Answer to Question #109529 in Electrical Engineering for Bharti

Question

Given a forward biased si diode with I = 1mA. If the diffusion capacitance is Cd =1microferad , what is the diffusion Length Lp? Assume that the doping of the P-side is much greater than that of the n-side . Use Dp = 13metersquare second.

Answer 1

In this problem, we will need the following:

$L_p$ - diffusion length for holes, m;

$D_p$ - diffusion constant for holes, $\text{m}^2/\text{s}$;

$C_D$ - diffusion capacitance, μF;

$\tau_p$ - meal life time for holes, s;

$V_T$ - volt equivalent of temperature $T$ (26 mV for 300 K);

$\eta=1$ for Ge and $\eta=2$ for SI - a constant, $\eta=I_0(e^{V/\eta V_T}-1).$

The diffusion length is

$$L_p=(D_p\tau_p)^{1/2}.\\ \space\\ C_D=\frac{\tau_p I}{\eta V_T},\\ \space\\ V_T=\frac{T}{11600.}$$

Therefore, combining all this, we get

$$L_P=\sqrt{\frac{D_pC_D\eta V_T}{I}}=\\ \space\\ =\sqrt{\frac{13\cdot(1\cdot10^{-6})\cdot2\cdot(300/11600)}{(1\cdot10^{-3})}}=0.0259\text{ m}.$$