DC TRANSIENT
ANALYSIS
1
SUB - TOPICS




NATURAL RESPONSE OF RL CIRCUIT
NATURAL RESPONSE OF RC CIRCUIT
STEP RESPONSE OF RL CIRCUIT
STEP RESPONSE OF RC CIRCUIT
2
OBJECTIVES


To investigate the behavior of currents and
voltages when energy is either released or
acquired by inductors and capacitors when
there is an abrupt change in dc current or
voltage source.
To do an analysis of natural response and step
response of RL and RC circuit.
3
FIRST – ORDER CIRCUITS



A circuit that contains only sources, resistor
and inductor is called and RL circuit.
A circuit that contains only sources, resistor
and capacitor is called an RC circuit.
RL and RC circuits are called first – order
circuits because their voltages and currents are
describe by first order differential equations.
4
R
R
i
i
L
–
+
An RL circuit
vs
–
+
Vs
C
An RC circuit
5
Review (conceptual)

Any first – order circuit can be reduced to a Thévenin (or
Norton) equivalent connected to either a single equivalent
inductor or capacitor.
RTh
-
RN
L
VTh
–
+
IN
C
In steady state, an inductor behave like a short circuit.
In steady state, a capacitor behaves like an open circuit.
6


The natural response of an RL and RC circuit
is its behavior (i.e., current and voltage ) when
stored energy in the inductor or capacitor is
released to the resistive part of the network
(containing no independent sources)
The steps response of an RL and RC circuits is
its behavior when a voltage or current source
step is applied to the circuit, or immediately
after a switch state is changed.
7
NATURAL RESPONSE OF AN
RL CIRCUIT

Consider the following circuit, for which the switch is
closed for t<0, and then opened at t = 0:
t=0
Is
Ro
i
L
+
R
V
–

The dc voltage V, has been supplying the RL circuit
with constant current for a long time
8
Solving for the circuit


For t ≤ 0, i(t) = Io
For t ≥ 0, the circuit reduce to
i
Io
Ro
L
+
R
v
–
Notation:
 0 is used to denote the time just prior to switching.

0 is used to denote the time immediately after switching.
9
Continue…

Applying KVL to the circuit:
v(t ) Ri (t ) 0
di(t )
L
Ri (t ) 0
dt
di(t )
L
Ri (t )
dt
di(t )
R
dt
i (t )
L
(1)
(2)
(3)
(4)
10
Continue

From equation (4), let say;
du
R
dv
(5)
u
L

Integrate both sides of equation (5);
i ( t ) du
R t
(6)
dv
i ( to ) u
L to
Where:
i(to) is the current corresponding to time to
i(t) ia the current corresponding to time t



11
Continue


Therefore,
i(t )
R
ln
t
i(0)
L
(7)
hence, the current is
i(t ) i(0)e
( R / L )t
I 0e
( R / L )t
12
Continue

From the Ohm’s law, the voltage across the
resistor R is:
v(t ) i(t ) R

I 0 Re
( R / L )t
And the power dissipated in the resistor is:
p vRi(t )
2
0
I Re
2( R / L )t
13
Continue

Energy absorb by the resistor is:
w
1 2
LI 0 (1 e
2
2( R / L )t
)
14
Time Constant, τ


Time constant, τ determines the rate at which the
current or voltage approaches zero.
Time constant,
L
R
(sec)
15

The expressions for current, voltage, power and
energy using time constant concept:
i (t )
I 0e
v(t )
t/
I 0 Re
2
0
t/
2t /
p
I Re
w
1 2
LI 0 (1 e
2
2t /
)
16
Switching time




For all transient cases, the following instants of
switching times are considered.
t = 0- , this is the time of switching between -∞ to 0 or
time before.
t = 0+ , this is the time of switching at the instant just
after time t = 0s (taken as initial value)
t = ∞ , this is the time of switching between t = 0+ to
∞ (taken as final value for step response)
17

The illustration of the different instance of switching
times is:
∞
-∞
t
0
t
0
18
Example

