Bipolar Junction Transistors Questions and Answers

Answer: D

The source impedance is reflected the other way, divided by the current gain, and adds to the intrinsic emitter resistance:

\[ R_{out} \approx 26 + \frac{10000}{100} = 126 \ \Omega \]

so about 126 ohms. The reflection works in both directions, dividing rather than multiplying when looking the other way, and that symmetry is what makes the follower a genuine impedance converter rather than merely a buffer.

Answer: A

For a given voltage across the emitter resistor, the base supplies only a beta-th of the current, so the apparent resistance at the base is beta times larger. The junction resistance is small, the collector current does not flow in the base, and bypassing would remove the effect entirely. Whenever a terminal handles a fraction of the current at the same voltage, the impedance seen there is scaled by the reciprocal of that fraction - a rule that covers transformers and followers alike.

Answer: B

A three-to-one spread means any circuit whose gain or bias depends on beta will vary threefold between devices, so the design must be arranged so that it does not - through emitter degeneration or overall feedback. The current figure is a measurement condition rather than a limit, and neither the frequency nor the voltage rating tracks the gain spread. A data sheet range is a design constraint rather than a nuisance, and reading it that way changes what circuit gets drawn.

Answer: C

This is the base, and both its thinness and its light doping exist so that injected carriers cross it without recombining. A channel belongs to a field effect transistor, a substrate is the wafer the device is built on, and a depletion layer is the carrier-free zone at a junction. Every one of the base's properties is chosen to serve transistor action, which is why the region is so much harder to make than either of the others. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: D

This is the Early effect, and it is what gives the output characteristics their slope and the device its finite output resistance. The Miller effect is the apparent multiplication of a bridging capacitance by voltage gain, the Zener effect is a tunnelling breakdown mechanism, and the Hall effect is the transverse voltage in a magnetic field. Effects named after people are worth attaching firmly to their mechanisms, since the names carry no clue.

Answer: A

This is the transition frequency, and it equals the low-frequency beta times the beta cut-off frequency, which is the quite different point where the gain has fallen by 3 dB. A corner frequency is a general term for a filter's break point and resonance belongs to tuned circuits. Two frequencies describing the same rolling-off gain, separated by a factor of beta, is exactly the kind of pair worth keeping distinct.

Answer: B

Even with an infinite load resistor the gain saturates at the transconductance times the device's output resistance, which is the Early voltage divided by the thermal voltage - a few thousand. The supply voltage limits the swing rather than the gain, beta does not appear in the expression, and base resistance affects bandwidth and noise. A limit that survives making the external components ideal is an intrinsic device limit, and that is what makes it worth knowing.

Answer: C

A practical collector resistor is a few kilohms while the device's output resistance is hundreds of kilohms, so the load dominates and the gain is set by it. Transconductance depends only on current, saturation is a matter of the operating point, and the base current stays a beta-th of the collector current. When two resistances appear in parallel, the smaller decides the result, which is why the intrinsic limit matters only in stages loaded by current sources.

Answer: D

Replacing the collector resistor with a current source, whose impedance is very high, lets the device's own output resistance set the gain - which is how the input stage of an operational amplifier reaches its enormous gain. A large emitter resistor reduces the gain deliberately, a follower has none to speak of, and common base has the same intrinsic limit but no current gain. Removing the element that dominates a limit is the way to reach the limit underneath it.

Answer: A

A Darlington feeds one transistor's emitter into the next one's base, so the overall current gain is roughly the product of the two - thousands rather than hundreds. Parallel devices share current, a complementary pair forms a push-pull stage, and cross-coupling makes a bistable. Cascading to multiply a gain is the obvious move once a single device's gain is insufficient, and the penalty is always somewhere else.