it's ok
Yes I'm sure it's correct,
it's coming from the expression of a Schottky diode (current density vs Voltage)
so it's :
J = a(Exp(-bV) -1)
because you want 0 current at bias of 0 Volts...
I already try many way to solve this equations, til now I'm fully stuck
Even numerically I...
ok but my problem is in the followings...
Thank you for your very quick answer and the welcome.
Yes I agree til there it's easy but it's the following that gives me troubles.
so
starting from :
y'' = a Exp(-b y) -a
y ' y'' = y ' a Exp(-b y) -a y '
after the integration on y on both...
Hi everybody,
How do I solve this differential equation ??:
y'' = a(Exp(-b*y)-1) ;
where a, b are constants
with the boundaries conditions :
y'(x=0)=-K1
y'(x=L)=0
without the constant term I can do
y''*y' = y' a Exp(-b y)
then integrate it
\ {1/2} (y')^2= {a/b}...