电工学原理及应用(Electrical Engineering)经典双语详解讲义 (4)

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电气工程及其自动化专业英语第一章课文翻译

电气工程及其自动化专业英语第一章课文翻译

第一章第一篇sectiongTwo variables u(t) and i(t) are the most basic concepts in an electric circuit,they characterize the various relationships in an electric circuitu(t)和i(t)这两个变量是电路中最基本的两个变量,它们刻划了电路的各种关系。

Charge and CurrentThe concept of electric charge is the underlying principle for explaining all electrical phenomena. Also, the most basic quantity in an electric circuit is the electric charge。

Charge is an electrical property of the atomic particles of which matter consists, measured in coulombs (C)。

电荷和电流电荷的概念是用来解释所有电气现象的基本概念.也即,电路中最基本的量是电荷.电荷是构成物质的原子微粒的电气属性,它是以库仑为单位来度量的。

We know from elementary physics that all matter is made of fundamental building blocks known as atoms and that each atom consists of electrons, protons, and neutrons. We also know that the charge e on an electron is negative and equal in magnitude to 1。

电路原理双语课件

电路原理双语课件

The transfer function can be written as
The transfer function H(ω) of a circuit is the frequency-dependent ratio of a phasor output Y(ω) (an element voltage or current) to a phasor input X(ω) (source voltage or current). Thus,
Vo 1/ jωC 1 H (ω ) = = = Vs R + 1/ jωC 1 + jω RC
The magnitude and phase of H(ω) are
Amplitude response 幅频特性
ω H= , φ = − tan 2 ω0 1 + (ω / ω 0 )
1
−1
where ω0 = 1/RC. At ω = 0, H = 1 and φ = 0. At ω = ∞, H = 0, and φ = − 90°. Also, at ω = ω0, H = 1/ 2 and φ = − 45°.
14.1 INTRODUCTION
In our sinusoidal circuit analysis, we have learned how to find voltages and currents in a circuit with a constant frequency source. If we let the amplitude of the sinusoidal source remain constant and vary the frequency, we obtain the circuit’s frequency response. The frequency response of a circuit is the variation in its behavior with change in signal frequency. The sinusoidal steady-state frequency responses of circuits are of significance in many applications, especially in communications and control systems.

电工学原理及应用(Electrical Engineering)经典双语详解讲义

电工学原理及应用(Electrical Engineering)经典双语详解讲义

Alternating Current
On the other hand, a current that varies with time, reversing direction periodically, is called alternating current, abbreviated as AC.

4Ω 2Ω 3Ω a 6Ω
10V
Congsidering current I, N=4,B=6,L=7
N=3,B=5,L=6
1.4 Kirchhoff’s current LAW (KCL)
The net(净,总的) current a node is zero.
I= 0


entering
• Electrical Engineering Subdivisions
• • • • • • • • Communication systems Computer systems Control systems Electromagnetics Electronics Photonics Power systems Signal processing
P1 P2 P3 P4 P5 90 20 120 90 80 0
功率平衡!电路中所有元件的功率之和为 0 ! 常用作对分析结果的检验准则。
•About Kirchhoff
古斯塔夫· 罗伯特· 基尔霍夫(Gustav Robert Kirchhoff )
——/wiki/Gustav_Kirchhoff
Box3
P3 U 3 I 3 60V ( 2)A 120W<0
U1 1
U2 2 U5

电工学原理及应用(Electrical Engineering)经典双语详解讲义 (5)

电工学原理及应用(Electrical Engineering)经典双语详解讲义 (5)

U d
U Ii
f
(1 A F)ri
2) 并联负反馈 使电路的输入电阻降低
ii
id
+
ui if
ri

ri f
1
ri A0 F
无负反馈时:
ri
ui id
+
有负反馈时:

rif
ui ii
ui id i f
1
1 AF
ri
6.对输出电阻的影响
1) 电压负反馈使电路的输出电阻降低
rof
1
ro A0 F
uo 电压负反馈具有稳定输出电压的作用
2. 并联电压负反馈
if RF
i1
+ ui –
R1 R2
id
– +

++
RL
uo –
设输入电压 ui 为正, 差值电流 id = i1 – if if 削弱了净输入电流(差 值电流) ——负反馈
反馈电流
if
uo Rf
取自输出电压——电压反馈
反馈信号与输入信号在输入端以电流的形式比
反之若比较的结果使净输入信号增大,因而输 出信号也增大,称为正反馈(positive feedback) 。
17.1.2 负反馈与正反馈的判别方法 反馈类型的判别步骤
1) 找出反馈网络(一般是电阻、电容) 2) 采用瞬时极性法判别正负反馈 3) 判别是交流或直流反馈? 4) 是负反馈!判断是何种类型的负反馈?
电压负反馈具有稳定输出电压的作用,
即有恒压输出特性,故输出电阻降低。
2) 电流负反馈使电路的输出电阻提高
rof (1 A0F )ro
电流负反馈具有稳定输出电流的作用,

