Topic: Analog Circuits
For the current mirror circuit shown below, if the emitter area of $Q_{2}$ is thrice of $Q_{1}$ the current I is
(The circuit is a basic BJT current mirror. $Q_1$ is diode-connected. $V_C = +10\text{ V}$, $R = 20\text{ k}\Omega$ in series with $Q_1$. The emitters of $Q_1$ and $Q_2$ are connected to $V_E = -10\text{ V}$. Current $I$ flows into the collector of $Q_2$.)
- A. 0.328 mA
- B. 2.955 mA
- C. 0.105 mA
- D. 0.012 mA
Answer: B. 2.955 mA
Explanation: Assuming the transistor is Ge with cut-off $V_{BE} = 0.3\text{ V}$ (as per official solution logic).
$V_{B1} = V_{E} + V_{BE1} = -10\text{ V} + 0.3\text{ V} = -9.7\text{ V}$.
The reference current $I_{1} = \frac{10 - (-9.7)}{20\text{ k}\Omega} = \frac{19.7}{20} = 0.985\text{ mA}$.
Assuming very high $\beta$, $I_{C1} \approx I_{1} = 0.985\text{ mA}$.
Since the emitter area of $Q_{2}$ is three times that of $Q_{1}$, $I_{C2} = 3 I_{C1} = 3 \times 0.985\text{ mA} = 2.955\text{ mA}$.