3 C H

Transcription

3 C H

CHAPTER 3 Resistive Network Analysis
C H A P TE R 3
R e s i s t i ve N e t w o r k An a l ys i s
本章的重點在於介紹各種電路分析的方法：
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Node voltage method.
Mesh current method.
Nodal and mesh analysis with controlled source.
Principles of superposition.
One-port network and equivalent circuits ＞＞ reduction of an
arbitrary circuit to equivalent circuit form （ Thevenin & Norton
equivalent circuit）
ë
Graphical and numerical techniques for analysis of nonlinear circuit
elements.
3. 1 T HE NODE V OLTAG E M E T HOD
The node voltage method is the most general method for analysis
of electrical circui ts.
This method is based on:
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Defining the voltage at each node as an independent variable.
One of the nodes is selected as reference node.
Each of the other node voltage is referenced to this reference
node.
Ohm’s law is applied between any two adjacent nodes in order
to determine the current flowing in each branch.
Each branch current is expressed in terms of one or more
node voltage.
下圖即是用來介紹如何利用 nodal voltage method 來定義 branch’
current:
i=
v a − vb
R
3-1
只要每一個 branch current 都已經清楚定義之後，就可以 Applying
KCL at each node:
∑i = 0
下圖即是介紹利用 KCL 的過程：
其中， By KCL: i1 − i2 − i3 = 0
va − vb vb − vc vb − vd
−
−
=0
R1
R2
R3
3-2
METHODOLOGY：Nodal voltage analysis method
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ë
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ë
Select a reference node （ usually ground ） . All other node
voltages will be referenced to this node.
Define the remaining n-1 node voltages as the independent
variables.
Apply KCL at each of the n-1 nodes, expressing each current
in terms of the adjacent node voltages.
Solve the linear system of n-1 equations in n-1 unknown.
Consider the circuit：
The reference（ ground） node is chosen to be the node at the
bottom of the circuit, and the direction of current flow is selected
arbitrarily（假設 i s 為正向電流）。
Applying KCL at node a: iS − i1 − i2 = 0
Applying KCL at node b: i2 − i3 = 0
Applying KCL at node c: i1 + i3 − iS = 0
(3.2)
(3.3)
(3.4)
The currents are expressed as functions of node voltage:
i1 =
va − vc
R1
(3.5)
3-3
i2 =
va − vb
R2
i3 =
vb − vc
R3
(3.6)
其中， eq.(3.4) 可以由 eq.(3.2)與 eq.(3.3)相減取得，此一結果正
驗證了： In a circuit containing n nodes, we can write at most n-1
independent equations 。
將 equation（ 3.5）與（ 3.6）代入（ 3.2）與（ 3.3）可以得到：
is −
v a va − v b
−
=0
R1
R2
va − v b vb
−
−=0
R2
R3
or
 1 + 1  va +  − 1 v b = is




 R1 R2 
 R2 
(3.7)
(3.8)
 − 1  va +  1 + 1 v b = 0 (3.9)




 R2 
 R 2 R3 
ððð Solve va and vb
E xa mple 3 . 1 No da l Ana ly s i s
P ro b le m :
Find all unknown current and voltages
G i ve n:
Source currents, resistor values.
I1 = 10mA 、 I2 = 50mA 、 R1 = 1KΩ 、 R2 = 2KΩ 、 R3 = 10KΩ 、 R4 = 2KΩ
3-4
F i nd :
All node voltages and branch current?
A nal ys i s :
The reference node is chosen to be the node at the bottom of the
circuit.
Applying KCL at node 1 and node 2:
At node 1 I1 −
At node 2
v1 − 0 v 1 − v 2 v1 − v 2
−
−
=0
R1
R2
R3
v1 − v 2 v1 − v 2 v 2 − 0
+
−
− I2 = 0
R2
R3
R4
1
1
1
 1 − 1  v 2 = I1
ððð  +
+
 v1 +  −

 R1 R2 R3 
 R2 R3 
1
1
1
 1
 1
−
+
+
−
v 1 + 
v 2 = −I2
 R2 R3 
 R2 R3 R3 
Solving v1 and v2
Solving branch currents.
3-5
E xa mple 3 . 2 No da l Ana ly s i s
P ro b le m :
Write the nodal equations and solve for the node voltages.
G i ve n:
Source currents, resistor values.
ia = 1mA 、 ib = 2mA 、 R1 = 1KΩ 、 R2 = 500Ω 、 R3 = 2.2KΩ 、 R 4 = 4. 7KΩ
F i nd :
All node voltages and branch current?
A nal ys i s :
circuit.
Applying KCL at node a and node b:
va v a − vb
−
=0
R1
R2
va − vb
vb vb
+ ib + −
−
=0
R2
R3 R4
ia −
Rewriting these equations to obtain a linear equation:
3-6
 1 + 1 v a +  − 1 vb = ia




