Flashcards in Chapter 8 - Photovoltaics Deck (96):

1

## What is the photovoltaic effect?

### Direct transformation of sunlight into electrical energy without moving parts.

2

## Draw the Feynman diagram of photovoltaics.

### See notes.

3

## How are photons described?

###
Energy = hf = hbar * omega

Momentum = p = hbar * k, with k = 2π/lambda = omega / c = 2πf / c.

4

## What kind of particle are photons?

### Bosons with angular momentum J = 1.

5

## How many polarization states exist for photons? What are they?

###
Three states exist.

1) Right circular polarization.

2) Left circular polarization.

3) Linear polarization.

6

## Derive the photon density of states.

### See notes.

7

## How does the photon density of states depend on the refractive index? What is the consequence of this?

### It goes as n^3. The consequence is that going from a volume with one refractive index to another with a different refractive index, some of the photons must be reflected at the interface.

8

## What is the absorber material in a solar cell?

### A semiconductor (eg. Si).

9

## How does the interactions between light and electron occur?

### Through electrical field. Basic principle is acceleration of an electron by the electrical field of the light.

10

## What determines the interaction strength?

### It is given by the dipole moment e*x in the absorbing medium.

11

## How is the interaction described quantum mechanically?

###
The excitation from initial state |i> to final state |f> is governed by the transition probability, W.

W is proportional to |E*|^2 = E^2e^2 |**|^2 which is proportional to I (intensity).*

*|^2 which is proportional to I (intensity).*

12

## How is the intensity of light described in the wave and particle models respectively?

###
Wave: I = 1/µ_0 |ExB| = 1/µ_0 * E_0 * B_0 = 1/(µ_0*c)*E_0^2 * sqrt(eps_0/µ_0) E^2.

Particle: hbar*omega * PHI, with PHI = photon flux density per area and time.

13

## Absorption of a photon is subject to three conserved quantities. Which?

###
Energy: E_f - E_i = hbar * omega

Momentum: p_f - p_i = hbar * k ≈ 0 (small momentum for photons)

Angular momentum: L_f - L_i = hbar.

14

## What are the consequences of angular momentum conservation in a) atoms and b) solids?

###
a) leads to dipole selection rule: ∆n = ±1

b) leads to creation of excitons.

15

## How is the complex dielectric function defned?

### eps = eps_1 + i*eps_2

16

## How is the complex refractive index defined, and what are the two parameters?

###
ñ = n + ik,

n = refractive index

k = absorption coefficient

17

## What happens to the speed of light in absorbing matter?

### It goes from c_0 to c_0/n, where c_0 is speed of light in vacuum and n is the refractive index.

18

## What happens to the wave vector in absorbing matter?

###
It goes from k = omega / c_0 to k = (omega / c_0)*ñ,

where ñ is the complex refractive index.

19

## How is the law of Lambert-Beer given?

###
From the expression of the electric field.

E = E_0 exp[i(kx - wt] = E_0 exp[i(omega/c_0 * (n+iK)x - wt)]

= E_0 exp[-omega/c_0 * kx] exp [i(omega/c_0 * nx-wt].

The law of Lambert-Beer is then given from the fact that I is proportional to E^2:

I(x) = I_0 * exp[-2 omega/c_0 * k x] = I_0 exp[-alpha x]

where alpha = 2 (omega/c_0) * k is the absorption coefficient.

20

## What is the consequence of the Lambert-Beer law, in terms of absorption length? How is this length for some materials, and why?

###
I(x) = I_0 * e^-(alpha*x)

Which means that 95% is absorbed in a length of x = 3/alpha.

For GaAs, as an direct semiconductor, this is about 1µm.

For c-Si, as an indirect semiconductor (so needs help from phonon to make the transition due to momentum conservation) this is about 10µm.

For a-Si this is about 1.5µm (no momentum conservation required since there is no periodic lattice, so all transitions are allowed).

21

## For normal incidence of light from the air (where n ≈ 1, and k ≈ 0), how is the reflectivity given?

### R = [(n-1)^2 + k^2]/[(n+1)^2 + k^2]

22

## For normal incidence of light from the air (where n ≈ 1, and k ≈ 0), how is the absorbance given?

###
A = (1-R) * (1-exp[alpha*d]),

where d is the absorbance thickness. Valid in the limit of strong absorption (alpha*d >> 1). Otherwise multiple reflections at the boundaries.

