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final state we left off on saturday

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David Lenfesty 2021-02-07 11:19:58 -07:00
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@ -370,53 +370,59 @@ ylabel('Current [A]');
\begin{subfigure}{0.33\textwidth} \begin{subfigure}{0.33\textwidth}
\centering \centering
\includegraphics[width=\textwidth]{q2_0V.png} \includegraphics[width=\textwidth]{q2_0V.png}
\caption{Fermi Functions at $0V$.} \caption{Fermi functions at $0V$.}
\label{fig:q2_0v} \label{fig:q2_0v}
\end{subfigure}% \end{subfigure}%
\begin{subfigure}{0.33\textwidth} \begin{subfigure}{0.33\textwidth}
\centering \centering
\includegraphics[width=\textwidth]{q2_0V05.png} \includegraphics[width=\textwidth]{q2_0V05.png}
\caption{Fermi Functions at $0.05V$.} \caption{Fermi functions at $0.05V$.}
\label{fig:q2_0v} \label{fig:q2_0v}
\end{subfigure}% \end{subfigure}%
\begin{subfigure}{0.33\textwidth} \begin{subfigure}{0.33\textwidth}
\centering \centering
\includegraphics[width=\textwidth]{q2_0V1.png} \includegraphics[width=\textwidth]{q2_0V1.png}
\caption{Fermi Functions at $0.1V$.} \caption{Fermi functions at $0.1V$.}
\label{fig:q2_0v} \label{fig:q2_0v}
\end{subfigure} \end{subfigure}
\begin{subfigure}{0.33\textwidth} \begin{subfigure}{0.33\textwidth}
\centering \centering
\includegraphics[width=\textwidth]{q2_0V2.png} \includegraphics[width=\textwidth]{q2_0V2.png}
\caption{Fermi Functions at $0.2V$.} \caption{Fermi functions at $0.2V$.}
\label{fig:q2_0v} \label{fig:q2_0v}
\end{subfigure}% \end{subfigure}%
\begin{subfigure}{0.33\textwidth} \begin{subfigure}{0.33\textwidth}
\centering \centering
\includegraphics[width=\textwidth]{q2_0V3.png} \includegraphics[width=\textwidth]{q2_0V3.png}
\caption{Fermi Functions at $0.3V$.} \caption{Fermi functions at $0.3V$.}
\label{fig:q2_0v} \label{fig:q2_0v}
\end{subfigure} \end{subfigure}
\caption{stuff and things} \caption{(a) depicts I-V curves at $V_G = 0.5$ V (higher curve) and $V_G = 0.25$ V.
(b) through (f) show Fermi functions and $D(E)$ at various drain voltages.}
\end{figure} \end{figure}
\subsection*{(b)} \subsection*{(b)}
We can see from these figures that the area under the $f_1(E + U) - f_2(E + U)$ curve As the drain voltage increases, we see that the overlapping area
gradually moves down in energy, this means that a smaller and smaller portion of it is under the "curve" of $f_1(E + U) - f_2(E+U)$ and $D(E)$ increases, up
above $E = 0eV$, which means there are fewer total energy levels through which current until the point where the drain only has electrons below the energy
can flow. From this we can see that with a high enough $V_D$ we will hit saturation, and levels available in the channel. The difference function also reaches its maximum height and it widens downward rather than upward.
be unable to pass more current. At this point, the overlapping area no longer
increases, and therefore the current does not increase either,
hitting {\em saturation current}.
% TODO re-word this %% WRONG EXPLANATION - kept for posterity :)
The self-consistent potential will decrease, which is what causes this shift in levels. % We can see from these figures that the area under the $f_1(E + U) - f_2(E + U)$ curve
% gradually moves down in energy. This means that a smaller and smaller portion of it is
% above $E = 0eV$, which means there are fewer total energy levels through which current
% can flow. From this we can see that with a high enough $V_D$ we will hit saturation, and
% be unable to pass more current. \\
\subsection*{(c)} \subsection*{(c)}
At $V_D = 0.3V$, the energy levels from approximately $0$ to $0.2$ $eV$ are being At $V_D = 0.3V$, the energy levels from approximately $0$ to $0.2$ $eV$ are being
used for electron transport. This is the range where the difference in the contact Fermi used for electron transport. This is the range where the difference in the contact Fermi
functions is more than 0 and the energy is more than 0. functions is more than 0 and the energy is more than 0. \\
The difference between the two contacts is the greatest at $0eV$, because this is the The difference between the two contacts is the greatest at $0eV$, because this is the
point at which their Fermi functions have the greatest difference, and thus will be making point at which their Fermi functions have the greatest difference, and thus will be making
the most "effort" to equalize the channel potential. the most "effort" to equalize the channel potential.
