engr-43_lec-04-3b_thevein-norton_part-b.ppt 1 bruce mayer, pe engineering-43: engineering circuit...
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ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt 3 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Recall Norton Equivalent Norton Equivalent Circuit for PART A i N = Norton Equivalent CURRENT Source R N = Norton Equivalent PARALLEL RESISTANCETRANSCRIPT
[email protected] • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PERegsitered Electrical & Mechanical Engineer
Engineering 43
Chp 5.4Chp 5.4Maximum Maximum
PowerPowerTransferTransfer
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ReCall ThReCall Théévenin Equivalentvenin EquivalentLINEAR C IRC U IT
M ay containindependent and
dependent sourceswith their co ntro ll ing
variablesPART A
LINEAR C IRC U ITM ay contain
independent anddependent so urces
with their contro ll ingvariablesPART B
a
b_Ov
i
LINEAR C IRC U IT
PART B
a
b_Ov
i
THR
THv
PART A
Thevenin Equivalent Circuit for PART A
vTH = Thevenin Equivalent VOLTAGE Source
RTH = Thevenin Equivalent SERIES RESISTANCE
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Recall Norton EquivalentRecall Norton Equivalent
Norton Equivalent Circuit for PART A
iN = Norton Equivalent CURRENT Source
RN = Norton Equivalent PARALLEL RESISTANCE
LINEAR C IRC U ITM ay contain
independent anddependent sources
with their co ntro l l ingvariablesPART A
LINEAR C IRC U ITM ay contain
independent anddependent sources
with their contro ll ingvariablesPART B
a
b_Ov
i
LINEAR C IRC U IT
PART B
a
b_Ov
i
NRNi
PART A
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ExampleExample Recognize Mixed sources
• Must Compute Open Circuit Voltage, VOC, and Short Circuit Current, ISC
The Open Ckt Voltage
Use V-Divider to Find VX
XV bV
SSX VVRR
RV32)2(
2
bXTH VVV
SSXb VaRVaVRV )3/41()(2
THV
For Vb Use KVL
SSSXX
bXTH
VVRaVRaVVVVV
)3/2)(21()2(
SbXTH VaRVVV341
Solve for VTH
The Short Ckt Current• Note that Shorting a-to-b
Results in a Single Large Node Now VTH = Vx - Vb
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example contExample cont Need to Find Vx
KCL at Single Node
Then RTH
SCISingle node
XV
022
2
RVVaV
RV
RVV sX
XXSx
aRVV S
X 243
Solving For Vx
KCL at Node-b for ISC
RVVaVI SX
XSC 2
SSC VaRRaRI
)2(441
3)2(4 aRR
IV
IVR
SC
TH
SC
OCTH
The Equivalent Circuit
a
b
R TH
V TH
SVaR
341
3)2(4 aRR
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical AnalysisNumerical Analysis Using Excel Spreadsheet
Short INDEPENDENT Sources to Find RTH
kRVR
X
OCTH
100 WHEN, PLOT AND FIND
X
XXTH Rk
RkRkR
44||4
And VOC by 12V Source and V-Divider for V across RX
X
XOC Rk
RV4
612
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Analysis - PlotNumerical Analysis - Plot
X
XOC
X
XXTH
RkRV
RkRkRkR
4612
44||4
THEVENIN EQUIVALENT EXAMPLE
Rx[kOhm] Voc[V] Rth[kOhm]0 12 0
0.1 11.8537 0.0975609760.2 11.7143 0.190476190.3 11.5814 0.2790697670.4 11.4545 0.3636363640.5 11.3333 0.4444444440.6 11.2174 0.521739130.7 11.1064 0.5957446810.8 11 0.6666666670.9 10.898 0.734693878