For the circuit below, find the expression of
io(t) and Vo(t). The switch was closed for a
long time, and at t = 0, the switch was opened.
2Ω
i0
t=0
20A
0.1Ω
+
i
2H 10Ω
L
40Ω
V
–
19
Solution :
Step 1:
Find τ for t > 0. Draw the equivalent circuit. The
switch is opened.
RT
(2 10 // 40) 10
So;
L
RT
2
0.2
10
sec
20
Step 2:
At t = 0- , time from -∞ to 0-, the switch was closed for
a long time.
2Ω
20A
0.1Ω
i (0-)
10Ω
40Ω
L
The inductor behave like a short circuit as it being
supplied for a long time by a dc current source. Current
20A thus flows through the short circuit until the switch
is opened.
Therefore;
iL(0-) = 20A
21
Step 3:
At the instant when the switch is opened, the time t = 0+,
2Ω
io(0+)
+
20A
2H 10Ω
iL(0+)
40Ω
vo(0+)
–
The current through the inductor remains the same (continuous).
Thus,
which is the initial current.
iL (0 )
iL (0 )
20 A
Only at this particular instant the value of the current through the
inductor is the same.
Since, there is no other supply in the circuit after the switch is
opened, this is the natural response case.
22
By using current division, the current in the 40Ω resistor
is:
10
io
iL
10 40
4A
So,
io (t )
5t
4e A
Using Ohm’s Law, the Vo is:
Vo (t )
4 40
160
So,
V0 (t )
160e
5t
23
NATURAL RESPONSE OF AN RC
CIRCUIT

Consider the following circuit, for which the switch is closed
for t < 0, and then opened at t = 0:
Vo
+
Ro
C
t=0
+
v
–
R
Notation:
 0- is used to denote the time just prior to switching
 0+ is used to denote the time immediately after switching.
24
Solving for the voltage (t ≥ 0)


For t ≤ 0, v(t) = Vo
For t > 0, the circuit reduces to
i
Vo
+
+
Ro
C
v
R
–
25
Continue

Applying KCL to the RC circuit:
iC
iR
dv(t )
C
dt
dv(t )
dt
(1)
0
v(t )
R
0
(2)
v(t )
RC
0
(3)
dv(t )
dt
v(t )
RC
(4)
dv(t )
v(t )
1
dt
RC
(5)
26
Continue

From equation (5), let say:
dx
x

(6)
Integrate both sides of equation (6):
v (t )
Vo

1
dy
RC
1
du
x
1
RC
t
0
dy
(7)
Therefore:
v(t )
ln
Vo
t
RC
(8)
27
Continue

Hence, the voltage is:
v(t ) v(0)e

t / RC
Voe
t / RC
Using Ohm’s law, the current is:
i(t )
v(t )
R
Vo
e
R
t / RC
28
Continue

The power dissipated in the resistor is:
p(t ) viR

2
o
V
e
R
2t / RC
The energy absorb by the resistor is:
w
1
2
CVo (1 e
2
2t / RC
)
29
Continue

The time constant for the RC circuit equal the product
of the resistance and capacitance,

Time constant,
RC
sec
30

The expressions for voltage, current, power and
energy using time constant concept:
v(t )
Vo e
t/
Vo t /
i (t )
e
R
Vo2 2t /
p (t )
e
R
1
2
w(t )
CVo (1 e
2
2t /
)
31

For the case of capacitor, two important observation
can be made,
1) capacitor behaves like an open circuit when being
supplied by dc source
(From, ic = Cdv/dt, when v is constant, dv/dt = 0.
When current in circuit is zero, the circuit is open
circuit.)
2) in capacitor, the voltage is continuous / stays the
same that is, Vc(0+) = Vc(0-)
32
Example
The switch has been in position a for a long time. At
Time t = 0, the switch moves to b. Find the expressions
for the vc(t), ic(t) and vo(t) and hence sketch them for t =
0 to t = 5τ.
5kΩ
a
b
18kΩ
t=0
90V
+
10kΩ
0.1μF
60kΩ
12kΩ
+
Vo
–
33
Solution
Step 1:
Find t for t > 5τ that is when the switch was at a. Draw
18kΩ
the equivalent circuit.
0.1μF
RT
(18k
RT C
12kΩ
60kΩ
12k ) // 60k
3
20 10
0.1 10
20k
6
2ms
34
Step 2:
At t = 0, the switch was at a. the capacitor behaves like
An open circuit as it is being supplied by a constant
source.
5kΩ
90V
+
10kΩ
+
Vc(0-)
–
10
vc (0 )
90 60V
15
35
Step 3:
At t = 0+, the instant when the switch is at b.
18kΩ
+
60V
0.1μF
–
60kΩ
12kΩ
+
Vo
–
The voltage across capacitor remains the same at this
particular instant.
vc(0+) = vc(0-) = 60V
36
Using voltage divider rule,
Vo (0 )
12
60
30
24V
Hence;
vc (t )
vo (t )
ic (t )
60e
500 t
24e
500 t
V
0.03e
V
500 t
A
37
Summary
No RL circuit
1
2
3
RC circuit
L
R
Inductor behaves like a
short circuit when being
supplied by dc source for a
long time
Inductor current is
continuous
iL(0+) = iL(0-)
RC
Capacitor behaves like an
open circuit when being
supplied by dc source for a
long time
Voltage across capacitor is
continuous
vC(0+) = vC(0-)
38
Step Response of RL Circuit

