Download Basic Concepts

Advanced Electrical Engineering
Advanced Electrical Engineering
Michael E. Auer
Basic Concepts
Michael E.Auer
02.05.2012
AEE01
Advanced Electrical Engineering
Chapter Content
Michael E.Auer
•
Introduction
•
Complex Numbers and Phasors
•
Circuit Theory Review
•
Methods of Network Analysis
•
Locus Diagrams
•
Circuit Element Variations
02.05.2012
AEE01
Advanced Electrical Engineering
Chapter Content
Michael E.Auer
•
Introduction
•
Complex Numbers and Phasors
•
Circuit Theory Review
•
Methods of Network Analysis
•
Locus Diagrams
•
Circuit Element Variations
02.05.2012
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This course:
Fundamentals of
Electro-Magnetism !!!
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Fundamental Forces of Nature
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The Electro-magnetic Spectrum
microprocessor
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ac power network
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Scientific progress,
but same fundamental laws
Example Lighting
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Light bulb
Fluorescent lamp
Incandescence is
the emission of light
from a hot object
due to its
temperature.
Fluoresce means to
emit radiation in
consequence to
incident radiation of a
shorter wavelength
02.05.2012
LED diode lamp
When a voltage is applied in a forward-biased
direction across an LED diode, current flows
through the junction and some of the streaming
electrons are captured by positive charges
(holes). Associated with each electron-hole
recombining act is the release of energy in the
form of a photon.
AEE01
Advanced Electrical Engineering
Chapter Content
Michael E.Auer
•
Introduction
•
Complex Numbers and Phasors
•
Circuit Theory Review
•
Methods of Network Analysis
•
Locus Diagrams
•
Circuit Element Variations
02.05.2012
AEE01
Advanced Electrical Engineering
Complex Numbers
Is useful to represent sinusoids as complex numbers.
z = x + jy
Rectangular coordinates
z = z ∠θ = z e jθ
Polar coordinates
j = −1
Re(z ) = x
Im( z ) = y
Relations based
on Euler’s Identity
e ± jθ = cos θ ± j sin θ
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Relations for
Complex
Numbers
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Phasor Domain
1. The phasor-analysis technique transforms equations from the time domain
to the phasor domain.
2. Integro-differential equations get converted into linear equations with no
sinusoidal functions.
3. After solving for the desired variable -such as a particular voltage or current
- in the phasor domain, conversion back to the time domain provides the same
solution that would have been obtained had the original integro-differential
equations been solved entirely in the time domain.
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Time Domain vs. Phasor Domain (1)
Rotating phasor
Stationary phasor
Phasor counterpart of
V0 ∠ φ
V0 ∠
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π
2
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Time Domain vs. Phasor Domain (2)
It is much easier to deal with
exponentials in the phasor
domain than sinusoidal
relations in the time domain.
Just need to track
magnitude/phase, knowing that
everything is at frequency ω.
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Phasor Relation for Resistors
Current through resistor
Time Domain
Frequency Domain
Time domain
i = I m cos (ω t + φ )
υ = iR = RI m cos (ω t + φ )
Phasor Domain
V = RI m ∠φ
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Phasor Relation for Inductors
Time domain
Phasor Domain
Time Domain
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Phasor Relation for Capacitors
Time domain
Phasor Domain
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Circuit Analysis in the Phasor Domain
Time Domain
Phasor Domain
Differential
Equations
Algebraic
Equations
complex !
Transformation
Solution of the
Differential
Equations
Solution of the
Algebraic
Equations
Solution
Solution
Retransformation
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Basic Approach
Transform the circuit to the phasor or frequency domain.
Solve the problem using circuit techniques (nodal analysis, mesh
analysis, superposition, etc.).
Transform the resulting phasor to the time domain.
1.
2.
3.
Time to Freq
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Solve
Equations in
Freq Domain
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Freq to Time
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Phasor Analysis: General Procedure
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Example: RL Circuit (1)
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Example: RL Circuit (2)
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Transformation Table
x means a travelling wave!
