BJT Biasing and Stability Questions and Answers

Answer: B

Multiplying supply by current:

\[ P = 12 \times 20\mathrm{m} = 240 \ \mathrm{mW} \]

giving 240 milliwatts, of which some is in the resistors and the rest in the device. At best a quarter of it reaches the load, so the useful output is around sixty milliwatts. Computing the power drawn before the power delivered puts the efficiency question in the right order.

Answer: C

Class A with a resistive load reaches at best a quarter of the input power:

\[ P_{out} \le 0.25 \times 240 = 60 \ \mathrm{mW} \]

so about 60 milliwatts. Transformer coupling raises the ceiling to fifty per cent by removing the standing drop across the collector resistor, which is exactly why output transformers persisted as long as they did. A theoretical efficiency limit is worth attaching to its coupling arrangement rather than to the class alone.

Answer: D

A transformer's winding has negligible resistance, so the collector sits at the supply and can swing to nearly twice it, doubling the available swing for the same current. Impedance matching is a real benefit but a separate one, the conduction angle is unchanged, and the quiescent current still has to flow. Removing a standing voltage drop is a recurring way to improve efficiency, and it is why choke and transformer loads appear wherever supply voltage is scarce.

Answer: A

The signal sees the collector resistor in parallel with any coupled load, and any bypassed emitter resistor as a short, so that combination sets the slope. The supply and total resistance define the direct load line, and the quiescent current locates the pivot rather than the slope. The device's output resistance is usually far higher and negligible. Two load lines with different slopes through one point is the standard picture, and knowing which resistances belong to each is the whole of drawing it.

Answer: B

Because the signal sees a smaller effective resistance, its load line pivots about the quiescent point with a steeper slope and reaches further in current for the same voltage change. The supply does not rise, the bypass capacitor maintains the emitter rather than adding drive, and saturation voltage changes are trivial. That two different load lines apply to the same stage under different conditions is easy to miss and changes the swing calculation.

Answer: C

Driven beyond its range, the output flattens at both extremes, against saturation on one side and cut-off on the other. A stage biased off-centre clips one side first, but sufficient overdrive reaches both. Nothing is permanent and the distortion is plainly visible. Symmetrical clipping is itself diagnostic: it says the operating point is roughly centred and the drive is simply too large.

Answer: D

Clipping on one side well before the other means the quiescent point is nearer one extreme of the load line than the other, so one direction of swing runs out first. Load impedance, bypass capacitance and source impedance all affect gain and response rather than producing this asymmetry. The shape of a distorted waveform carries information about its cause, and this is the clearest example of it.

Answer: A

The first two are correct and the third overstates the limit: a resistively loaded class A stage reaches at best twenty-five per cent, and only transformer coupling raises the ceiling to fifty. Quoting the transformer figure for a resistive stage is the standard error here, and it matters because the two differ by a factor of two. An efficiency limit belongs to a coupling arrangement as much as to a class of operation. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Answer: B

The current gain does not enter into it: the swing is bounded by the supply, the quiescent point and the alternating load, all of which are circuit quantities. Beta affects how much drive is needed to reach that swing but not how far it can go. Separating what limits an output from what is needed to reach it is a distinction worth keeping, because the remedies are quite different.

Answer: C

A larger emitter resistor gives better stabilisation, because a given current change produces a larger opposing voltage, but it also consumes supply voltage that would otherwise be available for the output to swing into. A common rule of thumb allots about a tenth of the supply to it. The other pairings describe real trades governed by other components. Any component whose benefit and cost both scale with its value has an optimum rather than a direction, and stating both sides is what makes the choice defensible.