chap 4 techniques of circuit analysis n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chap. 4 Techniques of Circuit Analysis PowerPoint Presentation
Download Presentation
Chap. 4 Techniques of Circuit Analysis

Loading in 2 Seconds...

play fullscreen
1 / 39

Chap. 4 Techniques of Circuit Analysis - PowerPoint PPT Presentation


  • 141 Views
  • Uploaded on

Chap. 4 Techniques of Circuit Analysis. C ontents. 4.1 Terminology 4.2 Introduction to the Node-Voltage Method 4.3 The Node-Voltage Method and Dependent Sources 4.4 The Node-Voltage Method: Some Special Cases 4.5 Introduction to the Mesh-Current Method

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Chap. 4 Techniques of Circuit Analysis' - hall-lee


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
chap 4 techniques of circuit analysis
Chap.4 Techniques of Circuit Analysis

Contents

4.1 Terminology

4.2 Introduction to the Node-Voltage Method

4.3 The Node-Voltage Method and Dependent Sources

4.4 The Node-Voltage Method: Some Special Cases

4.5 Introduction to the Mesh-Current Method

4.6 The Mesh-Current Method and Dependent Sources

4.7 The Mesh-Current Method: Some Special Cases

4.8 The Node-Voltage Method Versus the Mesh-Current Method

4.9 Source Transformations

4.10 Thévenin and Norton Equivalents

4.11 More on Deriving a Thévenin Equivalent

4.12 Maximum Power Transfer

4.13 Superposition

Objectives

1. 了解並能夠使用節點電壓法求解電路。

2. 了解並能夠使用網目電流法求解電路。

3. 對於特定電路能夠決定節點電壓法或網目電流法何者是較佳的求解方式。

4. 了解電源轉換,並能夠使用它來求解電路。

5. 了解戴維寧和諾頓等效電路的觀念,並能針對電路建立等效電路。

6. 了解電阻負載最大功率轉移之情況,並能計算滿足此情況之負載電阻值。

4 1 terminology
4.1 Terminology

節點

必要節點

路徑

分支

必要分支

迴路

網目

平面電路

Node

Essential node

Path

Branch

Essential branch

Loop

Mesh

Planar circuit

A point where two or more circuit elements join

A node where three or more circuit elements join

A trace of adjoining basic elements with no elementsincluded more than once

A path that connects two nodes

A path which connects two essential nodes without passingthrough an essential node

A path whose last node is the same as the starting node

A loop that does not enclose any other loops

A circuit that can be drawn on a plane with no crossingbranches

Nonplanar

Planar

ex 4 1 identifying node branch mesh and loop
EX4.1 Identifying Node, Branch, Mesh and Loop

Node

Essential node

Branch

Essential branch

Mesh

a, b, c, d, e, f, and g.

b, c, e, and g.

v1, v2, R1, R2, R3, R4, R5, R6, R7, and I .

v1 –R1 , R2 –R3 , v2 –R4 , R5, R6, R7, and I .

v1 –R1 –R5 –R3 –R2 , v2 –R2 –R3 –R6 –R4 , R5 –R7 –R6 , and R7 –I .

Find two paths that not loops or essential branches.

Find two loops that not meshes.

simultaneous equations how many
Simultaneous Equations—How Many?

電路中未知電流數= 分支數b。

若節點數= n,分支數 = b,則可套用KCL於(n-1)個節點上,或KVL於b-(n-1)個迴路或網目上,以列出方程式來求解。

若必要節點數= ne,必要分支數 = be ,亦可套用KCL於(ne -1)個必要節點上,或KVL於be -(ne -1)個迴路或網目上,以列出較少方程式來求解。

b:

c:

e:

Essential

nodes:

KCL

Meshes: KVL

節點電壓法使用(ne -1)條KCL方程式,而

網目電流法使用be -(ne -1)條KVL方程式來描述電路。

4 2 introduction to the node voltage method
4.2 Introduction to the Node-Voltage Method

節點電壓(node voltage) 為自參考點到非參考節點的電壓升。

節點電壓法之求解步驟:

1. 找出必要節點。

2. 選取有最多分支的節點為參考點。

3. 定義節點電壓。

4. 套用KCL,針對非參考點之節點列出方程式。

▼: reference node

Node 1:

Node 2:

ex 4 2 using the node voltage method
EX4.2 Using the Node-Voltage Method

1

a)

Node 1:

Branch currents:

b)

4 3 the node voltage method and dependent sources
4.3 TheNode-Voltage Method and Dependent Sources

當有相依電源時,需補上相依電源控制變數的限制方程式。

EX4.3 A Circuit with Dependent Source

1

2

Node 1:

Node 2:

相依電壓源控制變數之限制方程式:

8

4 4 the node voltage method some special cases
4.4 TheNode-Voltage Method: Some Special Cases

Case A.

