Day-16-1st-Order-Circuits1 - profmason.com

coalitionhihatElectronics - Devices

Oct 7, 2013 (4 years and 1 month ago)

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SERIES AND PARALLEL INDUCTORS

Now that the inductor has been added to our list of passive elements, it
is necessary to extend the powerful tool of series
-
parallel combination.
We need to know how to find the equivalent inductance of a series
-
connected or p
arallel
-
connected set of inductors found in practical
circuits.

Consider a series connection of N inductors, as shown in(a), with the
equivalent circuit shown in (b). The inductors have the same current
through them. Applying KVL to the loop,

v = v
1

+ v

2

+ v
3

+ ∙ ∙ ∙ + v
N

Substituting v k = L k di/dt results in


Where

L
eq

= L
1

+ L
2

+ L
3

+ ∙ ∙ ∙ + L
N

The equivalent inductance of series
-
connected inductors is the

sum of the individual inductances.

Inductors in series are combined in exactly the s
ame way as resistors
in series.

We now consider a parallel connection of N inductors, as shown in
(a), with the equivalent circuit in (b). The inductors have the same
voltage across them. Using KCL,

i = i
1

+ i
2

+ i
3

+ ∙ ∙ ∙ + i
N





From the definition:


Where


The initial current i(t 0 ) through L eq at t = t 0 is expected by KCL to be the sum of the inductor currents
at t 0 . Thus,

i(t 0 ) = i 1 (t 0 ) + i 2 (t 0 ) + ∙ ∙ ∙ + i N (t 0 )

The equivalent inductance of parallel inductors is the reciprocal

of the sum of the reciprocals of the
individual inductances.

Note that the inductors in parallel are combined in the same way as resistors in parallel.

Example: Given the collection of inductors shown below, find the value of the equivalent inductance:





Summary Table


1
st

Order Circuits

Now that we have considered the three passive elements (resistors, capacitors, and inductors) and one
active element (the op amp) individually, we are prepared to consider circuits that contain various
combinations o
f two or three of the passive elements.
First order circuits contain either
a
n

equivalent
resistor and capacitor or
a resistor and an inductor. These are called RC and RL circuits, respectively.

We carry out the analysis of RC and RL circuits by applyin
g Kirchhoff’s laws, as we did for resistive
circuits. The only difference is that applying Kirchhoff’s laws to purely resistive circuits results in
algebraic equations, while applying the laws to RC and RL circuits
produces differential equations.

The
di
fferential equations resulting from analyzing RC and RL circuits are of the first order. Hence, the
circuits are collectively known as first
-
order circuits.

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



SOURCE FREE RC CIRCUI
T

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. A circuit response is the manner in which
the circuit reacts to an excitation.

Consider a series comb
ination of a resistor and an initially charged capacitor, as
shown. (The re
sistor and capacitor may be the
equivalent resistance and equivalent
cap
acitance of combinations of re
sistors and capacitors.) Our objective is to
determine
the circuit response,
which, for pedagogic reasons, we assume to be the voltage v(t) across

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

v(0) = V
0

with the corresponding value of the energy stored as

w(0) = ½

CV
0
2

Applying KCL at the top node of the circuit,

i
C

+ i
R

= 0

By definition, i
C

= C dv/dt and i
R

= v/R. Thus,


This is a first
-
order differential equation, since only the first derivative of v is involved. To solve it, we
rearrange the terms as


Integrating both sides, we get


where ln A is the integration constant
. Thus,


Taking powers of e produces

v(t) = Ae


t/RC

But from the initial conditions, v(0) = A = V
0

. Hence,

v(t) = V
0

e


t/RC

This shows that the voltage response of the RC circuit is an exponential decay of the initial voltage. Since
the response is d
ue to the initial energy stored and the physical characteristics of the circuit and not due
to some external voltage or current source, it is called the natural response of the circuit.

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.

The natural response is illustrated graphically
at right. As t increases,
the
voltage decreases toward zero. The
rapidity with which the voltage
decreases is expressed

in terms of the time

constant, denoted by the
lower
case Greek letter tau, τ.

The time constant of a circuit is the time required for the response to decay by a

factor of 1/e or 36.8 percent of its initial value.

