GATE Electronics and Communications (EC) 2020 Solved Papers

Question : 61 of 65

Marks: +1, -0

In the voltage regulator shown below,

{V}_{1}

is the unregulated at

15 {V}

. Assume

{V}_{ {BE}}=0.7

{V}

and the base current is negligible for both the BJTs. If the regulated output

{V}_{0}

9 {V}

, the value of

{R}_{2}

. . . . . . Ω

Solution:

Voltage

V_{B}=V_{2}+V_{B E}

=3.3+0.7

V_{B}=4 V . . .(i)

∴ \;\; I=\;{9-4}/{1 K}

I=5 {mA}

Since base cement is negligible,

V_{B}=9 × \;{R_{2}}/{R_{1}+R_{2}}

4=\;{9 {R}_{2}}/{1 {K}+ {R}_{2}} ⇒ {R}_{2}=800 Ω

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