 R1 R2 
 R2 
 − 1  va +  1 + 1 + 1 vb = ib




 R2 
 R2 R3 R4 
ððð代入 ia、ib、R1、R2、R3、R4 可以得到
3va − 2vb = 1
− 2vz + 2. 67vb = 2
3 . 1 . 1 No da l Ana ly s i s w i t h Vo lt a ge S o ur c e s
前面的幾個例子中，出現在 circuit 中的 source 都是 current
source，如果 current source 換成 voltage source，則要如何利用 nodal
analysis？
ððð以 nodal analysis 來解答不僅沒有問題，而且會更容易！
以下圖所示的電路為例：
因 node a 已知，即 va =vs ，所以，只需要兩個 nodal equations （ at
node b and node c）：
vs − vb vb v b − vc
−
−
=0
R1
R2
R3
(node b)
3-7
vb − vc
vc
+ is −
=0
R3
R4
(node c)
Rewriting these equations:
 1 + 1 + 1 v b +  − 1  vc = vs




R1
 R1 R2 R3 
 R3 
(3.11)
 − 1  vb +  1 + 1  vc = iS




 R3 
 R3 R4 
E xa mple 3 . 4 No da l Ana ly s i s w i t h Vo lt a ge S o ur c e s
P ro b le m :
Find the node voltage:
G i ve n:
Source current and voltage, resistor values.
I = −2mA 、 V = 3V 、 R1 = 1KΩ 、 R2 = 2KΩ 、 R3 = 3KΩ
F i nd :
Node voltages?
A nal ys i s :
circuit.
Applying KCL at node a and node b:
3-8
I−
va − 0 va − v b
−
=0
R1
R2
va − vb vb − 3
−
=0
R2
R3
代入 I、 R1 、 R2
ððð
1.5 va − 0.5 vb = −2
− 0.5 va + 0. 833vb = 1
ððððððva =-1 . 1 6 7 V
vb =0.5V
3-9
3 . 2 T H E ME S H C U R R E N T ME T H O D
相較於之前所介紹的 node voltage method 係以 node voltage 作為
independent variables， mesh current method 則是以 mesh current 作
為 independent variables。
Mesh current method？
以下圖所示為例，先建立電流方向，並以該方向建立跨越電阻的電
壓降極性：
只要 around a mesh 的電流方向確定，即可 apply KVL around the
mesh（如下圖）：
.
METHODOLOGY：Mesh current method
ë
Define each mesh current consistently, we shall always define
mesh current C.W., for convenience.（ Using mesh currents as the
independent variables）。
ë
Applying KVL around each mesh, expressing each voltage in term
of one or more mesh current.
3-10
ë
Solve the resulting linear system of equation with mesh currents
as the independent variables.（ The number of equations is
equal to the number of meshes in the circuit.）
再次強調，在 mesh analysis 中，電流方向選擇的一致性是非常重
要的！為了避免產生困擾，在本課程所介紹的 mesh current method
中，一律將 mesh current 的方向設定為順時針向！
Consider a two-mesh circuit：
ððð The mesh current i1 a n d i2 are used to generate two
equations in the two unknowns.
For mesh 1:
v 2 = (i1 − i2 )R2
ððð The complete expression for mesh 1 is：
v s − i1R1 − (i1 − i2 )R2 = 0
3-11
For mesh 2:
v 2 = (i2 − i1)R2
ððð The complete expression for mesh 2 is：
(i2 − i1)R2 + i2R3 + i2R4 = 0
(3.15)
Combining the equations for the two meshes, we obtain the
following system of equations:
(R1 − R2)i1 − R 2i2 = vs
− R2i1 + (R 2 + R3 + R 4)i2 = 0
ððð Solve mesh current
i 1 and i 2
3-12
E xa mple 3 . 5 Me s h Ana ly s i s
P ro b le m :
Find the mesh currents:
G i ve n:
Source voltage, resistor values.
F i nd :
Mesh currents?
A nal ys i s :
Assume clockwise mesh currents i 1 and i 2 .
Write the mesh equation (KVL):
V1 − i1R1 − V 2 − (i1 − i2)R2 = 0
− (i2 − i1)R2 + V 2 − i2R3 − V 3 − i2R4 = 0
Rearranging the linear system of equations, we obtain:
……
ððð Solve mesh current i1 and i 2
3-13
E xa mple 3 . 6 Me s h Ana ly s i s
P ro b le m :
Write the mesh current equations:
G i ve n:
Source voltage, resistor values.
F i nd :
Mesh current equations?
A nal ys i s :
Assume clockwise mesh currents i 1 、 i 2 and i 3 .
Applying KVL around mesh 1: V1 − R1(i1 − i3) − R2(i1 − i2) = 0
Applying KVL around mesh 2: − R2(i2 − i1) − R3(i2 − i3 ) + V 2 = 0
Applying KVL around mesh 3: − R1(i3 − i1) − R4i3 − R3(i3 − i2) = 0
Rearranging these equations in standard forms:
(3 + 8)i1 − 8i2 − 3i3 = 12
− 8i1 + ( 6 + 8 )i2 − 6i3 = 6
− 3i1 − 6i2 + (3 + 6 + 4)i3 = 0
3-14
3 . 2 . 1 Me s h Ana ly s i s w i t h C ur r e nt S o ur c e s
之前所介紹（指 mesh current analysis）的電路中，似乎僅有 voltage
source，但是若電路中出現 current source 的話，要如何處理？
以下圖所示的電路為例，吾人還是和之前的分析一樣，把 mesh current 仙
界定清楚：
只是其中的 mesh current 必須滿足下列關係：
i1 − i2 = 2 A
(3.17)
至於跨越 current source 的電壓降？＞＞ vx
If the unknown voltage across the current source is vx ；
Application of KVL around mesh 1 and mesh 2:
10 − 5i1 − vx = 0
v x − 2i2 − 4i2 = 0
(3.18)
(3.19)
Substituting eq.(3.19) in eq.(3.18), and using eq.(3.17), we obtain
the system of equations:
5i1 + 6i2 = 10
− i1 + i2 = −2
ððð
i 1 =2 A
i2 =0A
3-15
E xa mple 3 . 7 Me s h Ana ly s i s w i t h C ur r e nt S o ur c e s
P ro b le m :
Find the mesh currents:
G i ve n:
Source current and voltage, resistor values.
F i nd :
Mesh currents?
A nal ys i s :
Assume clockwise mesh currents i 1 、 i 2 and i 3 .
For mesh 1: i1 = I
Applying KVL around mesh 2 and mesh 3:
− R2(i2 − i1) − R3(i2 − i3) + V = 0
− R1(i3 − i1) − R4i3 − R3(i3 − i2) = 0
ððð
i 2 =0 . 9 5 A
mesh 2
mesh 3
i3 =0.55A
3-16
3. 3 NO DAL AND ME S H ANALY S IS W ITH
C O N TR O L L E D S O U R C E S
當電路中出現 controlled source 時，怎麼辦？
ë
ë
Treat it as an ideal source and write the node and mesh
equations accordingly .
Write an equation relating the dependent source to one of the
circuit voltage or current（ constrain equation） .
ððð This constraint equation can be substi tuted in the set of
equations obtained by the techniques of nodal and mesh analysis.
ððð
unknown.
These equations can subsequently be solved for the
Consider（ A simplified model of a bipolar transistor amplifier）
Two nodes are easily recognized, and therefore nodal analysis is
chosen as the preferred method.
Applying KCL at node 1:
1
1
iS = v1
+ 
 RS Rb 
Applying KCL at node 2:
β ib +
v2
=0
Rc
Where i b is obtained by means of a simple current divider.
3-17
ib = is
1 / Rb
Rs
= is
1/ Rb + 1 / Rs
Rb + Rs
ððð
1
1
is = v 1 + 
 Rs Rb 
− βi s
ððð
Rs
v2
=
Rb + Rs Rc
Solve v1 and v2
E xa mple 3 . 8 Ana ly s i s w i t h D e pe nde nt S o ur c e s
P ro b le m :
Find the node voltages:
G i ve n:
Source current, resistor
relationship: v X = 2 × v 3 .
values,
F i nd :
Unknown node voltage v?
A nal ys i s :
Applying KCL to node v:
vx − v
v − v3
+I−
=0
R1
R3
Applying KCL to node v3 :
3-18
dependent
voltage
source
v − v3 v3
−
=0
R2
R3
Substituting the dependent source relationship into the first equation,
we obtain a system of equations in the two unknown v and v3 :
 1 + 1  v +  − 2 − 1 v 3 = I