电气工程及其自动化专业英语

电气工程及其自动化专业英语

专业英语电路基础characterize描绘…的特征,塑造人物,具有….的特征property 性质,财产equal in magnitude to 在数量(数量级)上等同于convert 转换converter 转换器time rate 时间变化率mathematically 从数学上来讲differen tiate v 区分,区别in honor of 为纪念某人name in honor of为纪念某人而以他命名electromotive force (e m f )电动势voltaic battery 伏打电池,化学电池an element 一个电器元件interpret 口译,解释,说明the potential at point a with respect to point b is点a关于点b的电势Potential difference/voltage 电势差/电压expend 花费,消耗instantaneous 瞬时的,促发的passive sign convention 关联参考方向the law of conservation of energy 能量守恒定律reference polarity 参考极性electron 电子electronic 电子的electric 电的,电动的time-varying 时变的constant-valued 常量的metal lic 金属的be due to 是因为,由于,归功于building block 模块Coulomb库伦,Ampere安培,joule焦耳,Volt伏特,Watt 瓦特,work 功变量u(t),i(t)是电路中最基本的概念。

他们描述了电路中的各种关系。

电荷量的概念是解释电现象的基本原理,电荷量也是电路中最基本的量。

电荷也是构成物质的原子的电器属性,量纲是库伦。

我们从初等物理可以得知所有物质是由基本组成部分原子组成,而原子又包括电子(electron),质子(proton)和中子(neutron)我们都知道电荷e是带负电的电子,在数量上等于1.60210*1019 C, 而质子携带同等电荷量的正电荷,相同数量的质子,电子使原子呈现电中性(neutrally charged)。

电气专业英语section 1-5中英

电气专业英语section 1-5中英

由于节点3连接有4条支路,所以选择其作为参考点, 用所示的接地符号标明。
The voltage between node 1 and the reference node 3 is identified as u1, and u2 is defined between node 2 and the reference.

每个非参考点相对于参考点的电压既是节点电压。
It is common practice to select polarities so that the node voltages are positive relative to the reference node.

选择极性时,通常使节点电压相对于参考点为正。

理解了电路基本定律(欧姆定律和基尔霍夫定律)后
we are now prepared to apply these laws to develop two powerful techniques for circuit analysis:

我们准备应用这些定律得出两种电路分析的有效方法
nodal analysis, which is based on a systematic application of Kirchhoff's current law (KCL),
0.2u1 1.2u2 2
(1-20)
Solve the Eq.(1-19) and Eq. (1-20) to obtain the unknown node voltage u1 and u2, then any current or power in the circuit may now be found.

电气工程专业英语unit6解读

电气工程专业英语unit6解读
Power system automation
This involves the application of automation technology in the power system, such as SCADA, EMS, and DMS, to achieve real-time monitoring and control of the power system.
Article structure
VS
Unit 6涉及的词汇主要包括电气工程相关术语、工业自动化和智能制造领域的专业词汇,以及一些科技类常用词汇。
短语总结
文章中出现的短语主要包括与电气工程和工业领域相关的短语,如"power system","automation control","intelligent manufacturing"等。这些短语对于理解文章内容非常重要,需要学生重点掌握。
Electrical engineering also contributes to the progress of other fields, such as computer science, robotics, and nanotechnology. Electrical engineers design and develop the electronic components and systems that are used in these fields, making it possible for them to advance.
Circuit analysis
Circuit analysis is a fundamental skill for electrical engineers. It involves using mathematical models to analyze the behavior of electrical circuits. Circuit analysis is essential for predicting circuit performance, optimizing circuit design, and troubleshooting circuit problems.

电工学原理及应用(Electrical Engineering)经典双语详解讲义

电工学原理及应用(Electrical Engineering)经典双语详解讲义

N

+
内电场被加 强,少子的漂 移加强,由于 少子数量很少, 形成很小的反 向电流。
PN 结加反向电压时,PN结变宽,反向电流较小, 反向电阻较大,PN结处于截止状态。 温度越高少子的数目越多,反向电流将随温度增加。
•二极管的符号与结构Fra bibliotekuDD 外形 iD 国标符号 D 美标符号
根据PN结构造面的特点,二极管分为点接触型、面接触 型、键型、合金型、扩散型、台面型、平面型、合金扩散型、 外延型、肖特基型等。
I D2
D1承受反向电压为-6 V
3
4mA
Ex.8.8 Assume ideal diodes, find the diode states for the circuits shown in Fig. 10.17.
D1 is on;
D2 is off;
D3 is off;D4 is on.
Electrical Engineering and Electronics II
Chapter 8 Diodes
4 Course Hours
Chapter 8 Diode
8.1 Basic Diode Concepts 8.2 Load-line analysis of diode circuits 8.3 Zener-diode voltage-regulator circuits 8.4 Ideal-diode model 8.5 Piecewise-linear diode models 8.6 Rectifier circuits 8.7 Wave-shaping Circuits 8.8 Linear Small-signal Equivalent Circuits
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How to get initial value
S t=0 C
Exa 4.2
R2 L
+ U -
R1
(a)
Knowns:before circuit changing,C、L have no energy; Find:the currents and voltages of all elements。
utransient C
E
New steady state
Old steady state
t
Why the transient response happens? Resistive circuit
K + E _ R t=0 I
I
No transient
•Resistor is a energy-consumption element, current is proportional to voltage, no transient response will happen even if changing source
+ u (0+) –C t=0+
uC (0 ) 8 iR (0 ) 1m A R2 8
R 8k 2

