What is the maximum wavelength of the light that produces voltage in a CdS
photovoltaic device? Discuss this in terms of the spectrum of sunlight as presented in Chapter 8.

Question

Quizplus · Accepted Answer

The Answer is From equation (9.3)

E = h c / \lambda ext { or } \lambda = h c / E

Using Planck's constant as

h = 4.136 imes 10 ^ { - 15 } \mathrm { eV } \cdot \mathrm { s }

, the speed of light as

\mathrm { c } = 3 imes 10 ^ { 8 } \mathrm {~m} / \mathrm { s }

, and the energy gap of CdS from Table

9.1

as

E = 2.42 \mathrm { eV }

then

\lambda = \frac { \left( 4.13 imes 10 ^ { - 15 } \mathrm { eV } \cdot \mathrm { s } ight) \left( 3 imes 10 ^ { 8 } \mathrm {~m} / \mathrm { s } ight) } { ( 2.42 \mathrm { eV } ) } = 5.13 imes 10 ^ { - 7 } \mathrm {~m} = 512 \mathrm {~nm}

From Figure

8.1

we see that this is near the peak in the spectrum of sunlight at sea level and only the small area to the left of the peak will contribute. This is in contrast to the case for pure Si where wavelengths less than about

1100 \mathrm {~nm}

will contribute and this represents nearly all the area under the spectrum.

What Is the Maximum Wavelength of the Light That Produces