Engineering 43
RC & RL
st
1 Order Ckts
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
C&L Summary
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Introduction  Transient Circuits
 In Circuits Which Contain Inductors &
Capacitors, Currents & Voltages
CanNOT Change Instantaneously
 Even The Application, Or
Removal, Of Constant
Sources Creates
Transient (TimeDependent) Behavior
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Basic Concept
 Inductors & Capacitors
Can Store Energy
 Under Certain Conditions This
Energy Can Be Released
 RATE of Energy Storage/Release
Depends on the parameters Of The
Circuit Connected To The Terminals Of
The Energy Storing Element
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
1st & 2nd Order Circuits
 FIRST ORDER CIRCUITS
• Circuits That Contain
ONE Energy Storing Element
– Either a Capacitor or an Inductor
 SECOND ORDER CIRCUITS
• Circuits With TWO Energy Storing
Elements in ANY Combination
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Circuits with L’s and/or C’s
 Up to this point we have considered DC
circuits which generate a System of
Linear Equations.
 We will now consider Circuits where the
Power Supply can Switched IN or OUT
Solve for 𝑣𝐶
Engineering-43: Engineering Circuit Analysis
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Solve for 𝑖𝐿
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
𝒗𝑪 𝒕 or 𝒊𝑳 𝒕 Solution Process
 For Either C or L the form of the Soln:
𝑥 𝑡 = 𝐾𝐼𝐶 𝑒 −𝑡
𝜏
+ 𝐾𝐹𝐶
• 𝐾𝐹𝐶 is a CONSTANT based on the FINAL
CONDITION (𝑡 → ∞) of The Circuit
• 𝐾𝐼𝐶 is CONSTANT based on 𝐾𝐹𝐶 and the
INITIAL CONDITION Of the Circuit:
𝑥 0− = 𝑥 0+
• 𝜏 is the TIME CONSTANT for the Circuit,
either 𝑅𝐶 or 𝐿 𝑅
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
𝒗𝑪 𝒕 or 𝒊𝑳 𝒕 Solution Process
 For Either C or L the form of the ODE:
𝑑𝑥 𝑡
𝜏
+ 𝑑𝑥 𝑡 = 𝑓 𝑡
𝑑𝑡
• 𝑓 𝑡 is the so-called “Forcing Function”
– In our case a either a Voltage or Current Src
• The ODE is derived from one of
– 𝑖𝐶 =
𝑑𝑣𝐶
𝐶
𝑑𝑡
– 𝑣𝐿 =
𝑑𝑖𝐿
𝐿
𝑑𝑡
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
𝒗𝑪 𝒕 or 𝒊𝑳 𝒕 Soln StepByStep
 Consider the circuit long before for the
switching event.
• This determines 𝒗𝑪 0− or 𝒊𝑳 0−
 Consider the circuit long after for the
switching event; i.e. as 𝑡 → ∞
• This Determines the FINAL Condition,
either 𝒗𝑪 ∞ or 𝒊𝑳 ∞
 Use any combination of KVL or KCL to
develop the ODE
Engineering-43: Engineering Circuit Analysis
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𝑑𝑥 𝑡
𝜏
𝑑𝑡
+ 𝑑𝑥 𝑡 = 𝑓 𝑡
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
𝒗𝑪 𝒕 or 𝒊𝑳 𝒕 Soln StepByStep
 Now the solution MATH → An important
property of LINEAR ODEs (see MTH6 or
ENGR25) is that the solution can be divided
into TWO pieces which are then
SuperPosed; to whit:
• Where
𝑥𝑡𝑜𝑡 = 𝑥𝑝 + 𝑥𝑐
– 𝑥𝑝 ≡ the particular Soln which is ANY Solution to the
ENTIRE ODE; usually found by (educated) Guessing
– 𝑥𝑐 ≡ the complementary Soln where 𝑓 𝑡 is
ARBITRARILY Set to ZERO
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
𝒗𝑪 𝒕 or 𝒊𝑳 𝒕 Soln StepByStep
 Find 𝑥𝑝 by Guessing
• For CONSTANT Powers Sources, 𝑉𝑠 or 𝐼𝑠 , 𝑥𝑝 is
usually one of: 𝑉𝑠 or 𝐼𝑠 or 𝑉𝑠 𝑅𝑒𝑞 , or 𝐼𝑠 ∙ 𝑅𝑒𝑞
 Find 𝑥𝑐 by Variable-Separation Calculus
which produces an ln
• 𝜏 and 𝐾𝐼𝐶 with 𝜏 now KNOWN
 Add 𝑥𝑝 and 𝑥𝑐 to obtain 𝑥𝑡𝑜𝑡
𝑥 𝑡 = 𝐾𝐼𝐶 𝑒 −𝑡
Engineering-43: Engineering Circuit Analysis
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𝜏
+ 𝐾𝐹𝐶
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
𝒗𝑪 𝒕 or 𝒊𝑳 𝒕 Soln StepByStep
 𝐾𝐹𝐶 is easily found by using one of
• A CAP is an OPEN to DC
• An IND is a SHORT to DC
 Insert into 𝑥𝑡𝑜𝑡 the value of 𝐾𝐹𝐶
 Now the hard part; Use the Initial Condition
• Find 𝑥𝑡𝑜𝑡 0− = 𝑥𝑡𝑜𝑡 0+ ; that is find one of
– 𝑣𝐶 0− = 𝑣𝐶 0+
– 𝑖𝐿 0− = 𝑖𝐿 0+
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Graphical Example
 Given
 The Ckt
Before Switching
 The Ckt After Switching
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
White Board Example
 Find 𝑣𝐶,𝑡𝑜𝑡 𝑡 for previous ckt
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Solutions Graphically
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
MATLAB Code for Plots
function CartPlot =
TempPlotXY(xU,yV)
% Bruce Mayer, PE
% ENGR25 * 10Sep12
% Function file = plotXY.m
clc % workspace window
%
% Make a nicely formatted XY
plot
plot(xU,yV, 'LineWidth', 3)
xlabel('t (mS)')
ylabel('v_C (volt)')
title('ENGR43 RC Transient RC
Circuit')
grid
Engineering-43: Engineering Circuit Analysis
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function CartPlot =
TempPlotXY(xU,yV)
% Bruce Mayer, PE
% ENGR25 * 10Sep12
% Function file = plotXY.m
clc % workspace window
%
% Make a nicely formatted XY
plot
plot(xU,yV, 'LineWidth', 3)
xlabel('t (µS)')
ylabel('i_L (µAmp)')
title('ENGR43 RL Transient RC
Circuit')
grid
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Complex Example  Slide-35
 Students should CAREFULLY Study:
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Circuits with L’s and/or C’s
 Conventional DC Analysis Using
Mathematical Models Requires The
Determination of (a Set of) Equations
That Represent the Circuit Response
 Example; In Node Or Loop Analysis Of
Resistive Circuits One Represents The
Circuit By A Set Of Algebraic Equations
Analysis
The Ckt
Engineering-43: Engineering Circuit Analysis
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The DC Math Model

