kasap solution manual, optoelectronics

Solutions for Optoelectronics and Photonics: Principles and Practices Chapter 4/5/6

4.8 Fabry-Perot optical resonator

a. Separation of the modes is,

∆υm = υ f =c

2L=

(3×108 m s-1)

2(0.5 m)= 3×108 Hz = 300 MHz.

The finesse is

F =

πR1/ 2

1− R=

π 0.991 / 2

1− 0.99= 312.6

and each mode width, spectral width, is

δυm =υ f

F=

3×108

312.6 = 9.6××××105 Hz = 960 kHz.

b Cavity mode nearest to the emission wavelength is

m =

2L

λ / n=

2(200×10−6)

(1300× 10−9)/ 3.7=1138.46 i.e. m = 1138.

Separation of the modes is,

∆υm = υ f =

c /n2L

=(3×108 m s-1) / 3.7

2(200×10−6 m)= 2.03×1011 Hz

The finesse is

F =

πR1/ 2

1− R=

π 0.81/ 2

1− 0.8 = 14.05

and each mode width, spectral width, is

δυm =υ f

F=

2.03× 1013

14.05 = 1.4×1010 Hz.

4.10 Threshold current and power output from a laser diode

a If Nph is the coherent radiation photon concentration, then only half of the photons, 1/2Nph, in the cavity would be moving towards the output face of the crystal at any instant. It takes ∆t = nL/c seconds for photons to cross the laser cavity length L.

Po = Energy flow per unit time in cavity towards face× Transmittance

=

hcλ

12 Nph( )dWL( )

∆t

Transmittance

=

hc

λ

12 Nph( )dWL( )

Lnc

1− R( )

=hc2NphdW

2nλ

1− R( )

and I =

Optical Power

Area=

Po

dW=

hc2Nph

2nλ

1− R( )


where R is the reflectance of the crystal face.

b Consider one round trip through the cavity. The length L is traversed twice and there is one reflection at each end. The overall attenuation of the coherent radiation after one-round trip is

RRexp[−α(2L)]

where R is the reflectance of the crystal end.

Equivalently we can represent this reduction by an effective or a total loss coefficient αt such that after one round trip, the reduction factor is

exp[−αt(2L)]

Equating the two,

RRexp[−α(2L)] = exp[−αt(2L)]

and rearranging,

αt = α +

1

2Lln

1

R2

c The reflectance is

R =

n −1

n +1

2

=3.5−1

3.5+1

2

= 0.309

The total loss coefficient is

αt = α +

1

2Lln

1

R2

=1000 m−1 +

1

60×10−6 m−1 ln1

0.3092

= 2.06×104 m-1.

∴ τ ph =

ncαt

=3.5

(3×108 m s−1)(2.06× 108 m−1)= 5.7×10-13 s (0.57 ps)

Coherent radiation is lost from the cavity after, on average, 0.57 ps.

For the above device, threshold current density Jth ≈ 500 A cm-2 and τsp ≈ 10 ps, d ≈ 0.25 µm,

From Jth =nthed

τ sp

we have, nth ≈Jthτ sp

ed≈

(500×104 A m-2)(10 ×10−9s)

(1.6×10−19C)(0.25×10−6m)≈ 1.25×1024 m-3 or 1.2×102×102×102×1018181818 cm-3

Now, the current density corresponding to I = 30 mA is

J = I/(WL) = (0.05 A)/[10×60×10-610-6 m2)] = 833×104 A m-2.

And, N ph ≈τ ph

edJ − J th( ) ≈

(5.7×10−13s)

(1.6×10−19C)(0.25×10−6m)(833−500)×104 A m-2

≈ 4.7×1×1×1×1000019191919 photons m-3

The optical power is

Po =hc 2N phdW

2nλ

1−R( )

= (6.62×10−34 J s)(3×108 m s-1)2(4.7×1019 m-3)(0.25×10−6 m)(10×10−6 m)

2(3.5)(1310×10−9 m)× 1− 0.309( )


≈ 0.00053 W or 0.53 mW.

Intensity = Optical Power / Area = Po / (dW)

= (0.00053)/[0.25×10×10-310-3 mm2)] = 223 W mm-2.

