Discussion
A semiconductor is irradiated with light such that carriers are uniformly generated throughout its volume. The semiconductor is N-type with \(N_{D} = 10^{19}/\text{cm}^{3}\). If the excess electron concentration in the steady state is \(\Delta_{n} = 10^{15}/\text{cm}^{2}\) and if \(\tau_{p}\) = 10 μs C minority carrier life time, then what is the generation rate due to irradiation ?
Answer: A
The generation rate is \(10^{20}\) e-h pairs/\(\text{cm}^{3}\)/s.
In the steady state, carriers are generated as fast as they recombine. With a uniform generation rate G and minority carrier lifetime \(\tau_{p}\), the steady excess concentration is \(\Delta n = \Delta p = G\tau_{p}\), since light creates electrons and holes in pairs.
\[ \begin{aligned} G &= \frac{\Delta n}{\tau_{p}} = \frac{10^{15}}{10 \times 10^{-6}} \\ &= \frac{10^{15}}{10^{-5}} = 10^{20} \end{aligned} \]
The unit is pairs per cm3 per second (the stem's 'cm2' for a volume concentration is a misprint for cm3). The donor concentration \(N_{D}\) only ensures that holes are the minority carriers; it does not enter the calculation. \(10^{10}\) comes from multiplying by the lifetime instead of dividing, and \(10^{14}\) and \(10^{8}\) from slips in the powers of ten.
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