Question 18.40: Design a series voltage regulator with the following specifi...

Refer to Fig. 18.30.
Selection of Zener diode

R_{L\,min} =\frac{V_{o}}{I_{L(max)}}=\frac{20}{50 × 10^{– 3}} = 400 Ω
V_{z} ≈\frac{V_{o}}{2}=\frac{20}{2} = 10 V

Hence, the Zener diode 0.5Z10 is chosen.

Since I_{R1} > I_{B2}, I_{R1} >\frac{I_{C2}}{β}, I_{R1} > \frac{10 × 10^{– 3}}{150}
I_{R1} > 66.7 μA
Let I_{R1} ≈ I_{R2} ≈ I_{R3} = 10 mA (neglecting I_{B2})
Let I_{C2} ≈ I_{E2} = 10 mA
So, the current flowing through the Zener,
I_{z} = I_{E2} + I_{R1} = 20 mA
P_{z} = V_{z}I_{z} = 10 × 20 × 10^{–3}
= 0.2 W < 0.5 W
Hence selection of 0.5WZ10 Zener diode is confirmed.

Selection of transistor Q_{1}
I_{E1} = I_{R1} + I_{R2} + I_{L}
= (10 + 10 + 50) mA = 70 mA
V_{i\,(max)} – V_{0} = 30 – 20 = 10 V

For transistor SL100, the ratings are
I_{C(max)} = 500 mA
V_{CE(max)} = 50 V
h_{FE} = 50 to 280
Hence, SL100 can be chosen for Q_{1}.
Selection of transistor Q_{2}
From the figure, V_{CE2(max)} + V_{z} = (V_{o} + V_{BE1})
Therefore, V_{CE2(max)} = (V_{o} + V_{BE1}) – V_{z} = 20.6 – 10 = 10.6 V
For transistor BC107, the ratings are
V_{CEO(max)} = 45 V
I_{C(max)} = 200 mA
h_{FE} = 125 – 300
Hence, transistor BC107 is selected for Q_{2}.
Selection of resistors R_{1}, R_{2} and R_{3}
V_{R1} = V_{o} – V_{z} = 20 – 10 = 10 V
R_{1} =\frac{V_{R1}}{I_{R1}}=\frac{10}{10 × 10^{– 3}} = 1 kΩ
V_{R2} = V_{o} – V_{R3} = 20 – 10.6 = 9.4 V
R_{2} =\frac{V_{R2}}{I_{R2}}=\frac{9.4}{10 × 10^{– 3}}= 940 Ω
V_{R3} = V_{z} + V_{BE2(sat)}
= 10 + 0.6 = 10.6 V
R_{3} =\frac{V_{R3}}{I_{R3}}=\frac{10.6}{10 × 10^{–3}} = 1060 Ω
Selection of resistor R_{4}
V_{B1} = V_{C 2} = V_{o} + V_{BE1} = 20 + 0.6 = 20.6 V
I_{B1} =\frac{I_{C1}}{β}=\frac{70 × 10^{– 3}}{50} = 1.4 mA

I_{R4} = I_{B1} + I_{C 2} = 11.4 mA

= \frac{30 – 20.6}{11.4 × 10^{– 3}} = 825 Ω

= \frac{22 – 20.6}{11.4 × 10^{– 3}} = 123 Ω

R_{4} =\frac{R_{4(max)} + R_{4(min)}}{2} = 474 Ω