47 lines
1.7 KiB
TeX
47 lines
1.7 KiB
TeX
\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}
|