This intensity is right at the crystal face over the optical cavity cross section. As the beam diverges, the intensity decreases away from the laser diode.

4.12 Laser diode efficiency

a The external quantum efficiency ηEQE, of a laser diode is defined as

ηEQE =Number of output photons from the diode per unit second( )Number of injected electrons into diode per unit second( )

∴ ηEQE =Optical Power / hυDiode Current / e

=Po /Eg

I / e=

ePo

IEg

The external differential quantum efficiency, ηEDQE, of a laser diode is defined as

ηEDQE =Increase in number of output photons from diode per unit second( )

Number of injected electrons into diode per unit second( )

∴ ηEDQE =(Change in Optical Power) / hυ

(Change Diode Current) / e=

∆Po /Eg

∆I / e=

e

Eg

dPo

dI

The external power efficiency, ηEPE, of the laser diode is defined by

ηEPE =Optical ouput power

Electical input power=

Po

IV=

Po

IV

eEg

eEg

=ePo

IEg

Eg

eV

∴ ηEPE =ePo

IEg

Eg

eV

= ηEQE

Eg

eV

b 670 nm laser diode

Eg ≈ hc/λ = (6.626×10-34)(3×108)/[(670×10-9)(1.6×10-19)] = 1.85 eV,

so that ηEQE =(1.6×10−19 C)(2×10−3 Js−1)

(80 ×10−3 A)(1.85 eV×1.6×10−19 eV/J)= 0.0135 or 1.35%

ηEDQE =e

Eg

dPo

dI

=

(1.6×10−19 C)

(1.85 eV×1.6×10−19 C)

3×10−3 − 2×10−3 Js−1

82×10−3 − 80 ×10−3 A

= 0.27 or 27%.

ηEPE =Po

IV=

2 ×10−3W

(80×10−3A )(2.3 V)= 0.011 or 1.1%

c 1310 nm laser diode

Eg ≈ hc/λ = (6.626×10-34)(3×108)/[(1310×10-9)(1.6×10-19)] = 0.9464 eV,

so that ηEQE =(1.6× 10−19 C)(3×10−3 Js−1)

(40×10−3 A)(0.9464 eV×1.6×10−19 J/eV)= 0.079 or 7.9%


∴ ηEDQE =e

Eg

dPo

dI

=

(1.6×10−19 C)

(0.9464 eV×1.6×10−19 C)

4 ×10−3 − 3×10−3 Js−1

45×10−3 − 40 ×10−3 A

= 0.21 or 21%.

and ηEPE =Po

IV=

3×10−3W

(40 ×10−3A )(1.4 V)= 0.054 or 5.4%

4.15 The SQW laser

The lowest energy levels with respect to the CB edge Ec in InGaAs are determined by the energy of an electron in a one-dimensional potential energy well

εn =h2n2

8me*d2

where n is a quantum number 1, 2, …, εn is the electron energy with respect to Ec in InGaAs, or εn = En − Ec.

Using d = 10×10-9 m, me* = 0.04me and n = 1 and 2, we find the following electron energy levels

n =1 ε1=εn =h2n2

8me* d2 =

(6.626×10−34 )2(1)2

8(0.04× 9.11×10−31)(10 ×10−9)2 = 1.51×10-20 J = 0.094 eV

n = 2 ε2 = 0.376 eV

Using d = 10×10-9 m, mh* = 0.44me and n = 1, the hole energy levels below Ev is

n =1 ′ ε n =h2n2

8mh* d2 =

(6.626×10−34 )2(1)2

8(0.44× 9.11×10−31)(10 ×10−9)2 = 1.37×10-21 J = 0.00855 eV

The wavelength light emission from the QW laser with Eg = 0.70 eV is

λQW =hc

Eg + ε1 + ′ ε 1=

(6.626×10−34)(3×108)

(0.70+ 0.094+ 0.00855)(1.602×10−19)= 1545××××10 -9 m (1545 nm)

The wavelength of emission from bulk InGaAs with Eg = 0.70 eV is

λg =hc

Eg

=(6.626×10−34 )(3×108)

(0.70)(1.602×10−19) = 1771××××10 -9 m (1771 nm)

The difference is λg − λQW = 1771 - 1545 = 226 nm.

