# Solutions to Selected Problems Exercise 2.1

 N 4 N

(a) Energy at earth’s mean distance/unit area/unit time = (rs/des)2 energy at sun’s surface/unit area/unit time

= (rs/deS)2 OTs4

Projected area of earth = nre2 Surface area of earth = 4nre2

Incoming energy = outgoing energy (rs/des)2 oTs4nre2 = oTe44nre2 Te = Ts (rs/des)1/2/21/2 = 289.38 K

(b) Trick Question!

The earth absorbs 0.7 of incident radiation but also only emits 0.7 as much. Hence the same radiation balance is maintained and:

Te = 289.38K

(c) The incoming radiation decreases to 0.7 but outgoing decreases to 0.6. Therefore the earth is hotter with such emission properties:

Te’ = (0.7 / 0.6 / 7 4 Te = 300.75 K

(d) This increase, 2°C, is roughly that predicted by a doubling of atmospheric CO2! The new value of emissivity is:

302.75 = (0.7 / є’ f74Te є = 0.584

A 2-3% reduction is therefore required.

Exercise 3.1

Energy from sun = fs oTs4/unit area

Energy re-radiated/unit time = oTc4/unit area Energy available/unit time/unit area = fsoTs4 – oTc4

Assume this available energy is converted at Carnot efficiency by a machine operating between a source at Tc and a sink at TA. Possible solar energy conversion efficiency is therefore:

n = [(1 – Tc 4 /(fsTs 4 ))](1 – Ta/Tc)

Solving iteratively:

For fs = 1 (full concentration), Tc must lie between TA (300 K) and TS (6000 K)

Try 3000 K, П3000 = 84.38% 2000 K, П2000 = 83.95% 2500 K, П2500 = 85.35%

2544 K, П2544 = 85.36%

For fs = 2.1646 x 10-5 (diffuse sunlight), Tc must lie between TA (300 K) and (fs)1/4 Ts (409 K)

Try 350 K, П350 = 6.64% 380 K, П380 = 5.40% 360 K, П360 = 6.69%

356 K, n356 = 6.72%

Exercise 4.1

(Quite difficult!).

The total rate of radiative recombination occurring in a cell can be found by integrating the Shockley-Van Roosbroeck Eq. (4.8) across the cell volume (thickness W). Assuming constant quasi-Fermi level separations:

RT = UT = (WA)(eqV/kT -1 )8nc(kT/ch)3 J n ae de

0 e£ -1

Under the conditions of the problem where a is zero for e < Ec/kT, this simplifies to:

The rate of photon emission, at least for thick cells can be found by the Shockley-Queisser approach that leads to Eqs. (3.9) and (3.10). This is an upper bound for thinner cells, so:

The difference is due to the photons that are recycled, i. e., generated but not emitted. This fraction is therefore given by:

f = (RT – re)/rt =(1 – re/rt )

Almost all are recycled if W >> (1/4n2a0). The problem was not intended to be any more difficult than this but some students have explored why negative values were predicted for small W.

These found f > (1 – 1/2n2), about 0.95. So, even when W is small, the fraction recycled stays large. This is due to the fact that photons generated by spontaneous radiative recombination are emitted in all directions, whereas Snell’s law only allows those emitted in a direction close to the normal to the cell’s surface to escape.

From the latter result, it can be deduced that the first result only provides a reasonably tight bound when W >> 1/(2n2a0).

An interesting follow-on problem that appears solvable is whether a cell of finite thickness has a higher efficiency than the infinite cell thickness case. See (Parrott 1993) and (Araujo and Marti 1994) where finite thickness is shown to have a definite advantage in two specific cases.

Exercise 4.3

(a) The “sky” radiation can be neglected since it corresponds to thermal genera­tion balanced by a corresponding recombination rate at thermal equilibrium. Hence:

Jsc = qfsN (Eg0,Ts)

2n(kT)3 7 e2de =qfs——£— h3c2 ege£-1

where Eg = 1.124 eV/kTs = 2.1739. From Eq. (C21):

в (0, Eg) – Eg2 Pa (-Eg)/2 + £gfti (-Eg) + e (-Eg)

From Eq. (C16):

exp( 2p)

2 j+1 [1 – (2 / 3 )j+1 exp(n)] with relative error equal to the second term times exp (-2.1739), i. e., 0.1137.

