Bipolar Junction Transistors Questions and Answers

Answer: A

The capacitor passes a burst of current at each edge, charging and then removing base charge quickly, while the resistor sets the steady drive. It does not reduce the steady current, offers no reverse protection, and cannot raise the available collector current. Giving a circuit a different response to transitions than to steady conditions is a recurring trick, and a capacitor across a resistor is its simplest form.

Answer: B

The emitter-base junction breaks down at only five to seven volts, because the emitter is heavily doped, and exceeding it degrades beta permanently even without destroying the device. Collector-base breakdown is the highest and collector-emitter falls between. Reverse voltage on the base-emitter junction is a common and easily overlooked hazard, and it is why a protective diode often appears across it.

Answer: C

Avalanche in the emitter junction damages the surface, so beta falls permanently even if the current was limited and the device still functions. That silent degradation is what makes the rating so treacherous: nothing fails outright, but a circuit's gain drifts and the cause is invisible. Any stress that degrades rather than destroys is worth designing against explicitly, because it will not announce itself.

Answer: D

Beta peaks at a moderate current: at low current recombination in the base takes a larger share, and at high current emitter crowding and high-level injection reduce the efficiency. The Early voltage and the thermal voltage are not current dependent in this way. That bell-shaped curve is why data sheets specify beta at a stated current, and why designs are placed near the peak where the parameter is least sensitive.

Answer: A

Longer carrier lifetime means less recombination in the base, so a larger fraction of the injected current reaches the collector and beta rises - typically half a per cent per degree. The falling base-emitter voltage is a separate temperature effect that also raises collector current, and leakage adds to the collector current rather than the base. Two independent mechanisms both pushing the current up is precisely why thermal runaway is a real risk without stabilising feedback.

Answer: B

Beta rises and the base-emitter drop falls, and both changes increase the collector current at a fixed bias - which, since more current means more heat, closes a positive feedback loop. Leakage rising is a third contributor in the same direction. That all the significant temperature effects push the same way is what makes bias stabilisation the central concern of the next topic rather than an afterthought.

Answer: C

Adding the rise across the junction-to-case path:

\[ T_{j} = 60 + (2 \times 3) = 66 \ \mathrm{^\circ C} \]

giving 66 degrees. The value 63 uses one watt and 126 doubles the case temperature. Thermal problems are all the same series circuit with temperature in place of voltage and power in place of current, which makes them easy once the analogy is accepted.

Answer: D

Interpolating along the derating line:

\[ \begin{aligned} P &= 10 \times \frac{150-75}{150-25} \\ &= 10 \times 0.6 = 6 \ \mathrm{W} \end{aligned} \]

so 6 watts. Five watts would be the midpoint of the temperature range rather than of the power range. A rating quoted at one temperature is nearly always a straight line to zero at the limit, and reading it that way is safer than assuming the headline figure applies.

Answer: A

Both are true and the second is the reason: beta can vary three to one across a production batch and drifts with temperature and operating current, so any performance figure proportional to it is unpredictable. The remedy is a circuit whose behaviour depends on resistor ratios instead, which is exactly what emitter degeneration and negative feedback achieve. Designing so that the least reliable parameter cancels out is the general lesson. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: B

The assertion is true but the reason is wrong: common base has substantial voltage gain; what it lacks is the inverting gain from input to output across the collector-base capacitance, so there is no Miller multiplication. The input is the emitter and the capacitance in question does not bridge it. A correct conclusion resting on a wrong mechanism is the most instructive error, because it survives every check except tracing the cause. Checking the units of each quantity catches a wrong formula before the arithmetic begins.