A slab of N-type Silicon (N_D = 10^15 cm^-3) is in equilibrium and kept at room

Question

A slab of N-type Silicon (N_D = 10^15 cm^-3) is in equilibrium and kept at room temperature under dark. What are the equilibrium carrier concentrations? If light falls on the semiconductor will it be still at equilibrium? Can it be at steady state? Find the steady-state carrier densities in the semiconductor under light and the net recombination rate. Assume uniform generation (e-h pairs) rate of 10^20 cm^3/s throughout the semiconductor volume and intrinsic carrier density in Si at room temperature is 10^10cm^-3; lifetime for electron and holes are tau_n = 100 ns; tau_p = 1 ns.

Explanation / Answer

Given , semiconductor type N-type, ND = 1015 cm-3

We know that for silicon intrinsic carrier concentration ni = 1.5 x 1010 cm-3

(a) Since material is n-type , electron will be majority carrier (no) ,

   At equilibrium , for majority carreier , majority carrier concentration will be equal to sum of intrinsic carrier concentration + dopoing concentration.

   so here no = ni + ND = 1.5 x 1010 cm-3 +   1015 cm-3 == 1 x 1015 cm-3

Minority carreir concentration (po) at equilibrium can be found by Mass Action law

no x p0 = ni2

So p0 = ni2 / no  = ( 1.5 x 1010 cm-3 )2 / 1 x 1015 cm-3 = 2.25 x 105 cm-3

So equilibrium carrier concentration no= 1 x 1015 cm-3 , po = 2.25 x 105 cm-3

(b) When light falls on semiconductor and if the energy of incident photon (light ) is greater than bandgap of n-typre silicon , then electron-hole pair will be generated,.

So total number of carriers will be increased due to light. But mojority carrier (here no) is high as compare to minority carrier (here po). So majority carrier will be relative unchanged but minority carrier (po) will keep increasing.

If keep shining light , then a time will come when minority carrier starts approaching intrinsic carrier concentration and semiconductor will no more n-type but become intrinsic semiconductor. So equilibrium of semiconductor from present state ( whare its n-type) will change to a new equilibrium state ( intrinsic semiconductor)

(c) Given that , e-h pair generation rate = 1020 cm-3/s = P (let say it)

( note in question its by mistake - sign is missing in power of cm)

. Tn = 100 ns , Tp = 1ns

So change in electron concentration (change in n ) = P* Tn = 1020 cm-3/s * 100 ns = 1013 cm-3

change in hole concentration (change in p ) = P* Tp = 1020 cm-3/s * 1 ns = 1011 cm-3

So total value of electron concentration = ( n at dark ) + (change in n due to light )

= 1015 cm-3   +   1013 cm-3   = = 1015 cm-3

So total value of hole concentration = ( p at dark ) + (change in p due to light )

= 2.25 x 105 cm-3   +   1011 cm-3   = =   1011 cm-3

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