The switch is closed at time t = 0.
i
Vs
+
R
+
t=0
L
v(t)
–

After switch is closed, using KVL
Vs
di
Ri (t ) L
dt
(1)
39
Continue

Rearrange the equation;
Ri (t ) Vs
L
di(t )
dt
Vs
R
i(t )
L
R
Vs
R
i
dt
L
R
di
R
di
dt
L
i(t ) Vs R
R
L
t
0
dv
i (t )
0
du
u (Vs R)
(2)
(3)
(4)
(5)
40
Continue

Therefore:
R
t
L

(5)
Hence, the current is;
i(t )

i (t ) (Vs R)
ln
I 0 (Vs R)
Vs
R
Io
Vs
e
R
( R / L )t
The voltage;
v(t ) (Vs I o R)e
( R / L )t
41
Example
The switch is closed for a long time at t = 0, the switch
opens. Find the expressions for iL(t) and vL(t).
t=0
10V
+
2Ω
3Ω
1/4H
42
Solution
Step 1:
Find τ for t > 0. The switch was opened. Draw the
equivalent circuit. Short circuit the voltage source.
3Ω
2Ω
1/4H
RT
(2 3)
L
RT
5
1
s
20
43
Continue
Step 2:
At t = 0-, the switch was closed. Draw the equivalent
circuit with 3Ω shorted and the inductor behaves like a
short circuit.
10V
+
2Ω
iL(0-)
iL (0 ) 10 / 2 5 A
44
Continue
Step 3:
At t = 0+, the instant switch was opened. The current in
inductor is continuous.
I0
iL (0 ) iL (0 ) 5 A
Step 4:
At t =∞, that is after a long time the switch has been left
opened. The inductor will once again be behaving like a
short circuit.
45
Continue
10V
Hence:
3Ω
2Ω
+
iL(∞)
iL ( ) Vs / RT
Vs
iL (t )
R
iL (t )
Io
2
2A
Vs
e
R
3e
20 t
( R / L )t
A
46
Continue

And the voltage is:
vL (t ) (Vs
vL (t )
I o R)e
15e
( R / L )t
20 t
V
47
Step Response of RL Circuit

The switch is closed at time t = 0
+
t=0
Is
R
C
i

vc(t)
–
From the circuit;
Is
dvc
C
dt
vc
R
(1)
48
Continue

Division of Equation (1) by C gives;
Is
C

vc
RC
(2)
Same mathematical techniques with RL, the voltage is:
vc (t )

