298 CHAPTER 6. BIPOLAR JUNCTION TRANSISTORS
Base
Emitter
x
z
XE = 2 μm
XB = 6 μm
WB = 100 nm
LBE = 0.5 μm
LBE
Figure 6.27: Figure for problem 6.35.
changing the current gain of the transistor. Note: You may ignore the vertical base
resistance in this problem.
(a) Write the expression for the minority charge,n(x, z), and integrated minority charge,
q(x)=
∫
n(x, z)dz(in cm−^2 ) in the base as a function of the varying emitter-base
potentialVBE(x).
(b) Derive the differential equation that relates the emitter base potentialVBE(x)toq(x)
in terms of the base resistance,RB, and recombination time,tN. What are the boundary
conditions?
(c) Using your results from (a) and (b), derive an expression forn(x, z),q(x),andthebase
currentIB(x).
(d) What is the total emitter current and base current for this transistor? Find the
expression for the current gain,β,intermsofRBandtN. Assume ideal emitter injection
efficiency.
Problem 6.36Consider an npn transistor where the base is open (figure 6.28). Assume
that theβof the transistor is not impacted by recombination in the base. Show that the
breakdown voltage, VCEO, in this configuration is reduced from the normal breakdown
voltage of the base-collector junction, VCBO. Derive an analytical expression for VCEO.
State all your assumptions.
Problem 6.37Consider a GaAs n-p-n transitor shown in figure 6.29. I make a mistake and
during the growth and insert a 10nm thick quantum well in the center of a 100 nm base.
The result of this mishap is to reduce the lifetime of the injected minority carriers to 0.1 ns