This implies that at t = τ the equation be
comes

V
0

e


τ/RC

= V
0

e


1

= 0.368 V
0

or

τ = RC

In terms of the time constant the equation can be written as

v(t) = V
0

e


t/τ

How many time constants are required for the final voltage to be 1% of the initial voltage?






Thus, it is customary to assume t
hat the capacitor is fully discharged (or charged) after five time
constants. In other words, it takes 5τ for the circuit to reach its final state or steady state when no
changes take place with time. Notice that for every time interval of τ, the voltage is

reduced by 36.8
percent of its previous value, v(t + τ) = v(t)/e = 0.368v(t), regardless of the value of t.

Observe that the smaller the time constant,
the more rapidly the voltage decreases, that
is, the faster the response. This is illustrated
at ri
ght. A circuit with a small time constant
gives a fast response in that it reaches the
steady state (or final state) quickly due to
quick dissipation of energy stored, whereas a
circuit with a large time constant gives a slow
response because it takes long
er to reach steady state. At any rate, whether the time constant is small
or large, the circuit reaches steady state in five time constants.

With the voltage v(t), we can find the current i
R

(t),


The power dissipated in the resistor is


The energy absorb
ed by the resistor up to time t is


Notice that as t → ∞ , w
R

( ∞ ) → ½ CV
0
2
, which is the same as w
C

(0), the energy initially stored in the
capacitor. The energy that was initially stored in the capacitor is eventually dissipated in the resistor.

In s
ummary:

T h e K e y t o W o r k i n g w i t h a S o u r c e
-

f r e e RC C i r c u i t i s

F i n d i n g :

1. The initial voltage v(0) = V
0

across the capacitor.

2. The time constant τ.

With these two items, we obtain the response as the capac
itor voltage v
C

(t) = v(t) = v(0)e


t/τ

. Once the
capacitor voltage is first obtained, other variables (capacitor current i
C

, resistor voltage v
R

, and resistor
current i R ) can be determined. In finding the time constant τ = RC, R is often the Thevenin

equivalent
resistance at the terminals of the capacitor; that is, we take out the capacitor C and find R = R
Th

at its
terminals.


Example:

Given the circuit below, find Vc(T) and Ic(t) given that Vc(0) = 10V.


1.


Find the time constant for the circuit





2.

Write an equation for Vc(t)





3.

Write an equation for Ic(t)








4.

Check with KVL








SOURCE FREE RL CIRCUIT

Consider the series connection of a resistor and an inductor, as shown. Our
goal is to determine the circuit response, which we will assu
me to be the
current i(t) through the inductor. We select the inductor current as the
response in order to take advantage of the idea that the inductor current
cannot change instantaneously. At t = 0, we assume that the inductor has an
initial current I
0

, or

i(0) = I
0

with the corresponding energy stored in the inductor as

w(0) = ½ LI
0
2

Applying KVL around the loop,

v
L

+ v
R

= 0

But v
L

= L di/dt and v
R

= iR. Thus,


Rearranging terms and integrating gives


Taking the powers of e, we have

i(t) = I
0

e


Rt/L

This shows that the natural response of the RL circuit is an
exponential decay of the initial current. The current response is
shown at right.

It is evident that the time constant for the RL circuit is


with τ again having the unit of seconds. Thu
s, the equation may be written as

i(t) = I
0

e


t/τ

With the current in, we can find the voltage across the resistor as

v
R

(t) = iR = I
0

Re


t/τ

The power dissipated in the resistor is

p = v R i = I
0
2

Re


2t/τ

The energy absorbed by the resistor is


Or


Note that as t → ∞ , w R ( ∞ ) → ½ LI
0
2
,, which is the same as w
L
(0), the initial energy stored in the
inductor. Again, the energy initially stored in the inductor is eventually dissipated in the resistor.

T h e K e y t o W o r k i n g w i t h a S
o u r c e
-

f r e e RL C i r c u i t i s t o F i n d :

1. The initial current i(0) = I
0

through the inductor.

2. The time constant τ of the circuit.

With the two items, we obtain the response as the inductor current i L (t) = i(t) = i(0)e


t/τ
. Onc
e we
determine the inductor current i L , other variables (inductor voltage v
L

, resistor voltage v
R

, and
resistor current i
R

) can be obtained. Note that in general, R is the Thevenin resistance at the terminals
of the inductor.

Example: For the circ
uit pictured below, find i(t) and VL(t) given that i(0) = 4 A.


1.