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Traveling Waves in the Phasor Domain
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Chapter Content
Michael E.Auer
•
Introduction
•
Complex Numbers and Phasors
•
Circuit Theory Review
•
Methods of Network Analysis
•
Locus Diagrams
•
Circuit Element Variations
02.05.2012
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Voltage Division
v1 = ii R1
and
v 2 = ii R2
Applying KVL to the loop,
v i = v1 + v 2 = ii (R1 + R2 )
and
vi
ii =
R1 + R2
Combining these yields the basic voltage division formula:
R1
v1 = v i
R1 + R2
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02.05.2012
R2
v2 = vi
R1 + R2
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Current Division
ii = i1 + i2
where
vi
i1 =
R1
vi
and i2 =
R2
Combining and solving for vi,
R1 R2
v i = ii
= ii
= ii (R1 || R2 )
1
1
R1 + R2
+
R1 R2
1
Combining these yields the basic current division formula:
R2
i1 = ii
R1 + R2
Michael E.Auer
02.05.2012
and
R1
i2 = ii
R1 + R2
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Thevenin and Norton Equivalent Circuits
Thévenin
Norton
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Thevenin Equivalent Voltage (1)
Applying KCL at the output node,
vo − vi vo
= G1 (v o − vi )+ GS v o
+
βi1 =
RS
R1
i1 = G1 (v o − v i )
Current i1 can be written as:
open circuit
Combining the previous equations
G1 (β +1)v i = [G1 (β +1) + GS ]v o
G1 (β +1)
R1RS
(β +1)RS
vi ×
=
vi
vo =
G1 (β +1) + GS
R1RS (β +1)RS + R1
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Thevenin Equivalent Voltage (2)
Using the given component values:
(β +1)RS
(50 +1)1 kΩ
vi =
v i = 0.718v i
vo =
(β +1)RS + R1
(50 +1)1 kΩ +1 kΩ
and
v th = 0.718v i
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Thevenin Equivalent Resistance
Applying KCL,
i x = −i1 − βi1 + G S v x
= G1v x + βG1v x + G S v x
= [G1 (β + 1) + G S ]v x
R1
1
vx
= RS
=
Rth =
β +1
G1 (β + 1) + G S
ix
Rth = RS
Michael E.Auer
R1
20 kΩ
= 1 kΩ
= 1 kΩ 392 Ω = 282 Ω
50 + 1
β +1
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Norton Equivalent Circuit
Applying KCL,
in = i1 + βi1
= G1v i + βG1vi
= G1 (β +1)v i
=
v i (β +1)
R1
short circuit
Short circuit at the output causes
zero current to flow through RS.
Rth is equal to Rth found earlier.
50 +1
vi
vi =
= (2.55 mS)v i
in =
20 kΩ
392 Ω
Michael E.Auer
02.05.2012
AEE01
Advanced Electrical Engineering
Final Thevenin and Norton Equivalent Circuits
Check of Results: Note that vth = inRth and this can be used to check
the calculations: inRth=(2.55 mS)vi(282 Ω) = 0.719vi, accurate within
round-off error.
While the two circuits are identical in terms of voltages and currents at the
output terminals, there is one difference between the two circuits. With no load
connected, the Norton circuit still dissipates power!
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Find Thevenin and Norton Equivalent Circuits!
Rth , vth , in
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Source Transformation (1)
• An equivalent circuit is one whose v-i characteristics are identical with the
original circuit.
• It is the process of replacing a voltage source vS in series with a resistor R
by a current source iS in parallel with a resistor R, or vice versa.
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02.05.2012
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Advanced Electrical Engineering
Source Transformation (2)
+
+
-
-
(a) Independent source transform
+
+
-
-
(b) Dependent source transform
Michael E.Auer
02.05.2012
•
The arrow of the
current source is
directed toward the
positive terminal of
the voltage source.
•
The source
transformation is not
possible when R = 0
for voltage source
and R = ∞ for current
source.
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Advanced Electrical Engineering
Source Transformation Example
Find io in the circuit shown below using source transformation.
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Superposition Theorem (1)
It states that the voltage across (or current through) an element in a
linear circuit is the algebraic sum of the voltage across (or currents
through) that element due to EACH independent source acting alone.
The principle of superposition helps us to analyze a linear circuit
with more than one independent source by calculating the
contribution of each independent source separately.
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Superposition Theorem (2)
We consider the effects of 8A and 20V one by one, then add the
two effects together for final vo.
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Superposition Theorem (3)
Steps to apply superposition principle
1. Turn off all independent sources except one source. Find
the output (voltage or current) due to that active source
using nodal or mesh analysis.
2. Repeat step 1 for each of the other independent sources.
3. Find the total contribution by adding algebraically all the
contributions due to the independent sources.
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02.05.2012
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Advanced Electrical Engineering
Superposition Theorem (4)
Two things have to be keep in mind:
1. When we say turn off all other independent sources:
 Independent voltage sources are replaced by 0 V (short circuit)
and
 Independent current sources are replaced by 0 A (open circuit).
2. Dependent sources are left intact because they are controlled by
circuit variables.
Michael E.Auer
02.05.2012
AEE01
Advanced Electrical Engineering
Superposition Theorem Example 1
Example 2
Use the superposition theorem to find v in the
circuit shown below.
3A is discarded
by open-circuit
6V is discarded
by short-circuit
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02.05.2012
AEE01
Advanced Electrical Engineering
Superposition Theorem Example 2
Use superposition to find vx in the circuit..
