Question

It is required to use a peak rectifier to design a dc power supply that provides an average dc output voltage of 12 V on which a maximum of plusminus 1-V ripple is allowed. The rectifier feeds a load of 200Ω. The rectifier is fed from the line voltage (120 V rms, 60 Hz) through a transformer. The diodes available have 0.7-V drop when conducting. If the designer opts for the half-wave circuit: Specify the rms voltage that must appear across the transformer secondary. Find the required value of the filter capacitor.

Answer 1

To design a dc power supply using a half-wave peak rectifier, we need to determine the rms voltage across the transformer secondary and the required value of the filter capacitor.

Step-by-step explanation:

To design a dc power supply using a half-wave peak rectifier, several steps need to be taken. First, we need to determine the rms voltage that must appear across the transformer secondary. We are given the line voltage, which is 120 V rms. To find the peak voltage, we can multiply the rms voltage by √2.

Next, we need to calculate the required value of the filter capacitor. The maximum allowed ripple is plus/minus 1 V, so the peak-to-peak ripple voltage is 2 V. The formula to calculate the required capacitance is C = I / (2 * f * Vripple), where I is the load current, f is the frequency, and Vripple is the ripple voltage. Plugging in the values, we can find the required capacitance.