Consider a p-type continuously graded bar i.e. nonuniformly doped bar as shown in the Fig. 1.

Fig. 1 Continuously graded p-type bar |

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,

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,

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,

But drift current can not 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.

__1.1 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 = x

_{1}and x = x_{2}as shown in the Fig. 2.Fig. 2 Graph of Concentration against distance x |

Let concentration of holes at x = x

_{1}is p = p_{1}
and concentration of holes at x = x

_{2}is p = p_{2}
This is due to nonuniform doping and is indicated in the graph shown in the Fig. 2.

There exists a potential difference between x

_{1}and x_{2}which is responsible to circulate drift current equal and opposite to diffusion current.
Let potential at x

_{1}= V_{1}
And potential at x

_{2}= V_{2}
From the equation (3) which indicates the the net current through the bar is zero, we can write,

According to Einstein'S relation, D

According to Einstein'S relation, D

_{p}/μ_{p}is V_{T}.
Now E is the electric field intensity generated internally. From the definition of electric field intensity we can write,

E = -dV/dx

Hence equation (5) modifies to,

To get the voltage generated between x

where V

From the equation (7) we can write,

Multiplying equations (8a) and (9a) we get,

Hence equation (5) modifies to,

To get the voltage generated between x

_{1}and x_{2}, integrate the equation (6),where V

_{21}is the potential difference between the two concentrations p_{1 }and p_{2}.**Note**: The potential difference depends on the concentrations and not on the distance between x_{1}and x_{2}.From the equation (7) we can write,

Similarly for n-type semiconductor bar which is nonuniformly doped, if
and are the concentrations of electrons at two different points, we can
write,

**Note**: There is change of sign of exponential term in equation (9a) and (9b) as compared to exponential term in equations (8a) and (8b). This is because diffusion current density in n-type bar is q dn/dx which is positive.

The product of the concentrations of electrons and holes is always constant. This proves the law of mass action.

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