4.16 A GaAs quantum well

The lowest energy levels with respect to the CB edge Ec in GaAs are determined by the energy of an electron in a one-dimensional potential energy well

εn =h2n2

8me*d2

where n is a quantum number 1, 2, …, εn is the electron energy with respect to Ec in GaAs, or εn = En − Ec. Thus,

n =1 ε1=εn =h2n2

8me* d2 =

(6.626×10−34 )2(1)2

8(0.07×9.11×10−31)(8×10−9)2 = 0.0839 eV

n = 2 ε2 = 0336 eV

n = 3 ε3 = 0.755 eV


Note: Whether ε3 is allowed depends on the depth of the QW and hence on the bandgap of the sandwiching semiconductor.

The hole energy levels below Ev is

′ ε n =h2n2

8mh* d2 =

(6.626×10−34)2(1)2

8(0.47× 9.11×10−31)(8×10−9)2 = 0.0125 eV

The wavelength of emission from bulk GaAs with Eg = 1.42 eV is

λg =hc

Eg

=(6.626×10−34 )(3×108)

(1.42)(1.602×10−19) = 874×10 -9 m or 874 nm.

Whereas from the GaAs QW, the wavelength is,

λQW =hc

Eg + ε1 + ′ ε 1=

(6.626×10−34)(3×108)

(1.42+ 0.0839+ 0.0125)(1.602×10−19)

= 818×10 -9 m or 818 nm.

The difference is λg − λQW = 874 − 818 =56 nm. 5.1 Bandgap and photodetection

a Given, λ = 600 nm, we need Eph = hυ = Eg so that,

Eg = hc/λ = (6.626×10-34 J s)(3×108 m s-1)/(600×10-9 m)

= 2.07 eV

b A = 5×10-2 cm2 and Ilight = 20×10-3 W/cm2. The received power is

P = AIlight = (5×10-2 cm2)(20×10-3 W/cm2) = 10-3 W

Nph = number of photons arriving per second = P/Eph

= (10-3 W)/(2.07×1.60218 ×10-19 J/eV)

= 2.9787×1015 Photons s-1.

= 2.9787×1015 EHP s-1.

c For GaAs, Eg = 1.42 eV and the corresponding wavelength is

λ = hc/ Eg = (6.626×10-34 J s)(3×108 m s-1)/(1.42 eV × 1.6 ×10-19 J/eV)= 873 nm

The wavelength of emitted radiation due to EHP recombination is therefore 873 nm.

It is not in the visible region (it is in the IR).

d For Si, Eg = 1.1 eV and the corresponding cut-off wavelength is,

λg = hc/ Eg = (6.626×10-34 J s)(3×108 m s-1)/(1.1 eV × 1.6 ×10-19 J/eV)= 1120 nm

Since the 873 nm wavelength is shorter than the cut-off wavelength of 1120 nm, the Si photodetector can detect the 873 nm radiation (Put differently, the photon energy corresponding to 873 nm, 1.42 eV, is larger than the Eg, 1.1 eV, of Si which mean that the Si photodetector can indeed detect the 873 nm radiation)

5.2 Absorption coefficient

a If Io is the intensity of incoming radiation (energy flowing per unit area per second), Ioexp(−αd) is the transmitted intensity through the specimen with thickness d (Figure 5.17) and thus Io[1− exp(−αd)] is the “absorbed” intensity. If Γph is the number of photons arriving per unit area per unit second (the


photon flux), then Γph = Io/hυ where hυ is the energy per photon. Thus the number of photons absorbed per unit volume per unit second of sample is

n ph =

AΓph

Ad=

Io 1− exp(−αd)[ ]hυd

=Io 1− exp(−αd)[ ]

dhυ

b For Ge, α ≈ 5.2 × 105 m-1 at 1.5 µm incident radiation (from figure 5.3).

∴ 1− exp(−αd) = 0.9

∴ d =1

αln

1

1− 0.9

=

1

5.2×105 ln1

1− 0.9

= 4.428×10−6 m = 4.428 µm.

