A power supply like that of Fig. 28.23 is supposed to deliver 22-V DC at a maximum current of 150 mA. The transformer's peak output voltage can charge the capacitor to a full \(22 \mathrm{V},\) and the primary is supplied with \(60-\mathrm{Hz}\) AC. What capacitance will ensure that the output voltage stays within \(3 \%\) of the rated \(22 \mathrm{V} ?\)

Short Answer

Expert verified
The capacitance needed to ensure that the output voltage stays within 3% of the rated 22V is about 0.0019 F or 1900 μF.

Step by step solution

01

Convert DC voltage to peak voltage.

The power supply is supposed to deliver a peak voltage of 22V DC. This means that the peak output voltage (\(V_{peak}\)) is 22V.
02

Define maximum current output.

The maximum current output (\(I_{max}\)) is given as 150 mA. Convert this to amps by dividing by 1000. So, \(I_{max}\) = 0.15 A.
03

Calculate the rated voltage.

The rated voltage is given as 22V.
04

Determine the ripple voltage.

The problem states that the output voltage should stay within 3% of the rated 22V, so the ripple voltage (\(V_{r}\)) is 3% of 22V DC. Calculate \(V_{r} = 0.03 * 22V = 0.66V.\)
05

Determine the frequency of the AC peak voltage.

The frequency (\(f\)) of the AC peak voltage is given as 60Hz.
06

Calculate the capacitance using the ripple voltage formula.

The ripple voltage formula in a power supply is given by \(V_{r} = I_{max}/ (2*f*C)\), where \(C\) is the required capacitance. Solving for \(C\), we get \(C = I_{max}/ (2*f*V_{r}) = 0.15A / (2*60Hz*0.66V)\)

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