Gv  i
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Ckt w/ L’s & C’s cont.
 When The Circuit Includes Inductors Or
Capacitors The Models Become Linear
Ordinary Differential Equations (ODEs)
 Thus Need ODE Tools In Order To
Analyze Circuits With Energy Storing
Elements
• Recall ODEs from
ENGR25
• See Math-4 for
More Info on ODEs
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
First Order Circuit Analysis
 A Method Based On Thévenin Will Be
Developed To Derive Mathematical
Models For Any Arbitrary Linear Circuit
With One Energy Storing Element
 This General Approach Can Be
Simplified In Some Special Cases
When The Form Of The Solution Can
Be Known BeforeHand
• Straight-Forward ParaMetric Solution
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Basic Concept
 Inductors & Capacitors
Can Store Energy
 Under Certain Conditions This
Energy Can Be Released
 RATE of Energy Storage/Release
Depends on the parameters Of The
Circuit Connected To The Terminals Of
The Energy Storing Element
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Example: Flash Circuit
 The Battery, VS,
Charges the Cap
To Prepare for a
Flash
 Moving the Switch
to the Right
“Triggers” The
Flash
• i.e., The Cap
Releases its Stored
Energy to the Lamp
Engineering-43: Engineering Circuit Analysis
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Say
“Cheese”
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Flash Ckt Transient Response
 The Voltage Across the Flash-Ckt Storage
Cap as a Function of TIME
 Note That the Discharge Time (the Flash) is
Much Less Than the Charge-Time
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
General Form of the Response
 Including the initial
conditions the model
equation for the
capacitor-voltage or
the inductor-current
will be shown to be of
the form
dxt 

