Problem — Current from an illuminated semiconductor

Problem

A semiconductor of area $S = 2\;\text{m}^2$ is struck by green light of intensity $|\vv I| = 1{,}2\;\text{kW/m}^2$. The energy gap is $E_\text{gap} = 1{,}2$ eV and each green photon has energy $E_f \approx 2{,}5$ eV. Assuming each photon frees one electron, what is the theoretical current produced? Compare with the real efficiency ($15$–$25\%$) of a photovoltaic cell.

Solution

Step 1 — Incident light power. Intensity times area gives the collected power:

$P = |\vv I|\cdot S = 1{,}2\cdot 10^{3}\cdot 2 = 2{,}4\cdot 10^{3}\;\text{W}$

Step 2 — Photon energy in joules.

$E_f = 2{,}5\;\text{eV} = 2{,}5\cdot 1{,}6\cdot 10^{-19} = 4{,}0\cdot 10^{-19}\;\text{J}$

Step 3 — Photons per second. Each photon carries $E_f$, so the number per second is

$\frac{N}{t} = \frac{P}{E_f} = \frac{2{,}4\cdot 10^{3}}{4{,}0\cdot 10^{-19}} = 6{,}0\cdot 10^{21}\;\text{photons/s}$

Step 4 — Ideal current. If every photon frees one electron of charge $e$:

$i = e\cdot \frac{N}{t} = 1{,}6\cdot 10^{-19}\cdot 6{,}0\cdot 10^{21} = 9{,}6\cdot 10^{2}\;\text{A}$

$\ev{i \approx 9{,}6\cdot 10^{2}\;\text{A (ideal)}}$

Comparison. In reality only $15$–$25\%$ of photons contribute (reflection, recombination, sub-gap photons or photons with excess energy lost as heat). The real current is therefore of order $i_\text{real} \approx 0{,}15\text{–}0{,}25\cdot 960 \approx 1{,}4\cdot 10^{2}\text{–}2{,}4\cdot 10^{2}$ A.

Collegamenti

Argomenti: Quantum physics Concetti: Photon · Photoelectric effect Competenze: Unit conversion · Symbolic setup