Bipolar Junction Transistors Questions and Answers

Answer: D

The correct matching is a-i, b-iv, c-ii, d-iii.

• BJT: a high current gain β is its characteristic figure of merit (i).
• JFET: the gate-channel junction is operated reverse biased (iv).
• Zener diode: used in reverse breakdown as a voltage regulator (ii).
• MOSFET: an insulated gate controls the drain current by voltage (iii).

The other codes give the BJT a reverse biased gate or a voltage-regulator role, or give the JFET the insulated gate, which belongs to the MOSFET; a JFET has no oxide layer between gate and channel.

Answer: B

The output is 2·5 V.

With \( V_{in} = 0 \) there is no forward bias on the base-emitter junction, so the transistor is cut off and draws no collector current. \( R_{C} \) and \( R_{L} \) then form a simple voltage divider across the 5 V supply:

\[ \begin{aligned} V_{o} &= 5 \times \frac{R_{L}}{R_{C} + R_{L}} \\ &= 5 \times \frac{10}{20} = 2.5\ \text{V} \end{aligned} \]

Zero would be the output of a saturated transistor, which needs base drive. The full 5 V would appear only with no load connected, and 4·55 V or 1·0 V do not follow from equal 10 kΩ resistors.

Answer: A

The correct statements are 1 and 2.

In the Ebers-Moll model the BJT is two back-to-back diodes, base-emitter and base-collector, with controlled current sources. When both junctions are forward biased the transistor is in saturation: it conducts heavily with only a small collector-emitter voltage, a low impedance state, like a closed switch (1). When both are reverse biased it is in cut-off: almost no current flows, a high impedance state, like an open switch (2).

Statements 3 and 4 swap these states, so every option containing either of them is wrong.

Answer: C

The current amplification factors are 99 and 49 respectively.

\[ \beta = \frac{\alpha}{1 - \alpha} \]

\[ \begin{aligned} \beta_{1} &= \frac{0.99}{0.01} = 99 \\ \beta_{2} &= \frac{0.98}{0.02} = 49 \end{aligned} \]

A small fall in alpha from 0.99 to 0.98 halves beta, because \(1 - \alpha\) doubles. '99 and 98' wrongly scales beta in proportion to alpha, and the options with 99 second reverse the two transistors.

Answer: A

The mutual conductance is about 0.23 mho.

For a bipolar transistor the small-signal transconductance depends only on the collector current and the thermal voltage:

\[ \begin{aligned} V_{T} &= \frac{kT}{q} \approx 25.9\ \text{mV at } 300\ \text{K} \\ g_{m} &= \frac{I_{C}}{V_{T}} = \frac{6 \times 10^{-3}}{25.9 \times 10^{-3}} \\ &\approx 0.23\ \text{S (mho)} \end{aligned} \]

The current gain h_FE = 120 is not needed for g_m; it would be used to find r_π = β/g_m, about 520 Ω. 0.02 mho is ten times too small, and 2.3 or 4.33 mho would need currents of tens of milliamperes.

Answer: D

For amplification the junctions are biased forward and reverse respectively.

In the active region the base-emitter junction is forward biased, so the emitter injects carriers into the thin base, and the base-collector junction is reverse biased, so its field sweeps those carriers into the collector. The collector current is then controlled by the base current and nearly independent of the collector voltage, which is what makes the transistor an amplifier.

Both junctions forward biased is saturation, both reverse biased is cut-off (the two states used in switching), and reverse-forward is the inverse active mode, which gives very poor gain.