GATE Electrical Engineering (EE) 2018 Solved Paper

Question : 47 of 65

Marks: +1, -0

In the circuit shown in the figure, the bipolar junction transistor (BJT) has a current gain

β=100

. The base-emitter voltage drop is a constant,

v_{B E}=0.7 {V}

. The value of the Thevenin's equivalent resistance

R_{ {tn}}

(in

Ω

) as shown in the figure is (up to 2 decimal places).

Solution:

KVL @input loop,

10.7-10 {k} i_{b}-0.7-1 {k} i_{e}=0

10.7-10 {k} i_{b}-0.7-1 {k}(1+β) i_{b}=0

10=111 i_{b}

i_{b}=\;{10}/{111} {m} {A}

{V}_{ {ab}}=1 {k} × i_{e}=1 {k}(1+β) i_{b}=\;{1010}/{111} {V}

For short circuit condition

10.7-10 {k} i_{b}-0.7=0

i_{b}=1 {mA}

{I}_{ {SC}}=i_{e}=(1+β) i_{b}=101 × 1=101 {mA}

{R}_{ {Th}}=\;{ {V}_{ {ab}}}/{ {I}_{ {SC}}}=\;{1010}/{111 × 101 {m}}=\;{1010 × 1000}/{111 × 101}=90.09 Ω

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