Bipolar Junction Transistors Questions and Answers

Answer: C

Most majority carriers from the emitter pass through the base region to the collector.

In a bipolar transistor in the active region, the emitter is heavily doped and the base is thin and lightly doped. Carriers injected from the emitter into the base diffuse across it, and because the base is so thin and has few carriers to recombine with, typically more than 95 to 99 percent of them reach the reverse-biased collector junction and are swept into the collector. That is why the collector current is nearly equal to the emitter current (\(\alpha \approx 1\)).

Only a small fraction recombine in the base, forming the base current; they do not recombine in the emitter or remain in the base.

Answer: B

The collector current is 1.5 mA.

The current gain \(\beta\) relates collector current to base current:

\[ \begin{aligned} I_{C} &= \beta I_{B} = 50 \times 30\ \mu\text{A} \\ &= 1500\ \mu\text{A} = 1.5\ \text{mA} \end{aligned} \]

15 mA and 150 mA come from slipping the decimal point by one or two places when converting microamperes to milliamperes, and 1.5 A treats the base current as 30 mA. A useful check is that small-signal transistors biased with tens of microamperes of base current carry collector currents of a few milliamperes.

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.

Answer: D

\(h_{FE}\) is the forward current transfer ratio.

In h-parameter notation the first subscript names the parameter (i input impedance, r reverse voltage ratio, f forward current ratio, o output admittance) and the second the configuration (e common emitter). Capital subscripts mean the DC value, so

\[ h_{FE} = \frac{I_{C}}{I_{B}} = \beta_{dc} \]

the DC current gain of the transistor in common emitter.

The reverse transfer parameter is \(h_{re}\), a voltage ratio rather than a current ratio. Output behaviour is described by \(h_{oe}\), an admittance. Voltage gain is a circuit quantity that depends on the load, not an h-parameter of the transistor.

Answer: C

For a transistor working as an amplifier, the emitter is forward biased and the collector is reverse biased.

This is the active region. Forward bias on the emitter-base junction injects carriers from the emitter into the thin base; reverse bias on the collector-base junction sets up a field that sweeps almost all of them into the collector, giving \(I_{C} \approx \beta I_{B}\). This holds whatever the configuration, CE, CB or CC.

Both junctions forward biased is saturation, where the transistor acts as a closed switch. Both reverse biased is cut-off, an open switch. Emitter reverse and collector forward is the inverse-active mode, which works but with very low gain and is not used for normal amplification.