julia one-liner:

using Roots; ripple = (V, RC, f) -> V * (1 - ℯ^(-find_zeros(t -> ℯ^(-t/RC) - cos(2π * f * t), [0.75/f, 1/f])[1] / RC))

We can evaluate the half-wave designs with $f = 60$ and full-wave designs with $f = 120$.

julia> ripple(170, 10e3 * 10e-6, 60)
24.02914656344006

julia> ripple(170, 10e3 * 10e-6, 120)
12.780834052882188

Conservative Discharge Time

If we skip solving for the intersection of the discharge curve to the AC power curve, we can produce an analytic expression for the ripple voltage. We can simply use the worst-case of the capacitor discharging the entire period of the AC wave to produce a conservative ripple estimate. Let $T = f^{-1}$:

$$\begin{align*} \Delta V &= V - V e^{-T/RC} \\ &= V (1 - e^{-T/RC}) \\ \frac{\Delta V}{V} &= 1 - e^{-T/RC} \end{align*}$$

ripple = (V, RC, f) -> V * (1 - ℯ^(-1 / (f * RC)))

julia> ripple(170, 10e3 * 10e-6, 60)
26.0981067685956

julia> ripple(170, 10e3 * 10e-6, 120)
13.59244951301504

Constant Current Assumption

It can be argued that the model above is innacurate for most loads [2, p. 33]. Switched loads (such as boost converters and digital logic) will draw pulses of current at frequencies higher than that of the AC mains. This results in roughly the same energy being drawn each AC cycle, meaning that we can model the scenario equivalently with a constant current load.

The constant current assumption works well with resistive loads also in well-designed ripple filters. The accuracy of the calculation above is also questionable for resistive loads due to component tolerance, and as long as the cutoff frequency of the ripple filter is much lower than the AC frequency, both models produce very similar results.

$$f_c << f \implies \frac{1}{2\pi f_c} >> \frac{1}{f} \implies RC >> T$$

This assumption holds true for the 33uF capacitor & 10k load (0.48 >> 0.017) Considering the worst-case of the capacitor discharging for the entire period of the AC wave:

$$\begin{align*} i(0) &= \frac{V}{R} \\ i(T) &= \frac{V}{R} e^{-T/RC} \approx \frac{V}{R} \end{align*}$$

Assuming $i(t) = k$ greatly simplifies the analysis. We can now use the admittance relationship to directly derive the ripple voltage:

$$\begin{align*} i(t) &= C \frac{dv_C(t)}{dt} \\ \frac{dv_C(t)}{dt} &= \frac{k}{C} \\ v_C(t) &= v_C(0) + \int_0^t \frac{k}{C} \; d\tau \\ &= -V + \frac{kt}{C} \\ v_R(t) &= V - \frac{kt}{C} \end{align*}$$

We can apply the conservative discharge estimate to produce our final ripple expression:

$$\begin{align*} \Delta V &= V - (V - \frac{kT}{C}) \\ &= \frac{kT}{C} \\ \end{align*}$$

Changing some variables around gives us the familiar ripple expression:

$$\Delta V = \frac{I_{load}}{f_{ripple} C}$$

Julia definition:

ripple = (I, f, C) -> I / (f * C)

Results:

julia> ripple(170 / 10e3, 10e-6, 60)
28.333333333333332

julia> ripple(170 / 10e3, 10e-6, 120)
14.166666666666666

Sources