山东大学本科电子技术模拟题带答案1

山东大学本科电子技术模拟题带答案1
山东大学本科电子技术模拟题带答案1

电子技术模拟题 2017年

一、单项选择题:在下列各题中,将正确的答案代码填入括号内

1. P 型 半 导 体 中 空 穴 数 量 远 比 电 子 多 得 多,因 此 该 半 导 体

应( )。

(a) 带 正 电

(b) 带 负 电 (c) 不 带 电

2. 电 路 如 图 所 示, 所 有 二 极 管 均 为 理 想 元 件,则 D 1、D 2、D 3的 工 作 状 态 为( )。

(a) D 1导 通,D 2、D 3 截 止 (b) D 1、D 2截 止 , D 3 导 通 (c) D 1、D 3截 止, D 2导 通 (d) D 1、D 2、D 3均 截 止

+

3. 电 路 如 图 所 示,二 极 管 D 为 理 想 元 件,输 入 信 号 u i 为 如 图 所 示 的 三 角 波 ,则 输 出 电 压 u O 的

最 大 值 为( )。

(a) 5 V (b) 10 V (c) 2 V (d) 7 V

D

u O

4. 晶 体 管 处 于 截 止 状 态 时, 集 电 结 和 发 射 结 的 偏 置 情 况 为( )。

(a) 发 射 结 反 偏,集 电 结 正 偏 (b) 发 射 结、集 电 结 均 反 偏 (c) 发 射 结、集 电 结 均 正 偏 (d) 发 射 结 正 偏,集

电 结 反 偏

5.某固定偏置单管放大电路的静态工作点Q 如图所示,欲使工作点移至Q ′需 使( )。

(a ) 偏置电阻R B 增大。 (b ) 集电极电阻R C 减小 (c ) 偏置电阻R B 减小

CE (V)

A

=40μ

6.采用差动放大电路是为了( )。

(A) 稳定电压放大倍数 (B)增强带负载能力 (C) 提高输入阻抗 (D )克服零点漂移

7. 电路如图所示,运算放大器的饱和电压为±12V ,设稳压管的稳定电压为6V , 正向压降为零,当输入电压u i =1V 时,则输出电压u O 等于( )。

(a) 12V (b) 6V (c)0V

O

u i

8.自 激 正 弦 振 荡 器 是 用 来 产 生 一 定 频 率 和 幅 度 的 正 弦 信 号 的 装 置,此 装 置 之 所 以 能 输 出 信 号 是 因 为( ) 。

(a) 有 外 加 输 入 信 号 (b) 满 足 了 自 激 振 荡 条 件

(c) 先 施 加 输 入 信 号 激 励振 荡 起 来,然 后 去 掉 输 入 信 号

9. 一 个 振 荡 器 要 能 够 产 生 正 弦 波 振 荡,电 路 的 组 成 必 须 包 含( )。

(a) 放 大 电 路,负 反 馈 电 路 (b) 负 反 馈 电 路、选 频 电 路

(c) 放 大 电 路、 正 反 馈 电 路、 选 频 电 路

10 .逻 辑 电 路 如 图 所 示,A=“0”,B=“1” 时,C 脉 冲 来 到 后 JK 触 发 器( )。

(a) 具 有 计 数 功 能 (b) 置“0” (c) 置“1” (d) 保 持 原 状 态

B

11 .分 析 某 时 序 逻 辑 电 路 的 状 态 表, 判 定 它 是( )。

(a) 移 位 寄 存 器

(b) 二 进 制 计 数 器 (c) 十 进 制 计 数 器

12.已知某半导体存储器的容量为1024×4,则其具有的地址线数应为( )。

(a) 4 (b) 8 (c) 10 (d) 2 13. T 形 电 阻 网 络 D/A 转 换 器 是 由( )组 成。

(a) T 形 电 阻 网 络 和 集 成 运 算 放 大 器 (b) T 形 电 阻 网 络 和 触 发 器 (c) T 形 电 阻 网 络 和 振 荡 器

二、计算题

放 大 电 路 如 图 所 示, 晶 体 管 的 电 流 放 大 系 数 β=50,

U BE .V =06 ,R B1=110 k Ω, R B2= 10 k Ω, R C = 6 k Ω,R E = 400 Ω,

R L = 6 k Ω, 要 求:(1) 计 算 静 态 工 作 点;(2) 画 出 微 变 等 效 电 路;(3) 计 算 电

压 放 大 倍 数。

u o 12V

三、分析表达题

电 路 如 图 所 示 ,R R R 2

45k

===10Ω,R 1k =1Ω , 求:

(1) R 3 ,R 6 的 阻 值。

(2) A u u U F O I ==/?

(3) 若 运 算 放 大 器 的 电 源 电 压 为±15V ,

输 入 电 压 u 1V =2, 则 输 出 电 压

u O =?

u O

+

-

四、分析计算题

整 流 滤 波 电 路 如 图 所 示,二 极 管 是 理 想 元 件,电 容

C F =500μ,负 载 电 阻R L k =5Ω , 开 关 S 1闭 合、S 2 断 开 时,直 流

电 压 表

()V 的 读 数 为1414.V ,求:

(1) 开 关 S 1 闭 合、S 2 断 开时, 直 流 电 流 表 ()A 的 读 数 ; (2) 开 关 S 1 断 开、S 2 闭 合 时, 直 流 电 流 表 ()A 的 读 数 ;

(3) 开 关 S 1、S 2 均 闭 合 时, 直 流 电 流 表()A 的 读 数。 (设 电 流 表 内 阻 为 零, 电 压 表 内 阻 为 无 穷 大 )。

u 1

-

+

五、 分析逻 辑 电 路

逻 辑 电 路 如 图 所 示,写 出 逻 辑 式, 化 简 之, 并 列 出 状 态 表。

A C

B

电子技术参考答案

三、 (1) R R R 312//k =≈1Ω,R R R 645//k ==5Ω (2) A u u U F O I ==/20

(3) 当u I V =2, 则u O 约 等 于 30V

要 点 A u u u u u U F

O I O2O1I

==- u R

R u u O I I 12110

=-=- u R

R u u O O1I 25410

=-= 故 A u u U F

I I

==2020 四、(1) 开 关 S 1闭 合、S 2 断 开 时 : ()A 的 读 数 为 零。

(2) 开 关 S 1 断 开、S 2闭合 时,U 214142

100=

=.

V V; U U O V V ==?=0909100

902..; ()A 的 读 数 为 U R O L mA mA =

=90

5

18, (3)开 关 S 1、S 2 均 闭 合 时,U U O V V ==?=12

121001202..; ()A 的 读 数 为

U R O L mA mA ==1205

24。 五、

F ABC A ABC B ABC C =++

=++()()ABC A B C =+++ABC A B C ()

ABC

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