For In0.53Ga0.47As, α ≈ 7.5 × 105 m-1 at 1.5 µm incident radiation (Figure 5.3).

d =1

7.5×105 ln1

1− 0.9

= 3.07×10−6m = 3.07 µm.

c The quantum efficiency is unity. Therefore the collected electrons per unit area per unit second is equal to the absorbed photons per unit area per unit second. So, the current density (current per unit area),

[ ] [ ]

hc

de

h

deJ ph

)exp(1)exp(1 αλυ

α −−=

−−= oo II

Given, Io = 100 µW mm-2 = 100 × 10-6 × 106 W m-2 = 100 W m-2,

81.108100.310626.6

9.0105.11001060218.1834

619

=×××

×××××= −

−−

phJ A m-2 =10.881 mA/cm2.

NOTE: We neglected any light reflection from the surface of the semiconductor crystal (100% efficient AR coating assumed).

5.3 Ge Photodiode

a At λ = 850×10-9 m, from the responsivity vs. wavelength curve we have R = 0.25 A/W. From the definitions of quantum efficiency (QE) η and responsivity we have,

%36.5)m10850()C1060218.1(

)A/W25.0()ms103(Js)10626.6(919

1834

=×××

××××== −−

−−

λη

e

hcR

Similarly, we can calculate quantum efficiency at other wavelengths. The results are summarized in the following table.

Wavelength, (nm) 850 1300 1550

Responsivity R, (A/W) 0.25 0.57 0.73

Quantum efficiency η, (%) 36.5 54.3 58.4

b Given, photocurrent Iph = Id = 0.3 µA = 0.3×10-6 A and area, A = 8×10-9 m2, the incident optical power,

Po = Iph/R = (0.3×10-6 A)/(0.73 A W-1) = 4.1096 × 10-7 W

Light intensity, Io = P0/A = (4.1096 ×10-7 W)/(8×10-9 m2) = 51.4 W m-2 or 5.14 mW cm-2.

c From semiconductor data under Selected Semiconductors, for most semiconductors dEg/dT is negative, Eg increases with decreasing temperature. Stated differently, α vs λ curve shifts towards shorter λ with decreasing T. The change in α with T means that the amount of optical power absorbed in the depletion region and hence the quantum efficiency will change with temperature. The peak


responsivity will shift to lower wavelengths with decreasing temperature. If maximum photogeneration requires a certain absorption depth and hence a certain αmax, then the same αmax will occur at a lower wavelength at lower temperatures. In Figure 5Q3, maximum responsivity corresponds to αmax which occurs at λmax at high T and at λ′max at lower T.

Figure 5Q3.

λ

Absortion coefficient = 1/δ

Low THigh T

λmaxλ′max

αmax

The absorption coeff icient depends on the temperature

d Dark current (∝ exp(−Eg/kT)) will be drastically reduced if we decrease the temperature. Reduction of dark current improves SNR.

e The RC time constant = 100 × (4×10-12) = 0.4 ns. The RC time constant is comparable to the rise time, 0.5 ns. Therefore, the speed of response depends on both the rise time and RC time constant. (It is not simply 0.4 ns + 0.5 ns.)

5.4 Si pin Photodiodes

a For type A, responsivity RA = 0.2 A/W at 450 nm wavelength light.

Given, intensity Io = 1 µW cm-2 = 10-8 W mm-2 and area A = 0.125 mm2.

Power, P0 = IoA = (10-8 W mm-2)× (12.5 mm2) = 1.25 × 10-7 W.

Photocurrent, Iph = RAP0 = (0.2 A/W)× (1.25 × 10-7 W) = 2.5 × 10-8 A = 25 nA.

Quantum efficiency, %1.55)m10450()C1060218.1(

)A/W2.0()ms103(Js)10626.6(919

1834

=×××

××××== −−

−−

λη

e

hcA

AR

For type B, responsivity RB = 0.12 A/W at 450 nm wavelength light.

Photocurrent, Iph = RBP0 = (0.12 A/W)× (1.25 × 10-9 W) = 1.5 × 10-10 A = 0.15 nA.

