BJT Biasing and Stability Questions and Answers

Answer: C

Both are true and the second explains the first: the product of current and voltage peaks near the middle of the load line, so that is where self-heating is worst. A stage near either extreme dissipates less, which is one reason class B output stages run cooler than class A ones. The operating point chosen for maximum swing is also the one of maximum dissipation, and both facts follow from the same curve. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: D

All three are correct, and together they describe both what makes the arrangement work and what it costs. The stiffness condition and the emitter feedback are the two mechanisms, and the loading penalty is the price. Any design rule expressed as a comparison - here divider resistance against reflected impedance - has a cost on the other side of the comparison, and stating both is what makes the rule usable. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Answer: A

The first two are correct and the third has the direction reversed: leakage roughly doubles rather than halves every ten degrees. All three of the real effects push the collector current up, which is precisely why they cannot be relied on to cancel and why stabilisation must be built into the circuit. A statement with the right mechanism and the wrong sign is the most easily overlooked kind of error. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Answer: B

The first is true and the second is not: collector feedback improves on fixed bias by opposing current changes through the base drive, but the divider-and-emitter arrangement is better still because its operating point contains no device parameter at all. Ranking the arrangements by stability factor rather than by whether they have feedback is what distinguishes them, since all three of the better ones have some. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Answer: C

The first is correct and the second is impossible: the leakage current is itself part of the collector current, so a change in it must produce at least an equal change in the total, putting a floor of one on the factor. Recognising that a figure of merit has a theoretical floor set by definition, rather than by engineering limits, prevents effort being spent chasing an unreachable target.

Answer: D

The pairing is A-P, B-Q, C-R, and it is a ranking by how much negative feedback each arrangement applies to the bias loop: none, some through the base, and a great deal through the emitter. Because the ordering follows from the mechanism, it can be reconstructed rather than memorised, which is more reliable than recalling three formulae.

Answer: A

The correct pairing is A-P, B-Q, C-R. Three components, three separable jobs, which is what makes the arrangement easy to design: the divider is chosen for the base voltage, the emitter resistor for the current, and the capacitor for the low-frequency limit. A circuit whose parameters can be set one at a time is far easier to work with than one where every choice interacts.

Answer: B

The order is 1, 2, 3, 4: decide what the stage should do, then work back through the components that determine it, and verify the assumption the calculation relied on. Doing the stiffness check last rather than first is deliberate, since it needs the emitter resistor and the beta range to evaluate. A design procedure that ends by testing its own assumption is one that will not surprise anybody later. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Answer: C

The first three are correctly paired and the fourth is not, so three match. The bypass capacitor has no effect on the stability factor at all, since it is an open circuit at direct current; what it improves is the signal gain. Attributing a direct-current benefit to a component that only acts on signals is the standard confusion in this topic, and it is worth checking every time.

Answer: D

The emitter current follows from the base voltage, and both resistors then drop the supply:

\[ \begin{aligned} I_{E} &= \frac{2.8-0.7}{1000} = 2.1 \ \mathrm{mA} \\ V_{CE} &= 12 - 2.1\mathrm{m}(4300) \approx 3.0 \ \mathrm{V} \end{aligned} \]

giving about 3 volts - usable but low, and a smaller collector resistor would centre the point. Reading a table means seeing which values combine and in what order. Checking the units of each quantity catches a wrong formula before the arithmetic begins.