Transcript Defining the Cyber Domain for Wireless Communications Security

Lesson 10: Thevenin’s Theorem
and Max Power Transfer
1
Learning Objectives
• State and explain Thèvenin's theorem.
• List the procedure for determining the Thèvenin equivalence
of an actual circuit from the standpoint of two terminals.
• Apply Thèvenin's Theorem to simplify a circuit for analysis.
• Analyze complex series-parallel circuits using Thèvenin's
theorem.
• Apply the Maximum Power Transfer theorem to solve
appropriate problems.
2
Thévenin’s Theorem
• Thévenin’s theorem greatly simplifies analysis of
complex circuits by allowing us to replace all of the
elements with a combination of just one voltage
source and one resistor.
• Thévenin’s theorem provides a simplified circuit that
provides the same response (voltage and current) at
the load terminals.
− This allows the response to be easily determined for
various load values.
3
Thévenin’s Theorem
• Any complex two-terminal circuit can be replaced by an
equivalent circuit consisting of a voltage source VTh and a
series resistor RTh.
Original Circuit
Thévenin Equivalent Circuit
• The Thévenin equivalent circuit provides an equivalence at the
terminals only.
− The internal construction and characteristics of the original network
and the Thévenin equivalent are usually quite different.
4
Thévenin’s Theorem
• ETh is the open circuit voltage at the terminals.
• RTh is the input or equivalent resistance at the
terminals when the sources are turned off.
Original Circuit
Thévenin Equivalent Circuit
5
Thévenin’s Theorem
5 Steps:
1. Remove the load and
2. Label the terminals a and b.
3. Solve for RTH by setting all sources to zero.
4. Solve for VTH by returning all sources to their
original position and finding the open-circuit
voltage between a and b.
5. Draw the new equivalent circuit.
6
Thévenin’s Theorem
Steps 1 & 2
Convert to a Thévenin circuit:
1. Identify and remove the load from the circuit.
2. Label the resulting open terminals.
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Thévenin’s Theorem
Step 3
3. Solve for RTH and isolate the resistance from the
source.
Set all sources to zero:
 Replace voltage sources with shorts.
 Replace current sources with opens.
8
Zeroing Sources
3. “Zeroing” a source means setting its value equal to
zero.
• Voltage sources – 0 V is equivalent to a short-circuit.
• Current sources – 0 A is equivalent to a open-circuit.
Voltage Sources Become
Short-Circuits
Current Sources Become
Open-Circuits
9
Thévenin’s Theorem
Step 3
• With the load disconnected, turn off all source.
• RTh is the equivalent resistance looking into the
“dead” circuit through terminals a-b.
Rth
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Thévenin’s Theorem
Step 3
3. Set all sources to zero, and calculate RTh .
1
RTH
1 
 1
 Rab  
   31
 80  60 40 
Remember,
calculate RTH from
the a and b
perspective!
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Thévenin’s Theorem
Step 4
4. Solve for VTH and then, as needed:
•
•
Calculate the voltage (VLD) across the RLD.
Calculate the current (ILD) through RLD.
(VDR)  ETH  VTH  Vab  V40 
R
40
VTH  E * 40  20V *
 4.44V
RT
40  80  60
VLD 
I LD
12
RLD
ETh
RTh  RLD
ETh

RTh  RLD
Thévenin’s Theorem
Step 5
5. REDRAW the circuit showing the Thèvenin
equivalents (VTH and RTH) with the load installed.
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Applying Thévenin Equivalent
• Repetitive solutions for various load resistances now
becomes easy with the transformed circuit.
14
Example Problem 1
Find the Thévenin equivalent circuit external to RLD. Determine
ILD and VLD when RLD = 2.5 Ω.
a
15Ω
9Ω
R
9
(VDR)  ETH  Vab  V9  E * 9  10V *
 6V
RT
6  9
I LD 
RTH = 18.6Ω
= 2.5Ω
I LD 
1/2. Remove the load label the terminals a and b.
3. Solve for RTH.
4. Solve for VTH.
5. Draw the new equivalent circuit.
RTH
b


