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Semiconductor Devices and Circuits Flashcards

7 cards from real BEE practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 7 Semiconductor Devices and Circuits flashcards as text
  1. What is the effect of increasing temperature on the reverse saturation current IS of a p-n junction diode?

    Answer: IS increases approximately doubling every 10°C

    Reverse saturation current approximately doubles for every 10°C rise in temperature due to increased minority carrier generation.

  2. In a common-emitter BJT amplifier, adding an unbypassed emitter resistor RE primarily:

    Answer: Stabilizes the Q-point and increases input impedance but reduces voltage gain

    An unbypassed RE provides negative feedback that stabilizes bias, raises input impedance, but reduces the voltage gain from −RC/re to approximately −RC/RE.

  3. The output resistance of an ideal op-amp is:

    Answer: Zero

    An ideal op-amp has zero output resistance, meaning it can source or sink any current without a voltage drop.

  4. Channel length modulation in a MOSFET is modeled in the saturation region by including which factor in the drain current equation?

    Answer: (1 + λVDS)

    The channel-length modulation factor (1 + λVDS) is multiplied into the saturation current to model the slight increase in ID with VDS.

  5. Which breakdown mechanism dominates in a heavily doped p-n junction with a thin depletion region at low reverse voltages?

    Answer: Zener (tunneling) breakdown

    In heavily doped, narrow depletion junctions, quantum mechanical tunneling (Zener effect) dominates at low reverse voltages (typically below 5–6 V).

  6. In a half-wave rectifier circuit, the peak inverse voltage (PIV) across the diode is equal to:

    Answer: Vpeak

    During the negative half-cycle, the entire peak secondary voltage appears across the reverse-biased diode, so PIV = Vpeak.

  7. A silicon BJT has VBE = 0.7 V at IC = 1 mA and T = 300 K. If IC is increased to 10 mA, the new VBE is approximately:

    Answer: 0.758 V

    ΔVBE = VT × ln(10) ≈ 26 mV × 2.303 ≈ 59.8 mV, so new VBE ≈ 0.7 + 0.058 ≈ 0.758 V.