BJT Biasing and Stability Questions and Answers

Answer: A

This is the quiescent point, often written as the Q point, and it is what the bias network exists to establish. Saturation and cut-off are the two extremes of the load line that the quiescent point must sit between, and pinch-off belongs to field effect transistors. A term for the resting condition is needed precisely because everything about a stage's behaviour is measured as an excursion from it.

Answer: B

This is the stability factor, ideally unity and at worst one plus beta. Form factor and ripple factor describe rectifier waveforms, and damping belongs to resonant and control systems. Because it is defined as a sensitivity, it compares arrangements directly in a way that quoting their nominal operating points never could.

Answer: C

Bypassing the emitter resistor does nothing for stability, since the capacitor is an open circuit at direct current - it restores the signal gain instead. The other three all add feedback or reduce beta's influence on the operating point. Sorting the components of a stage into those acting on the bias and those acting on the signal is the single most useful distinction in reading these circuits.

Answer: D

The thermal voltage rises rather than falls with temperature, being proportional to absolute temperature, and in any case its effect on the operating point is small. The other three are the standard trio, all pushing collector current up. A distractor with the correct mechanism but the wrong direction is worth checking against the underlying physics rather than against memory.

Answer: A

The collector cannot go above the supply, so the upward swing is the difference between the supply and the quiescent collector voltage. Saturation voltage limits the downward swing instead. Base current and capacitor reactance limit the drive and the low-frequency response rather than the swing. Because the two directions are limited by different things, both have to be computed before the available output is known. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: B

Whichever direction has less headroom clips first, so the undistorted output is set by the shorter of the two excursions. The gain hardly changes over a modest range of operating points, dissipation is a separate concern, and the load line's slope is fixed by the resistors. A limit set by the smaller of two quantities is always improved by balancing them, which is exactly what centring the operating point does. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: C

Class A means the device never cuts off, conducting for the whole cycle, which gives the lowest distortion and the worst efficiency, at best twenty-five per cent for a resistive load. Half-cycle conduction is class B, short pulses class C, and alternate conduction is the push-pull arrangement that class B usually adopts. The class is defined entirely by the conduction angle, and every other property follows from it. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Answer: D

The quiescent current flows continuously, so the device dissipates its maximum with no output at all, and efficiency rises only as the signal approaches full swing. Device size and bias current are consequences rather than causes, and no transformer is necessarily involved. A stage that consumes the same power whether or not it is doing anything is inefficient by construction, which is why class B displaced it in output stages. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: A

Dissipation is the quiescent power minus whatever is delivered to the load, so it is greatest with no signal and falls as the output rises. That is the reverse of the intuition most people bring to it, and it means a class A stage must be sized for its idle condition. A design whose worst case is the do-nothing condition is unusual enough to be worth remembering explicitly.

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.