 R1 R2 
 R1 R 2 
 − 1  v +  1 + 1 v 3 = 0




 R2 
 R 2 R3 
ððð
Solve v and v3
3-19
3 . 4 TH E P R IN C IP L E O F S U P E R P O S ITIO N
The principle of superposition applies to any linear system and
for a linear circuit may be stated as follows:
In a linear circuit containing N sources, each branch
voltage and current is the sum of N voltages and
currents each of which may be computed by setting all
but one source equal to zero and solving the circuit
containing that single source.
在多個電源電路中，某一支路電流或某一節點電壓等於各個電源單獨作用
於此網路時，在該支路的電流或該節點電壓的代數和。說穿了，重疊定理並非
一種深奧的理論，只不過是各個擊破而已！
以下圖所示的電路為例，該電路中包括兩個 sources：
i=
vb 1 + vb 2 vb1 vb 2
=
+
= ib1 + ib 2
R
R
R
The basic principles are used frequently in the analysis of circuits
（當電路中出現多個電壓源與電流源時）:
ððð Set a voltage source equal to zero
Replace it with a short circuit.
ððð Set a current source equal to zero
Replace it with a open circuit.
3-20
a 考慮第一個電源，移走其他電源，也就是將其他電源中的電壓源短路，
電流源斷路。
b.求出該電源對元件的效應（指電壓或電流而言）。
c.對電路中的每一個電源，重覆步驟 a 及 b 來處理。
d.電源分別計算完後，將所有求出的電流作相加減，極性方向相同為加，
否則為減。其所得的結果，即為全部電源對此元件的總效應。
3-21
E xa mple 3 . 9 P r i nc i ple o f S upe r po s i t i o n
P ro b le m :
Find the current i 2
Part 1: Zero the current source
i2 − V =
10 V
= 0. 909A
5Ω + 2Ω + 4Ω
Part 2: Zero the voltage source（ three parallel branches）
Using the current divider rule:
i2 − I =
ððð
1/ 6
× ( −2) = −0.909A
1/ 5 + 1/ 6
i = i2 − V + i2 − I = 0 A
3-22
3. 5 O NE - P O RT NE TW O RK S AND
RE Q UIVALE NT CIRCUITS
Electrical system 的各種表達方式：
從上述各種表達方式來看，不管是 LOAD 或 SOURCE，都可以將
它們看成是一種 a two-terminals device ，然後透過 i-v characteristics
來描述其性質，並表示如下：
This config uration is called a ONE-PORT network and is
useful for introducing the notation of equivalent circuits.
E xa mple 3 . 1 0 E qui v a le nt R e s i s t a nc e C a lc ula t i o n
P ro b le m :
Determine the source and load current
3-23
ððð Load circuit and equivalent load circuit.
REQ =
1
1
1
1
+
+
R1 R2 R3
ððð i =
vs
REQ
3 . 5 . 1 The v e ni n a nd No r t o n E qui v a le nt C i r c ui t s
這裡所要介紹的是一個在電路分析中非常重要的觀念，那就是「equivalent
circuit」。透過這個觀念，我們將可以把一個複雜的電路看成一個簡單且等值的
source 與 load circuits。
這個觀念下，有兩個非常重要的理論，分別是 Thevenin theorem 與
Norton theorem。
The Thevenin theorem:
As far as a load is concerned, any network composed of ideal
voltage and current sources, and of linear resistors, may be
represented by an equivalent circuit consisting of an ideal
voltage source, vT, in series with an equivalent resistance, R T.
在任一含有線性電阻和獨立電源的網路中，任何接於兩點間的電路，皆可
3-24
由戴維寧等效電壓 Eth 和戴維寧等效電阻 Rth 串聯而成。如下圖所示：
The Norton theorem:
As far as a load is concerned, any network composed of ideal
voltage and current sources, and of linear resistors, may be
represented by an equivalent circuit consisting of an ideal current
source, iN, in parallel with an equivalent resistance, R N.
在任一含有線性電阻和獨立電源的網路中，任何接於兩點間的電路，皆可
由諾頓等效電流 IN 和諾頓等效電阻 RN 並聯而成。其中，諾頓等效電流 IN 即為
在元件移去後兩端所能獲得的最大短路電流，而諾頓等效雷阻 RN 與戴維寧等
效電阻 Rth 的求法相同，即是將所有的電壓源短路，電流斷路後的等效電阻。
如下圖所示：
3 . 5 . 2 D e t e r mi na t i o n o f No r t o n o r The v e ni n
E qui v a le nt R e s i s t a nc e
Computation of equivalent resistance of a oneport network:
ë Remove the load.
ë Zero all independent voltage and current sources.