23

## For normal incidence of light from the air (where n ≈ 1, and k ≈ 0), how is the transmission given?

###
T = (1-R)^2*exp[-alpha*d])

where d is the absorbance thickness. Valid in the limit of strong absorption (alpha*d >> 1). Otherwise multiple reflections at the boundaries.

24

## For normal incidence of light form the air (where n ≈ 1, and k ≈ 0), how is then the average volume density of excited electrons generated by a photon flux per unit time given?

###
G = 1/d * A*PHI = (1-R) 1/d (1-exp[-alpha*d]) * PHI

where d is the absorbance thickness. Valid in the limit of strong absorption (alpha*d >> 1). Otherwise multiple reflections at the boundaries.

G = average generation rate, cm^-3 * s^-1

25

## Explain three different ways to reduce reflection (antireflection) from surface.

###
(i) Macroscopic texture by anisotropic etching (especially for monocrystalline Si). Here the rays are reflected not only back, but into another part of the surface, thus decreasing the total amount of reflected light. Here the trenches are a lot larger than the wavelength of light.

(ii) Nanotexturing with structures that are a lot smaller than the wavelength. This means that the refractive index changes gradually from the surface and down to the bulk. At any point, the refractive index is basically the same as the previous point. This means that there will be no reflection.

(iii) lambda/4 - antireflex coating (e.g. Si3N4 on Si). A material with a refractive index in between the air and the absorber with a thickness lambda/4 is placed on top of the absorber. This means that there will be destructive interference (total path lambda/2) for reflected wave at surface and lower surface. This is only optimal for one incident angle and wavelength.

26

## Derive the expression for the emission.

### See notes. (page 7)

27

## What is thermalization, and on which time scale does this happen?

### Thermalization is when excited electrons and holes with excess energy (that is the incoming light was more than required to excite the exciton) emit phonons (and thus heat). This happens on a time scale of 10^-13 s. (0.1 ps)

28

## Describe thermalization in molecules using energy-configuration diagram.

### The excitation is a vertical transition, because electron configuration can be changed much faster that the atom configuration. It will then lose energy by the emittance of phonons, and the reemitted light will be red-shifted (Stokes-shift).

29

## Describe thermalization in a semiconductor using band diagrams.

### See notes (page 9)

30

## Will thermalization primarily happen in the conduction band or the valence band? Why?

### In the conduction band. Because of the effective mass is smaller in CB, the curvature is greater, and the energy in the CB will be higher for the same k-vector than in the VB.

31

## What are the best electron-phonon coupling realized by?

### Longitudnial optical phonon modes near the Brillouin-zone center.

32

## Describe the generation and recombination of electrons and holes mathematically.

###
Number of electrons: n_0 + ∆n(t)

Number of holes: p_0 + ∆p(t)

dn/dt = dp/dt = G-R. When G = R, then dn/dt = 0 and we have steady state.

33

## Most solar cells are doped. What kind of doping?

### p-doping. This is because electrons have a smaller effective mass than holes.

34

## What is the efficiency of a solar cell determined by, in terms of recombination?

### The recombination time Tau_min of minority carriers.

35

## How can recombination occur?

### Either through emitting photons (radiative recombination) or through emitting phonons (nonradiative recombination)

36

## What does the probability of radiative recombination depend on?

### The dipole matrix element (same as for the excitation).

37

## What is the radiative recombination time?

###
It is given as:

tau = (hbar omega)/P,

where P is the probability of radiative recombination in classical radiation theory.

tau = 6πeps_0 hbar * (c^3)/(omega^3 * e^2 * x^2)

where x is the distance between electron and hole in the exciton.

38

## What can be said for the recombination time of high bandgap semiconductors?

### It will be short, since tau goes as 1/omega^3.

39

## Why is the recombination time so much longer in an indirect semiconductor?

### Because of momentum conservation (needs phonon to help transition from CB min to VB).

40

## What are typical recombination times for direct and indirect semiconductors?

###
For direct semiconductors with photon energy = 1 eV, and exciton radius 10Å we have tau = 5 ns.

For indirect semiconductors this is closer to µs.

41

## How can the electron hole pair non-radiatively recombine?

###
- Phonon cascade (as in thermalization - requires continiuum of real states in band gap)

- Multi-phonon emission

42

## Describe how a multi-phonon emission process works.

### A higher order process that occurs with a probability ∆^N, where ∆ is the electron-phonon coupling (∆

43

## What is the recombination of monomolecular recombination?

###
R = C_n * ∆n * N_D,

∆n is the density of minority carriers, N_D is the volume density of recombination centers (defects, impurities) C_n is the capture coefficient of N_D.