@ -427,8 +433,7 @@ ylabel('Current [A]');
drain current. The energy levels vanishing at $-0.4eV$ would mean that there is a drain current. The energy levels vanishing at $-0.4eV$ would mean that there is a
greater number of energy levels that would be able to be used for conduction. There greater number of energy levels that would be able to be used for conduction. There
would be a larger {\em area} where there is a non-zero difference in the two Fermi would be a larger {\em area} where there is a non-zero difference in the two Fermi
functions at the contacts and there are energy levels available in the channel. functions at the contacts and there are energy levels available in the channel. \\
This can be contrasted with material B, which would have {\em no} energy levels in This can be contrasted with material B, which would have {\em no} energy levels in
this "conduction" zone and thus no current would be able to flow at all. this "conduction" zone and thus no current would be able to flow at all.
@ -494,7 +499,7 @@ ylabel('Current [A]');
\begin{equation*} \begin{equation*}
I = 0, N = 0 I = 0, N = 0
\end{equation*} \end{equation*}
%
This is because the only allowed energy level in the channel is at a higher energy This is because the only allowed energy level in the channel is at a higher energy
level than exists in either of the contacts, thus there are no electrons that would level than exists in either of the contacts, thus there are no electrons that would
flow into the channel from either contact, thus no current {\em and} no electrons in the channel. flow into the channel from either contact, thus no current {\em and} no electrons in the channel.
@ -502,15 +507,15 @@ ylabel('Current [A]');
\subsubsection*{(ii)} \subsubsection*{(ii)}
\begin{equation*} \begin{equation*}
I = 608nA, N = 0.5 I = 608\:nA, N = 0.5
\end{equation*} \end{equation*}
%
Given that we are operating with a single energy level in the channel, we can use Given that we are operating with a single energy level in the channel, we can use
the equations 9 and 10 (provided in the assignment) directly. the equations 9 and 10 (provided in the assignment) directly.
\begin{equation*} \begin{equation*}
I = \frac{q}{\hbar} \cdot \frac{0.005eV \cdot 0.005eV}{0.005eV + 0.005eV} I = \frac{q}{\hbar} \cdot \frac{0.005eV \cdot 0.005eV}{0.005eV + 0.005eV}
\cdot [1 - 0] = 608nA \cdot [1 - 0] = 608\:nA
\end{equation*} \end{equation*}
\begin{equation*} \begin{equation*}
@ -520,15 +525,15 @@ ylabel('Current [A]');
\subsubsection*{(iii)} \subsubsection*{(iii)}
\begin{equation*} \begin{equation*}
I = 0A, N = 1 I = 0\:A, N = 1
\end{equation*} \end{equation*}
%
Given that we are operating with a single energy level in the channel, we can use Given that we are operating with a single energy level in the channel, we can use
the equations 9 and 10 (provided in the assignment) directly. the equations 9 and 10 (provided in the assignment) directly.
\begin{equation*} \begin{equation*}
I = \frac{q}{\hbar} \cdot \frac{0.005eV \cdot 0.005eV}{0.005eV + 0.005eV} I = \frac{q}{\hbar} \cdot \frac{0.005eV \cdot 0.005eV}{0.005eV + 0.005eV}
\cdot [1 - 1] = 0A \cdot [1 - 1] = 0\:A
\end{equation*} \end{equation*}
\begin{equation*} \begin{equation*}
@ -548,7 +553,7 @@ ylabel('Current [A]');
\subsubsection*{(ii)} \subsubsection*{(ii)}
For $f_1(E)$ the step point occurs at $E = \mu_1 = 0.25$ eV. Since $\mu_1 = \mu - qV_S$ and $\mu = 0$ eV, $V_S = -0.25$ V. For $f_1(E)$ the step point occurs at $E = \mu_1 = 0.25$ eV. Since $\mu_1 = \mu - qV_S$ and $\mu = 0$ eV, $V_S = -0.25$ V.