ë Compute the total resistance between load terminals, with the load
removed.
3-25
ë
ë
將欲探討的支路上的元件移走，成為開路。
將網路中所有的電壓源短路，電流源開路，求元件移走後網路兩端的
等效電阻 Rth。
RT = (R1 // R 2 ) + R3
3-26
E xa mple 3 . 11 The v e ni n E qui v a le nt R e s i s t a nc e
P ro b le m :
Find the Thevenin equivalent resistance
RT = [( (R1 // R2 ) + R3) // R4 ] + R5
E xa mple 3 . 1 2 The v e ni n E qui v a le nt R e s i s t a nc e
P ro b le m :
Find the Thevenin equivalent resistance
RT = ((R1 // R2 ) + R3) // R4
3-27
3 . 5 . 3 C o mput i ng t he The v e ni n Vo lt a ge
Thevenin voltage
v T is defined as:
The equivalent（ Thevenin） source voltage is equal to the opencircuit voltage present at the load terminals （ with the load
removed）。
也就是說，將電壓源、電流源置回原處，再應用各種解電路的方法求出元
件移走後，網路兩端的電壓，即為等效電壓 Eth。
By KVL:
v T = RT (0) + vOC = vOC
Computation of Thevenin voltage:
ë
ë
ë
ë
Remove the load, leaving the load terminals open-circuited.
Define the open-circuit voltage vo c across the load terminals.
Apply any preferred method (nodal analysis) to solve for voc .
The Thevenin voltage vT = voc .
To compute vo c , we disconnect the load, and immediately
observe that no current flows through R3 , since there is no closed
circuit connect at that branch. Therefore, vo c must be equal to the
voltage across R 2 .
3-28
ððð
v OC = vR 2 = vS
R2
R1 + R2
Consider the original circuit and its Thevenin equivalent:
iL = vS
R2
1
vT
=
R1 + R 2 (R3 + R1 // R 2 ) + RL RT + RL
3-29
E xa mple 3 . 1 3 The v e ni n E qui v a le nt Vo lt a ge
（ O pe n- C i r c ui t L o a d Vo lt a ge ）
P ro b le m :
Compute the open-circuit voltage, vOC
Since vb is equal to the reference voltage
va is equa l to the open-circuit voltage
By nodal analysis
Applying KCL at two nodes ( v and va )
12 − v v v − va
−
−
=0
R1
10
R2
v − v a va
−
=0
R2
R3
ððð Solve v and va
3-30
E xa mple 3 . 1 4 L o a d C ur r e nt C a lc ula t i o n by
Th e v e n i n E q u i v a l e n t M e t h o d
P ro b le m :
Compute the load current i by the Thevenin equivalent method.
Compute the Thevenin equivalent resistance
RT = R1 // R2
Compute the Thevenin equivalent voltage
By nodal analysis. Applying KCL at node va
v − va
va
+I−
=0
R1
R2
ððð v a = vOC = vT = 27 V
Assemble the Thevenin equivalent circuit
i=
vT
= 3A
RT + RL
3-31
3 . 5 . 4 C o mput i ng t he No r t o n C ur r e nt
Norton equivalent resistance = Thevenin equivalent resistance
Definition
Norton equivalent circuit is equal to the short-circuit current that
would flow were the load replaced by a short circuit.
將電壓源及電流源放回去，且將移去的負載元件兩端設定為短路，再應用
前述各種求解電路的方法，求出流過此短路線上的電流 iSC，此電流即為等效電
流 IN。
Where isc (flowing through the short circuit replacing the load) =
Norton current i N .
How to calculate i s c ?
以下圖所示的電路為例：
3-32
其中， Load 已經被短路取代了！並令 i 2 ＝ i s c
By mesh analysis method:
(R1 + R2 )i1 − R2iSC = vS
− R2i1 + (R2 + R3 )iSC = 0
By nodal analysis method:
vS − v
v
v
=
+
R1
R2 RS
ððð v = vS
R2R3
R1R3 + R2R3 + R1R2
ðððððð iSC =
v
vSR2
= iN =
R3
R1R3 + R2R3 + R1R2
METHODOLOGY: Computing the Norton Current
ë Replace the load with a short circuit.
ë Define the short circuit current, iSC, to be the Norton equivalent
current.