1 10.8 0.81.1 10.7059 0.8627450981.2 10.6154 0.9230769231.3 10.5283 0.9811320751.4 10.4444 1.0370370371.5 10.3636 1.0909090911.6 10.2857 1.1428571431.7 10.2105 1.1929824561.8 10.1379 1.241379311.9 10.0678 1.288135593
USING EXCEL
0
2
4
6
8
10
12
14
0 2 4 6 8 10
Rx[kOhm]
Voc[
V]
Voc[V]
Rth[kOhm]
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Analysis - LimitsNumerical Analysis - Limits
VRk
RV
kRkRkR
X
XOC
R
X
XTH
R
Lim
Lim
X
X
64
612
444
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Plot using MATLAB Script FilePlot using MATLAB Script File ENGR43_Chp5_Rth_Voc_Analysis_MATLAB
_0602.m
% ENGR43_Chp5_Rth_Voc_Analysis_MATLAB_0602.m% Bruce Mayer, PE% ENGR43 * 27Feb06%Rx = [0:0.1:20]'; %define the range of resistors to useVoc = 12-6*Rx./(Rx+4); %the formula for Voc. Notice "./"Rth = 4*Rx./(4+Rx); %formula for Thevenin resistance. plot(Rx,Voc,'bx', Rx,Rth,'mv')title('USING MATLAB'), grid, xlabel('Rx (kOhm)'), ylabel('Voc (V), Rth (kOhm)')legend('Voc [V]','Rth [kOhm]')
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
0 5 10 15 200
2
4
6
8
10
12USING MATLAB
Rx (kOhm)
Voc
(V
), R
th (
kOhm
)Voc [V]Rth [kOhm]
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Thevenin Theorem – General ViewThevenin Theorem – General View Typical Interpretation
The General View
LINEAR C IRC U ITM ay contain
independent anddependent sources
with their co ntro l l ingvariablesPART A
a
b_Ov
i
+-
R
2 R
- V X +
2 R
a V X
V TH
a
b
LINEAR C IRC U IT withALL independent
sources set to zeroPART A
a
b_Ov
i
THV
Looks Like Series
Resistance
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Thevenin General - CommentsThevenin General - Comments VTH Becomes the Sole Equivalent
Power Source/Sink for the “Part-A” (a.k.a. Driving) Circuit• It’s Value is Set to Maintain The
Open Ckt Voltage at vo
This Interpretation Applies Even When The Passive Elements Include INDUCTORS and CAPACITORS
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Amplifier Driving SpeakerAmplifier Driving Speaker Consider an Amplifier Circuit
connected to a Speaker
DrivingCircuita.k.a. the
“SOURCE”
Speakera.k.a. the
“LOAD”
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Circuit SimplificationCircuit Simplification Thévenin’s Equivalent Circuit Theorem Allows
Tremendous Simplification of the Amp Ckt
Thevenin +
RS
VS
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Maximum Power TransferMaximum Power Transfer Consider The Amp-Speaker Matching Issue
From PreAmp(voltage ) To speakers
+-
R T H
V T H
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Maximum Power Xfer ContMaximum Power Xfer Cont The Simplest Model for a
Speaker is to Consider it as a RESISTOR only
Since the “Load” Does the “Work” We Would like to Transfer the Maximum Amount of Power from the “Driving Ckt” to the Load• Anything Less Results in
Lost Energy in the Driving Ckt in the form of Heat
+-
R T H
VT H SPEAKERM O DEL
BASIC MODEL FOR THE ANALYSIS OF POWER TRANSFER
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Maximum Power TransferMaximum Power Transfer Consider Thevenin Equivalent
Ckt with Load RL
Find Load Pwr by V-Divider
For every choice of RL we have a different power.• How to find the MAXIMUM
Power value?
