Capacitor Filters

It can be seen from Figures 82 and 84 that the waveform v_out is not very smooth. For many applications it is desired to have a much smoother DC waveform, and so a filtering circuit is used. We will consider the filtered half-wave rectifier of Figure 86, and leave the filtered full-wave rectifiers up to you to work out (not hard-see lab).

**Figure 86:** Filtered Half-wave rectifier.
$\begin{figure} \begin{center} \epsfig{file=images/diodeimg22.eps}\end{center}\end{figure}$

The waveform produced by this filtered half-wave rectifier is shown in Figure 87, illustrating the ripple.

**Figure 87:** Filtered Half-wave rectified waveform.
$\begin{figure} \begin{center} \epsfig{file=images/diodeimg21.eps}\end{center}\end{figure}$

Here, ripple is defined as the difference between the maximum and minimum voltages on the waveform, Figure 88 (i.e. peak-to-peak).

**Figure 88:** Filtered Half-wave rectified waveform showing V_rpp and V_DC.
$\begin{figure} \begin{center} \epsfig{file=images/diodeimg23.eps}\end{center}\end{figure}$

The (peak-to-peak) ripple factor r is defined as

$\begin{displaymath}r = \frac{V_{rpp}}{V_{DC}} \end{displaymath}$

where V_rpp is the peak-to-peak ripple voltage and V_DC is the DC component of the ripple waveform. T is the period of the AC source voltage: T=1/f, $\omega = 2\pi f$ . For f=50 Hz (the frequency of the AC supply in Australia), T= 20 ms.

We now explain how to calculate (approximately) V_rpp and V_DC. Think of the ripple waveform as being approximated by a triangular waveform (Figure 89), so that

$\begin{displaymath}V_{DC} = V_{m(out)} - \frac{V_{rpp}}{2} \end{displaymath}$

(80)

using symmetry. Suppose that at the beginning of a cycle the capacitor is fully charged to V_m(out), and that the capacitor is large enough so that the time constant R_L C is much larger than T. The rate of change of v_out at the beginning of the cycle t=0 is

$\begin{displaymath}-\frac{V_{m(out)}}{R_L C} \end{displaymath}$

so that at time t=T the capacitor voltage has decreased by an amount

$\begin{displaymath}V_{rpp} = \frac{V_{m(out)}T}{R_L C} \end{displaymath}$

(81)

approximately (straight line approximation). This allows one to design C for a given load and desired ripple.

Exercise. Show that V_rpp in the case of a full-wave rectifier is given by

$\begin{displaymath}V_{rpp} = \frac{V_{m(out)}T}{2R_L C} . \end{displaymath}$

(82)

**Figure 89:** Approximate (triangular) filtered Half-wave rectified waveform.
$\begin{figure} \begin{center} \epsfig{file=images/diodeimg24.eps}\end{center}\end{figure}$

[ENGN2211 Home]

ANU Engineering - ENGN2211