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Electric Circuit
Chapter 3
(3.1, 3.2)
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Chapter 3
Applications of Resistive Circuits
3.1 Real Sources and Power Transfer
3.2 Amplifier Models
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3.1 Real Sources and Power Transfer
• Real sources (e.g. batteries and generators) differ from ideal
sources in two respects:
(1) Real sources cannot deliver unlimited amount of power
(2) Real sources dissipate power internally.
Real Voltage Source Model
𝑣 =𝑣 −𝑅 𝑖
𝑣 : source voltage (open-circuit voltage)
𝑅 : source resistance
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Real Current Source Model
𝑖 = 𝑖 − 𝑣/𝑅
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𝑖 : source current (short-circuit current)
𝑅 : source resistance
• The models represent only the terminal behavior of real sources.
• Neither 𝑣 nor 𝑖 remains constant at the terminals of a real source.
• No theoretical distinction between the two types of sources operating in
their linear regions.
• The practical distinction between the two types of sources emerges
when a variable load is connected.
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Loading Effects
𝑣 =𝑣 −𝑅 𝑖 =
𝑅
𝑣
𝑅 +𝑅
𝑅 << 𝑅  𝑣 ≈ 𝑣
Criteria for a good voltage source?
Good volage source : small internal resistance
𝑖 = 𝑖 − 𝑣/𝑅 =
𝑅
𝑖
𝑅 +𝑅
𝑅 ≫𝑅 𝑖≈𝑖
Criteria for a good current source?
Good current source : large internal resistance
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Example
A battery: 𝑣 = 6 𝑉 when 𝑖 = 0 & 𝑣 = 5.8 𝑉 when 𝑖 = 0.05 𝐴
Determine the condition under which it acts like an ideal voltage source
𝑅 =−
∆𝑣
6 − 5.8
=−
= 4Ω
∆𝑖
0 − 0.05
The model holds for 𝑖 ≪
𝑣
= 1.5 𝐴
𝑅
Ideal condition: 𝑣 ≈ 𝑣 if 𝑅 ≫ 𝑅 (4Ω)
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Example
Find the current 𝑖
𝑖=
40 + 20 𝑉
= 4𝑚𝐴
5 + 10 𝑘Ω
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Power Transfer and Efficiency
𝑖 = 𝑣 ⁄(𝑅 + 𝑅 )
𝑃 =𝑅 𝑖 =
𝑅
𝑣
(𝑅 + 𝑅 )
=
𝑃 =𝑅 𝑖 =
=
1
1 + 𝑅 ⁄𝑅
(
𝑣
)
𝑅
𝑅
𝑣
(𝑅 + 𝑅 )
𝑅 ⁄𝑅
1 + 𝑅 ⁄𝑅
(
𝑣
)
𝑅
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𝑑𝑃
(𝑅 + 𝑅 ) 𝑅 − 2(𝑅 + 𝑅 )𝑅
𝑅 −𝑅
=
𝑣 =
𝑣
𝑑𝑅
(𝑅 + 𝑅 )
(𝑅 + 𝑅 )
𝑃 is maximum
𝑃
=
= 0 (i.e. 𝑅 = 𝑅 )
𝑅
𝑣
(𝑅 + 𝑅 )
=
𝑣
4𝑅
Maximum power transfer theorem
If a source has fixed nonzero resistance
𝑅 , then maximum power transfer to a
load resistance requires 𝑅 = 𝑅 .
(此時 𝑃 = 𝑃 =
)
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Power Transfer Efficiency
𝐸𝑓𝑓 ≜
𝑃
𝑃 +𝑃
When there is maximum power transfer (i.e. 𝑃 = 𝑃
𝑅 =𝑅
𝑣=
𝑣
2
=
)
𝑃 =𝑃
𝐸𝑓𝑓 = 0.5 = 50%
• Electric utility companies would not strive for maximum power transfer. They
seek higher power transfer efficiency by making 𝑃 as small as possible.
• We want maximum power transfer in applications where voltage or current
signals are used to convey information rather than to deliver large amount
of power.
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3.2 Amplifier Models
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Amplifier:
An electronic device which enlarges the variation of an electrical signal.
Voltage Amplifier
Transfer curve:
vout(t) = Av vs(t)
Av: overall voltage amplification
Av > 1: typical (non-inverting) amplifier
Av < - 1: inverting amplifier (反相放大器)11
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CKT Model of Voltage Amplifier
source
amplifier
load
Input resistance Ri
- amplifier may draw current from the source
- vin may differ from vs
VCVS with gain 
- represent amplification of vin
- VCVS represent open-ckt output voltage vc =  vin
Output resistance Ro
- account for power dissipation within amplifier
- output Thevenin network
- Thevenin equivalent resistance looking back into the output terminals with
the input source suppressed
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Impact of Internal Resistance in Volt Amplifier
- consider Av = vout / vs
- note: vin  vs, vout  vc, due to
loading at input and output
Chain expansion
- vout / vs = (vin / vs)  (vc / vin)  (vout / vc)
vin / vs = Ri / (Rs + Ri)
vc / vin = 
vout / vc = RL / (Ro + RL)
- Av = vout / vs = [Ri / (Rs + Ri)]    [RL / (Ro + RL)]
– Note: Av< due to loading effects
Requirement for good voltage amplifier (small loading effect)
Ri >> Rs
Av  
Ro << RL
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Circuit Model for Current Amplifier
– Input current through input resistance Ri
– Current amplification by CCCS with gain 
– Short-circuit output current ic =  iin
– Output Norton network with Thévenin equivalent output resistance Ro
Chain Expansion:
iin / is
ic / iin
iout / ic
Ai = iout / is = [Rs / (Rs + Ri)]    [Ro / (Ro + RL)] < 
Requirement of good current amplifier
Ri << Rs
Ro >> RL
Ai  
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Amplifiers in Cascade (串接)
– output of an amplifier driving input of another
– to acquire larger gains not achievable with a single amplifier
Example: Cascade Volt Amplifier
v
medical transducer
w/ voltage signal
vs = 0.1 𝑠𝑖𝑛 𝜔t (V)
Rs = 8 k
max. current = 10 A
strip-chart recorder
5 k load
available amplifier
10 V scale
Ri = 24 k, Ro = 1 k,  = -10
 |Av| < 10, |vout| < 1 for one amp.
cascade to get larger |Av|
goal: check the suitability
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v
𝐴 =

𝑣out 𝑣in 𝑣c1 𝑣
𝑣
𝑣out
=
×
×
× c2 ×
𝑣
𝑣
𝑣in 𝑣c1 𝑣
𝑣c2
24
24
5
 ( 10) 
 ( 10) 
 60
8  24
1  24
1 5
𝑣
= 60𝑣 = 6 𝑠𝑖𝑛 𝜔 𝑡 (𝑉)
𝑣
𝑖in =
= 3.125 𝑠𝑖𝑛 𝜔 𝑡 (𝜇𝛢)
(8 + 24) kΩ
check : within  10V
Both are OK for the transducer
check 𝑖in < 10 μA
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Suggested exercises:
A.B. Carlson, “Circuits - Engineering concepts and analysis of linear
electric circuits” Brooks/Cole.
Problem 3.3, 3.9, 3.12, 3.14, 3.16, 3.19, 3.21
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