This blog is for the purpose of my class ELE431. This blog is used for discussion forum on any topics in Solid State Devices. Students may post any topics that is related to the subject.
there are 2 boundary in base that is;1-0+eg:nB(0+)=nB0(e^(qVBE/kT-1))2-xB-eg:nB(xB-)=nB0(e^(qVBC/kT-1))~~2007126643~~eeb4f
The boundries in base;In base,there are two boudary.1) 0+;ΔnB(0+)=nBo(eqVBE/kT - 1)2) Xb-;Δnb(Xb-)=nBo(eqBC/kT - 1)2007121889muizz
the 2 boundary in base that are1. 0+example:nB(0+)=nB0(exp^(qVbe/kT-1))2. xB-example:nB(xB-)=nB0(exp^(qVbc/kT-1))
the 2 boundaries in base that are1st . 0+:nB(0+)=nB0(exp^(qVbe/kT-1))2nd. xB-:nB(xB-)=nB0(exp^(qVbc/kT-1))~2007290826~~EEB4E~
Boundaries in base that is;For 0+,nB(0+)=nB0(e^(qVBE/kT-1))and xB-,nB(xB-)=nB0(e^(qVBC/kT-1))2007270442EEB4E
The two boudary in base.1) for 0+;ΔnB(0+)=nBo(eqVBE/kT - 1)2) for Xb-;Δnb(Xb-)=nBo(eqBC/kT - 1)mohd ikhwan2007290828eeb4e
The two boudary in base are:1. for 0+;ΔnB(0+)=nBo(eqVBE/kT - 1)2. for Xb-;Δnb(Xb-)=nBo(eqBC/kT - 1)2007121763eeb4f
the 2 boundaries in base that are1st . 0+:nB(0+)=nB0(exp^(qVbe/kT-1))2nd. xB-:nB(xB-)=nB0(exp^(qVbc/kT-1))*2007292004**EEB4E*
boundries in base...1) for 0+; ∆nB(0+)= nBo(exp VBE/kT-1)2) for Xb; ∆nB(Xb)= nBo(exp VBC/kT-12007126683 eeb4e
for base,there are 2 boundary conditions:1) ∆nB (0+)=nB0(eqVBE/kT-1)2) ∆nB (xB-)=nB0(eqVBC/kT-1)shahizawanis bt shamshuri2006687187EEB4E
for base,there are 2 boundary conditions:at O+:1) ∆nB (0+)=nB0(eqVBE/kT-1)at xB-2) ∆nB (xB-)=nB0(eqVBC/kT-1)shahizawanis bt shamshuri2006687187EEB4E
The boundries in base are;a) 0+;ΔnB(0+)=nBo(eqVBE/kT - 1)b) Xb-;Δnb(Xb-)=nBo(eqBC/kT - 1)>MUHAMAD IZUAN BIN IBRAHIM<>2007126589<>EEB4E<
Two boundaries in base that are:-at X= 0+nB(0+)=nB0(exp^(qVbe/kT-1))at X=B-nB(xB-)=nB0(exp^(qVbc/kT-1))/*MOHD UBAIDILLAH BIN SHAMSUDIN*//*2007126591*//*EEB4E*/
The boundary in base a0re has 2 0boundary:1) 0+eg:nB(02+)=nB0(e^(qVBE/kT-1))2) xB-eg:nB(xB-)=nB0(e^(qVBC/kT-1)) mohd khairul anuar bin awang2007121771eeb4e
At base, there are 2 boundries1. at 0+ nB(0+)=nB0(e^(qVBE/kT-1))2. at xB- nB(xB-)=nB0(e^(qVBC/kT-1))NUR ATHIRA BINTI MOHAMAD2006687745EEB4E
Post a Comment
19 comments:
there are 2 boundary in base that is;
1-0+
eg:nB(0+)=nB0(e^(qVBE/kT-1))
2-xB-
eg:nB(xB-)=nB0(e^(qVBC/kT-1))
~~2007126643~~
eeb4f
The boundries in base;
In base,there are two boudary.
