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ОТЧЕТЫ/

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МИНОБРНАУКИ РОССИИ
Санкт-Петербургский государственный
электротехнический университет
«ЛЭТИ» им. В.И. Ульянова (Ленина)
Кафедра Микрорадиоэлектроники и технологии радиоаппаратуры
(МИТ)
ОТЧЕТ
**по ИДЗ**
**по дисциплине «ОЭиР»**
****Тема: Исследование контактных явлений в структуре
металл-полупроводник****
****Вариант 14,6,3****
https://www.meme-arsenal.com/memes/4598e00877a721c55a46dc4aafb78719.jpg
------------------ -- -------------
Студент гр. 1181 Шишков Д.А.
Преподаватель Филипюк И.А
------------------ -- -------------
Санкт-Петербург
2023
Задание:
Для заданной пары металл-полупроводник оценить кинетические свойства
заданных материалов, рассчитать и построить энергетическую диаграмму и
вольт-амперную характеристику контакта в заданном диапазоне температур,
дать рекомендации по применению исследуемого контакта.
Таблица 1. Некоторые свойства металлов
<table>
<tbody>
<tr class="odd">
<td><p>No</p>
<p>ВАР.</p></td>
<td>Элемент</td>
<td>Структура</td>
<td>Атомная масса</td>
<td>Параметр решетки, Å</td>
<td>Плотность, г/см3</td>
<td>Удельное сопротивление, мкОм·см</td>
<td>Температура, К</td>
<td>Работа выхода φ, эВ</td>
<td></td>
<td></td>
</tr>
<tr class="even">
<td>Дебая (TD)</td>
<td>Ферми (TF·10<sup>-4</sup>)</td>
<td>плавления (Tпл)</td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
</tr>
<tr class="odd">
<td>14</td>
<td>Au</td>
<td>ГЦК</td>
<td>196.9</td>
<td>4.08</td>
<td>19.28</td>
<td>2.2</td>
<td>165</td>
<td>6.39</td>
<td>1337</td>
<td>4.58</td>
</tr>
</tbody>
</table>
1\) Определить класс симметрии заданных материалов, построить прямую и
обратную элементарные ячейки заданных материалов. Определить размеры
Зоны Бриллюэна в направлениях X, L, К.
Гранецентрированная Кубическая решётка
Формула симметрии: 3L~4~4L~3~6L~2~9PC
Класс симметрии: m3m
Так как формулы симметрии ГЦК и простой кубической решётки совпадают,
приведём на рис. 1-3 изображения осей, плоскостей и центра симметрии для
последнего
![](pictures/Pictures/100000010000015E00000147D7C1BAAC07F91159.png){width="3.4504in"
height="2.9512in"}
Рис. 1 Изображение осей симметрии кубической решётки
![](pictures/Pictures/10000001000001CC00000131EEA170A8799B8110.png){width="4.5437in"
height="3.0126in"}
Рис. 2 Изображение плоскостей симметрии куба
![](pictures/Pictures/10000001000000DC00000100F9B19430CB57B85C.png){width="2.0882in"
height="2.4307in"}
Рис. 3 Изображение центра симметрии куба
Базисные вектора:
$$a_{1} = \begin{bmatrix}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} \\
0 \\
\end{bmatrix}$$, $$a_{2} = \begin{bmatrix}
\frac{1}{\sqrt{2}} \\
0 \\
\frac{1}{\sqrt{2}} \\
\end{bmatrix}$$, $$a_{3} = \begin{bmatrix}
0 \\
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} \\
\end{bmatrix}$$, считая, что постоянная решётки = 1
Построим кристаллическую решётку по заданным векторам (рис. 4)
![](pictures/Pictures/1000000100000102000000EC388FD10724EFA862.png){width="2.6874in"
height="2.4583in"}
Рис. 4 ГЦК
Объём элементарной ячейки:
$$V = {|{\overrightarrow{a_{1}} \cdot \left\lbrack {\overrightarrow{a_{2}} \times a_{3}} \right\rbrack}|} = \frac{1}{\sqrt{2}}$$
Базисные вектора обратной решётки:
$${a_{i}^{\ast} = \frac{2\pi}{V}}\left\lbrack {a_{j} \times a_{k}} \right\rbrack,{i \neq j \neq k}$$
$$a_{1}^{\ast} = \begin{bmatrix}
{\sqrt{2}\pi} \\
{{- \sqrt{2}}\pi} \\
{{- \sqrt{2}}\pi} \\
\end{bmatrix}$$, $$a_{2}^{\ast} = \begin{bmatrix}
{{- \sqrt{2}}\pi} \\
{\sqrt{2}\pi} \\
{{- \sqrt{2}}\pi} \\
\end{bmatrix}$$, $$a_{3}^{\ast} = \begin{bmatrix}
{{- \sqrt{2}}\pi} \\
{{- \sqrt{2}}\pi} \\
{\sqrt{2}\pi} \\
\end{bmatrix}$$
Изобразим её на рис. 5
![](pictures/Pictures/10000001000002D0000003619A2670E62D0CE93C.png){width="4.128in"
height="3.6374in"}
Рис. 5 Обратная решётка ГЦК
Первая зона Бриллюэна (рис. 6):
![](pictures/Pictures/10000001000001C20000011D0C7FC81F7FC396D5.png){width="4.6874in"
height="2.9689in"}
Рис. 6 Первая зона Бриллюэна
Размеры зоны Бриллюэна по направлениям X, L, K:
$$X = \begin{bmatrix}
0 \\
0 \\
1 \\
\end{bmatrix}$$ - центр верхнего квадрата
$$L = \begin{bmatrix}
0.5 \\
0.5 \\
0.5 \\
\end{bmatrix}$$ - центр шестиугольника
$$K = \begin{bmatrix}
0.75 \\
0.75 \\
0 \\
\end{bmatrix}$$ - середина грани соединяющей два шестиугольника
2\) Определить концентрацию электронов для заданного металла из условия
касания зоны Бриллюэна и сферы Ферми и сделать суждение о применимости
теории свободных электронов.
В момент касания волновой вектор kф, соответствующий радиусу сферы
Ферми, равен волновому вектору kз, при котором в направлении,
перпендикулярном отражающим плоскостям, выполняется уравнение
Вульфа-Брэггов. Условия касания для двумерной модели можно записать в
виде $$k_{ф} = k_{з}$$ или, поскольку $${k = 2}\frac{\pi}{\lambda}$$, то
$$\lambda_{ф} = \lambda_{з}$$.
Объем сферы Ферми в пространстве импульсов равен
$$\frac{4}{3}\pi p_{ф}^{3}$$ , где $$p_{ф}$$ -- импульс электронов на
поверхности Ферми. С другой стороны, этот же объем равен
$$\frac{N}{2}\frac{h^{3}}{V}$$, где N -- число электронов в объеме V.
Таким образом:
$$\frac{4}{3}\pi{p_{ф}^{3} = \frac{N}{2}}\frac{h^{3}}{V}$$
Откуда находим:
$${p_{ф} = h}\left( \frac{3N}{8\pi V} \right)^{1/3}$$
Поскольку по соотношению де-Бройля $$\lambda = \frac{h}{p}$$, то
$$\lambda_{ф} = \left( \frac{8\pi V}{3N} \right)^{1/3}$$
Найдем теперь λз. Из всех граней первой зоны Бриллюэна для
гранецентрированной решетки ближе всего к началу координат находятся
грани, обусловленные отражением электронов от плоскостей {111}. Поэтому
сфера Ферми впервые коснется именно этих граней. Таким образом, для
определения условий касания сферы с первой зоной Бриллюэна необходимо
найти длину волны, при которой электроны взаимодействуют с плоскостями
{111}. Из уравнения Вульфа -- Бреггов
$${\mathit{n\lambda} = 2}d\mathit{sin\theta}$$ находим:
3
$${\lambda_{з} = 2}{d = 2}{\frac{a}{\sqrt{H^{2} + K^{2} + L^{2}}} = 2}\frac{a}{\sqrt{3}}$$
(здесь θ = 90°, так как в точке касания волновой вектор перпендикулярен
к плоскостям {111}; п = 1, так как λ--наибольшее)
Если в объеме V число атомов Na, то число элементарных ячеек будет
$$\frac{N_{a}}{4}$$, так как на одну ячейку гранецентрированной решетки
приходится четыре атома, тогда $${V = a^{3}}\frac{N_{a}}{4}$$
Откуда $$a = \left( \frac{4V}{N_{a}} \right)^{1/3}$$, следовательно
$${\lambda_{з} = \frac{2}{\sqrt{3}}}\left( \frac{4V}{N_{a}} \right)^{1/3}$$
Возводя обе части этого тождества в куб и произведя необходимые
сокращения, получаем:
$$\frac{N}{N_{a}} = \frac{\pi\sqrt{3}}{4} = 1.36$$
Список литературы:
Астанин В.В. Физика твёрдого тела