How to get initial value
•Exercise 2: Given by R1=4Ω, R2=6Ω, R3=3Ω, C=0.1µ F, L=1mH, US=36V, switch S is closed for a long time. Open the switch S when t=0, how to get the initial values of all elements?
•Homework
•P4.2 •P4.6 •P4.8
4.3 First-order RC Circuits

First-order circuit Only one (equivalent) capacitor or inductor is included in a linear circuit. Equivalent circuit of First-order circuit Two parts: one (equivalent) capacitor or inductor; a two terminal network with resistance and sources. L
Answer: vx =5V, ix = 1A
t>>0
Exercise 4.3 Find steady-state values of labeled currents and voltages for t>>0.
Answer: va =50V, ia = 2A
i1 = 2A, i2=1A, i3=1A
Solu (1)Brofore switch
uC (0 ), i L (0 )
uC (0 ) uC (0 ) 0
We know that uC (0 ) 0, i L (0 ) 0 By the law of circuit changing
L (0 ) L (0 ) 0
Example 4.2
S C R2
iC (0+ ) uC (0+) u2(0+_ )
+
+ U -
t=0 R1 (a) circuit L U
+
-
i1(0+ )
R1
+ + u _ 1(0+) _ uL(0+)
R2
iL(0+ )
(b) t = 0+ circiut
(2) By the circuit at t=0+, Find other unknowns
Energy
can not change instantly because of accumulating or decaying period.
1 2 Electric field energy(Wc Cu ) C 2 uC Change WC Charging or discharging
When t=0-, capacitor is considered as open circuit, we get equivalent circuit.
R 1 4k
8 uC (0 ) 12 8 V 48
12V
8k
u (0 ) C – t=0-

How to get initial value
Transients
•The time-varying currents and voltages resulting from the sudden application of sources, usually due to switching. •By writing circuit equations, we obtain integrodifferential equations. These equations can be converted to pure differential equations by differentiating w.r.t time. •The study of transients require us to solve differential equations .

N
or
N
C
4.3 First-order RC Circuits

According to Thevenin Law
N
L
or
N
C
iL + R U + uL L R U uC
iC C
-
-
பைடு நூலகம்

Differential equation of first-order RC circuit
iL + U R uL L + U R uC C
4.2 Initial state and steady state
t=0t=0
t=0+
t
Assume changing circuit when t=0, then t=0– is end point of old steady state; t=0+ is the start point of transient state.
R 1 4k 12V
R1 4k 12V 8k u (0–) C
K
t=0
i R
8k R 2 2mF
u C
8 uC (0 ) 12 8 V 48
u C ( 0 ) u C ( 0 ) 8V
substituting voltage source for uC(0+)
i (0+) R
gradually
K + _E R
uC
E C
uC
t
Energy
can not change instantly because of accumulating or decaying period.
1 2 Magnetic field energy (W LiL ) L 2 WL i Change
iC 、uL alter suddenly

How to get initial value
R 1 4k 12 V
K t=0 i R 8k R 2 2m F u C
Exercise 1: Assuming old circuit is in DC steady state before switch K is closed. how to get uC(0+),iR(0+)? Solution:
Answer: Uc(0)=12V, iL(0)=4A,iR(0)=2A
Conclusions
1. At the moment of changing circuit,uC、 iL can not alter suddenly, but ic, uL can change suddenly . 2. Before circuits changed, if energy-storage elements have no energy, just after circuits changed (t=0 +) the capacitance and inductance are viewed as short circuits and open circuit, respectively.
2. Replace inductances with short circuits.
3. Solve the remaining circuit using methods in chapter 2.
Example 4.1 Find steady-state values of vx and ix in this circuit for t>>0.
WL (0 ) WL (0 ) WC (0 ) WC (0 )
i L ( 0 ) i L ( 0 ) uC (0 ) uC (0 )
The law of changing circuit (换路定则)
From t=0–to t=0+,iL 、 uC can not change suddenly.
Electrical Engineering and Electronics II
Chapter 4 Transients
4 Course Hours
4.Transients
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