Spring 2016
EE 529 Homework #4 Solutions
1. (20%) An InGaAs photodetector for =
λ 1.3 mm has a responsivity of R ≡ 0.8 (A/W) , a
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specific detectivity D* =
7 ×10 cm ⋅ Hz1/2 ⋅ W -1 , a bandwidth of B = 2.5 GHz , and a
dynamic range DR = 60 dB. It has a circular active area of 80 µm in diameter. The total
resistance including the load is =
R 50 Ω . The dark current and the background radiation
current of the detector are not known.
(a) What is the NEP of this photodetector? What is the photocurrent at this power level?
(b) What is the saturation optical signal power for this photodetector? What is the
photocurrent at this power level?
(c) What is the SNR at the saturation power level of this photodetector? Does the
photodetector operate in the quantum or thermal regime at this power level?
(d) What is the risetime of the detector response to an impulse signal?
(a) Photodetector area A =
π(80 / 2) 2 mm 2 =
5.0 ×10−9 m 2
( A B)1/2
= 5 nW
D*
4 nA
i ph =⋅
R NEP =
=
NEP
P
log sat
60 dB
=
=
(b) DR 10
NEP
6
Psat 10
NEP 5 mW
=
=
i=
isat = R ⋅ Psat
= 4 mA
ph
(c) At the saturation level, isat  id , ib . Therefore, with =
R 50 Ω ,
= 3.2 ×10−12 A 2
in2,=
2eBisat
sh
4k BTB
= 8.3 ×10−13 A 2
R
The photodetector operate in the quantum regime at this optical power level.
in2 = in2, sh + in2,th = 4.0 ×10−12 A 2
in2=
,th
2
is2 = isat + 2eBisat ≈ 1.6 ×10−5 A 2
i2
SNR = s =4.0 ×106 =66 dB
in2
(d)
=
f3dB 0.886
=
B 2.215 GHz
0.35
tr = 158 ps
=
f3dB
2. (5%) Continue to work on the problem about solution-processed quantum dot
photoconductors discussed in Slide 4 of “Photodetectors and Solar Cells.” Please express
NEP in units of fW and noise current rms( in ) in units of pA.
rms(in )= R ⋅ NEP
Area of the detector from the figure in Slide 7 = 5 mm × 3 mm = 1.5 ×10−4 cm 2 .
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Spring 2016
Modulation
frequency (Hz)
D* ( 1013 jones)
R (A/W)
NEP (fW)
rms( in ) (pA)
1
5
10
30
50
100
400
0.7
1000
1.75
1.75
1.5
800
1.82
1.46
2
600
1.93
1.16
1.2
350
5.59
1.96
0.4
250
21.65
5.41
0.3
200
40.82
8.16
0.1
80
244.95
19.6
3. (20%) (a) Show that the space-charge effect can appear if the length of the photoconductor
considered in the exercise on Slide 8 of “Photodetectors and Solar Cells” is reduced to
=
l 10 mm while all other parameters remain unchanged. The photoconductor is biased at 2V.
Find the optical signal power range for which the device is free of the space charge effect. (b)
Find the gain and the responsivity of the photoconductor for the optical signal powers in both
cases: photoconductor free of the space charge effect and photoconductor limited by the
space charge effect. Use the spontaneous recombination time constant τs at the transition
point for the calculations. For the latter case, use the dark conductivity to calculate the
dielectric relaxation time.
(a) The dark conductivity is still =
σ0 0.136 Ω −1m −1 , but VSC in the dark is reduced to
σ0l 2
= 131 mV
(m e + m h )e
Therefore, the bias voltage of 2V > VSC in the dark. For the device to be free of the space
charge effect, we need VSC > 2V, this is possible when
(m e + m h )eVSC
=
σ
= 207.6 Ω -1m -1
l2
For this conductivity, the needed photogenerated carrier density is
σ
=
> 1.46 ×1019 m −3
N
e(m e + m h )
wd
(Alternatively, you can use e
V ≤ eN ⋅ wdl to calculate N, which gives the same result)
l
With this and the initial doping density of n0 ,
1
1
τs ≈
<
=0.8 ms
−11
B( N + n0 ) 8 ×10 (1.46 ×1013 + 1×1012 )
The corresponding photogeneration rate is
N
Ggen = > 1.825 ×1022 m −3s −1
τs
This requires an optical signal power of
10 ×10−6 ⋅100 × 10−6 ⋅1× 10−6
1.24
lwd
=
Ps
Ggen hn >
⋅1.825 ×1022 ⋅
⋅1.6=
×10−19 6.74 pW
0.632
0.85
he
Therefore, the device is free of the space charge effect if Ps > 6.74 pW
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Spring 2016
(b) For Ps > 6.74 pW=
,G