Quantum efficiency, %1.33)m10450()C1060218.1(

)A/W12.0()ms103(Js)10626.6(919

1834

=×××××××== −−

−−

λη

e

hcB

BR

b, c Photocurrent and quantum efficiency can be calculated for other wavelengths in the same way. The values are summarized in the following table.

Table 5Q4 Summarized values for photocurrent and quantum efficiency.

Wavelength, nm Type A Type B

Responsivity, A/W

Photocurrent, nA

Quantum efficiency, %

Responsivity, A/W

Photocurrent, nA

Quantum efficiency, %

450 0.20 25 55.1 0.12 15 33.1

700 0.46 57 81.5 0.46 57 81.5

1000 0.15 19 18.6 0.40 50 50

d Quantum efficiency depends on the wavelength and also on the device structure. Both devices use a Si crystal , but, at a given wavelength such as 450 nm, for A, QE = 55.1% and for B, QE = 33.1%.


ηmaximum =

hcReλ

=(6.62×10−34 J s)(3×108 m s-1)(0.78 A W-1)

(1.6×10−19 C)(1210×10−9 m) ≈ 0.80

5.6 Maximum QE

The relationship between the responsivity R and the quantum efficiency (QE) η is,

R =

ηeλhc

or η =

hc

e

Rλ

The QE is maximum when dη/dλ = 0, thus differentiating the above expression with respect to λ we have,

dηdλ

=hc

eλdRdλ

+hcR

e

d

dλ1

λ

= 0

hc

eλdRdλ

−hcR

e

1

λ2

= 0 ∴

dRdλ

−Rλ

= 0 i.e.

dRdλ

=Rλ

0

0.2

0.4

0.6

0.8

1

800 1000 1200 1400 1600 1800Wavelength(nm)

The responsivity of anInGaAspin photodiode

Responsivity(A/W)

60040020000

0

0.10.20.3

0.40.5

0.60.70.8

0.5 1 1.5 2Wavelength(µm)

The responsivity of a commercial Gepnjunction photodiode

Responsivity(A/W)Tangent through origin

00

0.1

0.2

0.3

0.4

0.5

0.6

200 400 600 800 1000 1200Wavelength(nm)

A B

The responsivity of two commercial Sipin photodiodes

Responsivity(A/W)

0

Figure 5Q6

We can find the maximum QE by drawing a tangent to the R vs. λ curve that passes through the origin as in the three examples below. The actual graphical values are listed in Table 5Q6. For example for the InGaAs pin photodiode, the maximum QE is

ηmaximum =

hcReλ

=(6.62×10−34 J s)(3×108 m s-1)(0.78 A W-1)

(1.6×10−19 C)(1210×10−9 m) ≈ 0.80

Table 5Q6

InGaAs pin Si-pin-A Si-pin-B Ge photodiode

λ (nm) 1210 700 810 1500

R (A/W) 0.78 0.46 0.57 0.71

Maximum QE 0.80 0.81 0.87 0.57

Maximum QE % 80% 81% 87% 57%


5.12 The APD and excess avalanche noise

a We can find the value of x by plotting F vs. M on a log-log plot, which is shown in Figure 5Q12. From the plot, the index x = 0.857. The fit shows that, F = 1.095M.8571 which agrees well with the equation, F ≈ Mx.

1

10

1 10

Exc

ess

nois

e fa

ctor

Multiplication

F = (1.095)M0.857

Excess noise factor F vs. Mfor a GE APD; from Scansen andKasap 1992.

Figure 5Q12

b Given, Ido = 0.5 µA, M = 6, B=500 MHz and x = 0.857. From Equation (12) §5.10, the SNR can be written as,

SNR=Signal PowerNoise Power

=M2I pho

2e Ido + I pho( )M2 +x B[ ]

∴ M2Ipho

2 − 2eM2 + xB SNR( )[ ]Ipho − 2eM2+ x B SNR( )Ido[ ]= 0 (2)

Solving this quadratic Equation (2) with a SNR = 1 for Ipho we find,

Ipho = 1.9665×10-8 A or 19.665 nA

If Po is the incident optical power, then by the definition of responsivity, R = Ipho/Po,

Po = Ipho/R = (1.9665×10-8 A)/(0.8 A/W) = 2.458×10-8 W or 24.58 nW.

c Solving this quadratic Equation (2) with a SNR = 10 for Ipho we find,

Ipho = 6.4832×10-8 A or 64.665 nA

The incident optical power,

Po = Ipho/R = (6.4832×10-8 A)/(0.8 A/W) = 8.104×10-8 W or 81.04 nW.

