BJT Biasing and Stability Questions and Answers

Answer: A

The emitter resistor is used for stabilization of the operating point.

If the collector current rises, for example because \(\beta\) or the temperature increases, the voltage across \(R_{E}\) rises with it. That reduces the base-emitter voltage, which reduces the base current and pulls the collector current back. This DC negative feedback keeps the Q-point steady against transistor variation and thermal runaway.

An emitter resistor does not bypass AC signals; the capacitor placed across it does that, so that the resistor stabilises DC without cutting the AC gain. It does not provide collector bias. Unbypassed, it actually reduces the gain, so 'higher gain' is the opposite of its effect.

Answer: C

The emitter resistor is there to provide thermal stability.

If temperature rises, the collector current tends to rise with it, because leakage current and beta increase. The extra current raises the voltage across the emitter resistor, which lowers the base-emitter voltage and pulls the current back. This negative feedback holds the Q-point steady and prevents thermal runaway.

An unbypassed emitter resistor actually reduces gain and raises the input impedance, so 'increase gain' and 'reduce input impedance' are the reverse of what happens. It does not significantly reduce the output impedance.

Answer: D

Biasing in class A mode fixes the operating point near the middle of the load line so that the device conducts for the full cycle. The key accepts both the improvement of stability and the reduction in the number of sources: a properly designed bias network, such as the voltage divider with an emitter resistor, keeps the operating point steady against changes in temperature and in transistor parameters, and it lets a single supply provide both the base bias and the collector supply, which removes the need for separate sources. Saving power is not the aim, because a class A stage draws current throughout the cycle and is the least efficient class. So both (B) and (C) hold, which is the last option. The answer is both (B) and (C).

Answer: D

For a fixed bias circuit the stability factor is \(S = \dfrac{\partial I_C}{\partial I_{CO}} = \beta + 1\), the fourth option. In fixed bias the base current is set by the supply and the base resistor, so any leakage current \(I_{CO}\) is amplified by the full factor \(1 + \beta\) at the collector, which makes this circuit the least stable.

The values \(1/(1+\beta)\) and \(1/(\beta - 1)\) are less than one, which would mean good stability, and this is true of potential-divider bias and not of fixed bias. \((\beta - 1)^2\) has no such origin.