Find the time constant for the circuit





2.

Write an equation for I (t)





3.

Write an equation for V
L
(t)








4.

Check with KVL








Example 2:

The switch in the circuit of Fig. 7.16 has
been closed for a long time. At t = 0, the switch is opened.
Calculate i(t) for t > 0.



1.


Find the initial current out of the 40 V supply




When t < 0, the switch is closed, and the inductor acts as a short
circuit to dc. The 16
-
Ohm resistor is short
-
ci
rcuited

To get i 1, we combine the 4
-
Ohm and 12
-
Ohm resistors in parallel to
get


Hence,


2.


Find the initial current across the inductor.






We obtain i(t) from i 1 using current division, by writing


Since the current through an inductor cannot chang
e instantaneously,

i(0) = i(0−) = 6 A

3.

Find the equivalent resistance seen by the inductor and the time constant.





When t > 0, the switch is open and the voltage source is
disconnected. We now have the RL circuit shown at right.
Combining the resistors,

we have

R eq = (12 + 4) || 16 = 8 Ohm

The time constant is


4.

Write the equation for the current as a function of time based on i0 and the time constant.



Thus,

i(t) = i(0)e


t/τ

= 6e


4t

A



Integrators and Differentiators

Capacitors and inductors poss
ess the following three special properties that make them very useful in
electric circuits:

1. The capacity to store energy makes them useful as temporary voltage or current sources. Thus, they
can be used for generating a large amount of current or volta
ge for a short period of time.

2. Capacitors oppose any abrupt change in voltage, while inductors oppose any abrupt change in
current. This property makes inductors useful for spark or arc suppression and for converting pulsating
dc voltage into relativel
y smooth dc voltage.

3. Capacitors and inductors are frequency sensitive. This property makes them useful for frequency
discrimination. The first two properties are put to use in dc circuits, while
the third one is taken advantage of in ac circuits.

I n
t e g r a t o r

Important op amp circuits that use energy
-
storage elements include
integrators and differentiators. These op amp circuits often involve resistors
and capacitors; inductors (coils) tend to be more bulky and expensive.

An integrator is an op
amp circuit whose output is proportional to the
integral of the input signal.


If the feedback resistor Rf in the familiar inverting amplifier of (a) is
replaced by a capacitor, we obtain an ideal integrator, as shown in (b). It is
interesting that we can obtain a mathematical representation of integration this way. At node a in Fig.

(b), iR = iC but


Substituting these
into iR = iC
, we obtain


Integrating both sides gives


To ensure that vo(0) = 0, it is always necessary to discharge the integrator’s capacitor prior to the
application of a signal. Assuming vo(0) = 0,


which show
s that the circuit in (b) provides an output voltage proportional to the integral of the input.
In practice, the op amp integrator requires a feedback resistor to reduce dc gain and prevent saturation.
Care
must be taken that the op amp operates within the linear range so that it does not saturate.

D i f f e r e n t i a t o r

A differentiator is an op amp circuit whose output is proportional to the rate
of change of the input signal.

If the input resistor is

replaced by a capacitor, the resulting circuit is a
differentiator, shown at right. Applying KCL at node a, iR = iC

But


Substituting these
into iR = iC
, we obtain


showing that the output is the derivative of the input. Differentiator circuits are ele
ctronically unstable
because any electrical noise within the circuit is exaggerated by the differentiator. For this reason, the
differentiator circuit is not as useful and popular as the integrator.

SWITCHING FUNCTIONS

Singularity functions (also called
switching functions) are very useful in circuit analysis. They serve as
good approximations to the switching signals that arise in circuits with switching operations. They are
helpful in the neat, compact description of some circuit phenomena, especially t
he step response of RC
or RL circuits to be discussed in the next sections.

Singularity functions are functions that either are discontinuous or have

discontinuous derivatives.

The three most widely used singularity functions in circuit analysis are the un
it step, the unit impulse,
and the unit ramp functions.

The unit step function u(t) is 0 for negative values of t and 1 for positive values of t.

In mathematical terms,

u(t) = 0, t < 0

u(t) = 1, t > 0

The unit step function is undefined at t = 0, where

it changes abruptly from 0 to 1. It is
dimensionless, like other mathematical functions such as sine and cosine. The unit
step function is shown at right.