во (-Eg) = 0.1137 + 0.0067

= 0.1207 (rel. error ~ 0.0007)

Px (-Eg) = 0.1137 + 0.0034

= 0.1171 (rel. error ~0.0004)

P2 (-Eg) = 0.1137 + 0.0017

= 0.1154 (rel. error ~ 0.0002)

.•.p2 (0, Eg) = 0.6552 (rel. error < 0.0001)

Jsc = 2,869 A/cm2 (fs = 1)

= 62.1 mA/cm2 (fs = 2.1646 x 10-5)

(b)

In the Shockley-Queisser approach:

In this case, eg = 1.124 eV/kTc = 43.478

Hence, во (-Eg) « ві (-Eg) « fa (-Eg) « exp (-Eg)

Voc = (kT/q) In [(Jsc + Jo)/Jo]

= 889.4 mV (fc = 1, f = 2.1646 x 10-5)

= 1,167.1 mV (fc = 1, fs = 1)

Note that the latter is larger than the bandgap! This means that the formulation is invalid in this region since non-degenerate carrier populations have been assumed. Also, once quasi-Fermi levels move into the bands, electron states at the bottom of the conduction band become more occupied than those at the top of the valence band. Stimulated emission increases and the effective absorption coefficient becomes negative.

For an infinitely thick cell, this prevents the progression of quasi-Fermi levels into bands. For thin cells, however, penetration can occur in principle (Parrott 1986).

Chemical Potential

At open-circuit, J = 0 so Eq. (4.19) becomes:

fcN(EG,~qV0c, Tc ) = fsN(EG,~,0,Ts) + (fc – fs )N(EG,<*>,0T )

= 5,242 (fc = fs)

= 0.1135 (fs = 2.1646 x 10-5 ,fc = 1)

Now,

\$2 (П, є) = є2 во (П-є)/2+єві(п-є)+в2 (П-є)

Note that ві and в2 are bounded as є – a n but во ^ This will limit є to a value less than n. Neglecting в1 and в2 terms gives:

в0(П-є) < 5.546 (fc = fs )

From Eq. (C.12), 1 – exp (n – є) > e-5.546

П – є < – 0.0039

Voc < Ec/q – 0.00010 V

Similarly, assuming в1 = в2 = в0 gives:

в0 (П – є) > 5.297

0.00010 V < (EG/q – Voc) < 0.00013 V (f = fs)

i. e., Voc = 1.1239 mV.

Following the same procedure in the case where fs << fc gives:

0.2334 < (Ec/q – Voc) < 0.2346

A tighter lower bound in this case can be obtained by assuming в1 = в2 = в0 = exp (n – є).

0.2346 < (Eq/q – Voc) < 0.2346

i. e., Voc = 889.4 mV.

This is a similar result to that from to the Shockley-Queisser approach.

As the two theories are more similar for smaller voltages, they will be more nearly similar near the maximum power voltage than near open-circuit. Hence, the Shockley-Queisser theory, obeying the traditional solar cell law but having a higher Voc, will have a lower fill factor.

(c) At 300 suns, the same approach gives similar values in both cases:

Voc (Shockley-Queisser) = 1.1049 V Voc (Chemical Potential) = 1.0364 V

Exercise 5.1

A close-up of 5.2 is shown in Fig. F.1.

n

n – 1

Fig. F.1 : Close-up of central cells in Fig. 5.2 showing the radiatively emitted current components.

As well as the currents shown in Fig. F.1, we have the most important current, that generated by sunlight and the background thermal radiation:

S = qA[ fsN(EGn, EG(n+1 ),°,TS ) + IV(EGn, EG(n+1 ),°,TA )]

The current-voltage relationship for cell n is therefore given by:

In = S – (A + B + C) + D + E

where

(A + B + C) = 2qAfcN ( Eg„ , -, qVn, Tc )

D = qAfcN ( eg( n+1 ),-,qVn+1 ,Tc)

E = qAfcN (EGn>™ >qVn-1 ,Tc)

Note that, in the case on a monolithic tandem cell (all fabricated on one piece of semiconductor material of varying composition), components B, D, C and E would all be enhanced by a factor of n2, where n is the refractive index. Rather than the performance limit in this case being a few percentage points below that of the filtered tandem cell limits of Table 5.1, it will lie a few percentage units below.

Exercise 5.2

The optimum operating voltage for each cell is given by the solution of Eq. (5.3). The easiest way to solve it is probably just, in each case, insert the given value of E = Eg, and vary Vm between 0 and Ec/q until the value giving zero is found. Following this procedure gives:

(a) Eg = 0.25 eV, Vm = 6.0 mV

(b) Eg = 0.50 eV, Vm = 158.9 mV

(c) Eg = 2.00 eV, Vm = 1.517 V

These points should lie on the iiopt versus E line for the diffuse case shown in Figs. 6.8 and 7.3.

The reason these voltages are so small for the small values of E is that the light coming from the sun is very weak at these energies for the diffuse case, while the light emitting ability of the cell is comparatively high. Only a small chemical potential enhancement is feasible, otherwise the cell will emit more light than it absorbs.

