BEE Semiconductor Devices and Electronics 3 — Questions and Answers
Question 1: In a CMOS inverter, static power dissipation is nearly zero because:
- Both transistors are always in saturation
- Only one transistor conducts at a time in steady state (Correct answer)
- The supply voltage is very low
- Both transistors have identical threshold voltages
Correct answer: Only one transistor conducts at a time in steady state
In steady state, either the PMOS or NMOS is OFF, creating no DC path from VDD to GND and eliminating static power.
Question 2: The built-in potential (contact potential) of a p-n junction at thermal equilibrium is caused by:
- Applied external voltage
- Diffusion of majority carriers and resulting charge separation (Correct answer)
- Generation-recombination in the depletion region
- Minority carrier injection
Correct answer: Diffusion of majority carriers and resulting charge separation
Diffusion of holes into the n-side and electrons into the p-side creates charge separation, establishing the built-in potential.
Question 3: Hall effect measurements in a semiconductor are used to determine:
- Bandgap energy and dielectric constant
- Carrier concentration and carrier type (Correct answer)
- Junction capacitance and built-in potential
- Minority carrier lifetime and diffusivity
Correct answer: Carrier concentration and carrier type
The Hall effect reveals carrier type (n or p) from the sign of the Hall voltage and carrier density from its magnitude.
Question 4: Which breakdown mechanism dominates in heavily doped p-n junctions at low reverse voltages?
- Avalanche breakdown
- Zener (tunneling) breakdown (Correct answer)
- Thermal breakdown
- Punch-through
Correct answer: Zener (tunneling) breakdown
Zener breakdown (band-to-band tunneling) occurs in heavily doped junctions with narrow depletion regions at voltages typically below 5 V.
Question 5: The transconductance of a BJT in the active region is given by gm =
- IC / VT (Correct answer)
- IB / β
- VCE / IC
- IC / VCC
Correct answer: IC / VT
For a BJT, gm = IC/VT where IC is collector current and VT = kT/q ≈ 26 mV at room temperature.
Question 6: In an LED, light is emitted due to:
- Blackbody radiation from resistive heating
- Radiative recombination of electron-hole pairs (Correct answer)
- Photon absorption increasing carrier energy
- Zener breakdown in the depletion region
Correct answer: Radiative recombination of electron-hole pairs
In an LED, forward-biased minority carriers recombine radiatively, releasing photons with energy equal to the bandgap.
Question 7: For an ideal op-amp, which of the following is true?
- Finite input impedance, zero output impedance, finite gain
- Infinite input impedance, zero output impedance, infinite open-loop gain (Correct answer)
- Zero input impedance, infinite output impedance, finite gain
- Infinite input and output impedance, unity gain
Correct answer: Infinite input impedance, zero output impedance, infinite open-loop gain
An ideal op-amp has infinite input impedance, zero output impedance, and infinite open-loop differential gain.
BEE Semiconductor Devices and Electronics 3
In a CMOS inverter, static power dissipation is nearly zero because: