Question

A semiconductor, in thermal equilibrium, has a hole concentration of p0 = 2x1016 cm-3. The minority carrier life time = 3x10-7 s. (Assume, ni = 1010 cm-3)

(a) Determine the thermal equilibrium recombination rate of electrons.
(b) Determine the recombination rate of electrons if an excess electron concentration of În = 1013 cm-3 exists.
(c) Calculate the change in the recombination rate when excess electron concentration exists (compared to thermal equilibrium).

Answer 1

`1.67* 10^(9)\ \text{cm}^(-3)\text{s}^(-1)`

`3.33* 10^(19)\ \text{cm}^(-3)\text{s}^(-1)`

`3.33* 10^(19)\ \text{cm}^(-3)\text{s}^(-1)`

Step-by-step explanation:

`p_0` = Hole concentration =
`2* 10^(16)\ \text{cm}^(-3)`

`n_i` = Intrinsic concentration =
`10^(10)\ \text{cm}^(-3)`

`\tau_(n0)` = Minority carrier life time =
`3* 10^(-6)\ \text{s}`

`\delta n` = Excess concentration of electrons =
`10^(13)\ \text{cm}^(-3)`

Majority carrier electron concentration is given by

`n_0=(n_i^2)/(p_0)\\\Rightarrow n_0=((10^(10))^2)/(2* 10^(16))\\\Rightarrow n_0=5000\ \text{cm}^(-3)`

Recombination rate is given by

`R_(n0)=(n_0)/(\tau_(n0))\\\Rightarrow R_(n0)=(5000)/(3* 10^(-6))\\\Rightarrow R_(n0)=1.67* 10^(9)\ \text{cm}^(-3)\text{s}^(-1)`

The recombination rate is
`1.67* 10^(9)\ \text{cm}^(-3)\text{s}^(-1)`

Recombination rate is given by

`R_n=(\delta_n)/(\tau_(n0))\\\Rightarrow R_n=(10^(13))/(3* 10^(-7))\\\Rightarrow R_n=3.33* 10^(19)\ \text{cm}^(-3)\text{s}^(-1)`

The recombination rate is
`3.33* 10^(19)\ \text{cm}^(-3)\text{s}^(-1)`

Change in the recombination rate is

`\Delta R_n=3.33* 10^(19)-1.67* 10^(9)\\\Rightarrow \Delta R_n=3.33* 10^(19)\ \text{cm}^(-3)\text{s}^(-1)`

The change in the recombination rate is
`3.33* 10^(19)\ \text{cm}^(-3)\text{s}^(-1)`