44

## What is the density of minority carriers given for in steady state?

###
Steady state: dn/dt = 0

∆n = G/(C_n * N_D)

45

## What is the minority carrier lifetime?

###
R = ∆n/tau, which means that

tau = 1/(C_n * N_D)

46

## What do we take into account when using the "detailed balance" principle?

### The thermal emission of captured carriers, that is the thermal excitation of electrons from the defect level to conduction band (and holes from defect level to valence band).

47

## What are typical recombination centers?

### Intrinsic defects (vacancies, dangling bonds) or impurities (Fe, Au etc.)

48

## What is bimolecular recombination and what is the recombination rate ?

###
When electrons in the conduction band recombine with mobile holes in the valance band (interaction with both electron and hole simultaneously). The recbomination rate is given as

R = C* ∆n ∆p, when ∆n and ∆p are much larger than intrinsic carrier densities.

49

## What is the minority carrier density for bimolecular recombination in steady state?

###
Given the recombination rate R = C* ∆n ∆p, and using that ∆n = ∆p, we get

dn/dt = G - R = G - C* ∆n^2 => ∆n = sqrt(G/C*)

50

## What is the minority carrier lifetime for bimolecular recombination?

### tau_min = 1/(C*∆p)

51

## What is Auger recombination? Show schematically.

###
Auger recombiation is when the energy and momentum of the recombining electron-hole pair are transferred to a third electron or hole, which returns to its ground state by thermalization.

See p. 13 of Chapter 8.

52

## How are the recombination rates for Auger recombination for heavily p-doped and n-doped materials respectively?

###
For heavily p-doped:

R = C_A * ∆n * p_0^2

For heavily n-doped:

R = C_A * ∆p * n_0^2

53

## What is the steady state density of minority carriers for Auger recombination?

###
dn/dt = G - R = G - c_a * ∆n * p_0^2

=> ∆n = G/(c_A*p_0^2)

54

## What is the minority carrier lifetime for Auger recombination?

### 1/(C_A*p_0^2)

55

## When are Auger recombination mainly relevant?

### In highly doped regions, for example around contacts.

56

## What is surface recombination?

### Recombination at intermediate level due to dangling bonds at the surface (or contact with other materials).

57

## How is the surface defects' influence on minority carrier lifetime described?

###
By the surface recombination velocity.

S = R_surface / ∆n_bulk

R_surface = recombination rate at the surface.

∆n_bulk = optically excited density of minority carriers in bulk.

58

## What are typical values for the surface recombination velocity?

### 1 cm/s for passivated Si, up to 10^6.

59

## Where do high values of surface recombination velocity occur?

### In unpassivated areas, such at contacts (interfaces between semiconductors and metals). So-called metal-induced gap states (MIGS).

60

## How does one avoid metal-induces gap states (MIGS)?

### By using a back surface field.

61

## What are typical values of exciton binding energies?

### 10 meV.

62

## How do we obtain an efficient and complete separation of charge?

### With an internal electric field generated by a p/n-, p/i/n- or Schottky diode.

63

## What is the requirement for a solar cell to operated as a diffusion cell?

### That the minority carrier diffusion length, L_n, is much larger than the absorption length (1/alpha).

64

## What is meant with a diffusion cell?

### That the carrier transport mainly happens with no electric field in most of the cell.

65

## What is meant with a drift cell?

### That there is an electric field to drive the carrier transport.

66

## Sketch the band diagram of a crystalline Si pn-diffusion cell.

### See notes.

67

## How are the band structures in real space calculated for solar cells?

### Via the Poisson equations.

68

## How is the built-in voltage defined in a pn-diffusion cell?

### eU_bi = Eg - Ea - Ed, where Ea is the height from the valence band up to the back contact and Ed is the height from the front contact up to the CB.

69

## What is the purpose of a back surface field?

### It surpresses minority carrier diffusion into the back contact.

70

## What is the requirement for good efficiency of a pn-diffusion cell?

### That the mean free path of the carriers are much larger than the spatial extention of the absorber.

71

## Sketch the band diagram of a thin film drift cell.

### See notes.

72

## What is an advantage of drift cells over diffusion cells?

### Since the electric field in the absorber is non-zero, we can get good efficiencies also when the absorber length is larger than the mean free path.

