We wish to build a 135 V, 20 A, DC power supply using a single phase bridge rectifier and an inductive filter. The peak to peak current ripple should be about 10%. If a 60 Hz AC source is available; Calculate the peak to peak current ripple. 

Explanation:

DC Power Supply Using Single Phase Bridge Rectifier and Inductive Filter

Problem Statement: We aim to design a DC power supply capable of delivering 135 V and 20 A using a single-phase bridge rectifier with an inductive filter. The peak-to-peak current ripple in the output should be approximately 10%. The available AC source operates at 60 Hz. The task is to calculate the peak-to-peak current ripple in the system.

Key Formula and Concepts:

The peak-to-peak current ripple in the output of a DC power supply using an inductive filter can be calculated using the following equation:

ΔI_pp = (ΔV × T) / L

Where:

• ΔI_pp: Peak-to-peak current ripple
• ΔV: Ripple voltage across the filter
• T: Time period of the AC waveform = 1 / f
• L: Inductance of the filter

In this case, the ripple percentage is specified as 10%. Therefore, the peak-to-peak current ripple can be calculated as:

ΔI_pp = 0.1 × I

Where:

• I: DC output current, which is 20 A

Calculation:

Given data:

• DC output current (I): 20 A
• Ripple percentage: 10% (0.1 × I)

Substituting the values into the formula for peak-to-peak current ripple:

ΔI_pp = 0.1 × I

ΔI_pp = 0.1 × 20 A

ΔI_pp = 2 A

Hence, the peak-to-peak current ripple is 2 A.