當節點電壓值 v1 = 100 V時,其KCL方程式不需列出,

只需節點2 之KCL方程式。

Node 2:

9

4 4 case b
4.4 Case B

Supernode (超節點)

1

Case B.

2

3

當電壓源兩端為必要節點且皆非參考點時,可引進一自定未知電流(如圖中之i),然後於方程式求解過程中將它消去。

Node 2:

Node 3:

( +

The Concept of a Supernode

當電壓源兩端為必要節點且皆非參考點時,可將其兩端節點合併為超節點,而此超節點也符合克希荷夫電流定律(KCL)。

10

4 4 case b contd
4.4 Case B Contd.

Supernode

Node 1:

Supernode:

電壓源限制:

相依電源控制變數:

11

node voltage analysis of the amplifier circuit
Node-Voltage Analysis of the Amplifier Circuit

Node a:

Supernode:

相依電源控制變數:

電壓源限制:

12

4 5 introduction to the mesh current method
4.5 Introduction to the Mesh-Current Method

網目電流(mesh current) :

只存在於網目周圍的假想電流,可能無法以安培計量測。

實際的分支電流可由網目電流加減組合表示而成。

網目電流法可用be- (ne -1)條KVL方程式來描述電路。

[7-(4-1)] = 4

網目電流法之求解步驟:

1. 定義網目電流。

2. 套用KVL於各網目上建立 be- (ne -1)個聯立方程式。

3. 求解網目電流。

4. 由網目電流求解分支電流。

13

evolution of the mesh current technique
Evolution of the Mesh-Current Technique

KCL:

KVL:

以網目電流表示分支電流:

將(ne -1)條KCL方程式帶入be- (ne -1)條KVL方程式

可去除(ne -1)個分支電流未知數

完全

相同

指定網目電流,直接列出be- (ne -1)條KVL方程式

網目KVL方程式

14

ex 4 4 using the mesh current method
EX4.4 Using the Mesh-CurrentMethod

b-(n-1)=7-(5-1)=3

a)

Mesh a:

Mesh b:

Mesh c:

b)

15

4 6 the mesh current method and dependent sources
4.6 TheMesh-Current Method and Dependent Sources

當有相依電源時,需補上相依電源控制變數的限制方程式。

EX4.5 A Circuit with Dependent Source

Find the power dissipated in the 4 resistor.

b-(n-1)=6-(4-1)=3

Mesh 1:

Mesh 2:

Mesh 3:

相依電源控制變數:

16

4 7 the mesh current method some special cases
4.7 TheMesh-Current Method: Some Special Cases

Case A.

當電流源僅有一個網目電流通過時,其KVL方程式不需列出,

直接指定該網目電流。

Mesh 1:

Mesh 2:

Mesh 3:

17

4 7 case b
4.7 Case B

Case B.

當電流源有兩個網目電流通過時,可引進一自定未知電壓(如圖中之v),然後於方程式求解過程中將它消去。

Mesh a:

Mesh c:

( +

The Concept of a Supermesh

當電流源有兩個網目電流通過時,可將其兩網目合併為超網目,而此超網目也符合克希荷夫電壓定律(KVL)。

18

4 7 case b contd
4.7 Case B Contd.

Case B.

Supermesh:

電流源限制:

Mesh b:

19

mesh current analysis of the amplifier circuit
Mesh-Current Analysis of the Amplifier Circuit

Supermesh:

電流源限制:

Mesh b:

相依電源控制變數:

20

4 8 the node voltage method versus the mesh current method
4.8 TheNode-Voltage Method Versus the Mesh-Current Method

使用節點電壓法或網目電流法考慮因素:

◆ 何者聯立方程式數目較少?

◆ 有超節點嗎?考慮節點電壓法。

◆ 有超網目嗎?考慮網目電流法。

◆ 要求解的電路部分適用何者解法?

