### Add solutions to first three homework assignments.

master Danila Fedorin 8 months ago
parent
commit
526a7c50bc
7 changed files with 178 additions and 1 deletions
1. 12
HW1.tex
2. 46
HW2.tex
3. 121
HW3.tex
4. BIN
Q1HW3.png
5. BIN
Q2.png
6. BIN
Q2.xcf
7. BIN
Q3.png

#### 12 HW1.tex View File

 @ -58,7 +58,10 @@ Not quite sure what this question means, but I have a few thoughts:  193nm light, and focused on refining the technique.  \item We started to perform multiple lithography (and maybe etch)  steps for a single layer, which made it possible to  halve (or further reduce) the minimum pitch.  halve (or further reduce) the minimum pitch. Exposing  photoresist more than once also made it possible (from what I can tell) to use all the special  techniques (off-axis illumination, immersion, RET), which otherwise constrain masks to being only  horizontal or only vertical.  \item Self-aligned mutli-patterning techniques cannot really lay down  holes'' in lines; these holes have to be added later.  As a result, layouts of modern CPUs are very regular, @ -71,7 +74,14 @@ Not quite sure what this question means, but I have a few thoughts: From the Rosetta Stone of Lithography'' it looks like with true double patterning, the smallest we can get is the 10nm node (50nm pitch).   However, the Breakfast Bytes article, right after saying 50nm is the smallest pitch we can get with double patterning, brings up SADP, which is also double patterning, but can go as low as 40nm. In the Rosetta Stone, however, 40nm seems to correspond to Higher-order pitch division', and not double patterning, so I still think 50nm pitch / 10nm node is the answer here.   \section*{Q7} We use tin plasma! Apparently, tin is fairly efficient'' at converting laser light into EUV.   \end{document}

#### 46 HW2.tex View File

 @ -0,0 +1,46 @@ \documentclass{article} \usepackage[margin=1in]{geometry} \usepackage{graphicx} \title{Homework 2} \begin{document} \maketitle \section*{Q1} The current scales linearly with oxide capacitance per unit area, $C_{ox}$. Doubling the thickness of the insulator is akin to doubling the distance between the two plates, which halves $C_{ox}$. Thus, current would be halved as well.   \section*{Q2} This occurs, by definition, at the threshold voltage, $V_t$.   \section*{Q3} \begin{figure}[h]  \centering  \includegraphics[width=0.8\linewidth]{Q3.png}  \label{fig:iv}  \caption{} \end{figure}   \section*{Q4} The depletion region is larger because of the potential on the drain. Since (for an NMOS transistor) the source is tied to the lowest potential, to drive current through the transistor, we need to apply potential to the drain. Doing so pushes more carriers into the depletion region, causing it to grow.   \section*{Q5} The potential is $V_{gs} - V_t$, which is also written as $V_{GT}$ in the book.   \section*{Q6} The transistor in the picture likely suffers from velocity saturation. I think so because for each step in gate voltage, the current increases by the same amount. However, our simple models predict that this should be a quadratic increase. This difference in behavior is typically caused by velocity saturation.   \section*{Q7} The transistor in the picture likely suffers from impact ionization. I think so because at high drain-source voltages, the current starts to "bend upwards", increasing more than it is expected to past the saturation point. This can be caused by "hot" electrons producing electron/hole pairs on impact with the substrate atoms. These pairs serve as carriers, thereby contributing to increased current.   \end{document}