For $f_2(E + U)$ the step point occurs at $E = \mu_2 = -0.25$ eV. Since $\mu_1 = \mu - qV_D$ and $\mu = 0$ eV, $V_D = 0.25$ V. For $f_2(E)$ the step point occurs at $E = \mu_2 = -0.25$ eV. Since $\mu_1 = \mu - qV_D$ and $\mu = 0$ eV, $V_D = 0.25$ V.
This assumes $U = 0$ eV. This assumes $U = 0$ eV.
\subsection*{(c)} \subsection*{(c)}
@ -572,7 +577,7 @@ ylabel('Current [A]');
I = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot [10^4 \cdot 0.2 - 10^4 \cdot -0.1] I = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot [10^4 \cdot 0.2 - 10^4 \cdot -0.1]
\end{equation*} \end{equation*}
\begin{equation*} \begin{equation*}
I = 1.8mA I = 1.8\:mA
\end{equation*} \end{equation*}
\subsubsection*{(iii)} \subsubsection*{(iii)}
@ -596,59 +601,54 @@ ylabel('Current [A]');
Using equation (5) in the assignment and plugging in the given values we obtain: Using equation (5) in the assignment and plugging in the given values we obtain:
\begin{equation*}
\begin{aligned}
U = -q[\frac{C_SV_S+C_GV_G+C_DV_D}{C_E}] + \frac{q^2(N-N_0)}{C_E} \\
U = -\alpha_SV_S - \alpha_GV_G - \alpha_DV_D + U_0(N - N_0)\:\:[eV] \\
U = -0*0 - 0.5*0\:eV - 0.5*0.6\:eV + 0.25\:eV*(0.325 - 0)\:\:[eV] \\
U = -0.21875\:eV
\end{aligned}
\end{equation*}
\begin{align*}
U &= -q[\frac{C_SV_S+C_GV_G+C_DV_D}{C_E}] + \frac{q^2(N-N_0)}{C_E} \\
U &= -\alpha_SV_S - \alpha_GV_G - \alpha_DV_D + U_0(N - N_0)\:\:eV \\
U &= -0 \cdot 0 - 0.5 \cdot 0\:eV - 0.5 \cdot 0.6\:eV + 0.25\:eV\cdot(0.325 - 0)\:\:eV \\
U &= -0.21875\:eV
\end{align*}
%
Then, starting with Equation \ref{eq:N_new} and plugging in $D(E-U) = \delta(E - \varepsilon)$, we obtain Then, starting with Equation \ref{eq:N_new} and plugging in $D(E-U) = \delta(E - \varepsilon)$, we obtain
\begin{equation*} \begin{equation*}
N = \frac{\gamma_1f_1(\varepsilon + U) + \gamma_2f_2(\varepsilon + U)}{\gamma_1 + \gamma_2} N = \frac{\gamma_1f_1(\varepsilon + U) + \gamma_2f_2(\varepsilon + U)}{\gamma_1 + \gamma_2}
\end{equation*} \end{equation*}
Recalling that the expressions for the contact Fermi functions are (with $\mu = 0$ eV): Recalling that the expressions for the contact Fermi functions are (with $\mu = 0$ eV):
\begin{equation*}
\begin{aligned}
f_1(\varepsilon + U) = \frac{1}{1 + e^{frac{\varepsilon + U + qV_S}{k_BT}}} = \frac{1}{1 + e^{frac{0.2 - 0.21875 + 0}{0.025}}} = 0.679 \\
f_2(\varepsilon + U) = \frac{1}{1 + e^{frac{\varepsilon + U + qV_D}{k_BT}}} = \frac{1}{1 + e^{frac{0.2 - 0.21875 + 0.6}{0.025}}} \simeq 0
\end{aligned}
\end{equation*}
Then, solving for $\gamma_2$:
\begin{equation*}
\begin{aligned}
\gamma_2 = \gamma_2\frac{N - f_1(\varepsilon + U)}{f_2(\varepsilon + U) - N} = 5.45 \times 10^-3 \: eV
\end{aligned}
\end{equation*}
\begin{align*}