Find the power dissipated in the 300 resistor in the following circuit.

網目電流法:

節點電壓法:

be-(ne-1) = 8-(4-1) = 5

ne-1 = 4-1 = 3

21

ex 4 6 understanding the node voltage method v s mesh current method
EX4.6 Understanding the Node-Voltage Method v.s. Mesh-Current Method

a)

Supernode:

b)

Node a:

Node b & 相依電源控制變數:

電壓源限制式與相依電源控制變數:

Node c:

Node 2:

22

ex 4 7 comparing the node voltage and mesh current method s
EX4.7 Comparing the Node-Voltage and Mesh-Current Methods

a)

Node o:

Node a:

Node b:

節點電壓法:

ne-1 = 4-1 = 3

相依電源

控制變數:

網目電流法:

be-(ne-1) = 6-(4-1) = 3

b)

Supermesh:

2個電流源限制式與相依電源控制變數:

23

4 9 source transformations
4.9 Source Transformations

電源轉換法(source transformation):電壓源子電路可與電流源子電路互換。

24

special source transformation techniques
Special Source Transformation Techniques

對端點a, b 而言,

並聯於電壓源之電阻Rp以及

串聯於電流源之電阻Rs是沒有作用的。

ex 4 9 using special source transformation techniques
EX4.9 Using Special Source Transformation Techniques
  • Use source transformations
  • to find the voltage vo.
  • b) Find the power developed by the 250-V source.
  • c) Find the power developed by the 8-A source.

+

-

b)

(參考原圖)

a)

(supplied 2800W)

c)

27

(參考原圖)

(supplied 480W)

4 10 th venin and norton equivalents
4.10 Thévenin and Norton Equivalents

戴維寧(Thévenin)和諾頓(Norton)等效(Equivalent)電路

適用於任何線性(Linear)電路。

Thévenin equivalent circuit

戴維寧等效電路(Thévenin equivalent circuit):

一個獨立電壓源VTh串聯一個電阻RTh,用以取代任何由電源和電阻器構成之電路。

求解VTh:令a, b端點開路,求出由a, b端看進去的電壓值。

求解RTh:令a, b端點短路,求出流經a, b端電流isc,而

RTh = VTh / isc。

另解RTh:令所有獨立電壓源短路,所有獨立電流源開路,求解由a, b端點看進去的等效電阻值。

28

finding a th venin equivalent
Finding a Thévenin Equivalent

a, b 端開路,求解VTh = v1 = vab。

a, b 端短路,求解isc。

29

the norton equivalents
The Norton Equivalents

諾頓等效電路(Norton equivalent circuit):

一個獨立電流源IN並聯一個諾頓等效電阻RN,用以取代任何由電源和電阻器構成之電路。

諾頓等效電路可直接從戴維寧等效電路,施以電源轉換求得。

即 IN = isc= VTh / RTh,RN= RTh。

注意電流方向與電壓極性

30

4 11 more on deriving a th venin equivalent
4.11 More on Deriving a Thévenin Equivalent

求解RTh亦可先將所有獨立電源拿掉(電壓源短路且電流源開路),然後由端點往回看求出等效電阻值。

若電路有相依電源,首先拿走所有獨立電源,然後在a, b端掛上一個測試用電壓源或電流源,該測試電源之電壓除以電流就是RTh。

32

ex 4 11 finding the th venin equivalent using a test source
EX4.11 Finding theThévenin Equivalent Using a Test Source

vT為測試電壓源,vT/ iT就是RTh。

33

4 12 maximum power transfer
4.12 Maximum Power Transfer

The derivative is zero and

pis maximized when

CONDITION FOR MAXIMUM POWER TRANSFER

(最大功率轉移時的條件)

最大功率轉移量

35

ex 4 12 calculating the condition for maximum power transfer
EX4.12 Calculating the Condition for Maximum Power Transfer

Also,

The percentage of the source power delivered to the load is

36

4 13 superposition
4.13 Superposition

重疊原理(superposition):在線性系統中,將各別獨立電源造成之響應相加,就可以得到總響應。

1)

37

4 13 contd
4.13 Contd.

v3

v4

2)

將電壓源驅動之電流i’和電流源驅

動之電流i”相加,可得總電流。

38