Question

The one-dimensional Poisson equation for the surface space charge region at the drain is
1. \(\frac{{{\partial ^2}{\rm{\Psi }}}}{{\partial \;{x^2}}} = - \frac{q}{{{\varepsilon _s}}}\left( {N_D^ + + N_A^ - } \right)\)
2. \(\frac{{{\partial ^2}{\rm{\Psi }}}}{{\partial {x^2}}} = \frac{{ - q}}{{{\epsilon_S}}}\left( {N_D^ + + N_A^ - + p + n} \right)\)
3. \(\frac{{{\partial ^2}{\rm{\Psi }}}}{{\partial {x^2}}} = \frac{{ - q}}{{{\epsilon_S}}}\left( {N_D^ + - N_A^ - + p - n} \right)$\)
4. \(\frac{{{\partial ^2}{\rm{\Psi }}}}{{\partial {x^2}}} = \frac{{ - q}}{{{\epsilon_S}}}\left( {p - n} \right)\)

Answer 1

Correct Answer - Option 3 : \(\frac{{{\partial ^2}{\rm{\Psi }}}}{{\partial {x^2}}} = \frac{{ - q}}{{{\epsilon_S}}}\left( {N_D^ + - N_A^ - + p - n} \right)$\)

Gauss Law for Electrostatics states that:

\(\nabla .E = \frac{{{\rho _v}}}{\epsilon}\) 

ρ_v = Volumetric charge density

ϵ = Permittivity of the material

The fact that the curl of E is zero (the electrostatic electric field intensity E is a conservative field) allowed us to define the electric field in terms of the electric scalar potential as:

\(E = - \nabla V\) 

Substituting this in the above equation, we get:

\(\nabla .E = - \nabla .\left( {\nabla V} \right) = \frac{{{\rho _v}}}{\epsilon}\) 

Replacing ∇.(∇ V) = ∇²V, the above Equation becomes,

\({\nabla ^2}V = - \frac{{{\rho _v}}}{\epsilon}\) 

In Cartesian coordinates, this can be written as:

\(\frac{{{\partial ^2}V}}{{\partial {x^2}}} + \frac{{{\partial ^2}V}}{{\partial {y^2}}} + \frac{{{\partial ^2}V}}{{\partial {z^2}}} = - \frac{{{\rho _v}}}{\epsilon}\) 

This is called the Poisson’s Equation in three dimensions.

Application:

When a voltage is applied across the source-drain contacts, the MOS structure is in a nonequilibrium condition forming a depletion region (or space charge region).

For doped semiconductor junctions with N_a acceptors and N_d donors, space charge region is created by the diffusion of electrons from n-side to p-side and diffusion of holes from p-side to n-side leaving behind uncompensated charge carriers N_d⁺ and N_a^–  respectively.

The net positive charge = q (N_d⁺ + p)

Net negative charge = – q (N_a^– + n)

The net charge becomes the sum of the positive and the negative charges, i.e.

ρ_v = q (N_d⁺ + p – (N_a^– + n))

ρ_v = q (N_d⁺ + p – N_a^– – n)      ---(1)

The potential distribution ψ (x) in the semiconductor is now determined from a solution of the one-dimensional Poisson’s Equation as:

\({\nabla ^2}\psi \left( x \right) = - \frac{{{\rho _v}\left( x \right)}}{\epsilon}\) 

Using Equation (1), we get:

\({\nabla ^2}\psi \left( x \right) = - \frac{q}{\epsilon}\left( {N_d^ + + p - N_a^ - - n} \right)\)