f_1(\varepsilon + U) = \frac{1}{1 + e^{(\varepsilon + U + qV_S)/k_BT}} = \frac{1}{1 + e^{(0.2 - 0.21875 + 0)/0.025}} = 0.679 \\
f_2(\varepsilon + U) = \frac{1}{1 + e^{()\varepsilon + U + qV_D)/k_BT}} = \frac{1}{1 + e^{(0.2 - 0.21875 + 0.6)/0.025}} \simeq 0
\end{align*}
Then, solving for $\gamma_2$:
\begin{align*}
\gamma_2 = \gamma_2\frac{N - f_1(\varepsilon + U)}{f_2(\varepsilon + U) - N} = 5.45 \times 10^{-3} \: eV
\end{align*}
\subsection*{(e)} \subsection*{(e)}
Assuming the density of states for each molecule can be modeled by $D(E) = \delta(E-\varepsilon)$, Assuming the density of states for each molecule can be modeled by $D(E) = \delta(E-\varepsilon)$,
Equation (12) in the assignment is valid here. Thus the current will be maximized when $[f_1(E) - f_2(E)]|_{E=\varepsilon}$ is maximized. Equation (12) in the assignment is valid here. Thus the current will be maximized when $[f_1(E) - f_2(E)]|_{E=\varepsilon}$ is maximized.
Solving $[f_1(E) - f_2(E)]|_{E=\varepsilon}$ for each molecule, we obtain: Solving $[f_1(E) - f_2(E)]|_{E=\varepsilon}$ for each molecule, we obtain:\\
Molecule A: Molecule A:
\begin{equation*} \begin{equation*}
[f_{1}(E) - f_{2}(E)]|_{E=\varepsilon_A} = [f_{1}(E) - f_{2}(E)]|_{E=\varepsilon_A} =
[\frac{1}{1 + e^{frac{\varepsilon_A}{k_BT_1}}} - [\frac{1}{1 + e^{frac{\varepsilon_A}{k_BT_2}}}] \frac{1}{1 + e^{\frac{\varepsilon_A}{k_BT_1}}} - \frac{1}{1 + e^{\frac{\varepsilon_A}{k_BT_2}}}
= [\frac{1}{1 + e^{frac{0}{0.024}}} - [\frac{1}{1 + e^{frac{0}{0.027}}}] = 0 = \frac{1}{1 + e^{\frac{0}{0.024}}} - \frac{1}{1 + e^{\frac{0}{0.027}}} = 0
\end{equation*} \end{equation*}
Molecule B: Molecule B:
\begin{equation*} \begin{equation*}
[f_{1}(E) - f_{2}(E)]|_{E=\varepsilon_B} = [f_{1}(E) - f_{2}(E)]|_{E=\varepsilon_B} =
[\frac{1}{1 + e^{frac{\varepsilon_B}{k_BT_1}}} - [\frac{1}{1 + e^{frac{\varepsilon_B}{k_BT_2}}}] \frac{1}{1 + e^{\frac{\varepsilon_B}{k_BT_1}}} - \frac{1}{1 + e^{\frac{\varepsilon_B}{k_BT_2}}}
= [\frac{1}{1 + e^{frac{-0.05}{0.024}}} - [\frac{1}{1 + e^{frac{-0.05}{0.027}}}] = 0.02493 = \frac{1}{1 + e^{\frac{-0.05}{0.024}}} - \frac{1}{1 + e^{\frac{-0.05}{0.027}}} = 0.02493
\end{equation*} \end{equation*}
Molecule C: Molecule C:
\begin{equation*} \begin{equation*}
[f_{1}(E) - f_{2}(E)]|_{E=\varepsilon_C} = [f_{1}(E) - f_{2}(E)]|_{E=\varepsilon_C} =
[\frac{1}{1 + e^{frac{\varepsilon_C}{k_BT_1}}} - [\frac{1}{1 + e^{frac{\varepsilon_C}{k_BT_2}}}] \frac{1}{1 + e^{\frac{\varepsilon_C}{k_BT_1}}} - \frac{1}{1 + e^{\frac{\varepsilon_C}{k_BT_2}}}
= [\frac{1}{1 + e^{frac{-0.1}{0.024}}} - [\frac{1}{1 + e^{frac{-0.1}{0.027}}}] = 0.00877 = \frac{1}{1 + e^{\frac{-0.1}{0.024}}} - \frac{1}{1 + e^{\frac{-0.1}{0.027}}} = 0.00877
\end{equation*} \end{equation*}
Molecule B should be chosen. Molecule B should be chosen.
