In a Zener regulated power supply of a Zener diode with $V_{2} = 6 V$ is used for regulation. The load current is to be $4 m A$ and the unregulated input is $10 V$ . The value of series resistor $R_{S}$ is (Assume $I_{z} = 5 I_{L}$ ):

Question

Verified by Toppr

Answer 1

Given,
$v_{z} = 6 v$
Load current $= 4 m A = I_{L}$
unregulated input $= 10 V$
To find: value of series resistance $R_{s} = ?$
Solution:
$I = I_{Z} + I_{L}$
$I_{Z} = 5 I_{L}$ (Given)
$I_{Z} = 5 \times 4 = 20 m A$
$I = 20 + 4 = 24 m A$

Voltage drop across $R_{s}$ is
$V_{s} = 10 - 6 = 4 V$
Therefore $R_{s} = \frac{4}{2 4 m A} = 167 r$

i.e., $R_{s}$ is greater than $100 r$ .

Answer 2

Given,

v_{z} = 6 v

Load current

= 4 m A = I_{L}

unregulated input

= 10 V

To find: value of series resistance

R_{s} = ?

Solution:

I = I_{Z} + I_{L}

I_{Z} = 5 I_{L}

(Given)

I_{Z} = 5 \times 4 = 20 m A

I = 20 + 4 = 24 m A

Voltage drop across

R_{s}

is

V_{s} = 10 - 6 = 4 V

Therefore

R_{s} = \frac{4}{2 4 m A} = 167 r

i.e.,

R_{s}

is greater than

100 r

.