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  1. My question is during equilibrium is achieved under no biasing, why it is that in fermi level diagram representation fermi levels are raised in p type nd lowered in n type. Is it not possible we increase the fermi level of n type than p type.

  2. If we finally look at the picture, it still confuses me and other readers too that fermi level then shown seems like fermi level of p type (Evp) and fermi level of n type (Ecn) are still closer as they were individually were.. Some text says to see the scaling of energy levels of p and n level. But i am not satisfied.

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You can find the position of Fermi level in two cases: $$n_0=n_i\exp\left[\frac{E_F-E_{Fi}}{k_BT}\right]$$ $$p_0=n_i\exp\left[-\frac{E_F-E_{Fi}}{k_BT}\right]$$ Rearranging the two relations, We get:

$$\rightarrow E_F-E_{Fi}=k_BT\ln\left(\frac{n_0}{n_i}\right)$$ $$\rightarrow E_{Fi}-E_{F}=k_BT\ln\left(\frac{p_0}{n_i}\right)$$

For an $n$-type semiconductor, $n_0>n_i$ and $E_F>E_{Fi}$. The Fermi level for an $n$-type semiconductor is above $E_{Fi}$. For a $p-type$ semiconductor $p_0>n_i$ and from the second equation we see that $E_{Fi}>E_F$. The Fermi level for a $p$-type semiconductor is below $E_{Fi}$.


You can find the position for the fermi level explicitly thus can see whether it's how close it's to conduction or valence band.

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  • $\begingroup$ My question is basically related to energy band diagram perspective...if you can answer that will be great. I am just aksing for the visual representation the doubt i am getting $\endgroup$ Jun 20 '21 at 5:30

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