要求每天阅读一篇技术文档,不需要记下来,只是能看懂就好。。后发现,这就是专业英语的课程资料。

----------------------------------------------------------------------------------------------------------------------

At the most basic level, a signal amplifier does exactly what you expect – it makes a signal
bigger! However, the way in which it is done does vary with the design of the actual amplifier,
the type of signal, and the reason why we want to enlarge the signal. [1] We can illustrate this by
considering the common example of a “Hi-Fi” audio system.
In a typical modern Hi-Fi: system, the signals will come from a unit like a CD player, FM
tuner, or a Tape/Minidisk unit. The signals they produce have typical levels of the order of
100 mV or so when the music is moderately loud. This is a reasonably large voltage, easy to
detect with something like an oscilloscope or a voltmeter. However, the actual power levels of
these signals are quite modest. Typically, these sources can only provide currents of a few
milliamps, which by P=VI means powers of just a few milliwatts. A typical loudspeaker will
require between a few Watts and perhaps over 100 Watts to produce loud sound. Hence we will
require some form of Power Amplifier (PA) to “boost” the signal power level from the source
and make it big enough to play the music.
Fig. 1.1 shows four examples of simple analog amplifier stages using various types of
device. In each case the a.c. voltage gain will usually be approximated by

provided that the actual device has an inherent gain large enough to be controlled by the resistor
values chosen.Note the negative sign in expression 1-1 which indicates that the examples all
invert the signal pattern when amplifying. [2] In practice, gains of the order of up to hundred are
possible from simple circuits like this, although it is usually a good idea to keep the voltage gain
below this. Note also that vacuum state devices tend to be called “valves” in the UK and “tubes”
in the USA.

Many practical amplifiers chain together a series of analog amplifier stages to obtain a high
overall voltage gain. For example, a PA system might start with voltages of the order of 0.1 mV
from microphones, and boost this to perhaps 10 to 100 V to drive loudspeakers. This requires an
overall voltage gain of 109, so a number of voltage gain stages will be required.
In many cases we wish to amplify the current signal level as well as the voltage. The
example we can consider here is the signal required to drive the loudspeakers in a “Hi-Fi” system.
These will tend to have a typical input impedance of the order of 8 Ohms. So to drive, say, 100
Watts into such a loudspeaker load we have to simultaneously provide a voltage of 28 Vrms and
3.5 Arms. Taking the example of a microphone as an initial source again a typical source
impedance will be around 100 Ohms. Hence the microphone will provide just 1 nA when
producing 0.1 mV. This means that to take this and drive 100 W into a loudspeaker the amplifier
system must amplify the signal current by a factor of over 109 at the same time as boosting the
voltage by a similar amount. [3] This means that the overall power gain required is 1018 – i.e. 180
dB!
This high overall power gain is one reason it is common to spread the amplifying function
into separately boxed pre- and power-amplifiers. The signal levels inside power amplifiers are so
much larger than these weak inputs that even the slightest ‘leakage’ from the output back to the
input may cause problems. By putting the high-power (high current) and low power sections in
different boxes we can help protect the input signals from harm.

In practice, many devices which require high currents and powers tend to work on the basis
that it is the signal voltage which determines the level of response, and they then draw the current
they need in order to work. [4] For example, it is the convention with loudspeakers that the
volume of the sound should be set by the voltage applied to the speaker. Despite this, most
loudspeakers have an efficiency (the effectiveness with which electrical power is converted into
acoustical power) which is highly frequency dependent. To a large extent this arises as a natural
consequence of the physical properties of loudspeakers. We won’t worry about the details here,
but as a result a loudspeaker’s input impedance usually varies in quite a complicated manner with
the frequency. (Sometimes also with the input level.)
Fig. 1.2 shows a typical example. In this case, the loudspeaker has an impedance of around
12 Ohms at 150 Hz and 5 Ohms at 1 kHz. So over twice the current will be required to play the
same output level at 1 kHz than is required at 150 Hz. The power amplifier has no way to “know
in advance” what kind of loudspeaker you will use, so simply adopts the convention of asserting
a voltage level to indicate the required signal level at each frequency in the signal and supplying
whatever current the loudspeaker then requires.
This kind of behavior is quite common in electronic systems. It means that, in information
terms, the signal pattern is determined by the way the voltage varies with time, and ideally the
current required is then drawn. Although the above is based on a high-power example, a similar
situation can arise when a sensor is able to generate a voltage in response to an input stimulus but
can only supply a very limited current. In these situations we require either a current amplifier or

a buffer. These devices are quite similar, and in each case we are using some form of gain device
and circuit to increase the signal current level. However, a current amplifier always tries to
multiply the current by a set amount. Hence it is similar in action to a voltage amplifier which
always tries to multiply the signal current by a set amount. The buffer differs from the current
amplifier as it sets out to provide whatever current level is demanded from it in order to maintain
the signal voltage told to assert. Hence it will have a higher current gain when connected to a
more demanding load.

specialized English for automation-Lesson 1 Analog Amplifiers的更多相关文章

  1. specialized English for automation-Lesson 2 Basic Circuits of Operational Amplifiers

    排版有点乱.... ========================================================================= Operational Ampl ...

