DC Circuits: First-Order...

EENG223: CIRCUIT THEORY I

DC Circuits:

First-Order Circuits Hasan Demirel


• Introduction

• The Source-Free RC Circuit

• The Source-Free RL Circuit

• Step Response of an RC Circuit

• Step Response of an RL Circuit

First-Order Circuits


• A first-order circuit can only contain one energy storage element (a capacitor or an inductor).

• The circuit will also contain resistance.

• So there are two types of first-order circuits: RC circuit RL circuit

• A first-order circuit is characterized by a first-order differential equation.

First-Order Circuits: Introduction


• A source-free circuit is one where all independent sources have been disconnected from the circuit after some switch action.

• The voltages and currents in the circuit typically will have some transient response due to initial conditions (initial capacitor voltages and initial inductor currents).

• We will begin by analyzing source-free circuits as they are

the simplest type. Later we will analyze circuits that also contain sources after the initial switch action.

First-Order Circuits: The Source-Free Circuits


• A source-free RC circuit occurs when its dc source is suddenly disconnected.

• The energy already stored in the capacitor is released to the resistors.

First-Order Circuits: The Source-Free RC Circuits

V0

• Since the capacitor is initially charged, we can assume that at time t=0, the initial voltage is:

• Then the energy stored:

• Applying KCL at the top node:

• By definition, iC =C dv/dt and iR = v/R. Thus,



V0

• This is a first-order differential equation, since only the first derivative of v is involved.

• Rearranging the terms:

• Integrating both sides:

• ln A is the integration constant. Thus

• Taking powers of e produces:

• From the initial conditions: v(0)=A=V0

• The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excitation.


• General form of the Differential Equations (DE) and the response for a 1st-order source-free circuit:


In general, a first-order D.E. has the form:

00)(1

tfortxdt

dx

Solving this DE (as we did with the RC circuit) yields:

0)0()(

tforextx

t

here τ= (Greek letter “Tau”) = time constant(in seconds)


• Notes concerning τ:


So, for an RC circuit: RC

1) For the Source-Free RC circuit the DE is: 00)(1

tfortvRCdt

dv

2) τ is related to the rate of exponential decay in a circuit as shown below.

3) It is typically easier to sketch a response in terms of multiples of τ than to be concerning with scaling of the graph.


First-Order Circuits: The Source-Free RC Circuits Ex. 7.1: In Fig. 7.5, let vC(0)= 15 V. Find vC , vx and ix for t>0.

Solution

• Equivalent Circuit for the above circuit can be generated:



• Equivalent Resistance seen by a Capacitor For the RC circuit in the previous example, it was determined that

τ= RC. But what value of R should be used in circuits with multiple resistors?

In general, a first-order RC circuit has the following time constant:

where REQ is the Thevenin resistance seen by the capacitor.

More specifically, REQ = R (seen from the terminals of the capacitor for t>0 with independent sources killed.)

CREQ


First-Order Circuits: The Source-Free RC Circuits Ex. : Refer to the circuit below. Let vC(0)= 45 V. DeterminevC , vx and io for t≥0.

Solution

• Time constant τ :

• Then:

• Consider Req seen from the capacitor.

12818

612eqR

sCReq 43

112

V45)0()( 25.04 t

t

CC eevtv

V15453

1)(

84

4)( 25.025.0 tt

Cx eetvtv

V75.38

4515

8

)()()( 25.0

25.025.0t

tt

Cxo e

eetvtvti


First-Order Circuits: The Source-Free RC Circuits Ex. 7.2: The switch in the circuit below has been closed for a long time, and it is

opened at t= 0. Find v(t) for t≥0. Calculate the initial energy stored in the capacitor.

Solution

• For t<0 the switch is closed; the capacitor is an open circuit to dc, as represented in Fig. (a).

• For t>0 the switch is opened, and we have the RC circuit shown in Fig. (b).


• Then:

• The initial energy stored in the capacitor:


First-Order Circuits: The Source-Free RC Circuits Ex. : If the switch in Fig. below opens at t= 0, find v(t) for t≥0 and wC(0).

Solution

• For t>0 the switch is opened, and we have the RC circuit shown in Fig. (b).


• Then:

• The initial energy stored in the capacitor:

• For t<0 the switch is closed; the capacitor is an open circuit to dc as shown in Fig. (a).

V8)0(

0V82463

3)(

0

Vv

tfortv

C

C

(a)

(b)

316

412eqR

sCReq 5.06

13

V8)0()( 25.0 t

t

C eevtv

J33.5)8(6

1

2

1)0(

2

1)0( 22

CCvwC


First-Order Circuits: The Source-Free RC Circuits Ex. : The switch in the circuit shown had been closed for a long time and then opened at time t = 0.

a) Determine an expression for v(t). b) Graph v(t) versus t. c) How long will it take for the capacitor to

completely discharge? d) Determine the capacitor voltage at time

t=100ms.

e) Determine the time at which the capacitor voltage is 10V.


• A source-free RL circuit occurs when its dc source is suddenly disconnected.

• The energy already stored in the inductor is released to the resistors.

First-Order Circuits: The Source-Free RL Circuits

• At time, t=0 , the intuctor has the initial current:

• Then the energy stored:

• We can apply KVL around the loop above :

• By definition, vL =L di/dt and vR = Ri. Thus,

I0

t=0



• This is a first-order differential equation, since only the first derivative of i is involved.

• Rearranging the terms and integrating:

• Then:

• Taking powers of e produces:

• Time constant for RL circuit becomes:

The natural response of the RL circuit is an exponential decay of the initial current.

