Potential Variation in a Continuously Graded Semiconductor

Potential Variation in a Continuously Graded Semiconductor

• Consider a p‒type continuously graded bar i.e. nonuniformly doped bar as shown in Fig. 1.18.1.

• No external voltage is applied to the bar. The bar is open circuited. As no external voltage is applied and it is open circuited, net current through the bar is zero.

• But due to the nonuniform doping there exists a diffusion current as holes move from high concentration to low concentration area. Hence there exists a diffusion current density of,

Jp = ‒q Dp dp/dx

         ………. (1.18.1)

• But as bar is open circuited, net current through it is zero.

• This means there exists one more internal current which is equal to diffusion current but in opposite direction to it. This is a drift current flowing in the bar in opposite direction to that of diffusion current.

• The current density of this current is,

 J_p = p qμ_pE           ……..(1.18.2)

• But drift current cannot exist without a potential difference and applied voltage to the bar is zero. So externally E is zero.

• This indicates that the E required for the circulation of drift current gets generated internally.

• This indicates that nonuniform doping of bar results in the induced voltage.

• And as net current through the bar is zero we can write for open circuited bar,

   ‒ q D_p dp/dx + p q μ_p E = 0

          ………….(1.18.3)

1. Expression for the Potential Difference

• To derive the expression for the potential difference between any two points of a nonuniformly doped bar, consider the two points at a distance of x = x₁ and x = x₂ as shown in Fig. 1.18.2.

• Let concentration of holes at x = x₁ is p = P₁ and concentration of holes at x = x₂ is p = p₂.

• This is due to nonuniform doping and is indicated in the graph shown in Fig. 1.18.2.

• There exists a potential difference between x₁ and x₂ which is responsible to circulate drift current equal and opposite to diffusion current.

Let

 potential at x₁ = v₁ and potential at x₂ = v₂

From the equation (1.18.3) which indicates that net current through the bar is zero, we can write,

 p q μ_p E = q D_p dp/dx

pE = D_p/μ_p . dp/dx

         …………. (1.18.4)

According to Einstein's relation, D_p/μ_p is V_T.

PE = V_T. dp/dx

i.e. E = V_T/P . dp/dx

         …………. (1.18.5)

• Now E is the electric field intensity generated internally. From the definition of electric field intensity we can write,

 E = ‒ dV/dx

hence equation (1.18.5) modifies to,

‒ dv/dx = V_T/P . dp/dx

i.e. dV = ‒ V_T/P dp

             ………. (1.18.6)

• To get the voltage generated between x₁ and x₂, integrate the equation (1.18.6),

 _V1∫^V2 dV = ‒V_{T  P1}∫^P2 1/P dp

i.e. V₂ ‒ V₁ = − V_T [In p]_P2^P2

V₂₁ = V_T In P₁/P₂

         ……..(1.18.7)

where V₂₁ is the potential difference between the two concentrations P₁ and P₂.

Key Point: The potential difference depends on the concentrations and not on the distance between x₁ and x₂.

2. Law of Junction

• From the equation (1.18.7) we can write,

In P₁/P₂ = V₂₁/V_T

i.e.  P₁/P₂ = e^V21/VT

P₁ = P₂ e^V21/VT

and

P₂ = P₁ e^‒V21/VT

           ……….(1.18.8)

• Similarly for n‒type semiconductor bar which is nonuniformly doped, if n₁ and n₂ concentrations of electrons at two different points, we can write,

n₁ = n₂ e^‒V21/VT

and

n₂ = n₁ e^V21/VT

           ……….(1.18.9)

Key Point: There is change of sign of exponential term in equation (1.18.9) as compared to exponential term in equations (1.18.8). This is because diffusion current density in n‒type bar is q D_n dn/dx which is positive.

• The equations (1.18.8) and (1.18.9) represent law of junction.

• Multiplying equations (1.18.8) and (1.18.9) we get,

i.e.

n₁ P₁ = P₂ e^V21/VT n₂ e^‒V21/VT

n₁ P₁ = n₂ P₂

• The product of the concentrations of electrons and holes is always constant.

This proves the law of mass action.

Review Questions

1. Derive the expression for the potential difference existing in a continuously graded semiconductor.

2. Explain the law of junction. Hence prove the law of mass action from it.