Prof. Jasprit Singh
Fall 2000
EECS 320
Solutions to Homework 7
Problem 1 An optical power density of 1W/cm2 is incident on a GaAs
sample. The photon energy is 2.0 eV and there is no reection from the surface.
Calculate the excess electron-hole carrier densities at the surface and 0.5 m
from the surface. The e-h recombination time is 10;8 s.
Solution
The optical absorption coecient at h! = 2.0 eV is (you may use Fig. 3.11
also)
4
1=2
(h! = 2:0 eV) = 5:7 10 (22::00 ; 1:43) = 2:15 104 cm;1
The generation rate at the surface is
4 ;1 )(1W cm;2 )
GL = Phop!(0) = (2:15(2:010 1cm
:6 10;19 J)
= 6:72 1022 cm;3 s;1
The excess carrier density is
n = p = GL n = GL p = (6:72 1022 cm;3 s;1 )(10;8)
= 6:72 1014 cm;3
The optical power density at a depth of 0.5 m is
Pop (x = 0:5m) = Pop (0) exp (;x)
= 0:34W cm;2
The e-h pair generation rate is now
GL = 2:28 1022 cm;3 s;1
and the excess charge density is
n = p = 2:28 1014 cm;3
Problem 2 Consider a long Si p-n junction with a reverse bias of 1 V at
1
300 K. The diode has the following parameters:
Diode area,
p-side doping,
n-side doping,
Electron diusion coecient,
Hole diusion coecient,
Electron minority carrier lifetime,
Hole minority carrier lifetime,
Optical absorption coecient,
Optical power density,
Photon energy,
A
Na
Nd
Dn
Dp
n
p
Pop
h!
=
=
=
=
=
=
=
=
=
=
1 cm2
3 1017 cm;3
1017 cm;3
12 cm2 =s
8 cm2 =s
10;7 s
10;7 s
103 cm;1
10 W=cm2
1:7 eV
Calculate the photocurrent in the diode.
Solution
We will assume that the carrier generation rate is constant. This is a good
approximation for Si where is quite small. We need to nd the values of
GL ; W; Lp and Ln to nd the photocurrent.
The carrier generation rate is
(103 cm;1 )(10 W cm;2 )
op
GL = P
=
h!
(1:7 1:6 10;19J)
22
= 3:68 10 cm;3 s;1
The built-in voltage in the diode is
Vbi = kB T `n nnn
p
= 0:85 eV
The depletion width is
1=2
W = 2(Vbie+ Vr ) NNa +NNd
= 1:8 10;5 cm
a d
The diusion lengths are
Lp = (Dp p )1=2 = 8:94 10;4 cm
Ln = (Dn n )1=2 = 10:95 10;4 cm
The photocurrent is now
IL = eGL(Lp + Ln + W )A
= (1:6 10;19 C)(3:68 1022 cm;3 s;1 ) 8:94 10;4 + 10:95 10;4 + 1:8 10;5 cm (1 cm2 )
= 11:8 A
2
Problem 3 Consider a long Si p-n junction solar cell with an area of 4 cm2
at 300 K. The solar cell has the following parameters:
n-type doping,
Nd = 1018 cm;3
p-type doping,
Na = 3 1017 cm;3
Electron diusion coecient,
Dn = 15 cm2 =s
Hole diusion coecient,
Dp = 7:5 cm2 =s
Electron minority carrier lifetime, n = 10;7 s
Hole minority carrier lifetime,
p = 10;7 s
Photocurrent,
IL = 1:0 A
Diode ideality factor,
m = 1:25
Calculate the open circuit voltage of the diode. If the ll factor is 0.75, calculate
the maximum power output.
Solution
The open circuit voltage is given by
Voc = mkeB T `n IIL
o
The prefactor Io is given by
Io = eA DLp pn + DLn np
p
n
= (1:6 10;19 C)(4 cm2 )
(7:5 cm2 =s)(2:25 102 cm;3 ) + (15 cm2 =s)(7:5 102 cm;3 )
(2:73 10;3 cm)
(1:22 10;3 cm)
= 6:3 10;12 A
The open circuit voltage is now
0
Voc = (1:25)(0:026V ) `n 6:3 1:10
;12
= 0:84 V
The maximum power output is
Pout = IL Voc F = (1:0A)(0:84V )(0:75)
= 0:63 W
Problem 4 Consider a GaAs p-n+ junction LED with the following parameters at 300 K:
Electron diusion coecient,
Dn = 25 cm2 =s
Hole diusion coecient,
Dp = 12 cm2 =s
n-side doping,
Nd = 5 1017 cm;3
p-side doping,
Na = 1016 cm;3
Electron minority carrier lifetime, n = 10 ns
Hole minority carrier lifetime,
p = 10 ns
3
Calculate the injection eciency of the LED assuming no trap-related recombination.
Solution
The injection eciency for the GaAs pn+ LED is given by
inj =
The diusion lengths are
Ln =
Lp =
Also
Dn npo
Ln
Dp pno
Dn npo
Ln
LP
+
p
Dn n = 1:12 10;3 cm
Dp p = 1:1 10;3 cm
p
2
6 ;3 )2
npo = Nni = (1:84(101610cmcm
= 3:39 10;4 cm;3
;3 )
a
6 cm;3 )2
2
;6
pno = Nni = (1(5:84101017 cm
;3 ) = 6:77 10
inj = 0:9903
d
Problem 5 The diode in Problem 4 is to be used to generate an optical
power of 1 mW. The diode area is 1 mm2 and the external radiative eciency
is 20%. Calculate the forward bias voltage required.
Solution
The current in the diode is (relevant for photon emission)
In = AeDLn npo exp
n
and the photons generated per second are
Iph = Ien ext = ADnLnpo ext exp
n
The optical power is
eV
kB T ; 1
eV
kB T
(Pop A) = Iph h! = ADn nLpo ext h! exp
n
eV
kB T
If this power is 1 mW, the external voltage is
(Pop A)Ln
V = kBeT `n AD
!
n npo ext h
;3 W)(1:12 10;3 cm)
(10
= 0:026 `n (10;4 cm2 )(25 cm2 =s)(3:39 10;4 cm;3 )(0:2)(1:43 1:6 10;19 J)
= 1:165 V
4