EE6604 Continuous Assessment I
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EE6604 Continuous Assessment I
1. The diagram below shows the dopant concentration of the emitter and base of a npn polysilicon emitter bipolar transistor. The base is made of silicon.
Na (cm-3)
6x1018
1017 LINEAR PLOT
Collector
BASE
xe -0. 1m 0 0 WB xb
Given data:
T = 300K kB = 1.38 x 10-23 J/K ni = 1.4x1010 cm-3
nB = 103 cm2/Vs pE = 60 cm2/Vs
WB = 0.2 m AE = 1 m2 SP = 5x105 cm/s
Ege = 40meV in emitter Egb = 0eV throughout base
i. Calculate the base Gummel number, GB and provide its unit.
ii. What is the internal electric field (in V/m) in the base at xb = WB/3?
iii. Find the collector current density of this transistor as a function of VBE .
iv. What is the value of the transconductance for VBE = 0.8V?
v. Calculate the emitter Gummel number, GE .
vi. What is the common emitter current gain β and the common base current gain α for this bipolar transistor?
vii. Calculate the small signal input resistance for VBE = 0.8V.
viii. Find the base current density of this transistor as a function of VBE .
ix. From breakdown electrical measurement, BVCEO = 3.56V and BVCBO = 12V. Determine the value of the empirical parameter m.
x. What is the threshold value of M for avalanche breakdown to occur in the BVCEO circuit?
2. The Early voltage of a silicon germanium base bipolar transistor VA (SiGe) is given by the following equation:
VA (SiGe) = exp ))| - 1 VA (Si)
Here, VA (Si) is the Early voltage of a bipolar transistor with the same emitter, base, collector layer thicknesses and dopant profiles. ∆Eg, SiGe is the total bandgap narrowing within the base region of the silicon germanium base bipolar transistor and T is the absolute temperature. Suppose the Early voltage of the silicon germanium base bipolar transistor is to be a factor of 2 higher than that of the silicon bipolar transistor at 300K,
determine the value of ∆Eg, SiGe required in eV by using the graph paper provided. For the ∆Εg, SiGe calculated above, what is the ratio VA (SiGe)/VA (Si) at 355K? A computer algorithm should NOT be used for solving this problem.
2022-09-24