Discussion
For the following transistor circuit assume \(V_{BE} = 0.7\) V, \(\beta_{dc} = 100\); then the current flowing through \(R_{E}\) is :
Answer: C
The emitter current is about 4.3 mA.
Go round the base-emitter loop from ground, through \(R_{B}\), across the base-emitter junction and through \(R_{E}\) to the -5.0 V supply:
\[ \begin{aligned} 5.0 - 0.7 &= I_{B}R_{B} + I_{E}R_{E} \\ I_{B} &= \frac{I_{E}}{\beta + 1} = \frac{I_{E}}{101} \end{aligned} \]
\[ \begin{aligned} 4.3 &= I_{E}\left(1.0 + \frac{1.0}{101}\right)\ \text{k}\Omega \\ I_{E} &= \frac{4.3}{1.0099} \approx 4.26\ \text{mA} \end{aligned} \]
which rounds to 4.3 mA. The transistor is in the active region: the collector sits at about 15 - 6.3 = 8.7 V, well above the emitter. 4.0 mA and 3.6 mA come from subtracting the drop across \(R_{B}\) as if \(I_{B}\) were much larger, and 8.0 mA would need the full 15 V across the emitter loop.
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