Note: Although the SNR has gone up by a factor of 10, the required increase in the incident optical power is only a factor of 3.3.

6.3 Solar cell driving a load

a The solar cell is used under an illumination of 1 kW m-2. The short circuit current has to be scale up by 1000/600 = 1.67. Figure 6Q3-2 shows the solar cell characteristics scaled by a factor 1.67 along the current axis. The load line for R = 20 Ω and its intersection with the solar cell I−V characteristics at P which is the operating point P. Thus,

I′ ≈ 22.5 mA and V′ ≈ 0.45 V

The power delivered to the load is

Pout = Ι′V′ = (22.5×10-3)(0.45V) = 0.0101W, or 10.1 mW.

This is not the maximum power available from the solar cell. The input sun-light power is


Pin = (Light Intensity)(Surface Area) = (1000 W m-2)(4 cm2 × 10-4 m2/cm2) = 0.4 W

The efficiency is

η = 100Pout

Pin

= 1000.010

0.4= 2.50

0 which is poor.

b Point M on Figure 6Q3-2 is probably close to the maximum efficiency point, I′ ≈ 23.5 mA and V′ ≈ 0.44 V. The load should be R = 18.7 Ω, close to the 20 Ω load. At 600 W m-2 illumination, the load has to be about 30 Ω as in Figure 6.8 (b). Thus, the maximum efficiency requires the load R to be decreased as the light intensity is increased. The fill factor is

FF =ImVm

IscVoc

=(23.5 mA)(0.44 V)

(27 mA)(0.50 V)≈ 0.78

c The solar cell is used under an illumination of 400 W m-2. The short circuit current has to be scale up by 400/600 = 0.67. Figure 6Q3-2 shows the solar cell characteristics scaled by a factor 0.67 along the current axis. Suppose we have N identical cells in series, and the voltage across the calculator is Vcalculator. The current taken by the calculator is 3 mA in the voltage range 3 to 4 V and the calculator stops working when Vcalculator < 3 V. The cells are in series so each has the same current and equal to 3 mA, marked as I′ in Figure 6Q-2. The voltage across one cell will be V′ = Vcalculator/N. which is marked in Figure 6Q3-2. V′ = 0.46 V. Minimum number of solar cells in series = N = 3 / 0.46 = 6.5 or 7 cells , since you must choose the nearest higher integer.

If we want the calculator continue to work under low intensity levels, then we can connect more cells in series until we reach about 4 V; N = 4 V / 0.46 V = 8.7 or 9 cells in series.

The easiest estimate for the minimum required light intensity is the following: The calculator will stop working when the light intensity cannot provide energy for the solar cell to deliver the 3 mA calculator current. The short circuit current at 400 W m-2 is 11 mA in Figure 6Q3-2. Thus

Minimum light intensity=3 mA

11 mA400 W m−2=109 W m-2

V′V

I (mA)

0.60.4

?0

0

?0

The load line for R = 20 (I-V for the load)

I-V for a solar cell under anillumination of 1000 Wm-2.

P

0.2

I′

?0

M

V′V

I (mA)

0.60.4

?0

0

?0

I−V for a solar cell under anillumination of 400 W m-2.

