The silicon sample with unit cross-sectional area shown below is in thermal equilibrium. The following information is given: T = 300 K electronic charge = 1.6 × 10-19 C, thermal voltage = 26 mV and electron mobility = 1350 cm2 / V-s
The magnitude of the electron of the electron drift current density at x = 0.5 μm is
Answer : Option AExplaination / Solution:
Electron drift current density
The amplifier circuit shown below uses a silicon transistor. The capacitors Cc and CE can be assumed to be short at signal frequency and effect of output resistance r0 can be ignored. If CE is disconnected from the circuit, which one of the following statements is true
Answer : Option AExplaination / Solution:
The equivalent circuit of given amplifier circuit (when CE is connected, RE is short-circuited)
For a N -point FET algorithm N = 2m which one of the following statements is TRUE ?
Answer : Option DExplaination / Solution:
For an N-point FET algorithm butterfly operates on one pair of samples and involves two complex addition and one complex multiplication.
In a uniformly doped BJT, assume that NE, NB and NC are the emitter, base and collector doping in atoms/cm3, respectively. If the emitter injection efficiency of the BJT is close unity, which one of the following condition is TRUE
Answer : Option BExplaination / Solution:
Emitter injection efficiency is given as
Compared to a p-n junction with NA = ND = 10-14/cm3, which one of the following statements is TRUE for a p-n junction with NA = ND = 10-20/cm3?
Answer : Option CExplaination / Solution:
Reverse bias breakdown or Zener effect occurs in highly doped PN junction through tunneling mechanism. In a highly doped PN junction, the conduction and valence bands on opposite sides of the junction are sufficiently close during reverse bias that electron may tunnel directly from the valence band on the p-side into the conduction band on n-side.
Breakdown voltage
So, breakdown voltage decreases as concentration increases
Depletion capacitance
Depletion capacitance increases as concentration increases
Q10.25 persons are in a room, 15 of them play hockey, 17 of them football and 10 of them play both hockey and football. Then the number of persons playing neither hockey nor football is ;
Answer : Option DExplaination / Solution:
Number of people who play hockey n(A) = 15
Number of people who play football n(B) = 17
Persons who play both hockey and football n(A∩B) = 10
Persons who play either hockey or football or both :
n(AUB) = n(A) + n(B) - n(A∩B)
= 15 + 17 - 10 = 22
Thus people who play neither hockey nor football = 25 - 22 = 3
Total Question/Mark :
Scored Mark :
Mark for Correct Answer : 1
Mark for Wrong Answer : -0.5
Mark for Left Answer : 0