tµ
s
1+ h 
e 
tµ
tr 
e 
l2
= 58.8 ps , therefore G
= 1.4 ×107
µ eV
e
G e
=
R =h
6.1× 106 A/W
hν
τ
ε
For Ps < 6.74 pW , G = s , where τd = varies with the optical signal power. The dark
τd
σ
ε
= 858 ps .
conductivity is =
σ0 0.136 Ω −1m −1 , therefore τ0d=
s0
τs
G=
= 9.3 ×105
0
τd
e
R =G0 he
=4.0 ×105 A/W
hν
e
t=
tr
4. (20%) In this problem, we compare the performance of Si and GaAs p-i-n photodiodes that
have the same physical structures for optical detection at 850 nm wavelength. Both have the
same i-region thickness di= 3 mm and the same active-area diameter 2=
r 40 mm . Both are
reverse-biased at 3V. Take RL + Rs = 50 Ω and C p = 0 for both devices.
(a) At 300K and under 3V reverse bias, Si has the following parameters: α = 7 ×104 m −1 ,
ve = 8 ×104 m/s , v=
3.2 ×104 m/s , and=
ε 11.8ε 0 . Find the 3-dB cutoff frequency, f3dB ,
h
and the internal bandwidth-efficiency product, f3dB ηi , for the Si p-i-n photodiode.
(b) At 300K and under 3V reverse bias, GaAs has the following parameters: α = 1×106 m −1 ,
v=
1.2 ×105 m/s , v=
1.7 ×104 m/s , and=
ε 13.18ε 0 . Find the 3-dB cutoff frequency,
e
h
f3dB , and the internal bandwidth-efficiency product, f3dB ηi , for the GaAs p-i-n
photodiode.
(c) Compare the performance of these two devices. Which one has higher f3dB ηi ? Why?
(Note: In this problem, the bias is not high enough to achieve saturation velocities, and the
hole velocity in GaAs is slower which limits its bandwidth. In general, high-speed p-i-n
photodiodes are operated with saturation velocities for both electrons and holes. GaAs has
higher saturation velocities, and GaAs p-i-n photodiodes are typically faster than Si p-i-n
photiodes.)
(a) For the Si p-i-n photodiode,
1 e
1  3 ×10−6 3 ×10−6 
ttr=
+
=
(ttr + ttrh =
)

 65.2 ps
2
2  8 ×104 3.2 ×104 
pr 2
ε
=
43.7 pF
R = RL + Rs = 50 Ω , Ci =
di
τ RC= RCi= 2.19 ps
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Spring 2016
0.443
f3dB ≈
1/2
 ttr2 + (2.78t RC ) 2 
ηi =
1 − e −α d i =
19%
=
6.77 GHz
f3dB ηi =1.29 GHz
(b) For the GaAs p-i-n photodiode,
1 e
1  3 × 10−6
3 × 10−6 
ttr=
(ttr + ttrh =
)
+
=

