Small Signal Amplifiers Questions and Answers

Answer: C

Multiplying transconductance by load resistance:

\[ |A_{v}| = 0.08 \times 1500 = 120 \]

giving 120. The transconductance of eighty millisiemens corresponds to two milliamps of collector current, so the same answer follows from dividing 1500 by the thirteen-ohm intrinsic emitter resistance. Two routes to one answer is a useful check, and it also shows that the two forms of the gain expression really are the same statement.

Answer: D

The two resistances appear in parallel:

\[ \begin{aligned} 1.5 \ \mathrm{k} \| 100 \ \mathrm{k} \\ &\approx 1.48 \ \mathrm{k\Omega} \end{aligned} \]

so about 1.48 kilohms, within two per cent of the load resistor alone. That is why the device's output resistance is normally ignored in a resistively loaded stage, and why it becomes the dominant term as soon as the load is replaced by a current source.

Answer: A

Two resistances in parallel are dominated by the smaller, so the device's output resistance matters only once the external load becomes comparable with it - which happens with a current-source load or an active load. Operating current, bypassing and frequency change other things. Knowing the condition under which a neglected term becomes significant is what lets the same analysis be reused in a quite different circuit without surprises.

Answer: B

An active load is a transistor, usually configured as a current source, standing in for the collector resistor so that the load impedance is very high. The gain then rises towards the device's intrinsic limit. It is not a signal-dependent impedance, not a matching resistor, and not a shaping capacitor. Replacing a passive component with a device that behaves better in the one respect that matters is a recurring move in integrated design, and this is its commonest form.

Answer: C

Integrated resistors are long strips of lightly doped material, so a hundred kilohms occupies an area a transistor delivering the same impedance does not need. Transistors are not universally smaller, dissipation is comparable at the same current, and large resistors can be made, just wastefully. What is cheap in a technology shapes its circuits, and area is the currency in integration.

Answer: D

A current source needs a volt or so across it to stay in its useful region, so the output cannot swing as close to the supply rail as it could with a resistor. The gain rises rather than falls, does not become beta-dependent, and the arrangement works down to direct current. Trading headroom for gain is the exchange, and it is affordable precisely because integrated stages are usually followed by more gain rather than by a load.

Answer: A

Two matched devices sharing a temperature drift together, so the difference between their inputs is amplified while the common drift cancels. Its gain per device is no higher than a single stage's, it needs a tail current source, and single-rail operation is unrelated. Arranging a circuit so that the imperfection appears as a common-mode quantity and the signal as a differential one is the foundation of every precision analogue front end.

Answer: B

Gain is proportional to the effective load, so the ratio of gains gives the ratio of loads:

\[ \frac{2000}{40} = 50 \]

so the device's output resistance is about fifty times the old collector resistor, which is now the only thing limiting the gain. Reading a hidden parameter from the change a circuit modification produces is often quicker than measuring it directly.

Answer: C

Each stage's load is halved by the next stage's input impedance if the two are equal, so each gain falls to about seventy-five:

\[ (-75) \times (-75) \approx 5600 \]

and with lighter loading the figure rises towards about eleven thousand, the product of one loaded and one unloaded stage. The naive 22500 assumes no loading at all. Two inversions give a positive result, which is worth noting before the magnitude is considered.

Answer: D

Only the stages that drive another stage suffer interstage loading, so the final one sees whatever external load the design intends, which may be much lighter. Operating current, bypassing and input impedance vary by design rather than by position. Recognising that loading is a property of the interface rather than of the stage explains why the stages of a cascade often differ in measured gain despite being identical on paper.