P

0.2

I′ 3 mA

Figure 6Q3-1 Figure 6Q3-2

6.4 Open circuit voltage

The short circuit current is the photocurrent so that at

Isc2 = Isc1

I2I1

= 50 mA( ) 50 W m−2

100 W m−2

= 25 mA

Assuming n = 1, the new open circuit voltage is

Voc2 = Voc1 +

nkBT

eln

I2I1

= 0.55+1 0.0259( )ln 0.5( )= 0.508 V


Assuming n = 2, the new open circuit voltage is

Voc2 = Voc1 +

nkBT

eln

I2I1

= 0.55+ 2 0.0259( )ln 0.5( )= 0.467 V

6.5 Shunt resistance .

a Figure 6Q5-1 shows the equivalent circuit with the series resistance removed. The currents flowing into node A sum to zero (Kirchoff’s current law). Currents into a node are positive and those leaving a node are negative. Thus,

−I + Iph − Id − (V/Rp) = 0

∴ I = − I ph + Id +V

Rp

= − I ph + Io exp(eV

nkBT) −1

+

V

Rp

which is the required equation.

A′

Iph Rp RLV

IIph

Id

Solar cell Load

B

A

Figure 6Q5-1

b Plots of I vs. V for different values of Rp are shown in Figure 6Q5-2.

Rp =

Rp = 1000 Ω

Rp = 100 Ω

Voltage (V)

Current (mA)

Figure 6Q5-2

Conclusion: Lower Rp decreases the OC voltage and reduces the available output power.

6.6 Series connected solar cells

Consider two cells in series. Then there are two Rs in series and two pn junctions in series. If I is the total current through the devices then the voltage across one pn junction is

Vd =V − I(2Rs )

2

Thus, I = − Iph + Io expVd

nVT

−1

= − Iph + Io exp

V − I(2Rs )

2nVT

−1

∴ I ≈ − Iph + Io expV − I(2Rs )

2nVT

since Vd > nVT


Rearranging we find the following:

2 Cells in series:

V = 2nVT lnI + Iph

Io

+ 2Rs I

1 Cell: V = nVT lnI + Iph

Io

+ Rs I

Figure 6Q6-1 shows the I-V characteristics for 1 cell and 2 cells in series. Notice that the OC voltage for the series cells is twice the OC voltage for one cell.

1 cell

Iph = 10 mA

2 cells in series

Current in mA

Voltage

Figure 6Q6-1

Since Power = IV we can plot the power delivered as a function of current as shown in Figure 6Q6-2. Maximum power corresponds to

1 cell: I ≈ 7.7 mA, V ≈ 0.3 V, P ≈ 2.3 W

2 cell: I ≈ 7.7 mA, V ≈ 0.58 V, P ≈ 4.5 W

Power in mW

Current in mA

2 cells in series

1 cell

4.5 mW

2.3 mW

-7.7 mA

Figure 6Q6-2

6.7 Series connected solar cells

A

Iph1

V

Id1

Rs2

RL

I

Two different solar cells in series

Iph2

Id2

Rs1

Iph2

Iph2

V1

V2

I

Figure 6Q7-1


The equivalent circuit is shown in Figure 6Q7-1. The current through both the devices has to be the same. Thus, for cell 1

I = − Iph1 + Io1 expVd1

n1VT

−1

= −Iph1 + Io1 exp

V1 − IRs1

n1VT

−1

∴ V1 − IRs1

n1VT

= lnI + Iph1

Io1

−1

∴ V1 = n1VT lnI + I ph1

Io1

−1

+ IRs1

For cell 2 I = − Iph1 + Io2 expVd 2

n2VT

−1

= −Iph 2 + Io2 exp

V2 − IRs2

n2VT

−1

∴ V2 = n2VT lnI + I ph 2

Io2

+1

+ IRs2

But V = V1 + V2

∴ V = n1VT lnI + Iph1

Io1

+1

+ IRs1 + n2VT ln

I + I ph2

Io2

+ 1

+ IRs 2

We can now substitute Io1 = 25×10-6 mA, n1 = 1.5, Rs1 = 10 Ω, Io2 = 1×10-7 mA, n2 = 1, Rs2 = 50 Ω and then plot V vs. I (rather I vs. V since we can calculate V from I) for each cell and the two cells in series as in Figure 6Q7-2. Notice that the short circuit is determined by the smallest Isc cell. The total open circuit voltage is the sum of the two.

