Question

A p *n junction is one in which the doping concentration in the p region is much greater than that in the n region. In such a junction, the forward current is mostly due to hole injection across the junction. Show that I = Ip = For the specific case in which ND = 1016 /cm3, Dp = 10 cm2 / s, Lrp= 10 mu m, and A = 104 mu m2, find Is and the voltage V obtained when I = 0.5 mA. Assume operation at 300 K where ni = 1.5 Times 10 10/cm3.

Answer 1

The voltage is 0.742 V.

Explanation:

              $I = Aqn_{i} ^2[\frac{D_{p} }{L_{p} N_{d}} + \frac{D_{n} }{L_{n} N_{A} }](e^\frac{v}{v_{T} } - 1)$

        $I_{p} = Aq n_{i} ^2 \frac{D_{p} }{L_{p} N_{D} } (e^\frac{v}{v_{T} } -1)$ and  $I_{n} = Aqn_{i} ^2 \frac{D_{n} }{L_{n} N_{A} } (e^\frac{v}{v_{T} } - 1)$

For P+ - n junction, Na >>Nd, thus Ip>>In and I = $I_{p} = Aq n_{i} ^2 \frac{D_{p} }{L_{p} N_{D} } (e^\frac{v}{v_{T} } -1)$

For this case using,  $Is = Aqn_{i} ^2 \frac{D_{p} }{L_{p} N_{D} }$

                   = 10^4 $\times$ 10^-8 $\times$ 1.6 $\times$ 10^-19 $\times$ (1.5 $\times$ 10^10)^2 $\times$ (10 / 10$\times$10^-4$\times$10^17)

                   = 3.6 $\times$ 10^-16 A

               $I = Is(e^\frac{v}{v_{T} } -1)$

              ⇒ 1 $\times$ 10^-3 = 3.6 $\times$ 10^-16 $\times$ [e^(v/25.9$\times$10^-3) - 1] = 1 $\times$ 10^-3

                             V = 0.742 V