Bipolar Junction Transistors Questions and Answers

Answer: C

With the emitter resistance dominant, the gain is set by the collector and emitter resistors, both of which are far more predictable than the operating current or beta. The transconductance is unchanged, bias stability is a related but separate benefit, and the input impedance rises rather than falls. Whenever a performance figure can be made a ratio of passive components, it becomes as good as those components, which is the whole appeal of feedback.

Answer: D

The hybrid-pi model is a linearised picture valid for small excursions about a chosen bias point, with a transconductance, an input resistance and an output resistance. Switching, thermal and breakdown behaviour all need other models. Replacing a non-linear device by a linear model about one point is the standard way to make amplifier analysis tractable, and the model's validity is exactly as wide as the signal is small. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Answer: A

The finite slope of the output characteristics appears as a resistance from collector to emitter, and it caps the gain a stage can reach even with an infinite load. Transconductance follows from the operating current, input resistance from beta and transconductance together, and base spreading resistance is a physical bulk resistance. Every element of a small-signal model corresponds to one physical effect, and tracing that correspondence is what makes the model memorable.

Answer: B

Base spreading resistance works with the junction capacitances to form a low-pass network, so it limits bandwidth and adds noise. It has little effect on the direct current gain, and neither voltage rating nor saturation voltage depends on it. It is also why a thin base, which helps transistor action, hurts the high-frequency noise figure - a genuine conflict the device designer has to resolve rather than avoid.

Answer: C

The transition frequency is where the common-emitter current gain has fallen to unity, and it equals the low-frequency beta times the beta cut-off frequency. The 3 dB point of the voltage gain depends on the circuit rather than the device. The remaining options describe nothing standard. A gain-bandwidth product expressed as a single frequency lets two devices be compared regardless of the gain each is used at.

Answer: D

Dividing the transition frequency by the low-frequency gain:

\[ \begin{aligned} f_{\beta} &= \frac{300 \ \mathrm{MHz}}{100} \\ &= 3 \ \mathrm{MHz} \end{aligned} \]

giving 3 MHz, beyond which the current gain starts to fall. The other options misplace a decade. The reciprocal relation between gain and bandwidth is what the transition frequency encodes, and it means a stage run at lower gain automatically has more bandwidth available.

Answer: A

At high frequencies the base-emitter capacitance takes an increasing share of the drive current, so less of it produces collector current. Base thickness is fixed, leakage is a temperature effect, and the Early effect is a voltage effect. A gain that falls with frequency is nearly always a capacitance diverting something, and finding which capacitance and where it goes is the whole of high-frequency analysis.

Answer: B

Because the collector swings in antiphase to the base, the collector-base capacitance appears at the input multiplied by one plus the voltage gain, which usually dominates the stage's bandwidth. The emitter resistance is indeed multiplied at the base, but that is the impedance-reflection rule rather than the Miller effect. A small capacitance across an inverting gain becomes a large one, which is why the cascode arrangement, with no voltage gain at the lower device, exists at all.

Answer: C

In saturation the collector-emitter voltage is a few tenths of a volt, so the product of current and voltage is small and the device stays cool. Beta is irrelevant there, since the current is set by the external circuit. Off-state leakage is a cut-off property, and heavy drive actually slows turn-off because the stored charge must be removed. A switch wants low voltage when on and low current when off, and saturation delivers the first of those. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Answer: D

Overdrive fills the base with excess stored charge, and that charge must be removed before the device can begin to turn off, which appears as storage delay. The saturation voltage actually falls slightly with more drive, and neither leakage nor breakdown voltage is affected. The remedy is a Schottky clamp or a speed-up capacitor, both of which prevent the device from saturating deeply - a good example of designing around a limit rather than accepting it.