Question

A half-wave rectifier is needed to supply 15-V de to a load that draws an average current of 250 mA. The peak-to-peak rip- ple is required to be 0.2 V or less. What is the minimum value allowed for the smooth- ing capacitance? If a full-wave rectifier is needed?

Answer 1

The minimum value of the smoothing capacitance in a half-wave rectifier circuit is 0.0625 F. For a full-wave rectifier, the minimum value would be 0.03125 F.

Step-by-step explanation:

In a half-wave rectifier circuit, the smoothing capacitance is used to reduce the ripple voltage. To determine the minimum value of the smoothing capacitance, we can use the formula:

Cmin = (Iavg * T)/(∆V)

Where:

Cmin is the minimum required capacitance

Iavg is the average current

T is the time period of the input voltage

∆V is the maximum allowable ripple voltage

Substituting the given values, we have:

Cmin = (0.25 A * 1/20 Hz)/(0.2 V) = 0.0625 F

If a full-wave rectifier is needed, the minimum value for the smoothing capacitance would be half of the value calculated for the half-wave rectifier, which is 0.03125 F.

Answer 2

To calculate the minimum value for the smoothing capacitance in a half-wave rectifier circuit, you can use the formula Cmin = (Iavg * T) / Vp, where Iavg is the average current, T is the period, and Vp is the peak voltage.

Step-by-step explanation:

In a half-wave rectifier circuit, the smoothing capacitance is used to reduce the ripple voltage. The minimum value allowed for the smoothing capacitance can be calculated using the formula:

Cmin = (Iavg * T) / Vp

Where Cmin is the minimum capacitance, Iavg is the average current, T is the period of the waveform, and Vp is the peak voltage. Given that the average current is 250 mA, the period is 1/f (where f is the frequency), and the peak-to-peak ripple voltage is 0.2 V, you can substitute these values and solve for Cmin.