 xt   f t 
dt
 This is the General Eqn
 Now By Linear
Differential Eqn Theorem
(SuperPosition) Let
Engineering-43: Engineering Circuit Analysis
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 xp(t)  ANY Solution to the
General ODE
• Called the “Particular”
Solution
 xc(t)  The Solution to the
General Eqn with f(t) =0
• Called the “Complementary
Solution” or the “Natural”
(unforced) Response
• i.e., xc is the Soln to the
“Homogenous” Eqn
dxt 

 xt   0
dt
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
1st Order Response Eqns
 Given xp and xc
the Total Solution
to the ODE
xt   x p t   xc t 
 Consider the Case
Where the Forcing
Function is a Constant
• f(t) = A
 Now Solve the ODE in
Two Parts

dx p t 
 x p t   A
dt
dxc t 

 xc t   0
dt
 For the Particular Soln,
Notice that a x t   K
p
1
CONSTANT
Fits the Eqn: and so
dx p t 
dt
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
0
1st Order Response Eqns cont
 Sub Into the General
(Particular) Eqn xp and
dxp/dt
0  K1  A
or
K1  A
 Next, Divide the
Homogeneous (RHS=0)
Eqn by ∙xc(t) to yield
dxc t  dt
1

xc t 

Engineering-43: Engineering Circuit Analysis
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 Next Separate the
Variables & Integrate
 x t  dx t      dt
1
1
c
c
 Recognize LHS as a
Natural Log; so
ln xc t  
t

c
where c  const
 Next Take “e” to
The Power of the
LHS & RHS
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
1st Order Response Eqns cont
 Then
xc t   e t  c  ec e t 
xc t   K 2e t 
 Note that Units of TIME
CONSTANT, , are Sec
 Thus the Solution for a
Constant Forcing Fcn
xt   x p t   xc t 
xt   K1  K 2e
Engineering-43: Engineering Circuit Analysis
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t / 
 For This Solution
Examine Extreme
Cases
• t =0
• t→∞
x0  K1  K 2
xt     K1  K 2e   K1
 The Latter Case (K1) is
Called the Steady-State
Response
• All Time-Dependent
Behavior has dissipated
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Effect of the Time Constant
Tangent reaches x-axis in one
time constant
Decreases 63.2%
after One Time
Constant time
Drops to 1.8% after 4 Time
Constants
Engineering-43: Engineering Circuit Analysis
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e 4  0.0183
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Large vs Small Time Constants
 Larger Time Constants Result in Longer
Decay Times
• The Circuit has a Sluggish Response
Quick to Steady-State
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
1st Order Ckt Solution Plan
1. Use some form of KVL/KCL, 𝑖 =
and 𝑣 =
𝑑𝑖
𝐿
𝑑𝑡
𝑑𝑣
𝐶
𝑑𝑡
to develop an ODE
2. Isolate the “0th” order term which
reveals the
•
Time Constant
•
Forcing Function
3. “EyeBall” the ODE to “Guesstimate the
Particular Solution, 𝑖𝑝 𝑡 or 𝑣𝑝 𝑡
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
1st Order Ckt Solution Plan
4. CHECK that the particular Solution
Satisfies the ODE
5. Separate the Variables in the
Homogeneous Equation and Integrate
to obtain the Complementary Solution
𝑖𝑐 𝑡 or 𝑣𝑐 𝑡
6. A the particular and complementary
solutions to obtain the total Solution;
𝑥𝑡𝑜𝑡 𝑡 = 𝑥𝑐 𝑡 + 𝑥𝑝 𝑡
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
1st Order Ckt Solution Plan
 The Solution Should include a Negative
Exponential with an UnKnown
PreFactor
7. Use the initial condition in the total
solution to determine the Prefactor
which completes the total solution
8. Check the total Solution for extreme
cases:
• t = 0+
Engineering-43: Engineering Circuit Analysis
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• t→∞
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Example  1st Order Ckt Soln
 For the Ckt Below Find 𝑖𝐿 𝑡 for 𝑡 > 0
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Time Constant Example
 Charging a Cap
vC  v S
RS
vS