Cell 1

Voltage

Current(mA)

Cell 2

Cell 1 in series with 2

Figure 6Q7-2

6.11 Maximum Power and Fill Factor

The current through a solar cell is I = − I ph + Io expV

nVT

so that the power is P = IV = − I phV + IoV expV

nVT

At maximum power dP/dV = 0 when V = Vm

dP

dV= − Iph + Io exp

V

nVT

+ Io

V

nVT

expV

nVT

∴ dP

dV= − Iph + Io exp

Vm

nVT

1+

Vm

nVT

= 0


∴ 1+Vm

nVT

exp

Vm

nVT

=

Iph

Io

and since Vm >> nVT,

∴ Vm

nVT

exp(Vm

nVT

) ≈Iph

Io

(1)

Consider the current at maximum power,

Im = − Iph + Io expVm

nVT

Substitute for the exponential term,

Im = − Iph + Io

nVT Iph

VmIo

∴ Im = − Iph 1−nVT

Vm

(2)

The product ImVm is the maximum power from which we can calculate the FF.

NOTE: We can obtain better expressions for the maximum power voltage Vm as follows:

We have to find Vm by solving,

f (Vm) =Vm

nVT

exp(Vm

nVT

) −Iph

Io

= 0 (3)

Let vm = Vm/(nVT), a = Iph/Io and voc = Voc/(nVT). From Example 6.3.2 we know that

voc =Voc

nVT

= lnI ph

Io

= ln a

Thus, Eq. (3) that we have to solve is f (vm ) = vm exp(vm) − a = 0 (4)

We can use Newton’s approximation. If xo is an approximate solution, then a better solution is

x1 = xo −f (xo )df

dx

xo

Let x =V/(nVT), Since a = Iph/Io which is something like (10 mA cm-2)/(10 nA cm-2) = 106, we take xo, the rough solution, to be such that Eq. (4) becomes

f(xo) ≈ vocexp(xo) − a = 0

so that xo = lna

voc

= voc − lnvoc

A better solution is when

x1 = xo −xoe

x o − a

(xo +1)exo=

xo(xo +1)ex o − xoex o + a

(xo + 1)ex o

∴ x1 =xo

2ex o + a

(xo +1)ex o

Substituting xo = voc − ln(voc) and a = exp(voc),


x1 =(voc − ln vm )2evoc −ln v m + evoc

(voc − lnvoc +1)ev oc − lnv oc

Thus, vm =(voc − ln vm )2evoc −ln vm + evoc

(voc − lnvoc +1)evoc − lnv oc (5)

As an example, let us apply this to Example 6.3.1 where Voc ≈ 0.480 V, taking n =1, voc = 18.46, vm = 15.72 and Vm = 0.41 V.

The fill factor FF is FF =ImVm

IscVoc

≈− Iph 1−

nVT

Vm

Vm

− IphVoc

=1−

1

vm

vm

voc

(6)

Questions 6.9 gives FF as FF ≈voc − ln(voc + 0.72)

voc + 2 (7)

Using Example 6.3.1 where Voc ≈ 0.480 V, voc = 18.46, vm = 15.72 leads to

FF = 0.80 from Eq. (6) whereas FF = 0.76 from Eq. (7).

6.12 Energy band diagram

a

Ec

Ev

Ec

EFp

M

EFn

eVo

p nE

o

Evnp

No illumination

VI

np

Eo–E

e(Vo–V)

eV

Ec

EFn

Ev

Ev

Ec

EFp

SCL

Illumination

Figure 6Q12-1

b

Ev

Ec

Ev

Ec

(i) (ii) Figure 6Q12-2

Possible energy band diagrams for a semiconductor that has a decreasing bandgap are shown in Figure 6Q11-2 (i) and (ii). The photogenerated electron "rolls” down Ec. The hole also "rolls" down its own energy hill either towards the right, (i), or left, (ii), depending on how Ec and Ev vary along the device. Obviously in Figure 6Q12-2(ii) there is a photogenerated current and a voltage across the device. It is possible, in principle, to fabricate a device that has the right bandgap variation and the right doping variation to give the energy band diagram in Figure 6Q12-2(ii) and hence function as a solar cell.

Documents

kasap solution manual, optoelectronics