EENG 2610: Circuit AnalysisClass 12: First-Order Circuits

Oluwayomi AdamoDepartment of Electrical EngineeringCollege of Engineering, University of North Texas

Transient Analysis Circuits in transition

Transient analysis: study of circuit behavior in transition phase. Transition phase is caused by a sudden change in circuit

Suddenly apply or remove voltage or current source, Open or close a switch in the circuit.

Transition is affected by capacitor or inductor in circuit No transient analysis is needed for resistive circuit network. Because capacitor and inductor can store energy, the circuit

response to a sudden change will go through a transition period before settling down to a steady-state value.

First-Order and Second-Order Circuits First-order circuits contain only a single capacitor or inductor Second-order circuits contain both a capacitor and an inductor Two techniques for transient analysis that we will learn:

Differential equation approach Step-by-step approach

Laplace transform method is a much simpler method for transient analysis – you will learn it in another course.

First-order differential equationSolution of transient analysis requires to solve a first-order differential equation: )()(

)(tftax

dt

tdx

Now we only consider f (t) is constant, that is, f (t) = A :

Ataxdt

tdx )(

)(

General solution to the first-order differential equation:

/21)( teKKtx

circuit theofconstant time:

,/

,/1

,)(

,)0(

1

1

21

aAK

a

Kx

KKx

)(txp

)()()( txtxtx cp

xc(t) is the solution to the homogeneous equation: 0)()(

taxdt

tdx

Solution:

)(txc

: particular integral solution (or forced response)

: complementary solution (or natural response)

A drop of 63.2%

%1%67.0)5( 22 KKxc

The Differential Equation Approach State-variable approach

Step 1: Find state-variables Voltage across the capacitor and current through inductor are

called state-variables. State-variables cannot change instantaneously.

Step 2: Find initial value of state variables at

Step 3: Write KCL equation for the voltage across the capacitor and/or KVL equation for the current through the inductor for

Step 4: Solve first-order differential equation

)0(usually 00 ttt

0tt

0t

Example 7.1: Calculate the currentAssume the switch has been in position 1 for a long time.

.0for )( tti

0t

Example 7.2: Find output voltage .0for )( ttvo 0t)(tiL

0t

)(tiL

Step-by-Step Approachfor First-Order Circuits

Step1: We assume a solution for the variable x(t) (either voltage or current) of the form

Step 2: Assuming that the original circuit has reached steady state before switch actions, draw this previous circuit with the capacitor replaced by an open circuit or the inductor replaced by a short circuit. Solve for the voltage across the capacitor, vC(0-), or the current through the inductor, iL(0-), prior to switch action.

Step 3: Voltage across a capacitor and the current through an inductor cannot change in zero time: vC(0+)=vC(0-), iL(0+)=iL(0-). Therefore, draw the circuit valid for t=0+ with the switches in their new positions. Replace a capacitor with a voltage source of value vC(0+) or an inductor with a current source of value iL(0+). Solve for the initial value of the variable x(0+).

/21)( teKKtx

Step 4: Assuming that steady state has been reached after the switches are thrown, draw the equivalent circuit, valid for , by replacing the capacitor by an open circuit or the inductor by a short circuit. Solve for the steady-state value of the variable

Step 5: Find Thevenin equivalent resistance at the terminals of the storage element by looking into the circuit from the terminals of the storage element. The time constant for a circuit containing a capacitor is , and for a circuit containing an inductor the time constant is

Step 6: Evaluate the constants K1, K2

in step 1 using the relations:

Then, we can find that the solution is

Important: This solution form applies only to a first-order circuit having constant, DC sources.

5t

)(x

ThR

CRThThRL /

1

21

)(

)0(

Kx

KKx

/)]()0([)()( texxxtx

If switch action occurs at any time t0, the step-by-step analysis yields the following equations:

1

210

)(

)(

Kx

KKtx

0/)(

0 for ,)]()([)()( 0 ttextxxtx tt

The function is essentially time-shifted by t0 seconds.

Example 7.3: The circuit is in steady state prior to time t=0, when the switch is closed. Calculate the current i(t) for t>0.

-0t

0t

t

THRThR Find