RS
 Now let
a
+
C
vc
_
dvC b
C
dt
 Use KCL at node-a
dvC vC  vS
C

0
dt
RS
dvC vC vS
C


dt RS RS
Engineering-43: Engineering Circuit Analysis
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• vC(t = 0 sec) = 0 V
• vS(t)= VS (a const)
 ReArrange the KCL Eqn
For the Homogenous
Case where Vs = 0
dvC dt
1

vC
CRS
 Thus the Time Constant
  CRS
dxc t  dt
1
   xc t   K 2 e t 
xc t 

Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Time Constant Example cont
 “Fully” Charged Criteria
 Charging a Cap
RS
• vC >0.99VS
a
e  t   0.01 or
t    ln 0.01  4.6
+
vS


C
OR
vc
_
b
 The Solution Can be
shown to be
t
vC (t )  VS  VS e



Or vC (t )  VS 1  e t 
Engineering-43: Engineering Circuit Analysis
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
  RS C
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Differential Eqn Approach
 Conditions for Using This Technique
• Circuit Contains ONE Energy Storing Device
• The Circuit Has Only CONSTANT,
INDEPENDENT Sources
• The Differential Equation For The Variable Of
Interest is SIMPLE To Obtain
– Normally by Using Basic Analysis Tools; e.g., KCL, KVL,
Thevenin, Norton, etc.
• The INITIAL CONDITION For The Differential
Equation is Known, Or Can Be Obtained Using
STEADY STATE Analysis Prior to Switching
– Based On: Cap is OPEN, Ind is SHORT
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Example
 Given the RC Ckt At
Right with
• Initial Condition (IC):
– v(0−) = VS/2
 Model t>0 using KCL at
v(t) after switch is made
vt   VS
dv
t   0
C
 Looks Like a Single
R
dt
E-Storage Ckt w/ a
 Find Time Constant; Put
Constant Forcing Fcn
Eqn into Std Form
• Assume Solution of Form
 Find v(t) for t>0

t
x(t )  K1  K 2 e , t  0

K1  x(); K1  K 2  x(0)
Engineering-43: Engineering Circuit Analysis
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• Multiply ODE by R
dv
RC t   vt   Vs
dt
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Example cont
 Compare Std-Form with
Model
• Const Force Fcn Model
dx
  x K 1
dt
• In This Case
dv
RC
(t )  v(t )  Vs
dt
 Next Check SteadyState (SS) Condition
• In SS the Time Derivative
goes to ZERO
Engineering-43: Engineering Circuit Analysis
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 Note: the SS condition
is Often Called the
“Final” Condition (FC)
 In This Case the FC
t    v(t )  Vs
confirms : K1  Vs
 Now Use IC to
Find K2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Example cont.2
 At t = 0+
• The Model Solution
v(0)  K1  K 2 e
0 
 K 2  v(0)  K1
v(0−)
 Recall The IC:
=
VS/2 = v(0+) for a Cap
• Then
K 2  v(0)  K1
K 2  VS 2  VS
K 2   VS 2
Engineering-43: Engineering Circuit Analysis
48
 The Total/General Soln

t
vt   K1  K 2e  , t  0 or
t



RC
v(t )  VS 1  0.5e 


 Check


v(0)  VS 1  0.5e  0  0.5Vs


v()  VS 1  0.5e    Vs
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx


 vR 
Inductor Exmpl
KVL
 Find i(t) Given
i (t )
• i(0−) = 0

vL

 Recognize Single
E-Storage Ckt w/ a
Constant Forcing Fcn
 To Find the ODE Use
KVL for Single-Loop ckt
• Assume Solution of Form

t
x(t )  K1  K 2 e  , t  0
di
VS  vR  vL  Ri (t )  L (t )
dt
K1  x(); K1  K 2  x(0)  Now Consider IC
 In This Case x→i

t
i(t )  K1  K 2e  ;
Engineering-43: Engineering Circuit Analysis
49
t 0
i (t )
• By Physics, The Current
Thru an Inductor Can NOT
Change instantaneously
 i (0  )  i (0 )  0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Inductor cont
 Casting ODE in
Standard form
VS
L di
(t )  i (t ) 
R dt
R
i (t )
 Next Using IC (t = 0+)
K 2  i ( 0  )  K1
 Recognize Time Const
 L R
K 2  0  VS R
K 2   VS R
 Also Note FC
t    i(t )  Vs R
 Thus
K1  Vs R
i(t )  K1  K 2e
Engineering-43: Engineering Circuit Analysis
50
 Thus the ODE Solution

