A system needs to be powered with 9 V d.c. source of maximum load current 100 mA. Design a circuit to supply power with the available domestic a.c. line. Assume any data required, but reasonably. Provide short circuit protection.

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The secondary output of step-down transformer is [latex]\sqrt{2}[/latex] times the output d.c. voltage required. Therefore, the step-down transformer is wound to have 230 V:13 V.
Given data: The d.c. output voltage = 9 V and Load current = 100 mA
The current rating is 1.5 times the maximum load current, i.e. 150 mA.
A bridge rectifier or full wave rectifier is used to get the pulsating d.c. output.

[latex]R_{L} =\frac{V_{d.c.}}{I_{L}}=\frac{9}{100 × 10^{– 3}}[/latex] = 90 Ω

A capacitor filter is used to remove the ripple and get a smooth output.

Ripple factor Γ = [latex]\frac{1}{4 \sqrt{3} f CR_{L}}[/latex]

Assume the ripple factor to be 0.03.

C = [latex]\frac{1}{4 \sqrt{3} × 50 × 0.03 × 90}[/latex] = 1069 μ F ≈ 1000 μ F

The short circuit resistance [latex]R_{sc}[/latex] connected with the series pass transistor is

Rsc = [latex]\frac{V_{BE}}{I_{limit}}=\frac{0.7}{150 × 10^{– 3}}[/latex] ≈ 4.7 Ω

Assume 7.6 V Zener diode in series with 1.5 kΩ.
The designed circuit is shown in Fig. 18.35.

Question 18.41: A system needs to be powered with 9 V d.c. source of maximum...

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