UConn ECE4211 HW#2 01/26/2016, Due 02/2/16, F. Jain, Popquiz#1 2/2/16 NAME_________
HW#1 Q. 11-Q13 to be solved as of part of this assignment.
Q.11. If the Schrodinger equation is solved for an atomic chain of length L (Fig. 7) but with finite potential barrier
(unlike Q. 10 where it is infinite), the electron energy as a funciton of momentum or wave vector is shown in Fig. 8.
V(x)
V0
-(a+b)
-b
a
0
a+b
x
Fig. 7 [Fig. 22 notes] . Potential in Kronig-Penney model.
We get a solution of electron energy E as a function of k (the momentum vector) of the form shown below.
Energy
E
Conduction
Band
Gap
Valence
Band
1
2
0 h/L
h/2L
-h/2a
h/2a
(Momentum in units of h/2L, a = lattice constant )
Fig. 8 Energy E- wavevector k (related to momentum) diagaram [Fig. 24b Notes].
(a) what is the minimum unit of momentum or wave vector (Fig. 8)?
HINT: momentum p=(h/2k=h/, as k=2/; /2=L for a chain of L atoms.
(b) will momentum be smaller or higher if the length of chain is doubled to 2L. Circle one.
(c) does the interatmic separation and lattice constant ‘a’ determine the forbidden gap E g. Circle one
Yes
No
Q.12. which of following expression is valid for density of states in a conduction band?
(a) Circle one
Eq. 14
or
Eq. 15.

2m
N(E)dE =
( 2e ) 3 / 2 E 1 / 2 dE
2
2

1
(14), N(E)dE =
1
2
2
(
2mh  3 / 2 1 / 2
) E dE
2
(15)
(b) Plot N(E) for conduction band as a function of E. N(E) on x-axis and E on y-axis.
(c) Label the three distributions f(E) as Fermi-Dirac (FD), Maxwell Boltzmann (MB) and Bose-Einstein (BE).
f (E) 
f (E) 
f (E) 
1
1 e
FD
MB
BE
1
FD
MB
BE
 Ae E / kT
FD
MB
BE
( E  E f / kT )
1
e
( E  E B / kT )
1
e
( E  E f / kT )
1
Q.13. Si is available in Single Crystalline, poly Crystalline, and amorphous forms.
(a) Is the energy band gap same for crystalline and poly-crystalline? Circle one
YES
NO
(b) Is the energy band gap of crystalline and amorphous Si same? Circle one
YES
NO
Q. 1 If energy density of states in conduction and valence band are known Q12(a), and Fermi-Dirac distribution f(E) is
known. Conduction band density of states N(E) =

1
2
2
(
2 me 3 / 2 1 / 2
) E dE
2
and f ( E ) 
1
e
( E  E f / kT )
 Ae E / kT
(a) What is the expression of concentration of electrons in conduction band? HINT-p 38 Notes.
n=
Ev
(b) If the hole concentration is p   N ( E )[1  f ( E )]dE (77), what is the significance of term [1-f(E)]

(c) Write the expression for majority hole concentration in the valence band.
p=
Q.2(a) . Pick the charge neutrality equation for n-Si. Circle one
Q.2(b) How would you calculate Fermi level Ef in n-Si.
Q2(c) How do we calculate Fermi level Ef in p-Si.
Q.3(a) Figure below shows three Fermi level locations. Label them as n-type, p-type, and intrinsic.
CB
CB
Ef
Ec=0
Ec
Ef = Eg/2
Eg
VB Top edge
Ef
Ev
Q.3(b) Circle the energy band diagram which represents the donor energy level ED with respect to
Conduction band edge EC (left side).
Q. 3(c) Circle the B and acceptor leverl with respect to VB.
Which of the two represent acceptor level EA
LEFT
ED
Ec
Ec
EA
Ev
ED
Ev
2
EA
Q4. Charge neutrality condition in an n-p junction is
. Assume ND+~ND
and NA-~NA.
(a) Derive expressions for xn0 and xp0 in terms of doping if the junction width Wo = xno +xpo,.
xn0 =
xp0 =
(b) If n-type side is more heavily doped with donor atoms than the conctratoin of acceptors on pside, the magnitude of xn0 will be higher than xp0. Circle one
TRUE
FALSE
⃗
Q5. Gauss’ Law states that the net outflow of displacement flux 𝐷 intersecting a closed surface
enclosing charge is equal to the enclosed charge density . Poisson’s equation is expressed in terms
⃗ = ,
of displacement vector D and charge density  as: ∇ ∙ 𝐷
.
(a) Write it in terms of potential V. HINT: D is related to electric field E and E is related to V.
(b) Label the charge density plots of Figs. 1 and 2 as either p-n or n-p junction. Here, p(x) and n(x)
are neglected in the junction.
=Charge density

