Electrical Power Systems Technology

Direct Current Power Systems 197

The average DC voltage output can be expressed as:

4.16Idc

Vdc = Vmax – ———

C

Sample Problem:

Given: the peak AC voltage applied to a power supply is 28.2 volts.

A 500 uF filter capacitor is used.

Find: the DC output voltage when 300 mA of load current flows.

Solution:

4.16 × Idc

Vdc = Vmax – ————

C

4.16 × 300 mA

= 28.2 – ——————

500 uF

Vdc = 25.7 volts

Again, the value of Vdc is for a full-wave filter-capacitor circuit with

a light load.

With a heavier load (lower resistance) connected to the filter circuit,

more current (Idc) is drawn. As Idc increases, Vdc decreases. However, if the

value of filter capacitor C is made larger, the value of Vdc becomes closer

to that of Vmax. The value of C for a full-wave rectifier can be determined

by using the equation:

2.4 × Idc

C = ————

Vr(rms)

Sample Problem:

Given: a power supply is rated at 950 mA of load current and maxi-

mum ripple of 0.5 volts (rms).

Find: the minimum value of capacitor filter needed for a filter using a full-

wave rectifier.

Solution:

2.4 × Idc

C = ————

Vr(rms)