t

R

t

Vs
L
1  e 


R 

Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Solution Process Summary
1. ReWrite ODE in Standard Form
•
Yields The Time-Constant, 
2. Analyze The Steady-State Behavior
•
Finds The Final Condition Constant, K1
3. Use the Initial Condition
•
Gives The Exponential PreFactor, K2
4. Check: Is The Solution Consistent
With the Extreme Cases
•
•
t = 0+
t→
Engineering-43: Engineering Circuit Analysis
51
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Solutions for f(t) ≠ Constant
1. Use KVL or KCL to Write the ODE
2. Perform Math Operations to Obtain a
CoEfficient of “1” for the “Zeroth” order
Term. This yields an Eqn of the form
dx
  x  f t 
dt
(yields  )
3. Find a Particular Solution, xp(t) to the
FULL ODE above
• This depends on f(t), and may require
“Educated Guessing”
Engineering-43: Engineering Circuit Analysis
52
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Solutions for f(t) ≠ Constant
4. Find the total solution by adding the
COMPLEMENTARY Solution, xc(t) to
previously determined xp(t).
xc(t) takes the form: xc t   Ke t 
• The Total Solution
at this Point →
xt   Ke
t 
 x p t 
5. Use the IC at t=0+ to find K; e.g.; IC = 7
7  Ke
0 
 x p 0  7  K  M
• Where M is just a NUMBER
Engineering-43: Engineering Circuit Analysis
53
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Capacitor Example
 For Ckt Below Find vo(t) for t>0 (note f(t) = const; 12V)
R1
R2
C
 Assume a Solution of
the Form for vc

t
vC (t )  K1  K 2 e  , t  0
 At t=0+ Apply KCL
dvC
vC
C
(t ) 
0
dt
R1  R2
K1  vC (); K1  K 2  vC (0 ) ( R  R )C dvC (t )  v  0
1
2
c
dt
Engineering-43: Engineering Circuit Analysis
54
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Cap Exmp cont
 Step1: By Inspection of
the ReGrouped KCL
Eqn Recognize 
  ( R1  R2 )C
 6k  0.1mF
 0.6 s
 Now Examine the Reln
Between vo and vC
• a V-Divider
2
1
vO (t ) 
vC (t )  vC (t )
24
3
Engineering-43: Engineering Circuit Analysis
55
 Step-2: Consider The
Steady-State
• In This Case After the
Switch Opens The
Energy Stored in the Cap
Will be Dissipated as
HEAT by the Resistors
vo (t  )  vC (t  )  0
 With xp = K1 = SSsoln
K1  vC ()  0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Cap Exmp cont
 Now The IC
 If the Switch is Closed for a Long Time before t =0, a
STEADY-STATE Condition Exists for NEGATIVE Times
Recall: Cap is
OPEN to DC
 vo(0−) by V-Divider
 Recall Reln Between vo
and vC for t ≥ 0
2
2

vO (0 ) 
12V  12V v (t )  3v (t ) 
C
O
3 2 4
9
6
8

v
(
0

)

v
(
0

)

12V  8V
vO (0 )  V
C
C
9
3
Engineering-43: Engineering Circuit Analysis
56
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Cap Exmp cont
 Step-3: Apply The IC
K1  K 2  vC (0  )
and K1  vC ()  0

 K 2  vC (0 )  8V
 Now have All the
Parameters needed To
Write The Solution

t
vC (t )  K1  K 2e  , t  0
vC (t )  0  8Ve

t
0.6 s
Engineering-43: Engineering Circuit Analysis
57
, t  0)
 Recall vo = (1/3)vC
8 
vo (t )   V e
3 

t
0.6 s
, t 0
 Note: For f(t)=Const,
puttingthe ODE in Std
Form yields  and K1 by
Inspection:
dvC
( R1  R2 )C
(t )  vc  0
dt

