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In p-n junction diode , the current I , can be expressed $I= I_0 exp \bigg[ \large\frac{eV}{K_{B}T}-1 \bigg]$ Where $I_0$ is called the satiation current , V is the voltage across the dicode and in positive for forward bias and negative for reverse bias and I is the current through the dicode , $K_B$ is the BoHZmann constant $(8.6 \times 10^{-5}e^v/k)$ and T is the absolute temperature . If for a given diode $I_0 = 5 \times 10^{-12}A $ and $T=300\;K$ , then what will be the increase in the current if the voltage across the diode is increased to $0.7\;A$?what will be the Dynamic resistance

$\begin{array}{1 1} 0.0336 \Omega \\ 0.336 \Omega \\ 3.36 \Omega \\ 0.00336 \Omega \end{array} $

1 Answer

Solution :
$V= 0.7\;V$
$\therefore \large\frac{eV}{K_BT}=\large\frac{1.6 \times 10^{-19} \times 0.6}{8.6 \times 10^{-5} \times 1.6 \times 10^{-19} \times 300}$
$\qquad= 27.14 $
$I=5 \times 10^{-12}\bigg[e^{27.14}-1\bigg]$
$\quad= 5 \times 10^{-12} [6.07 \times 10^{11}-1]$
$\qquad= 3.305\;A$
$\Delta I= 3.035 -0.063$
$\qquad= 2.972\;A$
$\Delta V= 0.7 -0.6 =0.1\;V$
Dynamic resistance $= \large\frac{\Delta V}{\Delta I}= \frac{0.1\;V}{2.972\;A}$
$\qquad= 0.336 \Omega$
answered Dec 31, 2016 by meena.p
 

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