dynamic circuits : first-order transientcktse.eie.polyu.edu.hk/eie201/3.transient.pdf · dynamic...

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Electronic Circuits 1

Dynamic circuits :First-order transient

Contents• Inductor and capacitor• Simple RC and RL circuits• Transient solutions

Prof. C.K. Tse: Dynamic circuits—Transient

2Prof. C.K. Tse: Dynamic circuits—Transient

Constitutive relation♦ An electrical element is defined by its relationship between v and i.

This is called constitutive relation. In general, we write

♦ For a resistor,♦ v = i R

♦ The constitutive relation of a resistor has no dependence upon time.

v f i i g v= =( ) ( ) or

+ v –

i


Capacitor and inductor

+ vc –

icThe constitutive relation of alinear capacitor is:

The constitutive relation of alinear inductor is:

i Cdvdtc

c=

v LdidtL

L=+ vL –

iL

C

L

where the proportionalityconstant C is capacitance(unit is farad or F)

where the proportionalityconstant L is inductance(unit is henry or H)


What happens if a circuit has C and/or L?

♦ The circuit becomes dynamic. That means:

♦ Its behaviour is a function of time.

♦ Its behaviour is described by a (set of) differentialequation(s).

♦ It has a transient response as well as a steady state.


Resistive circuits have no transient♦ Consider a resistive circuit.

♦ When the switch is turned on, thevoltage across R becomes Vimmediately (in zero time).

♦ v = V = i R for all t > 0

♦ i = V / R for all t > 0


A simple first-order RC circuit♦Let us consider a very simple dynamic circuit, whichcontains one capacitor.

♦After t = 0, the circuit is closed. So, we can easily write

♦and

♦Thus, we have

♦Thus, we have

♦If the initial condition is vC(0+) = 0, then A = –Vo.

♦Thus, the solution is

⇒

Vo for t>0


Transient response of the RC circuit♦Once we have the capacitor voltage, we can findanything.

♦Starting with

♦We can derive the current as

♦We see the solution typically has a TRANSIENTwhich dies out eventually, and as t tends to ∞, thesolution settles to a steady state.

time constant


A simple first-order RL circuit♦Consider a RL circuit.♦Before t = 0, the switch is closed (turned on).Current goes through the switch and nothing goesto R and L. Initially, iL(0–) = 0.♦At t = 0, the switch is opened. Current goes to Rand L.♦We know from KCL that Io = iR + iL for t > 0, i.e.,

♦The constitutive relations give♦Hence,♦ ⇒

♦The solution is

From the initial condition, wehave iL(0–) = 0. Continuity of theinductor current means thatiL(0+) = iL(0–) = 0. Hence,

A = –Io

Thus,


Transient response of the RL circuit♦ Starting with

♦ We can find vL(t):

time constant


Observation — first-order transients♦ First order transients are always like these:


Let’s do some math

0

5

x(t)

t

x(t) = 5(1 – e–t/τ)

0

5

x(t)

t

x(t) = 5 e–t/τ

0

6

x(t)

t1

x(t) = 1 + 5(1 – e–t/τ)

x(t) = 1 + 5 e–t/τ

0

5

x(t)

t

–2

x(t) = –2 + 7(1 – e–t/τ)

0

4

x(t)

t

–3

0

6

x(t)

t1

x(t) = –3 + 7 e–t/τ


General first-order solution

NO NEED TO SOLVE ANY EQUATION, just find

1. the starting point of capacitor voltage or inductor current2. the ending point of ………… ………. ……. ………. ……….3. the time constant τ


Finding τ

For the simple first-order RC circuit, τ = C R.For the simple first-order RL circuit, τ = L / R.

The problem is

Given a first-order circuit (which may look complicated),how to find the equivalent simple RC or RL circuit.


A quick way to find τSince the time constant is independent of the sources, we first of all set allsources to zero. That means, short-circuit all voltage sources and open-circuit all current sources. Then, reduce the circuit to

+–

R1

R2 C

R1

R2 C CR1 || R2

CeqReq Req Leqeither or

Example:

τ = C (R1 || R2)


Example 1 (boundary conditions given)Find vc(t) for t > 0 without solving anydifferential equation.

Step 1: initial point (given)

vc(0–) = 50 V is known (but not what we want).Continuity of cap voltage guarantees thatvc(0+) = vc(0–) = 50 V.

Step 2: final point (almost given)

vc(∞) = –20 V.