1) 0+;
ΔnB(0+)=nBo(eqVBE/kT - 1)
2) Xb-;
Δnb(Xb-)=nBo(eqBC/kT - 1)
2007121889
muizz
the 2 boundary in base that are
1. 0+
example:nB(0+)=nB0(exp^(qVbe/kT-1))
2. xB-
example:nB(xB-)=nB0(exp^(qVbc/kT-1))
the 2 boundaries in base that are
1st . 0+
:nB(0+)=nB0(exp^(qVbe/kT-1))
2nd. xB-
:nB(xB-)=nB0(exp^(qVbc/kT-1))
~2007290826~
~EEB4E~
the 2 boundaries in base that are
1st . 0+
:nB(0+)=nB0(exp^(qVbe/kT-1))
2nd. xB-
:nB(xB-)=nB0(exp^(qVbc/kT-1))
~2007290826~
~EEB4E~
the 2 boundaries in base that are
1st . 0+
:nB(0+)=nB0(exp^(qVbe/kT-1))
2nd. xB-
:nB(xB-)=nB0(exp^(qVbc/kT-1))
~2007290826~
~EEB4E~
Boundaries in base that is;
For 0+,nB(0+)=nB0(e^(qVBE/kT-1))
and xB-,nB(xB-)=nB0(e^(qVBC/kT-1))
2007270442
EEB4E
The two boudary in base.
1) for 0+;
ΔnB(0+)=nBo(eqVBE/kT - 1)
2) for Xb-;
Δnb(Xb-)=nBo(eqBC/kT - 1)
mohd ikhwan
2007290828
eeb4e
The two boudary in base are:
1. for 0+;
ΔnB(0+)=nBo(eqVBE/kT - 1)
2. for Xb-;
Δnb(Xb-)=nBo(eqBC/kT - 1)
2007121763
eeb4f
the 2 boundaries in base that are
1st . 0+
:nB(0+)=nB0(exp^(qVbe/kT-1))
2nd. xB-
:nB(xB-)=nB0(exp^(qVbc/kT-1))
*2007292004*
*EEB4E*
boundries in base...
1) for 0+;
∆nB(0+)= nBo(exp VBE/kT-1)
2) for Xb;
∆nB(Xb)= nBo(exp VBC/kT-1
2007126683
eeb4e
for base,there are 2 boundary conditions:
1) ∆nB (0+)=nB0(eqVBE/kT-1)
2) ∆nB (xB-)=nB0(eqVBC/kT-1)
shahizawanis bt shamshuri
2006687187
EEB4E
for base,there are 2 boundary conditions:
at O+:
1) ∆nB (0+)=nB0(eqVBE/kT-1)
at xB-
2) ∆nB (xB-)=nB0(eqVBC/kT-1)
shahizawanis bt shamshuri
2006687187
EEB4E
The boundries in base are;
a) 0+;
ΔnB(0+)=nBo(eqVBE/kT - 1)
b) Xb-;
Δnb(Xb-)=nBo(eqBC/kT - 1)
>MUHAMAD IZUAN BIN IBRAHIM<
>2007126589<
>EEB4E<
Two boundaries in base that are:-
at X= 0+
nB(0+)=nB0(exp^(qVbe/kT-1))
at X=B-
nB(xB-)=nB0(exp^(qVbc/kT-1))
/*MOHD UBAIDILLAH BIN SHAMSUDIN*/
/*2007126591*/
/*EEB4E*/
The boundary in base a0re has 2 0boundary:
1) 0+
eg:nB(02+)=nB0(e^(qVBE/kT-1))
2) xB-
eg:nB(xB-)=nB0(e^(qVBC/kT-1))
mohd khairul anuar bin awang
2007121771
eeb4e
At base, there are 2 boundries
1. at 0+
nB(0+)=nB0(e^(qVBE/kT-1))
2. at xB-
nB(xB-)=nB0(e^(qVBC/kT-1))
NUR ATHIRA BINTI MOHAMAD
2006687745
EEB4E
Post a Comment