Question

Design a FULL WAVE BRIDGE RECTIFIER circuit that will:

Take 120volts ac, 60 hz, sinusoidal waveform and convert it to a “regulated “dc value

giving 12 volts +, - 1 volt across a 2000-ohm output load resistor with no more than 2%

ripple voltage.

You may assume:

a. An ideal power transformer as discussed in class.

b. For hand computations, you must assume a diode given by Figure 4.8 page 185.

c. A filter capacitor sized per the textbook equation that will meet the ripple voltage

requirement above. See Equations 4.28 through 4.29(b) on page 216.

d. A Zener diode within a shunt regular circuit arrangement as shown in Figure 4.21

page 205.

208 Chapter 4 Diodes Power transformer Voltage regulator Diode Filter Load + ac line 120 V (rms)31 -60 Hz rectifier V, V m, Figure 4.22 Block diagram of a de power supply. selecting an appropriate turns ratio (N./N.) for the transformer, the designer can step the li voltage down to the value required to yield the particular de voltage output of the supply. Bet instance, a secondary voltage of 8-V rms may be appropriate for a de output of 5 V. This car be achieved with a 15:1 turns ratio. In addition to providing the appropriate sinusoidal amplitude for the de power supply the power transformer provides electrical isolation between the electronic equipment and the power-line circuit. This isolation minimizes the risk of electric shock to the equipment user. The diode rectifier converts the input sinusoid vs to a unipolar output, which can have the pulsating waveform indicated in Fig. 4.22. Although this waveform has a nonzero average or a de component, its pulsating nature makes it unsuitable as a de source for electronic circuits, hence the need for a filter. The variations in the magnitude of the rectifier output are considerably reduced by the filter block in Fig. 4.22. In this section we shall study a number of rectifier circuits and a simple implementation of the output filter. The output of the rectifier filter, though much more constant than without the filter, still contains a time-dependent component, known as ripple. To reduce the ripple and to stabilize the magnitude of the de output voltage against variations caused by changes in load current, a voltage regulator is employed. Such a regulator can be implemented using the zener shunt regulator configuration studied in Section 4.4. Alternatively, and much more commonly at present, an integrated-circuit regulator can be used. Figure 4 waveform 4.5.1 The Half-Wave Rectifier The half-wave rectifier utilizes alternate half-cycles of the input sinusoid. Figure 4.23(a) shows the circuit of a half-wave rectifier. This circuit was analyzed in Section 4.1 (see Fig.. assuming an ideal diode. Using the more realistic constant-voltage drop diode model, we obtain withsta appear negativ peak on Vo=0, (4.21a) (4.21b) It is us 50% Vo = Us - VpUVp The transfer characteristic represented by these equations is sketched in Fig. 4. V. = 0.7 V or 0.8 V. Figure 4.23(C) shows the output voltage obtained when the imp to use sinusoid. " rectifi be ju In selecting diodes for rectifier design, two important parameters must be speci current-handling capability required of the diode, determined by the largest current ins is expected to conduct, and the peak inverse voltage (PIV) that the diode must be circui ers must be specified: the est current the diode diode must be able to cire canna

Answer 1

VD censtant

VD censtant

12 오17 Vo ILV Jo o m 4 6