If the abrupt change occurs at t = t 0 (where t 0 > 0)
instead of t = 0, the unit step function bec
omes

u(t − t
0

) = 0, t < t
0

u(t − t
0

) = 1, t > t
0

which is the same as saying that u(t) is delayed by t 0 seconds, as shown in (a). To
get the equations above from the previous equations, we simply replace every t by t
− t
0

. If the change is
at t = − t
0

, the unit step function becomes

u(t + t
0

) = 0, t < − t
0

u(t + t
0

) = 1, t > − t
0

meaning that u(t) is advanced by t
0

seconds, as shown in
(b).

We use the step function to represent an abrupt change in voltage or current, like the c
hanges that
occur in the circuits of control systems and digital computers. For example, the voltage

v(t) = 0, t < t
0

v(t) = V
0

, t > t
0

may be expressed in terms of the unit step function as

v(t) = V
0

u(t − t
0

)

If we let t
0

= 0, then v(t)
is simply the step voltage V
0

u(t). A voltage source of V 0 u(t) is shown in Fig. (a); its
equivalent circuit is shown in (b). It is evident in Fig. (b)
that terminals a
-
b are short
-

circuited (v = 0) for t < 0
and that v = V
0

appears at the terminals
for t > 0.
Similarly, a current source of I


0

u(t) is shown in (a),
while its equivalent circuit is in (b). Notice that for t <
0, there is an open circuit (i = 0), and that i = I
0

flows
for t > 0.

The derivative of the unit step function u(t) is the unit

impulse function δ(t), which we write as


The unit impulse function

also known as the delta function

is shown at right.

The unit impulse function δ(t) is zero everywhere except at t = 0, where it is
undefined.

Impulsive currents and voltages occur in ele
ctric circuits as a result of switching
operations or impulsive sources. Although the unit impulse function is not
physically realizable (just like ideal sources, ideal resistors, etc.), it is a very useful mathematical tool.

The unit impulse may be regar
ded as an applied or resulting shock. It may be visualized as a very short
duration pulse of unit area. This may be expressed mathematically as


where t = 0


denotes the time just before t = 0 and t = 0
+

is the time
just after t = 0. For this reason, it

is customary to write 1 (denoting
unit area) beside the arrow that is used to symbolize the unit
impulse function, as the figure above. The unit area is known as the
strength of the impulse function. When an impulse function has a strength other than un
ity, the area of
the impulse is equal to its strength. For example, an impulse function 10δ(t) has an area of 10. The
figure at right shows the impulse functions 5δ(t + 2), 10δ(t), and − 4δ(t − 3).

To illustrate how the impulse function affects other func
tions, let us evaluate the integral


where a < t
0

< b. Since δ(t − t
0

) = 0 except at t = t
0

, the integrand is

zero except at t
0

. Thus,



This shows that when a function is integrated with the impulse function, we obtain the value of the
function at
the point where the impulse occurs. This is a highly useful property of the impulse function
known as the sampling or sifting property. The special case of the above equation is for t
0

= 0.


Integrating the unit step function u(t) results in the unit ramp

function r(t); we write



The unit ramp function is zero for negative values of t and has a unit slope for
positive values of t.

The figure at right shows the unit ramp function. In general, a ramp is a function
that changes at a constant rate.

The uni
t ramp function may be delayed or advanced as shown. For the delayed
unit ramp function,

r(t − t
0

) = 0, t ≤ t
0

r(t − t
0

) = t − t
0

, t ≥ t
0


and for the advanced unit ramp function,

r(t + t
0

) = 0, t ≤ t
0

r(t + t
0

) = t − t
0

, t ≥
-
t
0


We should keep in mind that the three singularity functions (impulse, step,
and ramp) are related by differentiation as


or by integration as


Although there are many more singularity functions, we are only interested in these three (the i
mpulse
function, the unit step function, and the ramp function) at this point.

Example:

Given v
C
(t) = [5 u(t) +6 r(t)] V, find i
C
(t) for the circuit shown.


1.


i
C
(t) = C d/dt v
C
(t)








Practical Integrator:



Sketch the input and output waveforms fo
r 1kHz sine wave, triangle wave, and square wave inputs.

Note that the 10MΩ resistor (a very large value) is not in the ideal integrator circuit. What is it
there for? (Think about what would happen if a small DC component was present in

the input
waveform. What would integrating this constant do after a short time?) What happens when it is
removed?