densité de porteurs en excès dans le volume de la diode

L'équation différentielle à résoudre est : d 2 p ^ dx 2 p ^ L p = 0 {d^2 {hat p}} over {dx^2} - hat p over { L_p }=0 ,

La solution est de la forme : p ^ ( x ) = A exp ( x L p ) + B exp ( x L p ) hat p( x )= A exp left( - {x over L_p} right) + B exp left({x over L_p} right) .

en x = 0 : p ^ = p ^ 0 en x = W N : p ^ ( W N ) = 0 } 0 = A exp ( W N L p ) + B exp ( W N L p ) p ^ 0 = A + B . matrix{ left none matrix{ "en"` x=0 `: ` hat p = {hat p}_0 ## "en" `x=W_N `: ` {hat p}{(W_N)} = 0 } right rbrace # `drarrow` 0= A exp left( - {W_N over L_p} right) + B exp left({W_N over L_p} right) `drarrow` {hat p}_0 = A + B . }
0 = p ^ 0 exp ( W N L p ) + B [ exp ( W N L p ) exp ( W N L p ) ] B = p ^ 0 exp ( W N L p ) exp ( W N L p ) exp ( W N L p ) . `drarrow` 0 = {hat p}_0 exp left( -{W_N over L_p} right) + B left[ exp left({W_N over L_p} right) - exp left(-{W_N over L_p} right) right] `drarrow` B = - {hat p}_0 {exp left(-{W_N over L_p} right) } over { exp left({W_N over L_p} right) -exp left(-{W_N over L_p} right) } .

et

A = p ^ 0 [ 1 exp ( W N L p ) exp ( W N L p ) exp ( W N L p ) ] = p ^ 0 exp ( W N L p ) exp ( W N L p ) exp ( W N L p ) . A = {hat p}_0 left [ 1 -{ {exp left(-{W_N over L_p} right) } over { exp left({W_N over L_p} right) -exp left(-{W_N over L_p} right) } } right] = {hat p}_0 {exp left({W_N over L_p} right) } over { exp left({W_N over L_p} right) -exp left(-{W_N over L_p} right) } .
p ^ ( x ) = p ^ 0 exp ( W N x L p ) exp ( W N x L p ) exp ( W N L p ) exp ( W N L p ) hat p(x)= {hat p}_0 { exp left({{W_N -x} over L_p} right) - exp left( -{{W_N -x} over L_p} right) } over { exp left({W_N over L_p} right) -exp left(-{W_N over L_p} right) }

Comme exp ( a ) exp ( a ) = 2 sh ( a ) exp ( a ) - exp( -a ) = 2 "sh" ( a ) , on peut écrire  : p ^ ( x ) = p 0 ^ sh ( W N x L p ) sh ( W N L p ) hat p ( x ) = hat p_0 { { "sh" left( { {W_N -x} over L_p }right)}} over {"sh" left( W_N over L_p right)} .