 1 
 Rab  
  15  18.6
1
1
  
6 9
ETh
RTh  RLD
6V
 284mA
18.16  2.5
RLD
RTh  RLD
2.5
 6V
 0.71V
18.6  2.5
VLD  ETh
VLD
OR VLD  I LD * RLD  284mA * 2.5  0.71V
15
Example Problem 2
Find the Thévenin equivalent circuit external to RLD
and determine ILD.
RTH  Rab  20k  50k  70k 
ETH  Vab  V50k   I S * R50 k  100 A * 50k   5V
RTH = 70kΩ
ETH = 5V
I LD 
ETh
RTh  RLD
I LD 
5V
 20  A
70k  180k
= 180kΩ
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Maximum Power Transfer
In some applications, the purpose of a circuit is to
provide maximum power to a load.
Some examples:
• Stereo amplifiers
• Radio transmitters
• Communications equipment
The question is: If you have a system, what load should
you connect to the system in order for the load to
receive the maximum power that the system can
deliver?
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Maximizing PLD
• How might we determine RLD such that PLD is
maximized?
2
PLD  I RLD
2
LD
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 VTh


 RLD
 RTh  RLD 
Maximum Power Transfer
Theorem
• Maximum power is transferred to the load when the
load resistance equals the Thévenin resistance as seen
from the load (RLD = RTh).
− When RLD = RTh, the source and load are said to be
matched.
19
Maximizing PLD
• As RLD increases, a higher percentage of the total
power is dissipated in the load resistor.
• But since the total resistance is increasing, the total
current is dropping, and a point is reached where the
total power dissipated by the entire circuit starts
dropping.
2
PLD  I RLD
2
LD
20
 VTh


 RLD
 RTh  RLD 
Maximum Power
• The max power happens when RLD = RTh , but what is the level
of power at this point?
− Showing the derivation, we get:
2
2
PLD  I LD RLD
2
 V Th
2
 VTh 
 VTh


 RTh
 RLD  
 RTh  RLD 
 RTh  RTh 
Rth
4R
2

Th
V
2
Th
4 RTh
 PMAX
• BE CAREFUL!!! Note that this is not true if RLD  RTh.
− If RLD  RTh then use:
PLD  I LD RLD
2
2
 VTh


 RLD
 RTh  RLD 
21
Maximum Power Transfer
Theorem
• The total power delivered by a supply such as ETh is
absorbed by both the Thévenin equivalent resistance
and the load resistance.
• Any power delivered by the source that does not get
to the load is lost to the Thévenin resistance.
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Example Problem 3
a) Find the Thévenin equivalent circuit to the left of terminals a-b.
b) Calculate the maximum power transfer to the load if RLD = RTH.
c) Determine the power dissipated by RLD for load resistances of 2 and 6.
b) When R LD = R TH
V
2
Th
4 RTh
 PMAX
PMAX 
a
4Ω
302 V
4*4
 56.25W
c) When R LD  R TH
1Ω
2
 VTh

PL  I L 2 * RL OR PL  
 RLD
R

R
 Th
LD 
For the 2 load and because we already
calculated VTH and R TH let's use:
12Ω
b
1
1 1 
a) RTH  Rab      1  4
 4 12 
R
12
(VDR)  ETH  Vab  V12  E * 12  40V *
 30V
RT
4  12
2
 VTh

 30V 
PL  
 RLD  
 * 2  50W
 4  2 
 RTh  RLD 
Now, for the 6 load, and just to show it works let's use:
PL   I L  * RLD
2
23
2
2
 30V 

 *6  54W
 4  6 
Example Problem 4
A stereo is rated for max output power of 150W per channel when RLD = 8Ω
a) Sketch the Thévenin Equivalent circuit.
b) What would the output power be with two 8Ω speakers as the load and
are connected in parallel to one of the channels?
b) We know that PMAX 
RTH = 8Ω
= 8Ω
V
2
Th
4 RTh
Rearrange the above and solve for VTH :
V 2TH  PMAX * 4 RTh
VTH  PMAX * 4 RTh  150W * 4*8  69.28V
Now, to caculate output power for the two 8Ω resistors in parallel,
calculate R L (8//8 = 4),
RTH = 8Ω
and because we already know VTH and R TH let's use:
= 8Ω//8 Ω
=4Ω
2
 VTh

 69.28V 
PL  
 RLD  
 * 4  133.3W
 8  4 
 RTh  RLD 
You should note, PL for this case is less than PMAX .
2
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Efficiency
• When maximum power is delivered to RLD, the
efficiency is a mere 50%.
− Remember that max power occurs when RLD = RTH.
pout PL
I L 2 RLD


 2
 100%
2
pin PS I L RTh  I L RLD
When R L =R TH
i 2 RTh
  2
100%  50%
2
i RTh  i RTh
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Efficiency
• Communication Circuits and Amplifiers:
− Max Power Transfer Is More Desirable Than High
Efficiency.
• Power Transmission (115 VAC 60 Hz Power ):
− High Efficiency Is More Desirable Than Max Power
Transfer.
− Load Resistance Kept Much Larger Than Internal
Resistance Of Voltage Source.
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QUESTIONS?
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