BJT Biasing and Stability Questions and Answers

Answer: C

The capacitor shorts the emitter resistor at signal frequencies, restoring the full gain while leaving the direct stabilisation intact. Input coupling is a different capacitor, the collector is not bypassed in a normal stage, and the stabilisation comes from the resistor the capacitor is removing. One component doing nothing for direct current and everything for signals is what lets bias stability and gain be optimised independently. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: D

The relevant resistance is the emitter resistor in parallel with the low impedance looking into the emitter, which is only tens of ohms - so the capacitor must be large. The collector resistor, the divider and the load all set other break points in the response. Because the resistance here is small, this is usually the capacitor that dominates a stage's low-frequency limit, which surprises people who size it from the emitter resistor alone.

Answer: A

The resistance at the emitter is tens of ohms against a coupling capacitor's tens of kilohms, so the same cut-off frequency demands a capacitance perhaps a thousand times larger. It passes only the signal current, not the quiescent current, and the charge and voltage arguments do not apply. A time constant is a product, so a small resistance always forces a large capacitance if the frequency is to be reached. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: B

Subtracting the junction drop and dividing:

\[ \begin{aligned} I_{B} &= \frac{12 - 0.7}{470 \ \mathrm{k}} \\ &\approx 24 \ \mathrm{\mu A} \end{aligned} \]

giving about 24 microamps. With a beta of 150 that is 3.6 mA of collector current, and with a beta of 300 it would be 7.2 mA - the very spread that makes fixed bias unusable in a linear stage. The calculation is easy; the problem is what it depends on. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Answer: C

Multiplying the base current by the gain:

\[ I_{C} = 100 \times 24 \ \mathrm{\mu A} = 2.4 \ \mathrm{mA} \]

so 2.4 mA. If beta were 300 the current would be 7.2 mA and the stage would very likely saturate. Working out both extremes of the beta range rather than the nominal value is what reveals whether a bias arrangement will survive a device change.

Answer: D

Applying the divider ratio:

\[ V_{B} = 12 \times \frac{10}{57} \approx 2.1 \ \mathrm{V} \]

giving about 2.1 volts, before allowing for the base current the divider also has to supply. The unloaded ratio is the starting point and the loading correction is small in a stiff divider, which is exactly the approximation stiffness buys.

Answer: A

The two arms appear in parallel to the base:

\[ \begin{aligned} R_{th} &= \frac{47 \times 10}{57} \\ &\approx 8.2 \ \mathrm{k\Omega} \end{aligned} \]

so about 8.2 kilohms. For a stiff divider this must be well below the impedance looking into the base, which with a 1.2 k emitter resistor and a beta of 100 is over 120 k - so the condition is comfortably met. Computing both numbers and comparing them is the whole stiffness check.

Answer: B

The emitter current is set by the base voltage and the emitter resistor, neither of which depends on beta:

\[ I_{E} = \frac{V_{B} - 0.7}{R_{E}} \]

so the change is under one per cent, arising only from the slightly reduced base current loading the divider. That near-immunity to beta is what the arrangement was chosen for, and it is why the same circuit works with any device in the family.

Answer: C

Adding the rise to the ambient:

\[ T_{j} = 40 + (5 \times 8) = 80 \ \mathrm{^\circ C} \]

giving 80 degrees, with reasonable margin below a 150 degree limit. Every thermal calculation is the same series circuit with power in place of current, and the ambient is the reference the whole rise sits on top of.

Answer: D

Fifty-five degrees is between five and six doublings:

\[ 1 \times 2^{5.5} \approx 45 \ \mathrm{\mu A} \]

so about 50 microamps. Amplified by beta in a common-emitter stage that becomes milliamps, which is why leakage matters so much more in hot power stages than on a cool bench. Exponential dependences must be evaluated rather than estimated linearly.