+-
S O U R CE
(L O A D )
R TH
V TH
R L
LV
THLTH
LL
L
LL V
RRRV
RVP
;
2
2
2 THLTH
LL V
RRRP
Consider PL as a
FUNCTION of RL and find the maximum of such a function have at left!• i.e., Take 1st Derivative
and Set to Zero
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Max Power Xfer contMax Power Xfer cont Find Max Power Condition Using Differential Calculus
Set The Derivative To Zero To Find MAX or MIN Points• For this Case Set To Zero
The NUMERATOR
32 2
LTH
LLTHTH
L
L
RRRRRV
dRdP
02 LLTH RRR
Solving for “Best” (max) Load
THL RR *
This is The Maximum Power Transfer Theorem• The load that maximizes the
power transfer for a circuit is equal to the Thevenin equivalent resistanceof the circuit
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Max Power QuantifiedMax Power Quantified By Calculus we
Know RL for PL,max
THL RR *
Recall the Power Transfer Eqn
2
2 THLTH
LL V
RRRP
Sub RTH for RL
2
2max, THTHTH
THL V
RRRP
2
22
2max, 42 THTH
THTH
TH
THL V
RRV
RRP
So Finally
TH
THL R
VP2
max, 41
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Max Pwr Xfer ExampleMax Pwr Xfer Example Determine RL for
Maximum Power Transfer
Need to Find RTH
• Notice This Ckt Contains Only INDEPENDENT Sources
Thus RTH BySource Deactivation
ab
kkkkRTH 6634 This is Then the RL For Max
Power Transfer
To Find the AMOUNT of Power Transferred Need the Thevenin Voltage
Then use RTH = 6kΩ along with VTH
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Max Pwr Xfer Example contMax Pwr Xfer Example cont To Find VTH Use Meshes The Eqns for Loops 1 & 2
Solving for I2
][10*6*4 21 VIkIkVOC
mAI 21
03*6*3 212 VIkIIk
][31
31
9][3
12 mAIkVI
Now Apply KVL for VOC
Recall
22 TH
LTH
LL V
RRRP
At Max: PL = PMX, RL = RTH
TH
THMX R
VP4
2
][625
6*4][100 2
mWkVPMX
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Max Pwr XferMax Pwr Xfer Determine RL and Max
Power Transferred Find Thevenin Equiv.
At This Terminal-Set Recall for Max Pwr Xfer
a
b
THL RR TH
THMX R
VP4
2
This is a MIXED Source Circuit• Analysis Proceeds More
Quickly if We start at c-d and Adjust for the 4kΩ at the end
c
d
Use Loop Analysis
Eqns for Loops 1 & 2mAI 41
022)(4 2'
12 kIkIIIk X
mAIII X 16426 1'
2
2I1I
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Max Pwr Xfer contMax Pwr Xfer cont The Controlling Variable
Now Short Ckt Current• The Added Wire Shorts
the 2k Resistor
Remember now the partition points
VkIVmAII
II
OC
X
82and4
so
2
12
2'
mAII SCX 40"
Then RTH
kmAV
IVRSC
OCTH 2
48
V8
c
d The RTH for ckt at a-b =
2kΩ+4kΩ; So kRL 6
][38][
6*482
max mWmWP
a
b
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Thevenin & Norton SummaryThevenin & Norton Summary Independent
Sources Only• RTH = RN by
Source Deactivation
• VTH – = VOC or
– = RN·ISC
• IN – = ISC or
– = VOC/RTH
Mixed INdep and Dep Srcs• Must Keep
Indep & dep Srcs Together in Driving Ckt
• VTH = VOC
• IN = ISC
• RTH = RN = VOC/ ISC
DEPENDENT Sources Only• Must Apply V
or I PROBE– Pick One, say
IP = 1.00 mA, then Calculate the other, say VP
• VTH = IN = 0
• RTH = RN = VP/ IP
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
WhiteBoard WorkWhiteBoard Work Let’s Work Problem
5.109
Find Pmax for Load RL
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
What’s an “Algorithm”What’s an “Algorithm” A postage stamp
issued by the USSR in 1983 to commemorate the 1200th anniversary of Muhammad al-Khowarizmi, after whom algorithms are named.