A 5.0 V stabilized power supply is required to be designed using a 12 V DC power supply as input source. The maximum power rating of zener diode is 2.0 W. The minimum value of resistance Rs in Ω connected in series with zener diode will be (a) 16.5 (b) 17.5 (c) 18.5 (d) 15.5

Answer: (b). Maximum Zener current: IZ,max = P/VZ = (2.0)/(5.0) = 0.4 A Drop across Rs: 12 - 5 = 7 V Rs,min = 7/0.4 = 17.5 Ω So the answer is (b) 17.5.

A 5.0 V stabilized power supply is required to be designed using a 12… | MHT CET 2023 (9 May, Shift 1)

Q44MHT CET 2023 · 9 MAY · SHIFT 1Medium

Special Purpose Diodes

A $5.0\text{ V}$ stabilized power supply is required to be designed using a $12\text{ V}$ DC power supply as input source. The maximum power rating of zener diode is $2.0\text{ W}$ . The minimum value of resistance $R_\text{s}$ in $\Omega$ connected in series with zener diode will be

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Answer: (b)

Maximum Zener current: $I_{Z,max} = \frac{P}{V_Z} = \frac{2.0}{5.0} = 0.4$ A
Drop across $R_s$ : $12 - 5 = 7$ V
$R_{s,min} = \frac{7}{0.4} = 17.5\ \Omega$
So the answer is (b) $17.5$ .

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