Question

Consider a silicon diode at T=300 K with Nd = 4 × 1016 cm?3 , Na = 1 × 1015 cm?3 , Dn = 25 cm2/s, Dp = 10 cm2/s, ?n = 5 × 10?7 s, ?p = 10?7 s, A = 10?3 cm2 , ni = 1.5 × 1010 cm?3 , r = 11.7. (1) Determine diffusion capacitance and junction capacitance at (a) Va = 0.4 V; (b) Va = 0.6 V. (2) At what voltage the two capacitances are equal? Hint: the solution cannot be found analytically, so you have to do it approximately, e.g. by trial and error, graphically, using matlab, etc

Answer #1

Design an ideal abrupt silicon PN-junction at 300 K such that
the donor impurity concentration Nd (in
cm?3) in n-side is Nd =
5×1015/cm3 and the acceptor impurity
concentration Na in the p-side is Na = 715
×1015/cm3.
Given: the diode area A = 2×10?3 cm2,
ni = 1010/cm3, ?n =
10?8 s and ?p = 10?7 s
Determine the following when a forward bias of 0.6 V is applied
to the diode:
1. What are the values (in ?m)...

Consider a Si pn diode with NA = 1018 cm−3 and ND =
1016 cm−3 . Assume the device has a cross-sectional area
of 2 µm2 . Calculate the forward and reverse current at +1.0 V and
-1.0 V respectively. (At room temperature)

Consider a pn+ junction with doping NA = 1016 and ND
= 1018 cm−3 at a forward bias of 0.4 V.
(a) Calculate the excess carrier distribution δn(xp) and δ(xn).
Which one is largest?
(b) Draw the excess carrier distribution.
(c) At what voltage is the carrier injection starting to reach
high level? (i.e. ∼ 10% of majority level)

Consider a silicon crystal that has been doped with 4 x 1016
cm-3 boron atoms and 9 x 1015 cm-3 phosphorous atoms. (a) What is
the overall charge of the silicon? (b) Is the material n-type or
p-type? Explain. (c) What are the equilibrium electron and hole
concentrations at room temperature (300K)? (d) What is the ionized
donor density in the crystal? (e) What is the neutral donor density
in the crystal? (f) Now assume the sample is heated to...

The electron concentration in a semiconductor is given by n =
1016((1-2x)/L) cm-3 for 0 ≤ x ≤ L, where L = 15 μm. The electron
mobility and diffusion coefficient are μn = 1200cm2/V-s and Dn =
30.9cm2/s. An electric field is applied such that the total
electron current density is a constant over the given range of x
and is Jn = -50 A/cm2 . (a) Determine the electron diffusion
current density (b) Determine the required electric field versus...

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