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# %%
import numpy as np
def dr(*x, p = 3):
print(np.round(x, p))
def vol(x):
return np.dot(x[0], np.cross(x[1], x[2]))
# %%
A_metal = 4.08
# a_general = np.array(((1/np.sqrt(2), 1/np.sqrt(2), 0), (1/np.sqrt(2), 0, 1/np.sqrt(2)), (0, 1/np.sqrt(2), 1/np.sqrt(2))))
a_metal = A_metal/2 * np.array(((0, 1, 1), (1, 0, 1), (1, 1, 0)))
dr(a_metal)
a = a_metal
V_metal = np.dot(a[0], np.cross(a[1], a[2]))
dr(V_metal, A_metal**3/4)
a_ = 2*np.pi/V_metal * np.array((np.cross(a[1],a[2]), np.cross(a[0], a[2]), np.cross(a[0], a[1])))
dr(a_)
dr(a_[0][0], 2*np.pi/A)
X = np.array((0, 0, 1)) * A
L = np.array((0.5, 0.5, 0.5)) * A
K = np.array((0.75, 0.75, 0)) * A
dr(X, L, K)
dr(vol(a_))
# %%
A_diel = 6.48
a_diel = a_general * A_diel
dr(a_diel)
a = a_diel
V_diel = np.dot(a[0], np.cross(a[1], a[2]))
dr(V_diel)
a_ = 2*np.pi/np.abs(V_diel) * np.array((np.cross(a[1],a[2]), np.cross(a[0], a[2]), np.cross(a[0], a[1])))
dr(a_)
# %%
a = a_general
dr(2*np.pi/(np.dot(a[0], np.cross(a[1], a[2])))*np.cross(a[1], a[2]))

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