Question

The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance is ᴿL = 4kΩ. The series resistance of the circuit is Rᵢ = 1kΩ. If the battery voltage V₈ varies from 8 V to 16 V, what are the minimum and maximum values of the current through Zener diode?
(A) 0.5 mA; 0.6 mA (B) 1 mA; 8.5 mA
(C) 1.5 mA; 8.5 mA (D) 0.5 mA; 8.5 mA

Answer 1

Answer:

b.1mA;8.5mA

Explanation:

kya Karo Yar

Answer 2

The minimum and maximum values of current through zener diode is

(D) 0.5 mA; 8.5 mA  

Explanation:

Given :

Zener breakdown voltage $V_{Z} =$ 6 V

Variable battery voltage = 8 V to 16 V

Load resistance $R_{L}$ = 4 kΩ

Series resistance $R_{i} =$ 1 kΩ

From, KCL equation,

⇒  $I_{R} = I_{Z}+ I_{L}$

For $I_{Zmin}$ we take $I_{R min}$ as a minimum, so input voltage is also minimum.

∴ $I_{Z min} = I_{Rmin} - I_{L}$

From current division rule

⇒ $I_{R min} = \frac{8-6}{1}$ = 2 mA

⇒ $I_{L} = \frac{V_{Z} }{R_{L} } = \frac{6}{4}$ = 1.5 mA

So, $I_{Zmin} = 2 - 1.5 = 0.5$ mA

For $I_{Zmax}$ we take $I_{Rmax}$ as a maximum, so input voltage is also maximum.

∴  $I_{Z max} = I_{Rmax} - I_{L}$

From current division rule,

⇒ $I_{Rmax} = \frac{16-6}{1} = 10$ mA

So, $I_{Zmax} = 10-1.5 = 8.5$ mA

Thus, the minimum and maximum values of current through zener diode is 0.5 mA and 8.5 mA respectively.