03myersd said:
Please
anyone? The report is due in at 3.30 today and I would really like to
have the analysis of the derivation included in it. Its not essential
but it would essentially round the whole thing off nicely.
Thanks
Here's the derivation; I hope it's not late
{L_p} = diffusion length of holes ({cm})
{D_p} = diffusion constant ({cm^2}/s)
{L_p} = average carrier life time (s)
Firstly, we find \Delta{p}
where \Delta{p} is the minority carrier concentration at the
edge of the depletion region (DP)
we know that the built-in voltage is given by
{V_{bi}} = \frac{kT}{q}\ln({\frac{{N_A}{N_D}}{n_{i^2}}})
Applying the law of mass action {n_{i^2}} =<br />
<br />
{n_{no}}\times{p_{no}}
we get
{V_{bi}} = \frac{kT}{q}\ln({\frac{{n_{no}}{p_{no}}}{n_{i^2}}})
Rearranging
p_{po} = p_{no}\exp(\frac{qV_{bi}}{kT}) => eqn.1
For non-equilibrium situation, i.e. when there's forward bias voltage
{V_f}
p_{p(0)} = p_{n(0)}\exp(\frac{q(V_{bi}-{V_f})}{kT}) => eqn.2
or
p_{po} = p_{no}\exp(\frac{qV}{kT})
where V = {V_{bi}}-V_f
assuming low injection level, i.e. {p_p}\approx{p_{po}}
eqn.1/eqn.2 by doing this we get \Delta{p}
therefore
\Delta{p} = p_{no}\exp(\frac{qV}{kT}-1)
Using the continuity equation, we get an expression for the current density
J_{p(x)} = q\frac{D_p}{L_p}\delta_{p(x)}
since \Delta{p} = \delta_{p(x=0)}
so \delta_{p(x=0)}= p_{no}\exp(\frac{qV}{kT}-1)
J_{p(x)} = q\frac{{D_p}{p_{no}}}{L_p}\exp(\frac{qV}{kT}-1)
Doing the same with electrons, we get a similar expression, hence
J_{total} = J_p + J_n = J_s\exp(\frac{qV}{kT}-1)
This is the standard way of expressing the diode equation. However, if we multiply the above expression by the cross-section area, we get the current I.
