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科技翻译---模拟放大器中英文

(2011-03-17 16:47:07)
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Analog Amplifiers

模拟放大器

 

一般情况下,信号放大器正如人们期望的一样工作——将信号放大。然而,信号的放大方式随着实际放大器的设计、信号的类型以及放大信号目的的不同而变化。这一点可以通过一个常见的高保真音响系统实例来加以说明。

 

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.

 

在典型的现代高保真系统中,信号是来自于CD播放器、调频收音机或磁带/小型磁盘机等设备。当音乐声大小适当时,这些设备产生信号的幅度大概在100 毫伏左右。这种信号幅度相当高,易于用示波器或电压表等仪器检测到。但是,这些信号的实际能量水平并不高。典型情况下,这些信号源只能提供毫安级的电流。根据公式PVI,其功率只有几十毫瓦。典型的扬声器需要几十瓦到数百瓦的功率才能产生足够大的声音。因此,我们需要某些形式的功率放大器(PA)来提高来自信号源的信号功率,使其足以播放音乐。

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 PVI 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.

 

 

图 1.1 为四个采用不同器件的简单模拟放大器的例子。在任意一种情况下,如果实际器件的固有增益都能被所选电阻控制,电压增益就近似为Av≈-R1/R2。注意,表达式1-1中的负号表明,示例中的电路在放大时改变了信号的极性。实际上,这种简单电路的增益可达100 左右,尽管将电压增益限制在100 以下通常更好一些。此外,真空管在英国被叫做valves,而在美国被称为tubes。

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 Fig. 1.1 Examples of voltage amplifiers Av≈-R1/R2 (1-1) 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.

 

许多实际的放大器将多个放大器级联起来,以获得较高的电压增益。例如,一个功率放大系统的输入是来自于麦克风的0.1毫伏的电压,将其放大到10 伏到100 伏才能推动扬声器。这就要求电压的总增益达到109,因此就需要很多放大器级联起来。

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.

 

在很多情况下,除了信号的电压以外,我们还要放大信号的电流。这里我们考虑的例子是高保真系统中用来驱动扬声器的信号,其典型的输入电阻约为8 欧姆。因此,要驱动100 瓦的扬声器负载,就要同时提供28 伏的电压和3.5 安的电流信号。仍以麦克风作为初始信号源为例,典型的源阻抗在100 欧左右。因此,麦克风在产生0.1 毫伏的信号时,提供的电流仅为1 纳安。这就表示要接受这种输入信号并去驱动100 瓦的扬声器,放大电路就必须将信号的电流和电压同时放大109倍。这也就意味着总的功率增益为1018,即180分贝。

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!

 

一般都将放大功能分散到单独设计的前级放大器和功率放大器中,其原因就在于功率增益很大。功率放大器中的信号幅度比微弱的输入信号大得多,即使输出的极微小的泄漏传输到输入端,都会引发一些问题。通过将大功率(大电流)和小功率放大电路分置在不同的单元中,就可以避免输入信号受到干扰。

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.

 

实际上,许多需要大电流和大功率的设备往往都在特定的条件下工作,即由信号的电压决定响应的幅度,继而由设备吸收其所需要的电流而工作。例如,扬声器的音量通常是由所加电压控制。此外,大多数扬声器的效率(电能被转换为声能的效能)基本上与频率无关。在很大程度上,这是由扬声器的物理特性所产生的自然结果。这里不必考虑具体的细节,但扬声器的输入阻抗随频率的不同而呈复杂的变化(有时也与输入信号的幅度有关)。

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.)

 

 

图 1.2 是一个典型的例子。此时,扬声器在150 赫兹时的阻抗为12 欧,1 千赫时的阻抗为5 欧。所以,要达到与150 赫兹时同样大小的输出信号,在频率为1 千赫时就需要两倍以上的电流。功率放大器不可能预先知道将会使用的扬声器类型,因此就简单地按常规情况处理,提供一定大小的电压,表示信号中任意频率下所需信号的幅度,并提供扬声器所需要的电流。

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.

 

这种特点在电子系统中很常见。用信息术语来说就是信号类型取决于电压随时间变化的情况,且在理想情况下就能吸收所需的电流。尽管上述情况是基于大功率的例子,但当传感器在输入的激励下做出响应而产生一定的电压,却只能提供有限的电流时,类似的情况也会出现。这时我们就需要一个电流放大器或缓冲器。这些装置非常相似,在各种情况下都可采用一定形式的增益装置或电路来提高信号电流的大小。不过,电流放大器总是设法对电流进行一定的放大。这与电压放大器的功能相似。缓冲器总是可以提供任何你所需要的电流,以便维持其标称电压保持不变。这就是它与电流放大器的不同之处。因此,缓冲器在连接到要求比较高的负载时就具有较高的电流增益。

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.

 

 

[1] However, the way in which this 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.

然而,信号的放大方式随着实际放大器的设计、信号的类型以及放大信号目的的不同而变化。

本句为强调句型,句中的定语从句in which this is done修饰主语the way,does 用于强调谓语vary。短语结构the design of the actual amplifier 与the type of signal及and the reason why we want to enlarge the signal 并列作为介词with 的宾语。why 引导的定语从句修饰the reason。

[2] Note the negative sign in expression 1-1 which indicates that the examples all invert the signal pattern when amplifying.

注意,表达式1-1 中的负号说明,示例中的电路在放大时改变了信号的极性。

句中which 引导的定语从句和介词短语in expression 1-1 修饰动词Note 的宾语the negative sign,that引导的宾语从句在定语从句中作宾语,副词all和分词结构when amplifying 修饰谓语动词invert。

[3] This means that to take this and drive 100W 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.

这就表示要接受这种输入信号并去驱动100 瓦的扬声器,放大电路就必须将信号的电流和电压同时放大109倍。

句中that 引导的从句作谓语动词means 的宾语,不定式短语to take…在宾语从句中作目的状语,by a factor…为从句中的方式状语。

[4] 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.

实际上,许多需要大电流和大功率的设备往往都在特定的条件下工作,即由信号的电压决定响应的幅度,继而由设备吸收其所需要的电流而工作。

句中which 引导的定语从句修饰主语many devices,介词短语on the basis 作状语,that 引导的定语从句修饰the basis,which 引导的从句定语修饰the signal voltage,they need 为省略了关系代词的定语从句,修饰draw 的宾语the current。

本稿由Translover提供,仅供学习和交流适用,不得用于任何商业用途。如有任何疑问,请通过与我联系。

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