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ECE 3144 Lecture 23
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
1
Method of finding Thevenin equivalent Circuit
Number
of method
1
2
If the circuit
contains:
Thevenin
equivalent circuit
Resistors and
independent sources
only (case 1)
a)
Resistors and
independent and
dependent sources
(case 2) or case 1
a)
b)
b)
c)
3
Resistors and
dependent sources
only.
a)
b)
c)
Connect an open circuit between terminals a and b. Find
voc = vab the voltage across the open circuit
Deactivate the independent sources (replace independent
voltage sources with short circuits and independent current
sources with open circuits). Find RTH by methods
introduced in Chapter 2.
Connect an open circuit between terminals a and b. Find
voc = vab the voltage across the open terminals.
Connect a short circuit between terminals a and b. Find isc,
the current directed from a to b in the short circuit.
Calculate RTH = voc/isc.
Note that voc = 0
Connect a 1-A current source from terminal b to terminal a.
Determine vab
Then RTH = vab/1.
2
Method of finding Norton equivalent Circuit
Number
of
method
1
2
If the circuit
contains:
Norton
equivalent circuit
Resistors and
independent sources
only (case 1)
a)
Resistors and
independent and
dependent sources
(case 2) or case 1
a)
b)
b)
c)
3
Resistors and
dependent sources
only.
a)
b)
c)
Connect a short circuit between terminals a and b. Find isc,
the current directed from a to b in the short circuit.
Deactivate the independent sources (replace independent
voltage sources with short circuits and independent current
sources with open circuits). Find RN=RTH by methods
introduced in Chapter 2.
Connect an open circuit between terminals a and b. Find
voc = vab the voltage across the open terminals.
Connect a short circuit between terminals a and b. Find isc,
the current directed from a to b in the short circuit.
Calculate RN = RTH = voc/isc.
Note that isc = 0
Connect a 1-A current source from terminal b to terminal a.
Determine vab
Then RN = RTH = vab/1.
3
Maximum Power Transfer
• Many applications of circuits require that the maximum power
available from a source be transferred to a load resistor RL.
• Consider the circuit network A, terminated with load RL. A can
represent any circuit network, say power utility systems. Power
utility systems are designed to transport the power to the load
RL with the greatest efficiency by reducing the losses on the
power lines and power sources themselves.
• How to calculate the maximum power efficiency? We know that
the general circuit A can be reduced to its Thevenin (Norton)
circuit.
i
A
4
Maximum Power Transfer
•
For the given general circuit, we wish to what is the power delivered to the
load resistor RL.
2
p  i RL
•
Since current i is
•
vTH
RTH  R L
The power delivered to RL is
i 
2
vTH
RL
p
( RTH  RL ) 2
•
Since VTH and RTH are fixed for a given source, the power delivered is a
function of load resistor RL. To find RL that maximizes the power, we
differentiate the power with respect to RL:
2
2
dp ( RTH  RL ) 2 vTH
 vTH
RL (2)( RTH  RL )

0
4
dRL
( RTH  RL )
( RTH  RL ) 2  2RL ( RTH  RL )  0
RTH  R L
=>
=>
5
•
Maximum Power transfer
Confirm that RTH= RL gives the maximum power transfer instead of the
minimum power transfer. We know that
– If RL = 0 => p = i2RL = 0 => p is minimum in this case
– if RL =  =>p = vi = 0 => p is also minimum in this case.
– So RTH=RL gives maximum power transfer.
The maximum power achieved when RTH = RL
p max
•
•
2
2
vTH
RL
vTH


2
4 RL
( RL  RL )
The maximum power transfer theorem states that the maximum power
delivered by a source represented by its Thevenin circuit (Norton circuit) is
attained when the load is equal to the Thevenin (Norton) resistance RTH (RN).
The maximum power delivered is v2TH /4RL for Thevenin equivalent source
The power attained as RL varies is shown as
Power vs. load resistance
p
p max
Where
x

4x
(1  x ) 2
RL
RTH
1
0.8
p / pmax
•
0.6
0.4
0.2
0
0
0.5
1
1.5
RL / RTH
2
2.5
3
6
Efficiency of power transfer
•
The efficiency of power transfer is defined as the ratio of the power delivered
to the load, pout, to the power supplied by the source, pin. Therefore we have
the efficiency  as
 
•
•
•
p out
p in
In the ideal source case, all the power supplied by the source is absorbed by
the load =>=1.
For practical sources we have discussed, the maximum power transfer
happened when RL = RTH.
For the maximum case, power supplied by the source pin and the power
absorbed by the load pout are
vTH
v 2 TH
pin  vTH i  vTH (
)
RTH  RL
2 RL
2
v
pout  TH
4 RL
•
 max 
pout
 50%
pin
Therefore only 50% efficiency can be achieved at maximum power transfer
conditions
7
Maximum power transfer for Norton
Equivalent circuits
• We may also use Norton circuit to represent circuit A
io
A
p  i RL
2
d2p
dR L
2
0
2
iN
R N2 R L

( R N  RL ) 2
=> Maximum power occurs when RN = RL = RTH
The maximum power delivered is
p max
i N2 RL

4
8
Examples
• Provided in the class.
9
Homework for Lecture 23
• Problems 4.70, 4.71, 4.72, 4.73
• Due March 18
10