Topic: Analog Electronics
For the CE (Common emitter) circuit shown, what will be the value of $I_E$ and $V_{CE}$?
$V_{CC}=10 \text{ V}$, $R_C=2 \text{ k}\Omega$, $V_{BE}=0.7 \text{ V}$
- A. 3 mA, 3 V
- B. 4 mA, 4 V
- C. 3.02 mA, 4.2 V
- D. 3.02 mA, 4 V
Answer: D. 3.02 mA, 4 V
Explanation: $I_B \times 220 \text{ k}\Omega = 4 - 0.7$
$I_B = 0.015 \text{ mA}$
Let, $V_{CE} = 4 \text{ V}$
$I_C = \frac{10-4}{2 \text{ k}\Omega} = 3 \text{ mA}$
$I_E = 3 \text{ mA} + 0.015 \text{ mA} = 3.015 \text{ mA} \approx 3.02 \text{ mA}$