diff --git a/PS1/doc.tex b/PS1/doc.tex index 3a1a634..2f7e5e6 100644 --- a/PS1/doc.tex +++ b/PS1/doc.tex @@ -370,53 +370,59 @@ ylabel('Current [A]'); \begin{subfigure}{0.33\textwidth} \centering \includegraphics[width=\textwidth]{q2_0V.png} - \caption{Fermi Functions at $0V$.} + \caption{Fermi functions at $0V$.} \label{fig:q2_0v} \end{subfigure}% \begin{subfigure}{0.33\textwidth} \centering \includegraphics[width=\textwidth]{q2_0V05.png} - \caption{Fermi Functions at $0.05V$.} + \caption{Fermi functions at $0.05V$.} \label{fig:q2_0v} \end{subfigure}% \begin{subfigure}{0.33\textwidth} \centering \includegraphics[width=\textwidth]{q2_0V1.png} - \caption{Fermi Functions at $0.1V$.} + \caption{Fermi functions at $0.1V$.} \label{fig:q2_0v} \end{subfigure} \begin{subfigure}{0.33\textwidth} \centering \includegraphics[width=\textwidth]{q2_0V2.png} - \caption{Fermi Functions at $0.2V$.} + \caption{Fermi functions at $0.2V$.} \label{fig:q2_0v} \end{subfigure}% \begin{subfigure}{0.33\textwidth} \centering \includegraphics[width=\textwidth]{q2_0V3.png} - \caption{Fermi Functions at $0.3V$.} + \caption{Fermi functions at $0.3V$.} \label{fig:q2_0v} \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} \subsection*{(b)} - 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. + As the drain voltage increases, we see that the overlapping area + under the "curve" of $f_1(E + U) - f_2(E+U)$ and $D(E)$ increases, up + until the point where the drain only has electrons below the energy + levels available in the channel. The difference function also reaches its maximum height and it widens downward rather than upward. + 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 - The self-consistent potential will decrease, which is what causes this shift in levels. + %% WRONG EXPLANATION - kept for posterity :) + % 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)} 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 - 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 point at which their Fermi functions have the greatest difference, and thus will be making 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 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 - 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 "conduction" zone and thus no current would be able to flow at all. @@ -494,7 +499,7 @@ ylabel('Current [A]'); \begin{equation*} I = 0, N = 0 \end{equation*} - + % 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 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)} \begin{equation*} - I = 608nA, N = 0.5 + I = 608\:nA, N = 0.5 \end{equation*} - + % 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. \begin{equation*} 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*} \begin{equation*} @@ -520,15 +525,15 @@ ylabel('Current [A]'); \subsubsection*{(iii)} \begin{equation*} - I = 0A, N = 1 + I = 0\:A, N = 1 \end{equation*} - + % 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. \begin{equation*} 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*} \begin{equation*} @@ -548,7 +553,7 @@ ylabel('Current [A]'); \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 $f_2(E + U)$ 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. \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] \end{equation*} \begin{equation*} - I = 1.8mA + I = 1.8\:mA \end{equation*} \subsubsection*{(iii)} @@ -596,59 +601,54 @@ ylabel('Current [A]'); 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 \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*} 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)} 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. - 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: \begin{equation*} [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{0}{0.024}}} - [\frac{1}{1 + e^{frac{0}{0.027}}}] = 0 + \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 \end{equation*} Molecule B: \begin{equation*} [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{-0.05}{0.024}}} - [\frac{1}{1 + e^{frac{-0.05}{0.027}}}] = 0.02493 + \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 \end{equation*} Molecule C: \begin{equation*} [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{-0.1}{0.024}}} - [\frac{1}{1 + e^{frac{-0.1}{0.027}}}] = 0.00877 + \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 \end{equation*} 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 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 $0eV$. - + % Since we know the voltages on the contacts, we can calculate the effective fermi level at each. \begin{equation*} - \mu_1 = \mu_0 - V_S = 0 - (-2V) = 2eV + \mu_1 = \mu_0 - V_S = 0 - (-2V) = 2\:eV \end{equation*} \begin{equation*} - \mu_2 = \mu_0 - V_D = 0 - 1V = -1eV + \mu_2 = \mu_0 - V_D = 0 - 1V = -1\:eV \end{equation*} - + % 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 $[-1eV, 0eV]$. @@ -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 \end{equation*} \begin{equation*} - I_A = 12.2mA + I_A = 12.2\:mA \end{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 \end{equation*} \begin{equation*} - I_A = 6.1mA + I_A = 6.1\:mA \end{equation*} - + % Material A should be chosen. \subsection*{(g)} Using Ohm's law we find the corresponding conductance: \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*} - + % We expect this result to be an integer multiple of $G_0 = 38.76$ $\mu S$, the quantum of conductance. We find \begin{equation*} - \frac{\sigma{max}}{G_0} = 3. + \frac{\sigma_{max}}{G_0} = 3. \end{equation*} - + % We conclude there are 3 levels.