dvc
dt
I s R (Vo
I s R)e
t / RC
And the current is:
i (t )
Is
Vo
e
R
t / RC
49
Example
The switch has been in position a for a long time. At t = 0,
the switch moves to b. Find Vc(t) for t > 0 and calculate its
value at t = 1s and t = 4s.
3kΩ
24V
+
5kΩ
a
+
Vc
–
4kΩ
b
t=0
+
30V
0.5mF
50
Solution
Step 1:
To find τ for t > 0, the switch is at b and short circuit the
voltage source.
4kΩ
0.5mF
RC
2s
51
Continue
Step 2:
The capacitor behaves like an open circuit as it is being
supplied by a constant dc source.
3kΩ
24V
+
5kΩ
+
Vc (0-)
–
From the circuit,
5
Vc (0 ) 24
15V
8
52
Continue
Step 3:
At t = 0+, the instant when the switch is just moves to b.
Voltage across capacitor remains the same.
Vc (0 ) Vc (0 ) 15V
Step 4:
At t = ∞, the capacitor again behaves like an open
circuit since it is being supplied by a constant source.
4kΩ
+
Vc(∞)
–
+
30V
Vc ( ) 30V
53
Continue
Step 5:
Hence,
Vc (t ) 30 (15 30)e
0.5t
30 15e
0.5t
V
At t = 1s, Vc(t) = 20.9V
At t = 4s, Vc(t) = 28 V
54
THE END
55
56
An Introduction To
Two – Port Networks
The University of Tennessee
Electrical and Computer Engineering
Knoxville, TN
wlg
Two Port Networks
Generalities:
The standard configuration of a two port:
I1
+
V1
_
Input
Port
I2
The Network
The network ?
The voltage and current convention ?
* notes
Output
Port
+
V2
_
Two Port Networks
Network Equations:
Impedance
Z parameters
Admittance
Y parameters
Transmission
A, B, C, D
parameters
* notes
V1 = z11I1 + z12I2
V2 = b11V1 - b12I1
V2 = z21I1 + z22I2
I2 = b21V1 – b22I1
I1 = y11V1 + y12V2
V1 = h11I1 + h12V2
I2 = y21V1 + y22V2
Hybrid
H parameters
I2 = h21I1 + h22V2
V1 = AV2 - BI2
I1 = g11V1 + g12I2
I1 = CV2 - DI2
V2 = g21V1 + g22I2
Two Port Networks
Z parameters:
V
1
z
11 I
1
V
1
z
12 I
2
V
2
z
21 I
1
z
* notes
V
2
22 I
2
I
2
0
I 0
1
I
2
0
I 0
1
z11 is the impedance seen looking into port 1
when port 2 is open.
z12 is a transfer impedance. It is the ratio of the
voltage at port 1 to the current at port 2 when
port 1 is open.
z21 is a transfer impedance. It is the ratio of the
voltage at port 2 to the current at port 1 when
port 2 is open.
z22 is the impedance seen looking into port 2
when port 1 is open.
Two Port Networks
Y parameters:
I
1
y
11 V
1
I
1
y
12 V
2
I
2
y
21 V
1
y
* notes
I
2
22 V
2
V
2
0
V 0
1
V
2
0
V 0
1
y11 is the admittance seen looking into port 1
when port 2 is shorted.
y12 is a transfer admittance. It is the ratio of the
current at port 1 to the voltage at port 2 when
port 1 is shorted.
y21 is a transfer impedance. It is the ratio of the
current at port 2 to the voltage at port 1 when
port 2 is shorted.
y22 is the admittance seen looking into port 2
when port 1 is shorted.
Two Port Networks
Z parameters:
Example 1
Given the following circuit. Determine the Z parameters.
I1
8
I2
10
+
V1
+
20
20
_
Find the Z parameters for the above network.
V2
_
Two Port Networks
Z parameters:
Example 1 (cont 1)
For z11:
For z22:
Z11 = 8 + 20||30 = 20
Z22 = 20||30 = 12
I1
For z12:
8
+
+
V1
V
1
z
12 I
2
V1
I2
10
20
V2
20
_
_
I 0
1
20 xI 2 x 20
20 30
8 xI 2
Therefore:
z12
8 xI 2
I2
8
=
z 21
Two Port Networks
Z parameters:
Example 1 (cont 2)
The Z parameter equations can be expressed in
matrix form as follows.
V1
z11
z12
I1
V2
z 21
z 22
I2
V1
V2
20 8 I 1
8 12 I 2
Two Port Networks
