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Mark Lundstrom 10/17/2014
ECE-‐305 Fall 2014 1
ECE 305 Homework SOLUTIONS: Week 8
Mark Lundstrom Purdue University
(10.31.14) 1) A silicon diode
is asymmetrically doped at N D =
10
19 cm-‐3 and N A = 1016 cm-‐3.
(Note
that at N D = 1019 the semiconductor is
on the edge of degeneracy, but
we can assume
that non-‐degenerate carrier statistics
are close enough for this
problem.) Answer the following
questions assuming room temperature.
Assume that the minority electron
and hole lifetimes are
! n = ! p = 10
"6 s. The lengths of the
N and P regions are
L = 500 µm and L >> xp ,xn . Assume
an “ideal diode” and answer the
following questions.
1a) Compute JD = ID A , the
diode current density at a
forward bias of VA = 0.5 V.
Solution: Since this in a
one-‐sided junction with N D >> N A
, essentially all of the
current is due to electrons
injected into the P-‐region.
JD = q
ni2
N A
DnLn
eqVA kBT !1( ) = J0 eqVA kBT !1( ) (*)
From Fig. 3.5 on p. 80 of
SDF, µn = 1248 cm
2 V-s at N A = 1016 .
Using the Einstein relation, we
find
Dn =
kBTq
µn = 0.026!1248 = 32.4 cm2 /s
The diffusion length is:
Ln = Dn! n = 32.4"10#6 = 57 µm
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Mark Lundstrom 10/17/2014
ECE-‐305 Fall 2014 2
HW8 solutions (continued): Since
Ln
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Mark Lundstrom 10/17/2014
ECE-‐305 Fall 2014 3
HW8 solutions (continued): 2) Consider
the diode in problem 1) above
and answer the following questions.
2a) If the diode is biased
such that JD = 10!6 A/cm2 , what is
VA ?
Solution:
Begin with JD = q
ni2
N A
DnLn
eqVA kBT !1( ) = J0 eqVA kBT !1( ) In
moderate forward bias, we can
drop the -‐1:
JD = J0eqVA kBT
so
VA =
kBTq
lnJDJ0
!"#
$%&
Putting in numbers:
VA = 0.026ln
10!6
9.1"10!12#$%
&'(= 0.302 V
VA = 0.302 V
2b) If the temperature
changes from 300 K to 301
K, how much does VA change?
Solution: Differentiate the
expression that we obtained above:
dVAdT
= ddT
kBTq
lnJDJ0
!"#
$%&
'()
*)
+,)
-)=
kBq
lnJDJ0
!"#
$%&+
kBTq
J0JD
ddT
JD J0`.1( )
dVAdT
=kBq
lnJDJ0
!"#
$%&'
kBTq
1J0
dJ0dT
!"#
$%&
dJ0dT
= ddT
qni
2
N A
DnLn
!
"#$
%&
The strongest part of the
temperature-‐dependence comes from the
exponential factor in ni
2 , so we can ignore the
temperature dependence of the
diffusion coefficient, the diffusion
length, and the effective
densities-‐of-‐states and write:
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Mark Lundstrom 10/17/2014
ECE-‐305 Fall 2014 4
HW8 solutions (continued):
J0 = Ke!EG kBT
then
1J0
dJ0dT
= 1Ke!EG kBT
Ke!EG kBTEG
kBT2
"
#$%
&'=
EGkBT
2
"
#$%
&',
which can be used to find
dVAdT
=kBq
lnJDJ0
!"#
$%&'
kBTq
1J0
dJ0dT
!"#
$%&=
kBq
lnJDJ0
!"#
$%&'
kBTq
EGkBT
2
!
"#$
%&
so
dVAdT
=kBq
lnJDJ0
!"#
$%&'
EG qT
!"#
$%&.
