Question on equilibrium and reverse bias at pn junction
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작성자 20180275 박종은(Pa… 댓글 2건 조회 1,366회 작성일 20-04-12 00:40본문
in different concentration?
2. At reverse bias at chapter 2, why is I(diff,p)~=0 and I(diff,n)~=0?
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양희원님의 댓글
양희원 작성일
I'm also curious about #1 issue that you uploaded.
For #2, since the external voltage has strengthened the field, the barrier rises even higher than that in equilibrium, thus prohibiting the flow of current. Therefore, the junction carries a negligible current under reverse bias, which means that |I(diff,p)| and |I(diff,n)| goes to zero.
Gyujun Jeong님의 댓글
Gyujun Jeong 작성일
For #1, Yes. The doping concentrations are different for n and p region for that figure.
If you derive the depletion region for the n and p type, it could be expressed like this:
x_n=sqrt(2 * (epsilon_s/q) * (Na/Nd) * (1/(Na+Nd)) * (delta V) )
x_p=sqrt(2 * (epsilon_s/q) * (Nd/Na) * (1/(Na+Nd)) * (delta V) )
Since the doping concentration is different for n-type and p-type, the Space Charge Region length should be different.
It can be simply understood by charge neutrality: Na*xn=Nd*xp.