# Derivation of relationship between carrier density and current

• Mar 15th 2011, 08:09 AM
jonm84
Derivation of relationship between carrier density and current
Hi there,

I'm new here so I'm not sure if this is in the best section, but here goes...

I'm trying to establish how a formula (relating the carrier lifetime in a semiconductor active region with the injected current) from a paper I have read has been derived. The formula is:

$n=\frac{1}{qV} \int_0^I{\tau} dI$ (1)

where n is the carrier density, q is the elementary charge, V is the volume of the semiconductor active area.

From another source (textbook) I have:

$\frac{1}{\tau} = \frac{\partial R}{\partial n}$ (2)

where

$R(n) = An + Bn^2 + Cn^3$ (3)

and also the injected current I is related to n as follows:

$I = qVR(n)$ (4)

I have a complete mental block on how whether I can derive the first equation from the following three - any help would be appreciated!
-----

Also, in the textbook, it uses equations 2,3 and 4 above to define the relationship between $\tau$ and I as being:

$\frac {1}{\tau^2} = A^2 + \frac{4B}{qV}I}$ (5)

I just keep going round in circles when I try to derive this from equations 2,3 and 4 :(
-----

Any help in how to derive (1) or (5) would be appreciated! I realise that this is a 'physicsy" type question being asked on a math forum, but I hope someone can give me some pointers. Cheers!
• Mar 15th 2011, 09:03 AM
zzzoak
From (4):

$
\displaylist
dI=qV \ \frac{dR}{dn}dn=qV \ \frac{1}{\tau} \ dn
$

$
\displaylist
dn=\frac{1}{qV} \ \tau dI
$

and integrating dn from 0 to n
dI from 0 to I
gives (1).
• Mar 15th 2011, 09:20 AM
jonm84
Quote:

Originally Posted by zzzoak
From (4):

$
\displaylist
dI=qV \ \frac{dR}{dn}dn=qV \ \frac{1}{\tau} \ dn
$

$
\displaylist
dn=\frac{1}{qV} \ \tau dI
$

and integrating dn from 0 to n
dI from 0 to I
gives (1).

Hey, thanks for your reply. I can sort of follow what you've done, in that I'm not sure how you made the first step, but I can follow it on after that...

Can you explain how you went from (4) to $dI=qV \ \frac{dR}{dn}dn$

Did you differentiate (4) w.r.t. dn?
• Mar 15th 2011, 12:09 PM
zzzoak
If we have function $
f(x)
$

then its differential is

$
\displaystyle
df=f'(x)dx= \frac{df}{dx}dx
$
• Mar 16th 2011, 02:03 AM
jonm84
Thanks - you've been a big help