Q127 / 180 · AIIMS 2019 (25 May) · Physics

The given transistor operates in saturation region then what should the be value of $V_{B B}$ $\left(R_{\text {out }}=200 \Omega, R_{\text {in }}=100 \mathrm{k} \Omega, V_{C C}=3\right.$ volt, $V_{B E}=0.7$ volt, $\left.V_{C E}=0, \beta=200\right)$ Figure

  • A
    4.1 volt
  • B
    7.5 volt
  • Correct answer
    C
    8.2 volt
  • D
    6.8 volt

Solution

$\begin{aligned} & \text { Here, } R_0=200 \Omega, R_i=100 \mathrm{k} \Omega, V_{c c}=3 \mathrm{~V} \\ & V_{B E}=0.7 \mathrm{~V}, V_{C E}=0, \beta=200 \\ & V_{C E}=V_{C C}-I_C R_O \Rightarrow 0=3-I_C \times 200 \\ & I_C=\frac{3}{200}=15 \mathrm{~mA} \\ & \beta=200=\frac{I_C}{I_B} \Rightarrow I_B=\frac{I_C}{\beta}=\frac{15 \mathrm{~mA}}{200}=75 \mu \mathrm{A} \\ & V_{B E}=V_{B B}-I_B R_i \\ & V_{B B}=0.7+75 \times 10^{-6} \times 100 \times 10^3 \\ & =0.7+7.5=8.2 \text { volt }\end{aligned}$

Hence, the correct answer is option: 'C'.

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