Doping measurements through DL capacitance
This is a nice problem from semiconductor devices theory.
The depletion layer (DL) capacitance of a one-sided p-n diode is often used to determine the doping profile of the lightly doped side. Consider an [tex]$n^+ - p$[/tex] diode, with a nonuniform p-side doping concentration [tex]$N_a(x)$[/tex]. The capacitance per unit area at bias voltage [tex]$V$[/tex] is by definition [tex]$C=\frac{\text{d}Q_d}{\text{d}V}$[/tex] ([tex]$Q_d$[/tex] being the DL charge per unit area) and, in terms of the DL width [tex]$X_d$[/tex], one can easily show that [tex]$C=\frac{\epsilon}{X_d}$[/tex] for any doping profile ([tex]$\epsilon$[/tex] being the dielectric constant of the semiconductor). Show that the doping concentration at the DL edge is given by
[tex]$ \displaystyle N_a(X_d)=\frac{2}{q \epsilon \frac{ \displaystyle \text{d}\left(\frac{1}{C^2}\right)}{\displaystyle \text{d}V}}[/tex].
The depletion layer (DL) capacitance of a one-sided p-n diode is often used to determine the doping profile of the lightly doped side. Consider an [tex]$n^+ - p$[/tex] diode, with a nonuniform p-side doping concentration [tex]$N_a(x)$[/tex]. The capacitance per unit area at bias voltage [tex]$V$[/tex] is by definition [tex]$C=\frac{\text{d}Q_d}{\text{d}V}$[/tex] ([tex]$Q_d$[/tex] being the DL charge per unit area) and, in terms of the DL width [tex]$X_d$[/tex], one can easily show that [tex]$C=\frac{\epsilon}{X_d}$[/tex] for any doping profile ([tex]$\epsilon$[/tex] being the dielectric constant of the semiconductor). Show that the doping concentration at the DL edge is given by
[tex]$ \displaystyle N_a(X_d)=\frac{2}{q \epsilon \frac{ \displaystyle \text{d}\left(\frac{1}{C^2}\right)}{\displaystyle \text{d}V}}[/tex].
Risposte
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BONUS QUESTION:
Generalize to the case of a p-n junction (i.e. find an equation for both [tex]$N_d(X_n)$[/tex] and [tex]$N_a(-X_p)$[/tex], where [tex]$X_n$[/tex] and [tex]$-X_p$[/tex] are the borders of the DL in the n-side and the p-side, resp.).
BONUS QUESTION:
Generalize to the case of a p-n junction (i.e. find an equation for both [tex]$N_d(X_n)$[/tex] and [tex]$N_a(-X_p)$[/tex], where [tex]$X_n$[/tex] and [tex]$-X_p$[/tex] are the borders of the DL in the n-side and the p-side, resp.).
By definition, the capacitance per unit area at bias voltage $V$ is
$C(V)=\frac{"d"Q_d(V)}{"d"V}$. (1)
It can be shown that, for any doping profile, one has
$C=\frac{\epsilon}{X_d}$. (2)
Moreover, in a $n^+ -p$ diode the DL charge per unit area can be approximated to
$Q_d\approx q N_a X_d$,
that leads to
$"d"Q_d=q N_a\cdot "d"X_d$. (3)
The first step is to divide both sides of (3) by the capacitance
$\frac{"d"Q_d}{C}=\frac{q N_a\cdot "d"X_d}{C}$. (4)
Using (1) and (2) into (4) we get
$"d"V=\frac{q N_a X_d\cdot "d"X_d}{\epsilon}$ (5)
and, since $2X_d\cdot "d"X_d="d"X_d^2$, we can rewrite (5) as follows
$"d"V=\frac{q\epsilon N_a\cdot "d"X_d^2}{2\epsilon^2}$. (6)
From (2) it follows that
$"d"C^{-2}=\frac{"d"X_d^2}{\epsilon^2}$,
hence (6) reads now as
$"d"V=\frac{q\epsilon N_a\cdot "d"C^{-2}}{2}$,
or, by expliciting the p-side doping concentration, as
$N_a=\frac{ 2 }{q\epsilon\frac{"d"C^{-2}}{dV}}$.
$C(V)=\frac{"d"Q_d(V)}{"d"V}$. (1)
It can be shown that, for any doping profile, one has
$C=\frac{\epsilon}{X_d}$. (2)
Moreover, in a $n^+ -p$ diode the DL charge per unit area can be approximated to
$Q_d\approx q N_a X_d$,
that leads to
$"d"Q_d=q N_a\cdot "d"X_d$. (3)
The first step is to divide both sides of (3) by the capacitance
$\frac{"d"Q_d}{C}=\frac{q N_a\cdot "d"X_d}{C}$. (4)
Using (1) and (2) into (4) we get
$"d"V=\frac{q N_a X_d\cdot "d"X_d}{\epsilon}$ (5)
and, since $2X_d\cdot "d"X_d="d"X_d^2$, we can rewrite (5) as follows
$"d"V=\frac{q\epsilon N_a\cdot "d"X_d^2}{2\epsilon^2}$. (6)
From (2) it follows that
$"d"C^{-2}=\frac{"d"X_d^2}{\epsilon^2}$,
hence (6) reads now as
$"d"V=\frac{q\epsilon N_a\cdot "d"C^{-2}}{2}$,
or, by expliciting the p-side doping concentration, as
$N_a=\frac{ 2 }{q\epsilon\frac{"d"C^{-2}}{dV}}$.
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