my question:

During forward biased, is it true that the minority concentration (e.g. hole in n-type) could be higher than the majority concentration of the same type (e.g. hole in p-type, which is the original doping NA)?


$p_n$ = minority carrier, holes concentration in n-type region.

$p_p$ = majority carrier, holes concentration in p-type region.

with no biased

$$ \frac{p_n}{p_p} = e^{-\frac{qV_bi}{kT}} $$

I can say, $p_n<p_p$.

when it is in forward biased

$$ \frac{p_n}{p_p} = e^{-\frac{qV_bi}{kT}}e^{\frac{qV_F}{kT}} $$ $p_n>p_p$ mathematically, even with small $V_F$, $p_n$ can easily larger than $p_p$

$p_p = N_A$.

$N_A$ is the original doping if full ionization is assumed.

I can understand that $p_n$ with forward biased > $p_n$ without biased, but could it be larger than $N_A$?

is it a correct understanding?

or I am missing something here?


1 Answer 1


The formula $$ \frac{p_n}{p_p} = e^{-\frac{qV_bi}{kT}}e^{\frac{qV_F}{kT}} $$ hold approximately only for relatively low forward biases (weak injection). It does definitely not hold anymore when $V_F\ge V_{bi}$ i.e., when the forward bias overcomes the built-in voltage of the junction. Thus an injected minority concentration $p_n>p_p$ is not possible.


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