 is charge
density
qND+
-xpo
- - -- - - -- - -- - - -- - -- -- - -- - - - ---- - --- -
qND+
Direction of
Electric Field
+++++++++++++++++++
+++++++++++++++++++
xno
-xno
x
++++++++
++++++++
++++++++
++++++++
++++++
qNA-
xpo
Direction of
Electric Field
qNA-
Junction Width Wo
Fig. 1 Circle one: p-n
x
- - - - ---- - --- -
OR
Junction Width
Fig. 2 Circle one: p-n
n-p
OR
n-p
Q6. (a) Write the expression for electric field 𝐸⃗ using Poisson’s equation Q5 (a)
E=
(b) Plot the electric field E for the above two 𝜌 plots in Fig. 3a and 3b.
Show
max Em
-xpo
E=Electric Field
xno
x
-xno
Wo=xpo+xno
Fig. 3(a)
E=Electric Field
Em
xpo
Wo=xpo+xno
E-field plot for Fig. 1
Fig. 3(b) E-field plot for Fig. 2.
3
x
Q7. (a) Draw the energy band diagram for n-p junction.
(b) The band bending or built-in voltage Vbi is given by: CIRCLE ONE
kT  p po  kT  N A N D 
kT  p po  kT  N A N D 
=
q
ln 
ln 

OR
=
q
ln 
ln 

q  pno  q  ni 
q  pno  q  ni2 
Q8. (a) The relationship between hole concentration p(-xpo) on p-side at the junction boundary -xpo
and hole concentration p(xno) on n-side are related at equilibrium is expressed by
p(-xpo)/ p(xno) = ?
(b) Under forward bias the relationship is: complete the equation
p(-xp)/ p(xn) = p(-xpf)/ p(xnf) = ?
(c) Under reverse bias: complete the equation
p(-xpr)/ p(xnr) = ?
Q9. Find the coefficient A and B for a p-n junction diode shown in Fig. 4 with a finite neutral nregion ln.
-xp0
xn0
ln
ppo
nno
ne
p
pe
p(x)
n
n(x)
pno
npo
0
-xp
xn
x
0
Vf
Fig. 4
Carrier distribution equation (53A) p(x) = A e
Boundary conditions (BCs) are:
OR
-(x- x n )
Lp
 Be
qV
pn at x = xn is pe = pno e kT
pn ( x = ln )  pno
qV
(x- x n )
Lp
for  > x > xn
f
(BC#1) p ( x = xn) = p = pe - pno = pno ( e kT - 1) , and BC#2 p ( x = ln ) = 0 .
Here, ln is the length of n-region. The n-region is finite (not ∞).
4
f
qV
HINT: BC#1 gives A  B = pno ( e kT - 1) , and BC#2 gives A e
f
-(ln - x n )
Lp
 Be
( ln - x n )
Lp
 0 . Find A, B.
Q10. (a)Expression of junction capacitance under equilibrium (Wo) , forward bias (Wf) and reveres
bias Wr.
(b)Write expressions for diffusion capacitance Cdiff under forward bias.
Q11. Draw the carrier distribution plots under
(a) forward bias Vf=0.7V and
(b) reverse bias of 2V for the p-n junction given below in Fig. 5.
HOLES
ELECTRONS
Carrier
Conc.
cm
-3
nn=1017
1020
1016
1012
Pp=1020 cm-3
108
10
p- region
n- region
4
100
p+
-xp0
-xn0
0
n
ND=1017cm-3
NA=1020cm-3
Fig. 5
5
Q12. (a) Draw the energy band diagram under forward bias of 0.7V for the device of Fig. 5.
HINT: Draw the Fermi level as a horizontal line under equilibrium; draw two vertical lines
representing the junction with width Wf; draw a new Fermi level for p-side which represents
forward bias that is p-side level is qVf below the n-side Fermi level; draw the p-Si conduction and
valence bands with respect to the new Fermi level; draw the n-Si CB and VB with respect to old
Fermi level.
(b) Draw the energy band diagram for the device of Fig. 5 under reverse bias of 2.0V?
HINT: Draw a new Fermi level for p-side which represents reverse bias that is p-side level is
qVr above the n-side Fermi level.
.
6