Putting numbers in:
dVAdT
= 1.38!10"23
1.6!10"19ln 10
"6
9.1!10"12#$%
&'(" 1.12
300#$%
&'(
dVAdT
= 1.00!10"3 " 3.73!10"3 = "2.7 !10"3
dVAdT
! "3 mV/K
So we need to lower the
applied bias about 3 millivolts
to keep the current constant as
the temperature increases 1 K
or 1 degree C. PN
junctions can be used as
thermometers because the diode
current depends sensitively on
temperature. A more careful
treatment would include the
temperature dependencies of the
bandgap, effective densities-‐of-‐states,
diffusion coefficient, etc., but the
result would be close to the
value obtained here.
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Mark Lundstrom 10/17/2014
ECE-‐305 Fall 2014 5
HW8 solutions (continued): 3) The
sketch below shows the carrier
concentrations in a PN junction
at room
temperature. Answer the following
questions.
3a) Is the diode forward or
reverse biased? Explain your
answer.
Solution: Forward biased because there
are excess electrons on the
P-‐side and excess holes on the
N-‐side.
3b) What is the acceptor
concentration on the P-‐side?
Solution: NA = 1016 cm-3
3c) What is the donor
concentration on the N-‐side?
Solution: ND = 1014 cm-3
3d) What is the intrinsic
carrier concentration?
Solution:
n0 p0 = ni2
On the P-‐side: n0 p0 = 1016 !107 =
1023 ni = 10
23 = 3.16!1011 cm-3
On the N-‐side: n0 p0 = 1014 !109 =
1023 ni = 10
23 = 3.16!1011 cm-3
ni = 3.16!10
11 cm-3
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Mark Lundstrom 10/17/2014
ECE-‐305 Fall 2014 6
HW8 solutions (continued): 3e)
Do low level injection conditions
apply?
Solution: YES.
On the P-‐side: !n "xp( ) = 1010
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Mark Lundstrom 10/17/2014
ECE-‐305 Fall 2014 7
HW8 solutions (continued):
4a) Is the diode forward
or reverse biased?
Solution: Forward biased because Fn
> Fp .
4b) What is the value
of the applied bias?
Solution:
qVA = Fn ! Fp
VA = +0.5 V
4c) What is the bandgap
of the semiconductor?
Solution: Reading from the graph:
EC ! EV = 1.25 eV
4d) What is the built-‐in
potential of the junction.
Solution: From the plot:
Vj =Vbi !VA = 0.25 V
Since: VA = +0.5 V
Vbi =Vj +VA = 0.75 V
Vbi = 0.75 V
5) Consider the Si diode of
prob. 1) and make one change.
The minority carrier lifetime
increases by a factor of 1000,
so ! n = ! p = 10
"3 s. The lengths of the
N and P regions
are still L = 500 µm and L >> xp
,xn . Assume an “ideal diode”
and compute the
forward-‐biased current density at VA =
0.6 V.
Solution: We recognize that the
minority carrier diffusion length
will change.
Ln = Dn! n = 32.4"10#3 = 1800 µm
Now Ln >> L , so the
long base diode of prob. 1)
has become a short base diode.
Instead of eqn. (*) in
prob. 1), we have to replace
the minority carrier diffusion length
by the length of the P-‐region:
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Mark Lundstrom 10/17/2014
ECE-‐305 Fall 2014 8
HW8 solutions (continued):
JD = q
ni2
N A
DnL
eqVA kBT !1( ) = J0 eqVA kBT !1( )
Write the saturation current density
as
J0 = q
ni2
N A
DnL
= qni
2
N A
DnLn
!"#
$#
%&#
'#
LnL
The term in brackets was
computed in prob. 1a), and so
was the diffusion length.
Using these results:
J0 = q
ni2
N A
DnLn
!"#
$#
%&#
'#
LnL
= 9.1(10)12{ }( 57500 = 1.04(10)12 A/cm2
The current becomes:
JD = J0 e
qVA kBT !1( ) = 1.04"10!12 e0.6/0.026 !1( ) = 1.04"10!12
1.05"1010 !1( ) = 1.1"10!2 JD 0.6 V( ) = 1.1!10"2 A/cm2
Note: The lifetime is
longer, so the current density
must be smaller – this is
a sanity check. Key point
to remember: Long base diode:
L >> Ln
JD = q
ni2
N A
DnLn
eqVA kBT !1( )
Short base diode, L
