RC and RL CircuitsFirst Order Circuits

Objective of LectureExplain the operation of a RC circuit in dc circuits

As the capacitor releases energy when there is: a transition in a unit step function voltage or current

source or a voltage or current source is switched out of the

circuit.

Explain the operation of a RL circuit in dc circuitAs the inductor releases energy when there is:

a transition in a unit step function voltage or current source

or a voltage or current source is switched out of the circuit.

Natural ResponseThe behavior of the circuit with no external sources

of excitation. There was a transition in the source in the circuit

where the unit step function changed from 1 to 0 at t ≤ 0s.

There is stored energy in the capacitor or inductor at time t = 0 s.

For t > 0 s, the stored energy is released Current flows through the circuit and voltages exist across

components in the circuit as the stored energy is released. The stored energy will decays to zero as time approaches

infinite, at which point the currents and voltages in the circuit become zero.

RC CircuitSuppose there is some charge on a capacitor

at time t = 0s. This charge could have been stored because a voltage or current source had been in the circuit at t<0s, but was switched off at t = 0s.

We can use the equations relating voltage and current to determine how the charge on the capacitor is removed as a function of time.The charge flows from one plate of the

capacitor through the resistor R to the other plate to neutralize the charge on the opposite plate of the capacitor.

Equations for RC Circuit

R

vi

dt

dvCi

ii

vv

RR

CC

RC

RC

0

ottCC

C

C

C

C

CC

CR

RC

VRC

tV

dtRCV

dV

RCdt

dV

V

RC

V

dt

dV

VV

R

V

dt

dVC

lnln

1

011

0

0

to

CR

t

oC

stCo

eR

VtItI

steVtV

RCVV

)()(

0 when )(

and If0

to

t

s

R

to

RRR

eCV

dttptw

eR

VIVtp

22

0

22

12

)()(

)(

Since the voltages are equal and the currents have the opposite sign, the power that is dissipated by the resistor is the power that is being released by the capacitor.

RL Circuits

RVIdt

dILV

II

VV

RR

LL

RL

RL

0

stLL

L

L

L

L

LL

RL

ItL

RI

dtL

R

I

dI

L

R

dt

dI

I

L

RI

dt

dI

RIdt

dIL

0lnln

01

0

0

t

oLR

t

oL

stLo

eRItVtV

steItI

R

LII

)()(

0 when )(

and If0

to

t

s

R

t

oRRR

eLI

dttptw

eRIIVtp

22

0

22

12

)()(

)(

Since the voltages are equal and the currents have the opposite sign, the power that is dissipated by the resistor is the power that is being released by the capacitor.

Initial ConditionCan be obtained by inserting a d.c. source to the

circuit for a time much longer than (t at least t = -5 )t .Capacitor

Vo is the voltage calculated by replacing the capacitor with a resistor with infinite resistance (an open circuit) after the voltage across the capacitor has reached a constant value (steady state).

Inductor Io is the current flowing through the inductor calculated by

replacing the inductor with a resistor with zero resistance (a short circuit) after the current flowing through the inductor has reached a constant value (steady-state).

PSpice

You can set the initial condition on a capacitor or inductor by doubling clicking on the part symbol. Then, enter a value for IC in the pop-up window that opens.

Time constant, tThe time required for the voltage across the

capacitor or current in the inductor to decay by a factor of 1/e or 36.8% of its initial value.

V(t) = 6V [1 - u(t)]

Example #1IL(t)

Example #1 (con’t)IL(t)

Example #1 (con’t)t < 0s

VL = 0VVR = 6VIL = IR = 2mA

Therefore,

Io = 2mA

Find the initial condition.

Io

Example #1 (con’t)

IL(t)

Example #1 (con’t)t > 0st = L/R = 10mH/3k /W =

3.33msIL = IR =Ioe-t/t = 2mA

e-(t/3.33ms)

VR = 3kW IR = 6V e-(t/3.33ms)

VL = L dIL/dt = -6V e-(t/3.33ms)

Note VR + VL = 0 V

IL(t)

Example #2

+

VC

_

Example #2 (con’t)

+

VC

_

Example #2 (con’t)Calculate the initial condition - the voltage on

the capacitor. Replace the capacitor with an open circuit and find the voltage across the two terminals. Note that in this circuit, current will flow through R3 so

there will be a voltage across C, but it will not be equal to the magnitude of the voltage source in the circuit.

Example #2 (con’t)The voltage across the capacitor is equal to

the voltage across the 12kW resistor.

VC = Vo = [12kW /15kW] 5V = 4V

Example #2 (con’t)

+

VC

_

+

VC

_

Example #2 (con’t)Further simplification of the circuit

+

VC

_

kkkkReq 31231 where

Example #2 (con’t)

0)()(

33.1)()(

33.1)(

)6/4(2)(

)(

)()(

4)(

)(

2mswhen t

6)2(3

6/)2(

6/)2(

6/)2(

6/)2(

/)2(

2

tItI

emAtItI

emAtI

emsVFdt

tdVCtI

tVtV

VetV

etVV

msFkCR

CR

msmstCR

msmstC

msmstCC

RC

msmstC

mst

mstCC

eq

eq

eq

eq

IC(t)

IR(t)

SummaryThe initial condition for:

the capacitor voltage (Vo) is determined by replacing the capacitor with an open circuit and then calculating the voltage across the terminals.

The inductor current (Io) is determined by replacing the inductor with a short circuit and then calculating the current flowing through the short.

The time constant for:an RC circuit is = t RCan RL circuit is = t L/R

The general equations for the natural response of:the voltage across a capacitor is the current through an inductor is

/)(

/)(

)(

)(o

o

ttoL

ttoC

eItI

eVtV