2A is discarded by
open-circuit
20 Ω
10 V
10V is discarded by
open-circuit
20 Ω
v1
+
−
4Ω
0.1v1
v2
2A
4Ω
0.1v2
(b)
(a)
Michael E.Auer
Dependant source
keep unchanged !!!
02.05.2012
AEE01
Advanced Electrical Engineering
Chapter Content
Michael E.Auer
•
Introduction
•
Complex Numbers and Phasors
•
Circuit Theory Review
•
Methods of Network Analysis
•
Locus Diagrams
•
Circuit Element Variations
02.05.2012
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Introduction
Things we need to know in solving any resistive circuit
with current and voltage sources only:
Number of equations
•
Ohm’s Law
b
•
Kirchhoff’s Current Laws (KCL)
n-1
•
Kirchhoff’s Voltage Laws (KVL)
b – (n-1)
mesh = independend loop
Number of branch currents and
branch voltages = 2b (variables)
Problem: Number of equations!
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Mesh Analysis (1)
1.
Mesh analysis provides a general procedure for analyzing
circuits using mesh currents as the circuit variables.
2.
Mesh analysis applies KVL to find unknown currents.
3.
A mesh is a loop which does not contain any other loops
within it (independent loop).
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Mesh Analysis (2)
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Mesh Analysis (3)
Example – circuit with independent voltage sources
Equations:
R1∙i1 + (i1 – i2) ∙ R3 = V1
R2 ∙ i2 + R3 ∙(i2 – i1) = -V2
reordered:
(R1+ R3) ∙ i1 - i2 ∙ R3 = V1
- R3 ∙ i1 + (R2 + R3)∙i2 = -V2
Note:
i1 and i2 are mesh current (imaginative, not measurable directly)
I1, I2 and I3 are branch current (real, measurable directly)
I1 = i1; I2 = i2; I3 = i1 - i2
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Mesh Analysis (4)
Formalization: Network equations by inspection.
− R3   i1   V1 
 ( R1 + R3 )

 ⋅   = 

( R2 + R3 )   i2   − V2 
 − R3
Impedance matrix
Excitation
Mesh currents
General rules:
1. Main diagonal: ring resistance of mesh n
2. Other elements: connection resistance between meshes n and m
• Sign depends on direction of mesh currents!
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Mesh Analysis (5)
Example: By inspection, write the mesh-current equations in matrix
form for the circuit below.
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Mesh Analysis Special Cases
dependent source
ideal voltage source
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Mesh Analysis
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Nodal Analysis (1)
It provides a general procedure for analyzing circuits using node
voltages as the circuit variables.
Example
3
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Nodal Analysis (2)
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Nodal Analysis (3)
Example
Apply KCL at
node 1 and 2
v1
G3
G1
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v2
G1∙v1 + (v1 – v2) ∙ G3 = 1A
G2 ∙ v2 + G3 ∙(v2 – v1) = - 4A
reordered:
(G1+ G3) ∙ v1 - v2 ∙ G3 = 1A
- G3 ∙ v1 + (G2 + G3)∙v2 = - 4A
G2
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Nodal Analysis (4)
Formalization: Network equations by inspection.
− G3   v1   1A 
 (G1 + G3 )

 ⋅   = 

(G2 + G3 )   v2   − 2A 
 − G3
Admittance matrix
Excitation
Node voltages
General rules:
1. Main diagonal: sum of connected admittances at node n
2. Other elements: connection admittances between nodes n and m
• Sign: negative!
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Nodal Analysis (5)
Example: By inspection, write the node-voltage equations in matrix
form for the circuit below.
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Nodal Analysis Special Case
dependent source
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AC Network Nodal Analysis
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Advanced Electrical Engineering
Chapter Content
Michael E.Auer
•
Introduction
•
Complex Numbers and Phasors
•
Circuit Theory Review
•
Methods of Network Analysis
•
Locus Diagrams
•
Circuit Element Variations
02.05.2012
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Phasor Inversion Example
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R-C-Circuit Locus
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Locus Inversion Theorems
1. The inversion of a straight line through the origin is
again a line through the origin.
2. The inversion of a straight line not through the origin is
a circle through the origin and vice versa.
3. The inversion of a circle that does not pass through the
origin is again a circle that does not pass through the
zero point
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Inversion of a Straight Line
through the origin
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not through the origin
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Inversion of a Circle not Through the Origin
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Locus of a Low-Pass Transfer Function
Ua
1
=
T ( jω ) =
U e 1 + jΩ
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mit
ω
Ω=
ωg
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Locus of a Parallel-Series Circuit (1)
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Locus of a Parallel-Series Circuit (2)
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Equivalent
Networks
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Dual Networks (1)
In other words, both circuits are described by the same pair of equations:
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Dual Networks (2)
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Dual Networks (3)
Rules for obtaining the dual of a planar
circuit, regardless of wether or not it is a
series-parallel network:
Rule1:
Insight of each mesh, including the infinite
region surrounding the circuit, place a
node.
