Question 18.39: Referring to Fig. 18.27, design a linear voltage regulator t...

Refer to Fig. 18.27. We know that
V_{i(min)} = V_{o} + 3V = 15 + 3 = 18V
Assuming the ripple voltage V_{r} = 2V (max), the input voltage is
V_{i} = V_{i(min)} + \frac{V_{r}}{2} = 18 + 1 = 19 V
Therefore, the input voltage, V_{i} = 19 V with a 2 V (max) ripple superimposed
Then V_{z} =\frac{V_{i}}{2}=\frac{19}{2} = 9.5 V (use the Zener diode 1N758 for 10 V)
Therefore, V_{z} = 10 V
I_{z} ≈ 20 mA

R_{1} =\frac{V_{i} – V_{z}}{I_{z}} =\frac{19 – 10}{20 × 10^{– 3}} = 450 Ω
Let I_{2} = I_{B(max)} = 50 μA

R_{2} =\frac{V_{o} – V_{z}}{I_{2}} =\frac{15 – 10}{50 × 10^{– 6}} = 100 kΩ

R_{3} =\frac{V_{z}}{I_{2}} =\frac{105}{50 × 10^{– 6}} = 200 kΩ

Select C_{1} = 50 μF.
Specification of transistor Q_{1}
V_{CE(max)} = V_{i(max)} = V_{i} + V_{r}/2 = 19 + 2/2 = 20 V
I_{E} = I_{L} = 50 mA
P = V_{CE} × I_{L} = (V_{i} – V_{o}) × I_{L}
= (16 – 15) × 50 × 10^{– 3} = 200 mW
Use the transistor 2N718 for Q_{1}.