Bipolar Junction Transistors Questions and Answers

Answer: A

The base must be thin compared with the diffusion length, so carriers injected from the emitter cross it and reach the collector before recombining - which is what makes the collector current nearly equal to the emitter current and the base current small. A thin base actually raises the base resistance, and capacitance and breakdown voltage are affected in ways that constrain rather than motivate the design. Transistor action is entirely a matter of carriers surviving the base, and every geometric choice serves that.

Answer: B

Light base doping means few majority carriers there to recombine with the injected ones, so the emitter injection efficiency is high and almost all the emitter current becomes collector current. The raised bulk resistance is an unwanted consequence rather than the aim, and the two remaining effects are undesirable. Doping asymmetry is what makes the emitter inject far more than the base does back into it, and that asymmetry is why the device is not symmetrical.

Answer: C

The active region needs the emitter junction forward biased to inject carriers and the collector junction reverse biased to collect them. Both forward is saturation, both reverse is cut-off, and the reversed arrangement is the inverse active region, which works but with a much poorer gain because the doping is wrong for it. Naming the four regions by their two junction biases is the compact way to hold them, and it makes the bias conditions in a circuit immediately readable.

Answer: D

In saturation both junctions are forward biased, the collector can no longer take all the carriers offered, and the collector-emitter voltage falls to a few tenths of a volt. The collector current then depends on the external circuit rather than on the base current, which is precisely what a switching application wants. Recognising saturation from the junction biases rather than from a current ratio is more reliable, because the ratio is what breaks down there.

Answer: A

Cut-off has both junctions reverse biased, so only leakage flows and the device is effectively open. Both forward is saturation and the mixed case is the active region. Equal base and collector currents would mean a beta of one, which describes no useful region at all. Since a switch spends its life in cut-off or saturation, these two regions matter far more in digital work than the active region between them.

Answer: B

Alpha is the ratio of collector to emitter current, so it is a little less than unity - typically 0.98 to 0.998. Collector over base is beta, and emitter over base is gamma, one more than beta. Because alpha is so close to one, small changes in it produce large changes in beta, which is why beta varies so widely between nominally identical devices and why no good design depends on its exact value.

Answer: C

Converting between the two gains:

\[ \beta = \frac{\alpha}{1-\alpha} = \frac{0.98}{0.02} = 49 \]

so 49. Ninety-eight mistakes alpha's decimals for the answer, and fifty would follow from an alpha of 0.9804. The steepness of this relation is the important part: raising alpha from 0.98 to 0.99 doubles beta to 99, which is why manufacturing spreads in alpha appear as enormous spreads in beta.

Answer: D

Inverting the relation:

\[ \alpha = \frac{\beta}{1+\beta} = \frac{99}{100} = 0.99 \]

giving 0.99. The value 1.01 is impossible, since alpha must be less than one for a device that cannot deliver more collector current than emitter current. Checking that a computed alpha lies below unity is a quick sanity test, and it catches the commonest algebraic slip in this pair of formulae.

Answer: A

Applying beta and then Kirchhoff's current law at the device:

\[ \begin{aligned} I_{C} &= 100 \times 20\mathrm{\mu} = 2 \ \mathrm{mA} \\ I_{E} &= 2 + 0.02 = 2.02 \ \mathrm{mA} \end{aligned} \]

so 2 mA and 2.02 mA. The emitter current is the largest of the three, since it supplies both the others - a check worth making, because an answer with the emitter current smallest has a sign error somewhere.

Answer: B

The emitter current is largest, because it supplies both the collector and the base: the three are related by \( I_{E} = I_{C} + I_{B} \). The collector current is very nearly as large, and the base current is smaller by a factor of beta. They are certainly not equal. Getting the ordering right is a useful check on any analysis, and it follows from current conservation alone rather than from any device parameter.