@ -657,16 +657,16 @@ ylabel('Current [A]');
Referring again to equation 2 in the assignment, we know that $D(E)$ is valid for all $E$, however Referring again to equation 2 in the assignment, we know that $D(E)$ is valid for all $E$, however
since $\gamma_1$ is dependant on the energy level, we can use it to reduce our limits. For material since $\gamma_1$ is dependant on the energy level, we can use it to reduce our limits. For material
A, this means we only care about $E$ above $0eV$, and for B, $E$ below $0eV$. A, this means we only care about $E$ above $0eV$, and for B, $E$ below $0eV$.
%
Since we know the voltages on the contacts, we can calculate the effective fermi level at each. Since we know the voltages on the contacts, we can calculate the effective fermi level at each.
\begin{equation*} \begin{equation*}
\mu_1 = \mu_0 - V_S = 0 - (-2V) = 2eV \mu_1 = \mu_0 - V_S = 0 - (-2V) = 2\:eV
\end{equation*} \end{equation*}
\begin{equation*} \begin{equation*}
\mu_2 = \mu_0 - V_D = 0 - 1V = -1eV \mu_2 = \mu_0 - V_D = 0 - 1V = -1\:eV
\end{equation*} \end{equation*}
%
With the fermi levels of each contact, we can adjust the limits of integration further for each With the fermi levels of each contact, we can adjust the limits of integration further for each
material. Material A can be evaluated on $[0eV, 2eV]$, and material B can be evaluated on $[-1eV, 0eV]$. material. Material A can be evaluated on $[0eV, 2eV]$, and material B can be evaluated on $[-1eV, 0eV]$.
@ -674,31 +674,31 @@ ylabel('Current [A]');
I_A = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot \int_{0}^{2} [1 - 0] \cdot 10^4 dE I_A = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot \int_{0}^{2} [1 - 0] \cdot 10^4 dE
\end{equation*} \end{equation*}
\begin{equation*} \begin{equation*}
I_A = 12.2mA I_A = 12.2\:mA
\end{equation*} \end{equation*}
\begin{equation*} \begin{equation*}
I_B = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot \int_{0}^{2} [1 - 0] \cdot 10^4 dE I_B = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot \int_{0}^{2} [1 - 0] \cdot 10^4 dE
\end{equation*} \end{equation*}
\begin{equation*} \begin{equation*}
I_A = 6.1mA I_A = 6.1\:mA
\end{equation*} \end{equation*}
%
Material A should be chosen. Material A should be chosen.
\subsection*{(g)} \subsection*{(g)}
Using Ohm's law we find the corresponding conductance: Using Ohm's law we find the corresponding conductance:
\begin{equation*} \begin{equation*}
\sigma{max} = \frac{I_{max}}{V} = \frac{500\:nA}{4.3\:mV} = 116\:\mu S. \sigma_{max} = \frac{I_{max}}{V} = \frac{500\:nA}{4.3\:mV} = 116\:\mu S.
\end{equation*} \end{equation*}
%
We expect this result to be an integer multiple of $G_0 = 38.76$ $\mu S$, the quantum of conductance. We find We expect this result to be an integer multiple of $G_0 = 38.76$ $\mu S$, the quantum of conductance. We find
\begin{equation*} \begin{equation*}
\frac{\sigma{max}}{G_0} = 3. \frac{\sigma_{max}}{G_0} = 3.
\end{equation*} \end{equation*}
%
We conclude there are 3 levels. We conclude there are 3 levels.