光电子学作业答案

Solving the above equation for V yields
I n Ln kT (3.59 103 A)(5.0 104 cm) V ln(1 ) (0.026) ln[1 2 2 ] 1.02V e AeDn n p (10 cm )(1.6 1019 C )(25cm2 / s)(4 104 cm3 )
The injection efficiency is (assuming no recombination due to trap)
(25)(4 104 ) Ln Ln 5.0 0.986 eDn n p eDp pn Dn n p Dp pn (25)(4 104 ) (12)(8 106 ) 5.0 3.46 Ln Lp Ln Lp eDn n p Dn n p
Solution: The electron minority carrier density in p-side is
ni2 (2 106 )2 4 3 np 4 10 cm Na 1016
And hole minority carrier density in n-side is given by
p n GL p (1021 cm3s1 )(5107 s) 51014 cm3
c)the photoconductivity is
e p[ n p ] 0.128( S / cm)
d)The gain of the photoconductive detector is p IL p Gph (1 ) I Lp ttr n where
Id (1.6 1019 C )[(1200cm2 / V s)(1015 cm3 ) (400cm2 / V s)(2.25 105 cm3 )](104 cm2 )(5 /100 104V / cm) 0.0096 A 9.6mA
b)The excess carrier density is
L L L2 (100 104 cm)2 ttr 1.67 108 s 2 n F nVB / L nVB (1200cm / V s)(5V )
So
p 5 107 400 Gph (1 ) (1 ) 40 ttr n 1.67 108 1200 p
Gerneration rate,
Calculation: a)dark current ; b)the excess concentration; c) the photoconductivity; d)the device gain. Solution: a)The dark current is given by
The impact ionization coefficient
imp imp 104 cm1
Calculation :The multiplication factor
Solution: The multiplication factor from Eqn.(2.66) is
Me 1 1 impWav 1 2 4 4 1 10 0.5 10
ni2 (2 106 )2 6 3 pn 8 10 cm Nd 5 1017
The diffusion length are
Ln Dn n [(25)(10 109 )]1 / 2 5.0m
Lp D p p [(12)(10 10 9 )]1 / 2 3.46 m
3.15 Condition: An AlGaAs/GaAs heterojunction LED at 300K with 17 3 Injection density for electrons, n 10 cm Injection density for holes, p 1017 cm3 Eg 1.43eV Bandgap of GaAs,
inj
3.9 The diode in problem 3.8 is used to generate an optical power of 1 mW , 2 the diode area is 1 m m and the external efficiency is 20％。
Calculation: the forward bias voltage required. Solution: The photons generated per second are
1 E p E g E 1.433eV 2 0.865m
for injection density of
'
n' 1018 cm 3
n E kT 0.0578 eV Nc
Shift of the peak position
E ' E 0.052eV
I d 0 FA e(n n0 p p0 ) AF e(nn0 p p0 ) AVB / L
n0
=
Nd 1015 cm3
p0
=
ni2 / Nd (1.5 1010 cm3 )2 /1015 cm3 2.25 105 cm3
so the dark current is
or 1 E p ' Eg E ' 1.4589eV 2 ' 0.850 m 0.020 m
3.18 Condition: a heterojunction LED based on GaAs at 300K with Bias current density, J 100 A cm2 The width of the active layer, d 0.5m Calculation: the 3db cutoff frequency of the diode. Solution: the 3db cutoff frequency is
In eI ph (1.6 1019 C )(4.43 1015 s 1 ) 3.59mA 0.986*0.2
ext
the current when the diode is forward bias is given by
In AeDn n p Ln [exp( eV ) 1] kT
I ph Power 103 15 1 4 . 43 10 s 19 (1.41)(1.6 10 )
Which are also expressed by
I ph In I I Tot injQropt ext e e e
So the current required to generate the photons is
p-side doping,
N a 1016 cm3
n 10ns
Electron minority carrier lifetime,
Hole minority carrier lifetime,
p 10ns
Calculation: injection efficiency of the LED assuming no recombination due to traps.
The photocurrent is
I L eAJ ph (0)[1 exp(W )] (1.6 1019 C )(104 cm2 )(3.911017 cm2 s 1 )[1 exp(104 1.0 104 )] 3.95 106 A
2.16 Condition: An avalanche photodetector with Avalanche region width, Wav 0.5m
1.6eV
104 cm1
A 104 cm2
Calculation: the prompt photocurrent of the device.
Solution: The photon flux incident on the detector is
0.1W / cm2 17 2 1 J ph (0) 3.91 10 cm s 19 1.6(1.6 10 J ) Pop
Calculation: the position of emission peak; the shift in the peak position 18 3 if the injection density increased to 10 cm
Solution: From the relationship
3.8 Condition: a GaAs p-n+ junction LED with Electron diffusion coefficient, Dn 25cm2 / s Hole diffusion coefficient, Dp 12cm2 / s n-side doping, N d 5 1017 cm3
3.12 Condition: a GaAs LED coupled to an optical fiber with Refractive index for the core layer, nr1 1.51
Refractive index for the cladding layer, n 1.47 Calculation: the maximum angle of acceptance for the fiber and the coupling efficiency for the diode
2.10 Condition: a silicon photoconductor at 300K Background doping, Nd 1015 cm3 Electron mobility, n 1200cm2 / V s Hole mobility, Electron lifetime,
Eg h hc


hc 1.24 0.87 m Eg Eg
for injection density of n 1017 cm3
E n kT 0.0058 eV Nc
the peak position is half of linewidth at the bandedge
2.13 Condition: a GaAs p-i-n detector with W 1.0 m Intrinsic layer width, Optical power density, P W / cm2 op 0.1 Photon energy, Absorption coefficient, Device area,
p 400cm2 / V s
n 106 s
p 5 107 s
A 104 cm2
L 100 m
Hole lifetime,
Detector area, Detector length, Bias voltage,
VB 5V
GL 1021 cm3s1
r2
Solution: The maximum angle of acceptance is calculated by
A sin1 (nr21 nr22 )1/ 2 ency is
fiber sin2 A 0.12
Example 3.7 θ=32.6o η=0.29