BJT Biasing and Small Signal Models Quiz Quiz

Assess your understanding of Bipolar Junction Transistor (BJT) biasing techniques and small signal models with these medium-difficulty questions. Explore key concepts such as bias stability, voltage divider circuits, small signal parameters, and related calculations commonly encountered in electronics engineering.

  1. Fixed Bias Circuit Evaluation

    In a fixed bias BJT configuration, what is the main drawback regarding the stability of the operating point, especially with temperature variations?

    1. Complicated circuit layout
    2. Requires high-frequency components
    3. High sensitivity to changes in transistor beta (β)
    4. Low input impedance

    Explanation: The primary disadvantage of a fixed bias circuit is its high sensitivity to changes in transistor beta, which can cause significant shifts in the operating point, especially as temperature varies. Complicated circuit layout is not correct because the fixed bias is relatively simple. Low input impedance is not its main issue; this is more typical of other configurations. The requirement for high-frequency components is not relevant, as fixed bias circuits are often used for low-frequency applications too.

  2. Voltage Divider Bias Stability

    Why is the voltage divider bias method generally preferred over fixed bias for BJT amplifiers, considering Q-point stability?

    1. It makes the circuit operate in cutoff mode
    2. It offers greater stability against beta variations
    3. It increases the base-emitter voltage drop
    4. It reduces circuit cost significantly

    Explanation: The voltage divider bias provides greater stability of the Q-point with respect to changes in transistor beta, largely because the base voltage is determined by resistor values, not beta. Operating the circuit in cutoff mode is not a purpose of this biasing technique. While cost can factor into design choices, reducing cost is not the main reason for choosing voltage divider bias. Increasing the base-emitter voltage drop is also not the objective; biasing aims to keep operating points stable.

  3. Small Signal Model Resistance Calculation

    When analyzing a NPN BJT in its small signal model, what does the resistance 'r_pi' (rπ) represent if the transistor is biased with a collector current Ic of 2 mA and operates at room temperature (assume Vt = 25 mV)?

    1. The output resistance looking into the collector, defined as 1 / hoe
    2. The collector resistance, given as Vce / Ic
    3. The base spreading resistance, a fixed value typically in ohms
    4. The dynamic resistance between base and emitter, calculated as β × Vt / Ic

    Explanation: In the small signal model, r_pi models the dynamic resistance seen between base and emitter and is calculated as β × Vt divided by Ic. The collector resistance given as Vce / Ic pertains to DC analysis, not to small signal resistance. The output resistance looking into the collector, defined as 1 / hoe, describes output characteristics, but not r_pi. The base spreading resistance refers to a much smaller, intrinsic resistance and is not the same as r_pi.

  4. Bypass Capacitor Function

    In a BJT amplifier using an emitter resistor with a bypass capacitor, what is the primary function of the bypass capacitor for small signal AC operation?

    1. To provide a low impedance path to ground for AC signals, maximizing voltage gain
    2. To filter power supply noise
    3. To protect the transistor from thermal runaway
    4. To block DC current flowing through the base

    Explanation: The bypass capacitor shorts the emitter resistor at AC, providing a low impedance path to ground and thus increasing voltage gain for small signals. Blocking DC current is not correct; the capacitor is intended to affect AC, not DC. Filtering power supply noise is not its purpose in this context. Protection against thermal runaway is achieved through the resistor itself, not the capacitor.

  5. Transconductance Calculation

    Given a BJT biased such that collector current Ic is 1 mA at room temperature, what is the approximate value of the transconductance (gm) in the small signal model (assume Vt ≈ 25 mV)?

    1. 0.04 S (siemens)
    2. 1 kΩ
    3. 25 mA
    4. 0.025 S

    Explanation: Transconductance gm is given by Ic divided by Vt. For Ic = 1 mA and Vt = 25 mV, gm = 0.001 / 0.025 = 0.04 S. 25 mA is simply a current value, not transconductance. 1 kΩ is a resistance, not the appropriate unit or value for gm in this context. 0.025 S is close but results from a different current value; the correct calculation gives 0.04 S.