K1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Thevenin/Norton Techniques
 Obtain The Thevenin Voltage Across The Capacitor,
Or The Norton Current Through The Inductor
Circuit
with
resistances
and
sources
a
Inductor
or
Capacitor
b
Representation of an arbitrary
circuit with one storage element
RTH a
Thévenin
VTH
Inductor
or
Capacitor


b
 With This approach can Analyze a SINGLE-LOOP,
or SINGLE-NODE Ckt to Find
• Time Constant using RTH
• Steady-State Final Condition using vTH (if vTH a constant)
Engineering-43: Engineering Circuit Analysis
58
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Thevenin Models for ODE
RTH a
VTH

iR

C
RTH a
ic
+
vc
_
b
Case 1.1
Voltage across capacitor
 KCL at node a
dvC
ic  C
dt
dv
v v
C C  C TH  0
dt
RTH
ic  iR  0
vC  vTH
iR 
RTH
dvC
 RTH C 
 vC  vTH
dt
Engineering-43: Engineering Circuit Analysis
59
VTH

 vR 


L
iL
b
vL

Case 1.2
Current through inductor
 KVL for Single Loop
vR  RTH iL
vR  vL  vTH
vL  L
 L

 RTH
diL
dt
L
diL
 RTH iL  vTH
dt
 diL
v

 iL  TH  I N
RTH
 dt
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Find ODE Soln By Thevenin
 Break Out the Energy Storage Device
(C or L) as the “Load” for a
Driving Circuit
 Analyze the Driving Ckt to Arrive at it’s
Thévenin (or Norton) Equivalent
 ReAttach The C or L Load
 Use KCL or KVL to arrive at ODE
 Put The ODE in Standard Form
Engineering-43: Engineering Circuit Analysis
60
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Find ODE Soln By Thevenin
 Recognize the Solution Parameters
 For Capacitor
•  = RTHC
• K1 (Final Condition) = vTH = vOC = xp
 For Inductor
•  = L/RTH
• K1 (Final Condition) = vTH/RTH = iSC = iN = xp
 In Both Cases Use the vC(0−) or iL(0−)
Initial Condition to Find Parameter K2
Engineering-43: Engineering Circuit Analysis
61
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Inductor Example
Find iO (t); t  0
6
6
iO (t )
6
3H

24V 
t 0
6
 The Variable Of Interest
Is The Inductor Current
 The Thévenin model
L diO
vTH
 iO 
RTH dt
RTH
Engineering-43: Engineering Circuit Analysis
62
 Since this Ckt has a
CONSTANT Forcing
Function of 0V (the 24V
source is switched OUT
at t = 0), Then The
Solution Is Of Thet Form

iO (t )  K1  K 2e  ; t  0
 Next Construct the
Thévenin Equivalent for
the Inductor
“Driving Circuit”
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Inductor Example cont
Thevenin for t>0
at inductor terminals
6
6

24V 
t 0
6
a
6
b
 From This Ckt Observe
vTH  0
RTH  6 6  6  6  10

L
3H

 0. 3s
RTH 10
Engineering-43: Engineering Circuit Analysis
63
 The f(t)=const ODE in
Standard Form
diO
0.3s
 iO  0 ; t  0
dt
 The Solution
Substituted into the
ODE at t > 0
t

 K 2  0t.3  

0.3 



0.3 
e    K1  K 2e   0
 0.3
 

 Combining the K2 terms
shows shows K1 = 0, thus
iO (t )  K 2e

t
0.3
;t 0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Inductor Example cont.2
Find iO (t); t  0
6
6
 Analyzing Ckt with 3H
Ind as Short Reveals
iO (t )
6
3H

24V 
t 0
6
 Now Find K2
 Assume Switch closed
for a Long Time
Before t = 0
24V 1  24V 

iO 
 
6 2  6  3 
16
 A; t  0
3
• The Reader Should
Verify the Above
 Then the Entire Solution
iO (t )  5.33 Ae

t
0.3 s
• Inductor is SHORT to DC
Engineering-43: Engineering Circuit Analysis
64
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
;t 0
Untangle to
find iO(0−)
 Be Faithful to Nodes
Engineering-43: Engineering Circuit Analysis
65
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
First Order Inductor Ciruit Transient Solution
5.5
diO
diO
 iO
0.3s   iO  0  
dt
dt 0.3s 
5.0
4.5
diO
 iO
diO