Exercise 6.1

The efficiency of a hot carrier cell is given by: П = Puse / Solar Input

where Puse is given by Eq. (6.8).

This becomes:

[ fsN(Ec,™,0,Ts ) + (fc – fs )N(Ec,™,0,Tc )- fcN(Ec,-,AtiH, TH VA^/Th

+ [fsE(EG~,0,Ts) + (fc – fs)E(EG,~0,Tc)- fcE(EG,~,AnH, TH )](1 – TC/TH )

fsE (0,-0,Ts)

 2n(kT) h3c22n(kT)4h3c 2

For the case of EG = 0 and A/iH a large negative, the corresponding terms become:

N (0,—,APH, t )=ЩН?–

When ApH is a large negative:

where the latter integral equals 2. Similarly:

E(0,~,AVH, TH ) = ln(kTH2 )4eAlH/kTH.6

Equation (6.12) follows from this.

Substituting numerical values and finding optimal values of AjiH gives:

TH = 4000K, fs = 1,ApH =-0.7485eV, n= 86.7%

TH = 3600K, fs = 2.1646x10_5, ApH =-3.1738eV, n = -22.8%!!

The negative value is a bit of a surprise since earlier calculations (Ross and Nozic 1982), although not calculating for EG = 0, suggest values close to 68% should be obtained.

Exercise 7.1

For mEG < E < (m + 1) EG, ji(E) = mqV, where m is the integral part of E/EG. As Eg ^ 0:

1(E) « qV (E/Eg) = aE/EG

If we define a temperature Tj such that:

(E – aE/EG)/kTc = E/kT,

the analysis will be greatly simplified. Rearranging a/EG = (1 – Tc/Ti)

Substituting into Eq. (7.3)

Dividing by fsE(0,^,0,Ts) gives:

П= (1 – Tc/Ti )[1 + (fc – fs)(TA/Ts)4 /fs

Which equals Eq. (7.4).

Note that something a little weird has happened here. One would expect a/EG to approach infinity as EG 0. However, the approach to zero must be in such a

way that mEG remains finite to satisfy the constraints on the problem.

Exercise 8.1

(a)

(INM / A )max = q[ fsN(EA, EB,0,Ts ) + (fc – fs )N(Ea, Eb,0,Ta ) – fcN(EA, EB,0,Tc )] For I21, parameters are:

Ea = 0.7 eV, Eb = 1.2 eV, Ts = 6000 K, Tg = Ta = 300 K, fc = 1, fs = 2.1646 x 10-5 Giving (I21/A)max = 28.9 mA/cm2 For I32, changed parameters are:

Ea = 1.2 eV, Eb = 1.9 eV Giving (I32/A)max = 30.2 mA/cm2 For I31, changed values are:

Ea = 1.9 eV, Eb = 3.1 eV

Giving (I31/A)max = 21.8 mA/cm2

(b) To find variation with chemical potential assuming the Shockley-Queisser approximation, a term given below is subtracted:

(Inm/A )dark = fcN (EA, EB,0, Tc )(e^NM /kTc -1)

This gives standard solar cell current-voltage type curves with open-circuit values of 0.47 eV, 0.94 eV and 1.61 eV.

(c) Curves for I21 and I32 are combined using the standard approach, e. g., Fig. 5.5 (Wenham et al. 1994) since there is a mismatch in the current, the curve will be a bit flattened from a normal solar cell curve with a short-circuit current density of

28.9 mA/cm2 and an open-circuit potential of 1.41 eV.

(d) The curve from (c) is combined with the I31 curve as in Fig. 5.2 of the above reference to give the final output curve, which again looks fairly similar to a single solar cell curve (no humps or bumps!) with a short-circuit voltage of 1.41 eV.

(e) From the curve (d), the maximum power output of 62.8 mW/cm2 at a potential of 1.276 eV is found, compared to the solar input calculated as 159.1 mW/cm2, corresponding to an efficiency of 39.5%.

References

Araujo GL and Marti A (1994), Absolute limiting efficiencies for photovoltaic energy conversion, Solar Energy Materials and Solar Cells 33: 213-240.

Parrott JE (1993), Radiative recombination and photon recycling in photovoltaic solar cells, Solar Energy Materials and Solar Cells 30: 221-231.

Parrott JE (1986), Self-consistent detailed balance treatment of the solar cell, IEE Proceedings 133: 314-318.

Wenham SR, Green MA and Watt ME (1994), Applied Photovoltaics, Bridge Printers, Sydney.

[1] = qAfs N(EG,<*>,0,Ts) + qA(fc _ fs)N(EG,~,0Tc)_ qAfcN(EG,<*>,qV, Tc )

(4.19)

The first term on the right is identical to that in the earlier Eq. (4.10) while the third term corresponds to the second. Noting that for (E – qV) >> kT,