Bipolar Junction Transistors Questions and Answers

Answer: A

Increasing the voltage gain is not among them - an unbypassed emitter resistor reduces the gain, and that reduction is precisely the price paid for the other three benefits. Bias stability, predictability and a higher input impedance all follow from the same local feedback. A component added for three good reasons usually costs something, and recognising which of the four listed effects runs the other way is the whole of this item. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: A

A bipolar junction transistor (BJT) has three regions and three terminals: the emitter, the base and the collector. The emitter supplies charge carriers, the thin base controls the flow, and the collector gathers the carriers.

The other lists name the terminals of other devices: anode, cathode and gate are the terminals of a thyristor (SCR), base 1, base 2 and emitter (not gate) belong to a unijunction transistor, and drain, source and gate are the terminals of a field effect transistor. The answer is option A.

Answer: C

The avalanche effect takes place in the breakdown region.

When the reverse voltage across a junction, such as the collector-base junction of a transistor, becomes large, carriers crossing the depletion region gain enough energy to knock electrons out of the lattice. These free more carriers in turn, and the current multiplies rapidly. This avalanche multiplication is what defines the breakdown region of a transistor's characteristics.

In the active, saturation and cut-off regions the junction voltages are far below the breakdown value, so no avalanche occurs.

Answer: C

The input and output are 180° out of phase.

In a common emitter amplifier, an increase in base voltage raises the base current and therefore the collector current. The larger collector current increases the drop across the collector load resistor, so the collector (output) voltage falls. The output therefore moves opposite to the input: a phase reversal of 180°.

0° is the phase relation of the common base and common collector (emitter follower) circuits, and 90° and 270° do not arise in a resistively loaded amplifier stage.

Answer: C

The insufficient bias is \(V_{BB} \lt 0.2\) V.

A germanium junction needs a forward voltage of about 0.2 to 0.3 V before appreciable current flows (a silicon junction needs about 0.7 V). A base supply below 0.2 V cannot overcome the emitter-base barrier, so the transistor stays cut off.

0.7 V, anything above 0.7 V, and the range between 0.2 V and 0.7 V all reach or exceed the germanium cut-in voltage, so they can provide the forward bias the question asks about.

Answer: A

The current gain is \(\beta = I_{C}/I_{B}\).

\(\beta\) is the common-emitter current gain: the ratio of collector current to base current. Typical values run from about 50 to several hundred.

\(I_{C}/I_{E}\) is \(\alpha\), the common-base current gain, which is slightly less than 1. The two are linked by \(\beta = \frac{\alpha}{1-\alpha}\). \(I_{B}/I_{C}\) and \(I_{B}/I_{E}\) are the reciprocal-type ratios and are much smaller than one, so neither is a gain.

Answer: C

The largest current flows in the emitter.

In a bipolar junction transistor the emitter supplies all the carriers. Most of them cross the thin base into the collector and a small fraction recombine in the base, so the emitter current divides into the other two: I_E = I_B + I_C.

The collector current is close to the emitter current but always a little smaller (about 0.95 to 0.99 of it), and the base current is the smallest of the three, typically one hundredth of the collector current. The emitter current is the sum, so it is the largest.

Answer: B

The total collector current is \(I_{c} = \alpha I_{E} + I_{CO}\).

In the common base configuration a fraction \(\alpha\) of the emitter current crosses the base and reaches the collector. In addition, the reverse-biased collector-base junction passes a small leakage current \(I_{CO}\) (also written \(I_{CBO}\)) of minority carriers, which flows even with the emitter open. The total collector current is the sum of the two.

\(I_{c} = \alpha I_{E}\) leaves out the leakage component, so it is not the total. \(\beta\) is the current gain relating the collector current to the base current in the common emitter configuration, so \(\beta I_{E}\) is wrong, and the first option has the terms and signs mixed up.

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.