73

## How is the mobility of holes in a-Si? What are the practical consequences of this?

### The mobility of holes is much lower than for electrons. This means the illumination must come through the p-contact in order to minimize the distance of generated holes from p-contact.

74

## Sketch the band-diagram of a MIS-Schottky cell.

### See notes.

75

## Sketch the setup of a dye-sensitized solar cell. What is another name for this kind of cell?

###
See notes for diagram.

Other name: Grötzel cell.

76

## Sketch the setup of an organic bulk heterojunction cell.

### See notes.

77

## When talking about semiconductors, what is the ideality factor?

### The ideality factor says something aobut the recombination in the space-charge region. For n = 1 we have no recombination, for n = 2 we have complete recombination.

78

## What is the current of a solar cell in the dark?

###
j = js (exp{eU/nkT}-1)

with js being the saturation current density for U -> ∞, n being the ideality factor.

js = e((Dn/Ln)*n_0,p + (Dp/Lp)*p_0,n)

n_0,p = equilibrium electron concentration on p-side

p_0,n = equilibrium hole concentration on n-side

79

## What is the current of a solar cell under illumination?

###
j = js(exp{eu/nkT}-1) - jph,

where jph is the photon induced current given by

jph = e(∆n*µn + ∆p*µp)E

80

## Draw a typical I-V characteristic for a solar cell in the dark and under illumination assuming an ideal diode. Draw all important values.

### See notes.

81

## What is the open circuit voltage?

### The voltage when the solar cell is not connected to anything.

82

## What is the short circuit current?

### The current when the solar cell is short circuited, that is when we don't have an external load.

83

## What is the fill factor?

###
The fill factor is the ratio of the area of the square we get from the I-V curve at the maximum power point to the area of the square of the open circuit voltage and short circuit current.

FF = Um*jm/Uoc*jsc,

where Um and jm are the voltage and current at the maximum power point.

84

## What is the efficiency of a solar cell?

###
eta = Pout/Pin = Umjm/Pin = Uocjsc*FF / Pin.

Pin is the incoming light intensity in W/m^2.

85

## How can we model a real diode?

### As an ideal diode in parallel with a parallel (shunt) resistance, which comes from defects in the absorber and in series with a series resistance which is the combined resistances of contacts, leads, wires etc.

86

## What is the extra term that is added to the current of a real solar cell?

### U-RsI/Rp, where Rs is the series resistance, Rp is the shunt resistance and I is the current.

87

## How does the values of the shunt resistance and the series resistance in a solar cell affect the performance of the solar cell?

###
- Too large vales of Rs decrease Isc and FF.

- Too small values of Rp decrease Uoc and FF.

For real systems: Rs ≥ of a few ohm enough. Rp ≤ 100 ohm have a negative effect.

88

## How can we deduce the series and shunt resistance in a solar cell?

### From the I-V characteristics at U = 0 and U = Uoc.

89

## When we estimate the maximum efficiency, which assumptions do we make?

###
i) All photons wit energy > Egap are absorbed, all photons below are transmitted.

ii) No nonradiative recombinations (at defects or interfaces)

iii) No radiative recombinations, or a radiation balance between the sun and the solar cell (detailed balance)

iv) Ideal diode: Rp = ∞, Rs = 0, n = 1.

90

## What can be said of the maximum efficiency dependence on the band gap? What is the optimum band gap?

###
For Eg -> 0, the efficiency goes to 0 (because Uoc ∞ the efficiency goes to 0 (no absorption).

Optimum band gap is 1.1 eV (Si!!)

91

## Under illumination, what is the electric field from the open circuit voltage?

###
eUoc = E_F,n - E_F,p

where E_F,n and E_F,p are the quasi-Fermi levels of the electrons and holes respectively.

92

## What are the main approaches for an efficiency over 31%?

###
- Multijunction cells

- third generation solar cells

93

## What tactics can be used to increase the efficiency?

###
- Include optical down and up conversion.

- Tandem solar cells

- Hot carrier cells

94

## What is down-conversion in solar cells?

### We use a dye-layer in front of the solar cell, which converts photons with energy 2*Eg into two photons of ≈ Eg. Instead of loosing the extra energy to thermalization, we now have two photons that can excite an exciton.

95

## What are tandem cells?

### These are cells where we have two solar cells with different Egap which are optically and electrically in series. They can be split into three terminal devices and two terminal devices, where we need a tunnel junction in between the diodes.

96