  2. specialized English for automation-Lesson 3 CMOS Logic Circuit

    CMOS logic is a newer technology, based on the use of complementary MOS transistors toperform logic ...

  3. 运放参数的详细解释和分析-part1,输入偏置电流和输入失调电流【转】

    一般运放的datasheet中会列出众多的运放参数,有些易于理解,我们常关注,有些可能会被忽略了.在接下来的一些主题里,将对每一个参数进行详细的说明和分析.力求在原理和对应用的影响上把运放参数阐述清楚 ...

  4. 转载:AAC文件解析及解码

    转自:http://blog.csdn.net/wlsfling/article/details/5876016 http://www.cnblogs.com/gaozehua/archive/201 ...

  5. Lesson 14 Do you speak English?

    Text I had an amusing experience last year. After I had left a small village in the south of France. ...

  6. Lesson 25 Do the English speak English?

    Text I arrived London at last. The railway station was big, black and dark. I did not know the way t ...

  7. AD8275 Driver Amplifiers For Analog-To-Digital Converters

    Driver Amplifiers For Analog-To-Digital Converters What amplifiers are used to drive analog-to-digit ...

  8. English trip EM2-LP-4B At School Teacher:Russell

    课上内容(Lesson) Where is Loki a student?  Loki is in Meten, BaobaoStreet, Chengdu. What is he studying? ...

  9. 2016.09.14,英语,《Using English at Work》全书笔记

    半个月时间,听完了ESLPod出品的<Using English at Work>,笔记和自己听的时候的备注列在下面.准备把每个语音里的快速阅读部分截取出来,放在手机里反复听. 下一阶段把 ...

随机推荐

  1. linux 如何查看防火墙是否开启

    service iptables status可以查看到iptables服务的当前状态.但是即使服务运行了,防火墙也不一定起作用,你还得看防火墙规则的设置 iptables -L在此说一下关于启动和关 ...

  2. Python扩展之类的魔术方法

    Python中类的魔术方法 在Python中以两个下划线开头的方法,__init__.__str__.__doc__.__new__等,被称为"魔术方法"(Magic method ...

  3. 正则表达式,以python为例

    转载需注明原文地址和作者两项内容. 正则表达式目的是能够快速处理字符串内容,主要用于找出指定的字符串,配合其他操作完成任务.使用正则表达式时要了解自己语言的特性,python中的正则表达式默认情况是贪 ...

  4. Python:笔记(7)——yield关键字

    Python:笔记(7)——yield关键字 yield与生成器 所谓生成器是一个函数,它可以生成一个值的序列,以便在迭代中使用.函数使用yield关键字可以定义生成器对象. 一个例子 我们调用该函数 ...

  5. CCF地铁修建

    问题描述 A市有n个交通枢纽,其中1号和n号非常重要,为了加强运输能力,A市决定在1号到n号枢纽间修建一条地铁. 地铁由很多段隧道组成,每段隧道连接两个交通枢纽.经过勘探,有m段隧道作为候选,两个交通 ...

  6. 2017-2018 ACM-ICPC, Asia Daejeon Regional Contest Solution

    A:Broadcast Stations 留坑. B:Connect3 题意:四个栈,每次放棋子只能放某个栈的栈顶,栈满不能放,现在给出(1, x) 表示黑子放在第x个栈的第一个位置,白子放在第b个栈 ...

  7. poj1228 Grandpa's Estate

    地址:http://poj.org/problem?id=1228 题目: Grandpa's Estate Time Limit: 1000MS   Memory Limit: 10000K Tot ...

  8. JAVA面试题整理(5)-数据库

    数据库 1.Oracle/mysql分页有什么优化 2.悲观锁.乐观锁 悲观锁(Pessimistic Concurrency Control,PCC):假定会发生并发冲突,屏蔽一切可能违反数据完整性 ...

  9. OpenVAS安装过程

    OpenVAS安装过程 安装过程 检查安装状况 命令行下输入opensav-check-setup,显示错误NO CA certificate file,并显示解决方法 创建证书 输入命令openva ...

  10. POJ 1840 Eqs(乱搞)题解

    思路:这题好像以前有类似的讲过,我们把等式移一下,变成 -(a1*x1^3 + a2*x2^3)== a3*x3^3 + a4*x4^3 + a5*x5^3,那么我们只要先预处理求出左边的答案,然后再 ...