dt 3s 10
dt
4.0
Current (A)
3.5
diO

dt
3.0
2.5
t 0
t 0
i(t) (A)
 iO 0  

3s 10
16 A 3 16 10 A
A


 17.78
3s 10
33 s
s
2.0
iO (t )  5.333 Ae
1.5

t
0.3 s
;t 0
1.0
0.5
0.0
-0.2
0.0
0.2
0.4
file = Engr44_Lec_06-1_Last_example_Fall03..xls
Engineering-43: Engineering Circuit Analysis
66
0.6
0.8
1.0
1.2
1.4
1.6
Time (s)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
1.8
WhiteBoard Work
 Let’s Work This
Problem
t=0
24V
4
2
+
4
6
8
(1/4)H
VC(t)
-
 Well, Maybe
NEXT time…
Engineering-43: Engineering Circuit Analysis
67
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
WhiteBoard Work
Engineering-43: Engineering Circuit Analysis
68
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Engineering 43
Appendix
DE Approach
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-43: Engineering Circuit Analysis
69
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Differential Eqn Approach cont
 Math Property
• When all Independent Sources Are CONSTANT,
then for ANY variable y(t); i.e., v(t) or i(t), in The
Circuit The Solution takes the Form
yt   K1  K 2e

t

 The Solution Strategy
• Use The DIFFERENTIAL EQUATION And The
FINAL & INITIAL Conditions To Find The
Parameters K1 and K2
Engineering-43: Engineering Circuit Analysis
70
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Differential Eqn Approach cont
 If the ODE for y is
Known to Take This
Form
dy
a1
 a0 y  A
dt
y (0  )  y0

dy
K
 2 e
dt

t



  a0  K 1  K 2 e 





 A


t
t

Engineering-43: Engineering Circuit Analysis
71
 K 2  t
a1  
e
 

 a1

   a0  K 2 e   a0 K1  A
 

 We Can Use This
Structure to Find The
Unknowns. If: t

y (t )  K1  K 2e  , t  0

 Then Sub Into ODE
 Equating the TRANSIENT
(exponential) and
CONSTANT Terms Find
t
a1
 a1


   a0  K 2 e  0   
a0
 

A
K1 
a0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Differential Eqn Approach cont
 Up to Now
A
yt    K 2e
a0
 So Finally

t
a1 a0
 Next Use the Initial
Condition
y (0  )  y 0
y (0)  K1  K 2 e 0  y0
K1  K 2  y0
or
K 2  y0  K1
Engineering-43: Engineering Circuit Analysis
72
A 
A
y t     y0  e
a0 
a0 

t
a1 a0
 If we Write the ODE in
Proper form We can
Determine By
Inspection  and K1
K1
dy
a1 dy
A
a1
 a0 y  A 
y
dt
a0 dt
a0

Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Inductor Example
 For The Ckt Shown
Find i1(t) for t>0
i1 (t )
 Assume Solution
of the Form t

i1 (t )  K1  K 2 e  , t  0

vL
K1  i1 (); K1  K 2  i1 (0)
 The Model for t>0 →
KVL on single-loop ckt
di1
L
 6  12i1 (t )  0
dt
 Rewrite In Std Form
Engineering-43: Engineering Circuit Analysis
73

2 di1
(t )  i1 (t )  0
18 dt
 Recognize Time Const
1
 S
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
L
Inductor Example cont
 Examine Std-Form Eqn
to Find K1
1 di1
(t )  i1 (t )  0
9 dt
 K1  0
 For Initial Conditions
Need the Inductor
Current for t<0
 Again Consider DC
(Steady-State)
Condition for t<0
Engineering-43: Engineering Circuit Analysis
74
 The SS Ckt Prior to
Switching
• Recall An Inductor is a
SHORT to DC
i1 (0)
 So
i1 (0) 
12V
 1A
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx
Inductor Example cont.1
 Now Use Step-3 To
Find K2 from IC
• Remember Current Thru
an Inductor Must be
Time-Continuous
 The Answer

i1 (t )  e
t
1
9
[ A], t  0
i1 (t )  e 9t [ A], t  0
i1 (0)  i1 (0)  K1  K 2
 1[ A]  K 2
0
• Recall that K1 was zero
 Construct From the
Parameters The ODE
Solution
Engineering-43: Engineering Circuit Analysis
75
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx