Fundamentals of Electrical Engineering Part 2Fundamentals of Electrical Engineering Part 2 Summersemester 2004 University Duisburg-Essen, Campus Duisburg Department of Electrical Engineering

1FEE 2

Fundamentals of Electrical Engineering Part 2

Summersemester 2004

University Duisburg-Essen, Campus DuisburgDepartment of Electrical Engineering

2FEE 2

Syllabus

8.Transient Response of Passive Circuits

9.Electric Fields

10.Magnetic Fields and Electromagnetic Induction

11.Semiconductor Devices

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8. Transient Response of Passive Circuits


8.1 Introduction

Up to now: ➢ The forced response of passive circuits was described. (Forced means that it is the response to an ongoing stimulus from outside, the so-called forcing function)➢ This leads to AC or DC circuit analysis made in the previous

chapters

Novel in this chapter: ➢ The same circuits may show a different behavior, the natural response.➢ This is the behavior immediately after some disturbance has occurred

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Example: Spring-Mass System8. Transient Response of Passive Circuits

1. Applying a sinusoidal driving force

The second is called natural, because both➔ the oscillating frequency and➔ the time needed to dampare determined solely by the system's components themselves and not by the external force. It is also called transient because it tends to die out.

Oscillating at the frequency of this force

2. Simple push of mass

Mass oscillates damped by friction and finally comes to rest.

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Transients in Electric Circuits8. Transient Response of Passive Circuits

Reason for transient processes

Energy loss in the resistances

Properties of transients in electric circuits

➔ After a cicuit is switched on, there is an initial period, where both forced and natural response are important➔ After some time, the transient has feded off leaving only the forced response remaining in the system.➔ In many cases, transients can easily be calculated, as there is only one energy storing element in the circuit (i.e. capacitor or inductor).➔ Such circuits are called „first-order circuits“, because they are described by first order differential equations.

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8.2 Transients in First-Order Circuits8. Transient Response of Passive Circuits

First, we use the following particular stimulus:

Somewhere in the circuit are two terminals marked “output“:=> find the output voltage v

out(t)

t = 0 t >> 0All previous transients are faded off, v

out is constant

Period of interest Final steady-state

The transient response indicates the speed at which a circuit may respond to changes of input and thus the operating speed of the circuit.

voltage input00

2

1

≥

<=

tvtv

vIN

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Step Resonse 18. Transient Response of Passive Circuits

}-

+v

1(t) v

IN(t) v

OUT(t)

v1(t)

V1

V2

t

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Step Resonse 28. Transient Response of Passive Circuits

Looking for: vOUT

(t)

Condition:Network contains any combination of resistors plus one capacitor or inductor

General solution:

A, B: voltagesτ: time constant } to be determined

Only valid for t > 0, or we say the validity of the functions starts at t = 0+.

0withe)(OUT >+=−

ττt

BAtv

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Step Response 38. Transient Response of Passive Circuits

At t = 0+ : vOUT

(t)

As t approaches infinity, vout

(t) itself approaches some final value.

vout

(t) may be discontinuous at t = 0. Two rules define the behavior of voltages over time.

T is also called the duration of the transient.

dvOUT

(t)dt

decreases =>

has decreased to 1/e of its initial valueAt t = τ : dv

OUT(t)

dt

decreases

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Rules for Voltages and Currents 18. Transient Response of Passive Circuits

Similarly, for an inductor we have v = L di / dt and by the same line of deduction, we get:

Rule 1: The voltage across a capacitor cannot change instantaneously.

Rule 2: The current through an inductor cannot change instantaneously.

possible. not is whichinfinite, become would thus and dd ously,instantane

change to werev voltage the If .dd is capacitor a through current The

itv

tvCi =

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Rules for Voltages and Currents 28. Transient Response of Passive Circuits

Two additional rules concern the dc steady state, i.e. after all transients have faded off. Here all derivations with respect to time are zero. Thus, we get:

Rule 3: In the dc steady state, the current through a capacitor is zero.

Rule 4: In the dc steady state, the voltage across an inductor is zero.

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Example8. Transient Response of Passive Circuits

}-

+v

1(t) v

OUT(t)

R

C

vout

(t) = 0 for t < 0, i.e. the capacitor is uncharged before the circuit is switched on.

><

=000

)(1 tVt

tv

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Solution 1

Node equation for vout

:

With the general form and

we insert

By comparing, we find A = V and τ = RC, with B being still to be found.


0d

)(d)()( OUTOUT1 =−−

ttvC

Rtvtv

)(1)(1d

)(d1OUT

OUT tvRC

tvRCt

tv=+⇒

τt

BAtv−

+= e)(OUTτ

τ

tBt

tv −−= e

d)(d OUT

RCV

RCB

RCAB tt

=++−−−

ττ

τee

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Solution 2

Initial condition: vout

(t = 0) = 0

Thus, we finally get

vout(t)

t


VABBAtv

−=−=⇒

=+==⇒ 0)0(OUT

0 fore1)(OUT >

−=

−tVtv RC

t

00,20,40,60,8

11,2

0

0,3

0,6

0,9

1,2

1,5

1,8

2,1

2,4

2,7 3

3,3

3,6

3,9

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In the beginning, the capacitor is uncharged and the voltage across its terminals is consequently zero.

Physical Meaning

When the voltage step occurs, the current (V-Vout

)/R charges the capacitor. The current is maximum in the beginning.With increasing voltage across the capacitor, the charge current decreases and finally vanishes.

Vout

approaches V, but never reaches it exactly. However, in the dc steady state we assume it has.


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1. Write a node equation or loop equation

2. Substitute the general solution obtaining two equations for three unknowns

3. Use the initial condition of the circuit (rules 1 and 3) to find the third unknown

Steps for Solution (Summary)

When we have an inductor instead of a capacitor, the procedure is quite similar except that now rules 2 and 4 are used instead of 1 and 3.

This approach works, as long as there is only one capacitor or inductor in the network.

Alternatively, one may use Thévenin or Norton equivalents.


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In general, the situation is only slightly different (s. example below)

Problems with Switch

t < 0 t > 0


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The constraint here is that the input current is zero for t > 0.

Solution 1

For t < 0 the circuit is in the dc steady state, with vout

= 0.

For t = 0- we have iL = V

0/R

1. By rule 2, i

L must have this value

for t = 0+, too.

With vout = L diL/dt and i

L = -v

out/R

2, we have

0/d

d2

OUTout =+RL

vt

v


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Knowing that vout → 0 for t → ∞, we find that

Solution 2

And by vout(0+)=-R2iL(0+)=-R2V0/R1 we yield

LtR

VRRv

2

e01

2OUT

−−=


LtR

Bv2

eOUT

−=

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Problem especially in digital systems

Response to a Rectangular Pulse 1

Equivalent to excitation by two subsequent step inputs


00,20,40,60,8

11,2

-2 -1,5 -1 -0,5 0 0,5 1 1,5 2 2,5 3

+ =

00,20,40,60,8

11,2

-2 -1,5 -1 -0,5 0 0,5 1 1,5 2 2,5 3

voltage input0

000

)( 0

>

<<<

=

TtTtV

ttvIN

00,20,40,60,8

11,2

-2 -1,5 -1 -0,5 0 0,5 1 1,5 2 2,5 3

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As a consequence, the response may be found by adding the responses to the two step inputs.


Example: RC-combination with impulse length T equalling the time constant τ = RC.


0

0,2

0,4

0,6

0,8

1

0 0,5 1 1,5 2 2,5 3

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t < 0 0 < t < T t > T


Response to v1

Response to v2

Sum vout


0

0

0

0

For T = RC

−

−RC

t

V e10

−

−RC

t

V e10

−

−RC

t

V e10

−

−RC

t

V e10

−−

−−

RCTt

V e10

−

−10

RCT

RCt

eeV

( )10 −−

eeV RCt

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If T << RC, then vout does not have much time to rise before it must start downward again.



If T >> RC, then vout almost reaches its final value, before the downward impulse begins.

00,20,40,60,8

11,2

0 0,25 0,5 0,75 1 1,25 1,5 1,75 2

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It can be seen from the graph that the output resembles the rectangular input more, when RC is much smaller than T. For RC → 0, the output voltage becomes an exact rectangular pulse.

Pulse Distortion


Pulse distortion (T ≈ RC or T << RC) must be often avoided in electronic circuits. (E.g. ideal amplifier)However, in most circuits capacitances are present and thus – to avoid pulse distortion - T must be chosen much greater than RC.

Shorter impulse length means more information per time.But

Much effort is made to reduce capacitances.

↓

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8.3 Higher-Order Circuits8. Transient Response of Passive Circuits

If more than one energy-storing element is present, the solution gets more complex.

n energy-storing elements → nth order liner differential equation

Those problems will not be in the scope of these lectures.

Solution:

Find the roots of a polynomial function.

Find the final value of the output voltage or current.

Documents

Fundamentals of Electrical Engineering Part 2Fundamentals of Electrical Engineering Part 2 Summersemester 2004 University Duisburg-Essen, Campus Duisburg Department of Electrical Engineering