I0

t=0


• General form of the Differential Equations (DE) and the response for a 1st-order source-free circuit:


In general, a first-order D.E. has the form:

00)(1

tfortxdt

dx

Solving this DE (as we did with the RL circuit) yields:

0)0()(

tforextx

t

Then:

Where:

0)0()( 0

tforeIeiti

tt

R

L



• Equivalent Resistance seen by an Inductor For the RL circuit , it was determined that τ= L/R. As with the RC circuit,

the value of R should actually be the equivalent (or Thevenin) resistance seen by the inductor.

In general, a first-order RL circuit has the following time constant:

where REQ is the Thevenin resistance seen by the inductor.

More specifically, REQ = R (seen from the terminals of the capacitor for t>0 with independent sources killed.)

EQR

L


First-Order Circuits: The Source-Free RL Circuits Ex. 7.3: Assuming that i(0) =10 A, calculate i(t) and ix(t) in the circuit below.

Solution

• Substituting Eq. (2) into Eq. (1) gives.

• Thevenin resistance at the inductor terminals. we insert a voltage source with v0=1 V. Applying KVL to the two loops results (1)

(2)


First-Order Circuits: The Source-Free RL Circuits Ex. 7.3: Assuming that i(0) =10 A, calculate i(t) and ix(t) in the circuit below.

Solution

• Time constant is:

• Hence,

• Thus, the current through the inductor is:


First-Order Circuits: The Source-Free RL Circuits Ex. 7.4: The switch in the circuit below has been closed for a long time. At t=0

the switch is opened. Calculate i(t) for t>0.

Solution

• When t<0 the switch is closed, and the inductor acts as a short circuit to dc,

• Using current division:

• Current through an inductor cannot change instantaneously,


First-Order Circuits: The Source-Free RL Circuits Ex. 7.4: The switch in the circuit below has been closed for a long time. At t=0

the switch is opened. Calculate i(t) for t>0.

Solution

When t>0 the switch is open and the voltage source is disconnected. We now have the source-free RL circuit in Fig. (b).

• The time constant is :

• Thus,


First-Order Circuits: The Source-Free RL Circuits Ex. : Determine an expression for i(t). Sketch i(t) versus t.


First-Order Circuits: Step Response of an RC Circuit

• Step Response (DC forcing functions) • Consider circuits having DC forcing functions for t > 0 (i.e., circuits that have

independent DC sources for t > 0).

• The general solution to a differential equation has two parts: • x(t) = xh+ xp = homogeneous solution + particular solution • or • x(t) = xn+ xf = natural solution + forced solution

• xn is due to the initial conditions in the circuit • and xf is due to the forcing functions (independent voltage and current

sources for t > 0).

• xf in general take on the “form” of the forcing functions, • So DC sources imply that the forced response function will be a constant(DC), • Sinusoidal sources imply that the forced response will be sinusoidal, etc.



• Step Response (DC forcing functions) • Since we are only considering DC forcing functions in this chapter, we assume

that : xf = B (constant). • Recall that a 1st-order source-free circuit had the form Ae-t/τ. Note that there

was a natural response only since there were no forcing functions (sources) for t > 0. So the natural response was

0/ tforAex t

n

• The complete response for 1st-order circuit with DC forcing functions therefore will have the form: x(t) = xf + xn

/)( tAeBtx

• The “Shortcut Method”: An easy way to find the constants B and A is to evaluate x(t) at 2 points. Two convenient points at t = 0 and t = ∞ since the circuit is under dc conditions at these two points. This approach is sometimes called the “shortcut method.”



• Step Response (DC forcing functions) • The “Shortcut Method” :

So, x(0) = B + Ae0= B + A And x(∞) = B + Ae-∞= B

• Complete response yields the following expression:

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

• The Shortcut Method- Procedure: The shortcut method will be the key method used to analyze 1st-order circuit with DC forcing functions:

1. Analyze the circuit at t = 0-: Find x(0-) = x(0+)

2. Analyze the circuit at t = ∞: Find x(∞) 3. Find τ= REQC or τ= L/REQ

4. Assume that x(t) has the form x(t) = x(∞)+[x(0) –x(∞)] e-t/τ using x(0) and x(∞)



• Step Response (DC forcing functions)

Notes: The “shortcut method” also works for source-free circuits, but x(∞) = B=0

since the circuit is dead at t = ∞. If variables other than vC or iL are needed, it is generally easiest to solve for vC or iL first and then use the result to find the desired variable.


First-Order Circuits: Step Response of an RC Circuit Ex. 7.10: The switch in Fig. Below has been in position A for a long time. At t=0

the switch moves to B. Determine v(t) for t>0 and calculate its value at t =1 s and 4 s.

Solution

• Voltage across the capacitor just before t=0. Capacitor is open circuit under dc conditions:

• Capacitor voltage cannot change instantaneously:

• For t>0 (switch to B). Thevenin Resistance connected to the capacitor:

• Time constant:


First-Order Circuits: Step Response of an RC Circuit Ex. 7.10: The switch in Fig. below has been in position A for a long time. At t=0

the switch moves to B. Determine v(t) for t>0 and calculate its value at t 1 s and 4 s.

Solution

• Since the capacitor acts like an open circuit to dc at steady state, v(∞) = 30 V. Thus,



Ex. : Find v(t) and i(t) for t ≥0.


First-Order Circuits: Step Response of an RL Circuit

Ex. : Find v(t) and i(t) for t ≥0.

DC Circuits: First-Order...

Documents

Transcript of DC Circuits: First-Order...