Question

A BJT biasing circuit is shown in the figure, where $V_{BE = 0.7V$ and $\beta = 100$. The Quiescent Point values of $V_{CE}$ and $I_{C}$ are respectively:} \includegraphics[width=0.4\linewidth]{42.png}

• A. 4.6V and 2.46mA
• B. 3.5V and 2.46mA
• C. 2.61V and 3.13mA
• D. 4.6 V and 3.13mA

Hint:

For BJT circuits, use KVL and Kirchhoff's current law (KCL) to find the quiescent point.

Answer 1

The Correct Option is A

\begin{align*} V_{th} &= \frac{12 \times 50}{100 + 50} = 4V
R_{th} &= 50 || 100 = 33.33 k\Omega
\text{KVL:} -4 + I_B (33.33k) + 0.7 + (1 + \beta) I_B \times 1k &= 0
I_B &= \frac{3.3}{33.33k + 101k} = 24.56 \mu A
I_C &= \beta I_B = 100 \times 24.56 \times 10^{-6} = 2.46 mA
I_E &= (1 + \beta) I_B = 2.47 mA
\text{KVL:} -12 + (2K) I_C + V_{CE} + I_E R_E &= 0
12 &= 2K \times I_C + V_{CE} + I_E \times 1k
V_{CE} &= 12 - (2k)(2.46m) - (2.47m)(1k) = 12 - 7.39 = 4.61 V
V_{CE} &= 4.61 V \end{align*} Hence the correct option is (A).