Step 3: time constant

The equivalent RC circuit is:

Thus, τ = CR.

Answer is:

vc(t) = –20 + 70 e–t/CR

0

50

t–20


Example 2 (non-trivial boundary conditions)Find v1(t) and v2(t) for t > 0 without solvingany differential equation.

+v1–

+v2–

C12F

C23F

i1

R = 1Ωt=0

i2Suppose v1(0–) = 5 V and v2(0–) = 2 V.Problem: how to find the final voltage values.

Form 7 solution:You considered the charge transfer ∆q from C1 to C2.

++++++ ++++

++∆q

q1 ⇒ q1 – ∆q q2 ⇒ q2 + ∆q

Use charge balanceand KVL equations tofind the final voltagevalues.Clumsy solution!


Example 2 (elegant solution)

We need not consider CHARGE!

Step 1: initial point (given)

v1(0–) = 5 V and v2(0–) = 2 V are known.Continuity of cap voltage guarantees thatv1(0+) = 5 V and v2(0+) = 2 V.

Step 2: final point (non-trivial)

C1: for all t

C2: for all t

After t>0, we have i1 = –i2, i.e.,

+v1–

+v2–

C12F

C23F

i1

R = 1Ωt=0

idvdt

idvdt

11

22

2

3

=

=

2 3 01 2dv

dtdvdt

+ =

⇒ 2v1(t) + 3v2(t) = K for all t > 0.

i2

Integration constant

At t = 0+, this equation means2*5 + 3*2 = K. Thus, K = 16.Thus,2v1(t) + 3v2(t) = 16 for t > 0.

At t =∞, we have v1(∞)=v2(∞)from KVL. Hence, 2v1(∞)+3v1(∞)=16⇒ v1(∞)=v2(∞)=16/5 V.


Example 2 (elegant solution)

Step 3: time constant

The circuit after t = 0 is

This can be reduced to

The time constant is

+v1–

+v2–

C12F

C23F

i1

R = 1Ω

i2

C1 C2

C1 +C2

= 6/5 F R = 1Ω

τ =

+× =

C CC C

R1 2

1 2

65

s


Example 2 (answer)

5V

16/5=3.2V

v1

t

2V

16/5=3.2V

v2

t

v t e t1

5 63 2 1 8( ) . . /= + − V v t e t2

5 62 1 2 1( ) . ( /= + − − ) V

We can also find the current by i t

v t v tR

e t( )( ) ( ) /=

−= −1 2 5 63 A


General procedure♦ Set up the differential equation(s) for the circuit in terms of capacitor

voltage(s) or inductor current(s).

♦ The rest is just Form 7 Applied Math!

♦ E.g.,

♦ Get the general solution.

♦ There should be n arbitrary constants for an nth-order circuit.

♦ Using initial conditions, find all the arbitrary constants.

d v

dtA

dvdt

Bv Cc cc

2

2+ + =

In the previous example:


Basic question 1♦Why must we choose capacitor voltage and inductor current as thevariable(s) for setting up differential equations?

♦ Never try to set differential equation in terms of other kinds of variables!

♦Answer:♦Capacitor voltages and inductor currents are guaranteed to be CONTINUOUSbefore and after the switching. So, it is always true that

dvdt

kv VRR o+ =

v v i iC C L L( ) ( ) ( ) ( )0 0 0 0− + − += = and


Basic question 1♦Then, why capacitor voltages and inductor currents are guaranteed to becontinuous?

♦Answer:♦Let’s try to prove it by contradiction. Suppose vc and iL are discontinuous at t = 0.That means,

♦Now, recall the constitutive relations.

♦Then, we have

♦So, capacitor voltages and inductor currents must not be discontinuous.

v v i iC C L L( ) ( ) ( ) ( )0 0 0 0− + − +≠ ≠ and

i Cdvdtc

c= v LdidtL

L=and

i vC L→ ∞ → ∞ and

which is not permitted in the physical world.

vc or iL

t

slope = ∞


Basic question 2♦How to get the differential equation systematically for any circuit?

♦For simple circuits (like the simple RC and RL circuits), we can get it by an adhoc procedure, as in the previous examples. But, if the circuit is big, it seemsrather difficult!

♦Hint:♦Graph theory. (We’ll look into details later.)

dynamic circuits : first-order transientcktse.eie.polyu.edu.hk/eie201/3.transient.pdf · dynamic...

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