Rule2:
Suppose two of this nodes, for example a and b,
are in adjacent meshes. Then there is at least
one element in the boundary common to these
two meshes. Place the dual of each common
element between nodes a and b.
Michael E.Auer
02.05.2012
AEE01
Advanced Electrical Engineering
I-V-Relations in Dual Nezworks
∑ iµ = 0
∑ vν = 0
v = vq
i = iq
µ
v = R ⋅i
v=L
ν
di
dt
i=C
i ′ = iq′
Given Network
i-v-Duality
Dual Network
∑µ uµ′ = 0 ∑ν iν′ = 0
dv
dt
i ′ = G ′ ⋅ u′
i′ = C′
du ′
dt
u′ = L′
di ′
dt
v' = Z 0 ⋅ i
und
i′ =
v
Z0
For example Z0 = 1Ω
u ′ = uq′
Duality relations for the basic network elements:
G′ =
Michael E.Auer
R
Z 02
C′ =
L
Z 02
L′ = C ⋅ Z 02
iq′ =
02.05.2012
vq
Z0
vq' = Z 0 ⋅ iq
AEE01
Advanced Electrical Engineering
Chapter Content
Michael E.Auer
•
Introduction
•
Complex Numbers and Phasors
•
Circuit Theory Review
•
Methods of Network Analysis
•
Locus Diagrams
•
Circuit Element Variations
02.05.2012
AEE01
Advanced Electrical Engineering
Circuit Element Variations
•
All electronic components have manufacturing tolerances.
o Resistors can be purchased with ± 10%, ± 5%, and
± 1% tolerance. (IC resistors are often ± 10% and more.)
o Capacitors can have asymmetrical tolerances such as +20%/-50%.
o Power supply voltages typically vary from 1% to 10%.
•
•
•
Device parameters will also vary with temperature and age.
Circuits must be designed to accommodate these variations.
Mainly Worst-case Analysis and Monte Carlo Analysis
(statistical) are used to examine the effects of component
parameter variations.
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Tolerance Modeling
•
For symmetrical parameter variations
Pnom(1 - ε) ≤ P ≤ Pnom(1 + ε)
•
For example, a 10 kΩ resistor with ±5% percent tolerance could take on
the following range of values:
10kΩ(1 - 0.05) ≤ R ≤ 10kΩ(1 + 0.05)
9500 Ω ≤ R ≤ 10500 Ω
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Numeric Precision
•
Michael E.Auer
Most circuit parameters vary from less than ± 1 % to
greater than ± 50%.
•
As a consequence, more than three significant digits is
meaningless.
•
Results should be represented with three significant digits:
2.03 mA, 5.72 V, 0.0436 µA, and so on.
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Circuit Analysis with Tolerances
•
Worst-case analysis
– Parameters are manipulated to produce the worst-case min and max
values of desired quantities.
– This can lead to over design since the worst-case combination of
parameters is rare.
– It may be less expensive to discard a rare failure than to design for 100%
yield.
•
Monte-Carlo analysis
– Parameters are randomly varied to generate a set of statistics for desired
outputs.
– The design can be optimized so that failures due to parameter variation
are less frequent than failures due to other mechanisms.
– In this way, the design difficulty is better managed than a worst-case
approach.
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AEE01
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Worst Case Analysis Example
Problem: Find the nominal and worstcase values for output voltage and
source current.
Solution:
•
Known Information and Given
Data: Circuit topology and values
in figure.
•
Nominal voltage solution:
Unknowns:
nom
O
V
Michael E.Auer
min
O
, V
max
O
, V
, I
nom
I
, I
min
I
, I
max
I
02.05.2012
nom
O
V
R1nom
=V
R1nom + R2nom
18kΩ
= 15V
= 5V
18kΩ + 36kΩ
nom
I
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Monte Carlo Analysis
•
•
•
Parameters are varied randomly and output statistics are gathered.
We use programs like MATLAB, Mathcad, SPICE, or a spreadsheet to
complete a statistically significant set of calculations.
For example, with Excel, a resistor with a nominal value Rnom and
tolerance ε can be expressed as:
R = Rnom (1+ 2ε ( RAND() − 0.5))
The RAND() function returns
random numbers uniformly
distributed between 0 and 1.
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Monte Carlo Analysis Results
VO (V)
Average
4.96
Nominal
5.00
Standard Deviation
0.30
Maximum
5.70
W/C Maximum
5.87
Minimum
4.37
W/C Minimum
4.20
Histogram of output voltage from 1000 case Monte Carlo simulation.
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