ë Apply any preferred method(e.g., nodal analysis) to solve for i SC.
ë The Norton current is iN= i SC .
3-33
E xa mple 3 . 1 5 No r t o n E qui v a le nt C i r c ui t
P ro b le m :
Determine the Norton current and the Norton equivalent.
Compute the Thevenin (Norton) equivalent resistance:
RT = RN = R1// R2 + R3 = 4Ω
Compute the Norton current:
By nodal analysis.
Applying KCL at node 1 and node 2:
I−
v1
−i=0
R1
i−
v2 v2
−
=0
R2 R3
(Node 1 )
(Node 2)
Substitute v2 -V=v1 in the first equation (node 1):
3-34
I−
v2 − V
−i = 0
R1
ððð Solve v2 and I
ðððððð iN = iSC =
v2
= 1. 5A
R3
3-35
3 . 5 . 5 S o ur c e Tr a ns f o r ma t i o ns
Source transformation is a procedure that may be very useful in
the computation of equivalent circuits.
在之前的 Thevenin 與 Norton 定理介紹中，我們得知 any one-port
network can be represented by a voltage sources in series with a
resistance, or by a current source in parallel with a resistance.
由上述結果中，我們可以進一步說：利用 vT = RTiN ，我們可以把呈
現 Thevenin equivalent form 的電路，換成 Norton equivalent form 。
而以下這個電路就成為看出 source transformation 效果的最好範例：
如此一來，因三個 resistors 是呈現並聯，使得 i SC 的計算更是直
接：
According to current divider:
iSC = iN =
1/ R 3
vS
1/ R1 + 1/ R2 + 1/ R3 R1
3-36
註：下左圖為 Thevenin equivalent form 電路。
註：下右圖則為 Norton equivalent form 電路。
E xa mple 3 . 1 6 S o ur c e Tr a ns f o r ma t i o ns
P ro b le m :
Compute the Norton equivalent circuit using source transformations
Using source transformation technique:
ððð
其中， the branch consisting of V1 and R1 can be replaced by V1 /R1
（ Norton current source ） and R1 （ Equivalent resistance ） . The
series branch between node a’ and b’ can be replaced by V 2 /R 3
（ Norton current source） and R 3 （ equivalent resistance） .
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ððð
The equivalent resistance （ each current source is replaced by an
open circuit）： RT = R1// R2 // R3 + R4 = 200Ω .
The Norton current iN =0.5 -0.025-0.5=-0.025A
3 . 5 . 6 E xpe r i me nt a l D e t e r mi na t i o n o f The v e ni n
a nd No r t o n E qui v a le nt s
雖然章節的標題是要透過實驗的方法來決定 Thevenin and Norton
equivalents ，但其真正的目的是要去探討 Thevenin 或 Norton
quivalent circuits 後面所代表的 complex 或 unknown circuits。
之前的探討中，吾人已知 Thevenin voltage v T 為 open-circuit
voltage，而 Norton current i N 為 short-circuit current，因此，只要 vT
與 i N 可以量出來， Thevenin resistance 即可求出：
3-38
RT =
vT
iN
但問題是 vT 與 i N 如何量測？
ë
ë
利用 ammeter 去量測 short-circuit current（相關於 Norton
current i N）。
利用 voltmeter 去量測 open-circuit voltage （相關於
Thevenin voltage v T ） .
其中，令人注意的是量測者本身的內部電阻 r m。
因內部電阻的存在，使得量出的 i SC 與 vOC 不是 ideal current，也不
是 ideal voltage；然而其間的關係為何？
rm 

iN =" iSC "  1+

 RT 
 RT 
v T =" v OC " 1 +

rm 

Where r m=0 fo r ideal ammeter ,r m=∞ for ideal voltmeter.
3-39
FOCUS ON MEASUREMENTS
Experimental Determination of Thevenin Equivalent Circuit
Measured vOC=6.5V
Measured i SC=3.75mA
rm=15Ω
The unknown circuit is replaced by its Thevenin equivalent, and is
connected to an ammeter for a measurement of the short-circuit
current, and then to a voltmeter for the measurement of the opencircuit voltage.
vT = vOC =6.5V
rm 

iN = iSC 1 +

 RT 
RT =
v T vOC
=
− rm
iN
iSC
3-40

3 C H

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