#### 121 HW3.tex View File

 @ -0,0 +1,121 @@ \documentclass{article} \usepackage[margin=1in]{geometry} \usepackage{graphicx} \usepackage{amsmath} \title{Homework 3} \begin{document} \maketitle \section*{Q1} Given the logical formula, we can follow the following process to convert it into strictly inverters, NOR, and NAND gates:   \begin{equation*}  \begin{aligned}  & \lnot((AB+C)D+E) \\  \Leftrightarrow & \lnot(\lnot\lnot(AB+C)D+E) & \text{(negation involutive)} \\  \Leftrightarrow & \lnot(\lnot(\lnot(AB+C)+\lnot D)+E) & \text{(DeMorgan's Laws)} \\  \Leftrightarrow & \lnot(\lnot(\lnot(\lnot\lnot AB+C)+\lnot D)+E) & \text{(negation involutive)} \\  \Leftrightarrow & \lnot(\lnot(\lnot(\lnot(\lnot A + \lnot B)+C)+\lnot D)+E) & \text{(DeMorgan's Laws)} \\  \end{aligned} \end{equation*}   This corresponds to the following circuit:   \begin{figure}[h]  \centering  \includegraphics[width=0.7\linewidth]{Q1HW3.png} \end{figure}   \pagebreak \section*{Q2} Making Scott's adjustment (adding a top-level 'not' to the formula in the assignment) yields the following:   \begin{figure}[h]  \centering  \includegraphics[width=0.7\linewidth]{Q2.png} \end{figure}   \pagebreak \section*{Q3} The book gives the following equation for determing the ideal number of stages:   \begin{equation*}  \begin{aligned}  N &= \log_{\rho}F \\  0 &= p_\text{inv} + \rho(1-\ln\rho)  \end{aligned} \end{equation*}   Where $p_\text{inv}$ is the intrinsic delay of an inverter. For $p_\text{inv} = 5$, we have $\rho = 6.14$. We then compute $F$:   \begin{equation*}  \begin{aligned}  & F &= GBH \\  & G &= 1 \\  & B &= 1 \\  & H &= 1000 \\  \Rightarrow & F &= 1000  \end{aligned} \end{equation*}   The ideal number of stages is then:   \begin{equation*}  \log_\rho F = 3.8 \approx 4 \end{equation*}   For all inverters, then, we get the following:   \begin{equation*}  \begin{aligned}  \hat{f} &= \sqrt{1000} \\  \text{sz}_4 &= 1000/\hat{f}^1 = 178 \\  \text{sz}_3 &= 1000/\hat{f}^2 = 31.6 \\  \text{sz}_2 &= 1000/\hat{f}^3 = 5.62 \\  \text{sz}_1 &= 1000/\hat{f}^4 = 1 \\  \end{aligned} \end{equation*}   \pagebreak \section*{Q4} First, to compute stage effort $\hat{f}$.   \begin{equation*}  \begin{aligned}  & F &= GBH \\  & G &= \left(\frac{4}{3}\right)\left(\frac{5}{3}\right)\left(\frac{5}{3}\right) \\  & B &= 3 \\  & H &= 1000 \\  \Rightarrow & F &= 11111  \end{aligned} \end{equation*}   Assuimng a $p_\text{invs}$ of 1, and thus $\rho = 3.59$, we get:   \begin{equation*}  \log_\rho F = 7.2 \approx 7 \end{equation*}   Since we currently have 3 stages, we should insert 4 inverters. It appears as though inserting inverters only at the end makes it too difficult for the first-stage NAND gate to drive the 3-branched NOR gates (we end up with an optimal size less than 1). Instead, I will insert two inverters right after the NAND2 gate, and two more inverters at the end. We can now compute gate sizes:   \begin{equation*}  \begin{aligned}  \hat{f} &= \sqrt{11111} \\  \text{sz}_7 &= 1000/\hat{f}^1 = 264 \\  \text{sz}_6 &= 1000/\hat{f}^2 = 69.8 \\  \text{sz}_5 &= 1000/\hat{f}^3 * \left(\frac{5}{3}\right) = 30.8 \\  \text{sz}_4 &= 1000/\hat{f}^4 * \left(\frac{5}{3}\right)\left(\frac{5}{3}\right) = 13.5 \\  \text{sz}_3 &= 1000/\hat{f}^5 * \left(\frac{5}{3}\right)\left(\frac{5}{3}\right)3 = 10.7 \\  \text{sz}_2 &= 1000/\hat{f}^6 * \left(\frac{5}{3}\right)\left(\frac{5}{3}\right)3 = 2.84 \\  \text{sz}_1 &= 1000/\hat{f}^7 * \left(\frac{5}{3}\right)\left(\frac{5}{3}\right)3 = 1 \\  \end{aligned} \end{equation*}   \end{document}`

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