First Order Circuit Analysis: Analyzing Circuits with Energy Storing Elements using ODEs, Slides of Electrical Circuit Analysis

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RC & RL

st

Order Ckts

C&L Summary

1 st^ & 2 nd^ 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

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

The Ckt The DC Math Model

Analysis

v i

G =

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

Basic Concept

• Inductors And Capacitors

Can Store Energy

• Under Certain Conditions This

Energy Can Be Released

• RATE OF ENERGY RELEASE Depends on the

parameters Of The Circuit Connected To The

Terminals Of The Energy Storing Element

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

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

 xp (t) ≡ ANY Solution to the General ODE

• Called the “Particular” Solution  xc (t) ≡ The Solution to the General Eqn with f(t) =
• Called the “Complementary Solution” or the “Natural” (unforced) Response
• i.e., xc is the Soln to the  This is the General Eqn “Homogenous” Eqn

 Now By Linear Differential Eqn Theorem (SuperPosition) Let

x ( ) t^ f ( ) t

dt

dx t

( )

• x ( ) t = 0

dt

dx t

τ

1 st^ Order Response Eqns cont

• Sub Into the General (Particular) Eqn x (^) p and dx (^) p /dt

K A

K A

1

1

or

 Next, Divide the Homogeneous (RHS=0) Eqn by xc (t) to yield

 Next Separate the Variables & Integrate

( ) ( ) τ

1 = − x t

dx t dt

c

c

( )

dx ( ) t dt ∫ (^) x (^) c t c ∫

τ

 Recognize LHS as a Natural Log; so

[ ( )]

where const

ln

c

c

t

xc t

τ

 Next Take “e” to The Power of the LHS & RHS

1 st^ Order Response Eqns cont

• Then

( )

( ) τ

τ τ

t c

t c c t c

x t K e

x t e e e

−

− + −

2

 Note that Units of TIME CONSTANT, τ, are Sec

 Thus the Solution for a Constant Forcing Fcn

 For This Solution Examine Extreme Cases

• t =
• t → ∞

 The Latter Case is Called the Steady-State Response

• All Time-Dependent Behavior has dissipated

( ) ( ) ( )

( ) 1 2^ t /τ

p c

x t K K e

x t x t x t

= +^ −

( )

( ) (^121)

x t K K e K

x K K

−∞

Large vs Small Time Constants

• Larger Time Constants Result in Longer Decay

Times

• The Circuit has a Sluggish Response

Quick to Steady-State

Time Constant Example

• Charging a Cap

 Use KCL at node-a

 Now let

• vC(t = 0 sec) = 0 V
• vS(t)= VS (a const)  Rearrange the KCL Eqn For the Homogenous Case where Vs = 0

 Thus the Time Constant

−

vS +

R (^) S (^) a

b

C

+ vc _

dt

C dvC

S

C S R

v − v

S

S S

C C

S

C C S

R

v

R

v

dt

dv

C

R

v v

dt

dv

C

C S

C v CR

dv dt 1 = −

τ = CR S

( )^ τ

τ c^ t c

c (^) x t K e x t

dx t dt = − ⇒ = − 2

1

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

Example

• Given the RC Ckt At Right with - Initial Condition (IC): - v(0−) = VS /
• Find v(t) for t>
• Looks Like a Single E-Storage Ckt w/ a Constant Forcing Fcn - Assume Solution of Form

 Model t>0 using KCL at v(t) after switch is made

( ); ( 0 )

( ) , 0 1 1 2

1 2 = ∞ + = +

= + >

−

K x K K x

x t K K e t

t τ

( ) 0

( )

• =

− t dt

dv C R

v t VS

 Find Time Constant; Put Eqn into Std Form

• Multiply ODE by R

dt t v t V^ s

dv RC ( )+ ( )=