Pho.rs: PN junction physics

A1 ^?? Substitute $j=0$ and $v=1/u$ into the equation $(1)$ and get the relationship between $\varphi'$, $u'$, and $u$.

Straight substitution of $p=n_i u$, $n = n_i v$, $\mu_p = \alpha \mu_n$ and $\mathcal{E}=-\varphi'$ gives us the equation
\[0 = \alpha \mu_n \left[-\varphi' e n_i u - k_\text{B} T n_i u' \right] + \mu_n \left[ - \varphi' e n_i v + k_\text{B} T n_i v' \right].\]With $vu=1$ and $v'u+u'v=0$ we have:
\[\frac{\varphi' e}{k_\text{B} T} \left[ \alpha u + \frac{1}{u} \right] + u' \left[ \alpha + \frac{1}{u^2} \right] = 0 \quad \Rightarrow \quad \frac{\varphi' e}{k_\text{B}T}u + u' = 0\]

Ответ: \[\frac{\varphi' e}{k_\text{B}T}u + u' = 0\]

A2 ^?? Integrate the equation obtained in A1 and find the potential difference $V_{bi}=\varphi(+\infty) - \varphi(-\infty)$ of the electric field at the pn junction. Express the answer in terms of $N_p$, $N_n$, $n_i$

Using $u'/u=(\ln u)'$ we obtain:
\[-\frac{\varphi' e}{k_\text{B}T} = \left( \ln u \right)' \quad \Rightarrow \quad \frac{V_{bi}e}{k_\text{B}T}=\ln \frac{u(-\infty)}{u(+\infty)}.\]We can find $u$ far away from the pn junction from the electroneutraility condition $(p-N_A)-(n-N_D) = 0$. In the p-silicon $N_p \gg n_i$ and
\[u-\frac{N_p}{n_i}-\frac{1}{u} =0 \quad \Rightarrow \quad u \simeq \frac{N_p}{n_i}\]In the n-silicon $N_n \gg n_i$ and
\[u+\frac{N_n}{n_i}-\frac{1}{u} =0 \quad \Rightarrow \quad u \simeq \frac{n_i}{N_n}.\]Finally
\[ V_{bi} \simeq \frac{k_\text{B} T}{e} \ln \frac{N_p N_n}{n_i^2}\]

Ответ: \[ V_{bi} \simeq \frac{k_\text{B} T}{e} \ln \frac{N_p N_n}{n_i^2}\]

A3 ^?? Write down Gauss' theorem and express $\varphi'$ by $u$ and $N_D-N_A$. Get the differential equation for $\Psi$ where $\Psi = \ln u$ as
\[L^2 \Psi'' = 2 \sinh \Psi - \frac{N_A-N_D}{n_i} \tag{2}\]

In the 1D case, the Guass theorem is $\mathcal{E}' = \rho/(\varepsilon \varepsilon_0)$. Substituting $\mathcal{E} = - \varphi'$ and $\rho = e \left(p - N_A - n + N_D \right)$ we have
\[-\varphi'' = \frac{en_i}{\varepsilon\varepsilon_0} \left( u - \frac{N_A}{n_i} - \frac{1}{u} + \frac{N_D}{n_i} \right).\]However $\varphi'=-\dfrac{k_\text{B}T}{e} (\ln u )' = - \frac{k_\text{B}T}{e} \Psi'$ then we have the following equation for the $\Psi$:
\[\frac{k_\text{B} T}{e}\Psi'' = \frac{e n_i}{\varepsilon\varepsilon_0} \left( e^{\Psi} - \frac{N_A - N_D}{n_i} - e^{-\Psi} \right) =\frac{en_i}{\varepsilon\varepsilon_0} \left( 2 \sinh \Psi - \frac{N_A - N_D}{n_i} \right)\]

Ответ: \[\frac{k_\text{B}T \varepsilon\varepsilon_0}{e^2 n_i} \Psi'' = 2 \sinh \Psi - \frac{N_A-N_D}{n_i}\]

A4 ^?? Find the analytic expression for $\delta \Psi$ of the previous equation in regions far from the pn junction.

Let $\Psi_0 = \operatorname{arcsinh} \frac{N_A-N_D}{2n_i}$, so
\[2 \sinh \left( \Psi_0 + \delta \Psi \right) \simeq \frac{N_A-N_D}{n_i}+2 \cosh \Psi_0 \cdot \delta \Psi \simeq \frac{N_A-N_D}{n_i} + \left|\frac{N_A-N_D}{n_i} \right| \delta \Psi.\]Also $\Psi'' = (\Psi_0 + \delta \Psi)'' = \delta \Psi''$, so
\[L^2 \delta \Psi'' = \left|\frac{N_A-N_D}{n_i} \right| \delta \Psi.\]From the pair of two exponential solutions, we have to choose the one that decays.

Ответ: For $x<0$
\[\Psi = \ln \frac{N_p}{n_i}-A_- \exp \left( \frac{x}{L}\sqrt{\frac{N_p}{n_i}} \right), \quad \Psi'=-\frac{A_-}{L}\sqrt{\frac{N_p}{n_i}} \exp \left( \frac{x}{L} \sqrt{\frac{N_p}{n_i}} \right)\]For $x>0$
\[\Psi = -\ln \frac{N_n}{n_i}+A_+ \exp \left( -\frac{x}{L}\sqrt{\frac{N_n}{n_i}} \right), \quad \Psi'=-\frac{A_+}{L}\sqrt{\frac{N_n}{n_i}} \exp \left( -\frac{x}{L} \sqrt{\frac{N_p}{n_i}} \right)\]

A5 ^?? Find the analytic expression for $\Psi$ for the domain, where $n_i|2 \sinh \Psi| \ll |N_A - N_D|$.

In this region we have
\[L^2\Psi'' = \frac{N_D-N_A}{n_i}.\]By integrating we obtain:
\[\Psi(x)' = \Psi'(0) + \int\limits_0^x \frac{N_D - N_A}{L^2 n_i} dx = \Psi'(0)+\frac{N_D -N_A}{L^2 n_i}x\]
\[\Psi(x) = \Psi(0) + x\Psi'(0) + \frac{N_D -N_A}{2L^2 n_i}x^2.\]

Ответ: For $x<0$:
\[\Psi(x) = \Psi(0) + x \Psi'(0) - \frac{N_p}{2L^2 n_i} x^2, \quad \Psi'(x) = \Psi'(0)-\frac{N_p}{L^2 n_i}x\]For $x>0$:
\[\Psi(x) = \Psi(0) + x \Psi'(0) + \frac{N_n}{2L^2 n_i} x^2, \quad \Psi'(x) = \Psi'(0)+\frac{N_n}{L^2 n_i}x.\]

A6 ^?? Use the program pn-junction.py to solve equation (2) numerically for $N_p=2 \cdot 10^5$, $N_n=10^5$. Collect some data and plot a graph $\Psi(0)/\ln(N_nN_p)$ vs. $\ln N_n/N_p$.

B1 ^?? Find the dependences of the electric field $\mathcal{E}(x)$ in the chosen model and draw its graph. What relation between $x_p$, $x_n$, $N_p$, and $N_n$ must be satisfied so that the field $\mathcal{E}$ goes to zero at the boundaries $-x_p$ and $x_n$ and to is continuous at the point $x=0$?

The electric field grows linearly from zero at the $-x_p$ to $-\mathcal{E}_\text{max}$ at the $x=0$. Then it decays linearly and extincts at the $x_n$. Let's write the Gauss theorem for the region $x \in [-x_p,0]$:
\[\mathcal{E}_\text{max}-0 = \frac{eN_p x_p}{\varepsilon_0}\]Similarly, for the range $x \in [0, x_n]$ we have:
\[\mathcal{E}_\text{max}-0 = \frac{eN_n x_n}{\varepsilon_0}\]Finally,
\[ N_p x_p = N_n x_n\]

Ответ: \[ N_A x_p = N_D x_n\]

B2 ^?? Express $x_p'$ through $x_p$, $V_{bi}$ and $U$. Substitute $dQ = -eSN_p dx_p'$, where $S$ is the area of the pn junction, into the expression for the capacitance $C$ and find the dependence of $C(U)$. Express the dependence $C(U)$ in terms of $S$, $N_{n,p}$.

The voltage $V_{bi}-U$ is the integral $\int (-\mathcal{E})\, dx$, so
\[V_{bi}-U = \frac{\mathcal{E}_\text{max} x_p}{2} +\frac{\mathcal{E}_\text{max} x_n}{2} \quad \Rightarrow \quad x_p^2 = \frac{\varepsilon_0(V_{bi}-U)}{e N_p \left[ 1+ \frac{N_p}{N_n} \right]} \]It means that
\[\frac{x_p'^2}{x_p^2} = \frac{V_{bi}-U}{V_{bi}} \quad \Rightarrow \quad d x_p' = -\frac{x_p dU}{\sqrt{V_{bi}(V_{bi} - U)}}.\]With the definition of the capacity $C$ we have
\[C = \frac{dQ}{dU} = eS \sqrt{
\frac{\varepsilon_0}{k_\text{B}T \ln \frac{N_nN_p}{n_i^2} -eU} \frac{N_p N_n}{N_p+N_n}
}\]

Ответ: \[C = eS\sqrt{
\frac{\varepsilon_0}{k_\text{B}T \ln \frac{N_nN_p}{n_i^2} -eU} \frac{N_p N_n}{N_p+N_n}
}\]

C1 ^?? Suppose that due to the presence of the voltage $U$, the concentrations of holes and electrons have changed with respect to the equilibrium $U=0$ as follows $u \to u+ \delta u$, $v \to v + \delta v$. Express the current density $j$ through $\delta u$, $\delta v$, their derivatives, and the field $\mathcal{E}$.

\[ j = \alpha \mu_n \left[\mathcal{E} e n_i \delta u - k_\text{B} T n_i \delta u' \right] + \mu_n \left[\mathcal{E} e n_i \delta v + k_\text{B} T n_i \delta v' \right] \]

C2 ^?? Write the stationarity equations for $\delta u$, $\delta v$ through their values and the values of their derivatives, considering that changes in these quantities over time can only be due to currents and recombination.

Current $J_p$ of holes:
\[J_p = \alpha \mu_n n_i \left[ \mathcal{E} (u+\delta u) - \frac{k_\text{B} T}{e} (u+\delta u)' \right] = \alpha \mu_n n_i \left[ \mathcal{E} \delta u - \frac{k_\text{B} T}{e} \delta u' \right]. \]So, the concentration $(u+ \delta u) n_i$ of holes at the given point doesn't change only if
\[ J_p'+R = 0.\]Similarly,
\[J_n = \mu_n n_i \left[ -\mathcal{E} \delta v - \frac{k_\text{B} T}{e} \delta v' \right], \quad J_n'+R=0.\]We have used that $J_p=J_n=0$ in the equilibrium and it's easu to check.

Ответ: \[\alpha \mu_n n_i \left[ \mathcal{E}' \delta u + \mathcal{E} \delta u' - \frac{k_\text{B}T}{e} \delta u'' \right] + \frac{\delta u v + \delta v u}{\tau_p (u+1) + \tau_n (v + 1)} n_i = 0\]\[\mu_n n_i \left[- \mathcal{E}' \delta v - \mathcal{E} \delta v' - \frac{k_\text{B}T}{e} \delta v'' \right] + \frac{\delta u v + \delta v u}{\tau_p (u+1) + \tau_n (v + 1)} n_i = 0\]

C3 ^?? Transform the stationarity conditions for points where $\mathcal{E}=0$ and find the form of the functions $\delta v$ and $\delta u$ in the range $x<-x_p$. Take into account $N_p, N_n \gg n_i$ and the smallness of $\delta u$, $\delta v$.

Since $\mathcal{E} = 0$ our stationary conditions are
\[
\begin{cases}
-\alpha \mu_n n_i \frac{k_\text{B} T}{e} \delta u'' + \frac{u \delta v + v \delta u}{\tau_p(u+1) + \tau_n(v+1)} n_i = 0\\
-\mu_n n_i \frac{k_\text{B} T}{e} \delta v'' + \frac{u \delta v + v \delta u}{\tau_p(u+1) + \tau_n(v+1)} n_i = 0
\end{cases} \quad \Rightarrow \quad \alpha \delta u '' = \delta v''\]

In the region $x<-x_p'$ we have that $v \ll 1 \ll u$, so
\[ - \mu_n \frac{k_\text{B} T}{e} \delta v'' + \frac{\delta v}{\tau_p} = 0.\]This equation leads to $\delta v$ in the form of $A e^{x/L_p}$, where $L_p=\sqrt{\mu_n \tau_pk_BT/e}$. Also, at the $x \to -\infty$ both $\delta u$ and $\delta v$ have to decay, so with $\alpha \delta u'' = \delta v''$ we have $\delta u = \delta v / \alpha +Bx + C$.

Ответ: \[ \delta v = A \exp \left[ x \sqrt{\frac{e}{\mu_n \tau_pk_BT}} \right], \quad \delta u = \frac{A}{\alpha} \exp \left[ x \sqrt{\frac{e}{\mu_n \tau_pk_BT}} \right] + Bx + C\]

C4 ^?? Assuming that the equation from A1 is satisfied in the region $x \in [-x_p', x_n']$, find the constants for $\delta u$ and $\delta v$ for the results of C3.

On the other side $x > x_n$ we have very similar situation:
\[
\begin{cases}
-\alpha \mu_n \frac{k_\text{B}T}{e} \delta u'' + \frac{\delta u}{\tau_n} = 0 \\
\alpha \delta u '' = \delta v''
\end{cases},
\]
so
\[\delta u = D \exp \left[ -x \sqrt{\frac{e}{\alpha \mu_n \tau_n k_\text{B} T}} \right], \quad \delta v = D \alpha \exp \left[ -x \sqrt{\frac{e}{\alpha \mu_n \tau_n k_\text{B} T}} \right] + Ex + F. \]Since $\delta u$ and $\delta v$ are small corrections they can influence only on the density of minor charge carriers. With the equation from A1
\[\frac{\Delta \varphi e}{k_\text{B}T} = -\Delta \ln u = \Delta \ln v\]we have
\[\frac{V_{bi}e}{k_\text{B}T}\ - \frac{Ue}{k_\text{B}T} = - \ln \frac{1+\frac{\alpha D N_n}{n_i} e^{-x_n/L_n}}{\frac{N_pN_n}{n_i^2}} = \ln \frac{\frac{N_pN_n}{n_i^2}}{1+\frac{AN_p}{n_i}e^{-x_p/L_p}}. \]We can easily find the current of minor charges, so in p-silicon:
\[J_n(x)=-\mu_n n_i \frac{k_\text{B}T}{e} \delta v' = -\mu_n n_i\frac{k_\text{B}T}{e} A L_p e^{x/L_p}.\]Since $J_p'=-R=J_n'$ and $J_p(-\infty)=ej$, we have
\[J_p(x) = ej + J_n(x) \quad \Rightarrow \quad ej =-\alpha \mu_n n_i \frac{k_\text{B} T}{e} B.\]The value for $E$ could be found similarly with $J_n(+\infty) = -ej$.

Ответ: \[Ae^{-x_p/L_p}= \frac{n_i}{N_p} \left(e^{Ue/k_\text{B}T}-1 \right) \quad D = e^{-x_n/L_n}\frac{n_i}{\alpha N_n} \left(e^{Ue/k_\text{B}T}-1 \right) \]\[B = -\frac{e^2 j}{\alpha \mu_n n_i k_\text{B} T}, \quad E = \frac{e^2j}{\mu_n n_i k_\text{B}T}\]

C5 ^?? Obtain the I-V characteristic of the pn junction in the form $j = f(U)$.

Since we use the approximation $uv=1$ in the region $[-x_p,x_n]$ we have the recombination rate $R$ is equal to zero their. It means that $J'_p=J'_n=0$, so $J_p(-x_p) = J_p(x_n)$ and $J_n(-x_p)=J_n(x_n)$ are the same equations:
\[
\begin{cases}
ej+J_n(-x_p)=J_p(x_n)\\
J_n(-x_p)=-ej + J_p(x_n)
\end{cases}
\]
Finally,
\[ej = J_p(x_n) - J_n(-x_n) = \alpha \mu_n n_i \frac{k_\text{B} T}{e}DL_n e^{x_n/L_n} + \mu_n n_i \frac{k_BT}{e} AL_p e^{x_p/L_p}\]

Ответ: \[j = \mu_n n_i^2 \frac{k_\text{B}T}{e^2} \left[ \frac{L_n}{ N_n} + \frac{L_p}{N_p} \right] \left(e^{Ue/k_\text{B}T}-1 \right) \]

PN junction physics

Решение