A BJT With doping concentrations of N_E = N_c = 10^18 cm^-3, and N_B = 10^16 cm^

Question

A BJT With doping concentrations of N_E = N_c = 10^18 cm^-3, and N_B = 10^16 cm^-3, with a very narrow base can potentially have a high current gain hut will most likely have a low output resistance. In a P-channel J-FE1, pinch-off occurs at the lower potential end of the channel. For n-Channel MOSFETs, increasing the doping level of the semiconductor substrate results in increasing V_T. Reduction of the gate oxide thickness reduces the threshold voltage of n-channel MOSFETs. Threshold voltage for ideal (enhancement mode, normally off) p-channel MOSFETs has a negative value. Too much positive trapped charge in the gate oxide of a p-channel MOS transistor can result in a transistor which is "ON" even with V_gs = OV(depletion mode transistor). Work function of Aluminum is smaller than the distance between conduction band of silicon and vacuum level, therefore, Aluminum contacts with n-type silicon result in Shuteye junctions. By reducing the doping level of the channel in a J-FET absolute value of pinch off voltage decreases. Using a high-k dielectric (dielectric with high dielectric constant) lot gate insulator reduces the threshold voltage for n-channel MOSFETs.

Explanation / Answer

g) True

h)True

i)True(vt propotional to Vt*ln(Na/Ni))

j)true(Vt inverserly popotinal to oxide thickness)

k)true

l)False(lot fo negative ion trapped at oxide cause pmos to be on even at Vgs= 0V)

m) true

o) False

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