Rectifiers and Smoothing Filters

1From the wall socket to a phone

A socket in a Rwandan home supplies alternating current at 230 V, reversing direction 50 times every second (REG design voltage; the Rwanda Grid Code fixes the nominal frequency at 50 Hz). A phone battery needs a steady 5 V of direct current, always in one direction. Every charger, radio, laptop adaptor and solar inverter must therefore turn a.c. into d.c. Simple adaptors do it in three steps: a transformer lowers the voltage, a rectifier made of diodes makes the current flow one way only, and a smoothing capacitor fills in the gaps. Modern switch-mode chargers rectify the mains first and then use a high-frequency circuit to step it down, but the rectifier and the smoothing capacitor are still inside. In Unit 4 you found the mean value of a half-cycle of a.c.; here you use it.

2The half-wave rectifier

Figure 3.1: A half-wave rectifier.

During each half-cycle in which the top of the secondary is positive, the diode is forward biased and current flows through the load. During the other half-cycle, the diode is reverse biased and no current flows. The output is a series of positive half-cycles separated by gaps (Figure 3.3(b)): it is unidirectional but far from steady.

• Peak output: $V_0-0.7\mathrm{V}V\_0\; -\; 0.7\backslash ,\backslash mathrm\{V\}$, where $V_0=\sqrt{2}{V}_{\text{rms}}V\_0\; =\; \backslash sqrt\{2\}\backslash ,V\_\{\backslash text\{rms\}\}$ is the secondary's peak p.d.
• Mean output over a whole cycle: half the mean of one half-cycle, $\frac{1}{2}\times \frac{2V_0}{\pi }=\frac{V_0}{\pi }\approx 0.318V_0\backslash frac\{1\}\{2\}\; \backslash times\; \backslash frac\{2V\_0\}\{\backslash pi\}\; =\; \backslash frac\{V\_0\}\{\backslash pi\}\; \backslash approx\; 0.318V\_0$ (ignoring the diode drop).
• The output repeats at the supply frequency, 50 Hz.
• When the diode is not conducting it must withstand the full reverse p.d. of the secondary: its peak inverse voltage (PIV) rating must exceed $V_{0}V\_0$.

3The full-wave bridge rectifier

• when A is positive, current flows A → D1 → + terminal → load → − terminal → D4 → B;
• when B is positive, current flows B → D2 → + terminal → load → − terminal → D3 → A.

Either way, the current goes through the load in the same direction, so the output has two pulses per cycle (Figure 3.3(c)): peak $V_0-1.4\mathrm{V}V\_0\; -\; 1.4\backslash ,\backslash mathrm\{V\}$, because two diodes conduct at a time; mean $\frac{2V_0}{\pi }\approx 0.637V_0\backslash frac\{2V\_0\}\{\backslash pi\}\; \backslash approx\; 0.637V\_0$, twice the half-wave value; and a pulse frequency of 100 Hz. Each diode's PIV must exceed $V_{0}V\_0$. An older design, the centre-tap full-wave rectifier, uses only two diodes with a transformer whose secondary has a connection at its midpoint; each diode then conducts on alternate half-cycles, but each must withstand a reverse p.d. of $2V_{0}2V\_0$.

Figure 3.3: Waveforms for a 50 Hz supply. In (d) the dashed curve is the unsmoothed output.

4Smoothing with a reservoir capacitor

A capacitor connected across the load (Figure 3.4) charges up to the peak p.d. while the diodes conduct. As the rectified p.d. falls after each peak, the diodes become reverse biased and the capacitor alone supplies the load, discharging through it: $V=V_{\text{peak}}{e}^{-\frac{t}{RC}}V\; =\; V\_\{\backslash text\{peak\}\}\backslash ,e^\{-\backslash frac\{t\}\{RC\}\}$. Before it has discharged much, the next peak arrives and recharges it. The output becomes a nearly steady p.d. with a small sawtooth ripple (Figure 3.3(d)).

Figure 3.4: A reservoir (smoothing) capacitor. An electrolytic capacitor must be connected with its + terminal to the positive line.

Estimating the ripple. If the ripple is small, the load current $II$ is nearly constant, and the capacitor supplies it for almost the whole time between peaks, $\mathrm{\Delta }t\approx \frac{1}{f_r}\backslash Delta\; t\; \backslash approx\; \backslash frac\{1\}\{f\_r\}$, where $f_{r}f\_r$ is the ripple frequency (50 Hz for half-wave, 100 Hz for full-wave on a 50 Hz supply). The charge lost is $\mathrm{\Delta }Q=I\mathrm{\Delta }t\backslash Delta\; Q\; =\; I\backslash ,\backslash Delta\; t$, so

$$\mathrm{\Delta }V=\frac{\mathrm{\Delta }Q}{C}\approx \frac{I}{f_{r}C}\backslash Delta\; V\; =\; \backslash frac\{\backslash Delta\; Q\}\{C\}\; \backslash approx\; \backslash frac\{I\}\{f\_rC\}$$

Conditions: $RC\gg \frac{1}{f_r}RC\; \backslash gg\; \backslash frac\{1\}\{f\_r\}$, so that the decay between peaks is nearly linear. A larger capacitor, a smaller load current or full-wave rectification all reduce the ripple. For a really steady output, a Zener regulator (Lesson 2) or a regulator IC follows the capacitor.

5Experiment 3.1: watching rectification

Use a low-voltage a.c. supply from the laboratory power pack (for example 6 V r.m.s.), never the mains directly. Connect a single diode and a 1 kΩ load (Figure 3.1) and display the p.d. across the load on an oscilloscope or a data logger, with the time base at 5 ms per division. Then build the bridge (or use a bridge package), and finally add 100 µF and then 1000 µF across the load, observing the polarity of the electrolytic capacitor. Record the peak p.d. and the ripple each time, and compare the ripple with $\frac{I}{f_{r}C}\backslash frac\{I\}\{f\_rC\}$.