 100 ps
2
2  1.2 ×105 1.7 ×104 
pr 2
ε
=
48.9 pF
R = RL + Rs = 50 Ω , Ci =
di
τ RC= RCi= 2.45 ps
0.443
=
f3dB ≈
4.42 GHz
1/2
 ttr2 + (2.78t RC ) 2 
1 − e −α d i =
95%
ηi =
f3dB ηi =4.2 GHz
(c) Comparing these two devices, we find that the bandwidth-efficiency product of the Si p-i-n
photodiode is significantly smaller than that of the GaAs p-i-n photodiode even though the Si
p-i-n photodiode has a higher 3-dB cutoff frequency under this bias condition. The reason is
clearly the smaller internal quantum efficiency ηi for the Si p-i-n photodiode due to the low
absorption coefficient of Si at 850 nm wavelength.
5. (15%) Under illumination, a GaAs solar cell with a dark saturation current I S = 0.5 nA
produces a short-circuit current I SC = 100 mA . Estimate the value of the maximum voltage
Vm and maximum current I m .
=
VOC
k BT  I ph 
 0.1 
=
≅ 0.5 V
ln   0.026 ln 
−9 
e
 0.5 × 10 
 IS 
Vm =
VOC −
k BT  eVm 
ln 1 +
 , solve for Vm .
e
 k BT 
 eV 
An approximate solution can be obtained by putting Vm = VOC in ln 1 + m  , which gives
 k BT 
Vm ≅ 0.42 V .
eV
eVm kBTm
| I m | I=
83.7 mA
=
e
s
k BT
6. (20%) (a) The intensity of light arriving at a point on
earth, where the solar latitude is α, can be
approximated by the Meinel and Meinel equation
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Spring 2016
−0.678
(kW m −2 ) . The solar latitude α is the angle between the sun rays and
I ≈ 1.353(0.7)(sin α )
the horizon as shown in the figure. Around March 22nd and September 23rd, the sun rays
arrive parallel to the plane of the equator. Plot intensity I vs. α. What is the maximum power
available for a solar panel of area 1 m2 if its power conversion efficiency is 20%? (b)
Manufacturer’s characterization tests on a particular Si p-n junction solar cell at 300K
specifies VOC = 0.45 V and I SC = 400 mA when illuminated directly with light of intensity 1
kW m-2. The fill factor of the solar cell is 0.73. This solar cell is to be used in a portable
equipment application near Eskimo Point (Nunavut, Canada) at a latitude (φ) of 63⁰.
Calculate VOC and the maximum available power when the solar cell is used at noon on
September 23rd, when the temperature is around -10⁰C. What is the maximum power this
solar cell can generate? (Note: Assume the dependence of the dark current I S on temperature
is only through the exponential dependence on temperature for n p 0 and pn 0 , the bandgap
energy Eg (= 1.1 eV) and the fill factor remain the same when reducing the temperature
from 300K to -10⁰C, the temperature dependence of the effective density of states at the band
edge is negligible compared to the exponential dependence.)
(a) The result of I vs. α is shown below. The maximum is at α =90α as expected. The curve is
broad, implying that there is still substantial intensity when α is a low angle.
At α =90α , I max = 1.353 ⋅ 0.7= 0.95 kW m −2
Therefore, at the best spot on Earth, the best one can do with this solar panel is
Pmax= 0.2 ⋅ 0.95 ⋅1= 0.19 kW= 190 W
0.4
k BT  I ph 
ln    0.45 = 0.0259 ln
 I=
1.1× 10−8 A
S
I
e
I
S
 S 
From the figure, φ + α = π / 2 . Therefore, at Eskimo Point, α =27α which gives a light
intensity of
(b) VOC ≅
I ≈ 1.353(0.7)(sin 27
o −0.678
)
=
0.736 kW m −2
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Spring 2016
0.736
The photocurrent at Eskimo
Point I ph (400
=
=
mA)
294 mA
1
L

L
I S eA  h pn 0 + e n p 0 
=
τe
 τh

 Eg 
ni2
ni2 2
, pn 0 =
, ni N C NV exp  −
=
 according to Mass action law.
p p0
nn 0
 k BT 
k BT = 0.0227 eV at -10⁰C (263K)
=
np0
Eg
Eg


I S (Eskimo Point)
exp  −
=
+

I S (Test)
 k B ⋅ (263K ) k B ⋅ (300 K ) 
I S (Eskimo Point)
1.1
1.1 

=
exp  −
+
−8

1.1×10
 0.0227 0.0259 
=
I S 2.77 ×10−11 A at Eskimo Point
0.294
ln
0.523 V
=
VOC 0.0227
=
2.77 ×10−11
=
Pmax (=
FF ) I phVOC (0.73)(0.294 A)(0.523 V)=112 mW
One can see that although VOC is increased due to the lower temperature, the sun light
intensity is reduced at higher latitude, which results in a lower I ph and therefore a reduced
maximum output power.
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