# Thread: Derivation of relationship between carrier density and current

1. ## 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!

2. 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).

3. 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?

4. If we have function $
f(x)
$

then its differential is

$
\displaystyle
df=f'(x)dx= \frac{df}{dx}dx
$

5. Thanks - you've been a big help