Z parameters:
Example 2 (problem 18.7 Alexander & Sadiku)
You are given the following circuit. Find the Z parameters.
I1
1
I2
4
+
+
+
V1
_
1
2
Vx
-
V2
2Vx
_
Two Port Networks
Z parameters:
V
1
z
11 I
1
Example 2 (continue p2)
I1
I
2
I1
Vx
3V x
2
I1
;
2V x
6
6V x
but V x
I1
I1
2
+
+
Vx
6
V1
2V x
1
2
Vx
-
V2
2Vx
_
I1
Other Answers
Z21 = -0.667
Substituting gives;
3 V1
I2
4
+
0
V1
Vx
1
1
V1
or
I1
z 11
5
3
Z12 = 0.222
Z22 = 1.111
_
Two Port Networks
Transmission parameters (A,B,C,D):
The defining equations are:
V1
A B
I1
C
V1
A
V2
I1
C
V2
I2 = 0
I2 = 0
D
V2
I2
B
V1
I2
V2 = 0
D
I1
I2
V2 = 0
Two Port Networks
Transmission parameters (A,B,C,D):
Example
Given the network below with assumed voltage polarities and
Current directions compatible with the A,B,C,D parameters.
I1
+
-I2
+
R1
V1
R2
_
V2
_
We can write the following equations.
V1 = (R1 + R2)I1 + R2I2
V2 = R2I1 + R2I2
It is not always possible to write 2 equations in terms of the V’s and I’s
Of the parameter set.
Two Port Networks
Transmission parameters (A,B,C,D):
Example (cont.)
V1 = (R1 + R2)I1 + R2I2
V2 = R2I1 + R2I2
From these equations we can directly evaluate the A,B,C,D parameters.
V1
A
V2
C
I1
V2
I2 = 0
I2 = 0
=
=
R1
R2
R2
1
R2
B
D
V1
=
V
=
0
I2
2
I1
I2
V2 = 0
=
R1
1
Later we will see how to interconnect two of these networks together for a final answer
* notes
Two Port Networks
The equations for the hybrid parameters are:
Hybrid Parameters:
* notes
V1
h11
h12
I1
I2
h21
h22
V2
h11
V1
I1
h21
I2
I1
V2 = 0
V2 = 0
h12
V1
V2
h22
I2
V2
I1 = 0
I1 = 0
Two Port Networks
The following is a popular model used to represent
a particular variety of transistors.
Hybrid Parameters:
I1
I2
K1
+
+
+
V1
K2V2
K3V1
_
_
V2
_
We can write the following equations:
V1
I2
* notes
K4
AI 1 BV 2
CI 1
V2
D
Two Port Networks
Hybrid Parameters:
V1
I2
AI 1 BV 2
CI 1
V2
D
We want to evaluate the H parameters from the above set of equations.
h11
h21
V1
I1
I2
I1
=
K1
V2 = 0
=
V2 = 0
K3
h12
V1
V2
h22
I2
V2
=
K2
I1 = 0
=
I1 = 0
1
K
4
Two Port Networks
Another example with hybrid parameters.
Hybrid Parameters:
Given the circuit below.
I1
-I2
+
The equations for the circuit are:
+
R1
V1
R2
V1 = (R1 + R2)I1 + R2I2
V2
_
V2 = R2I1 + R2I2
_
The H parameters are as follows.
h11
h21
V1
I1
I2
I1
V2=0
=
=
V2=0
R1
h12
V1
V2
-1
h22
I2
V2
I1=0
I1=0
=
1
=
1
R2
Two Port Networks
Modifying the two port network:
Earlier we found the z parameters of the following network.
I1
8
I2
10
+
V1
+
20
20
_
V1
V2
* notes
V2
_
20 8
8 12
I1
I2
Two Port Networks
Modifying the two port network:
We modify the network as shown be adding elements outside the two ports
I1
6
+
10 v
_
8
I2
10
+
V1
+
20
_
20
V2
_
We now have:
V1 = 10 - 6I1
V2 = - 4I2
4
Two Port Networks
Modifying the two port network:
We take a look at the original equations and the equations describing
the new port conditions.
V1
V2
20 8
8 12
I1
I2
So we have,
10 – 6I1 = 20I1 + 8I2
-4I2 = 8I1 + 12I2
* notes
V1 = 10 - 6I1
V2 = - 4I2
Two Port Networks
Modifying the two port network:
Rearranging the equations gives,
1
I1
26
8
10
I2
8
16
0
I1
I2
0.4545
-0.2273
Two Port Networks
Y Parameters and Beyond:
Given the following network.
I1
+
V1
I2
1
+
1
s
s
_
V2
_
1
(a) Find the Y parameters for the network.
(b) From the Y parameters find the z parameters
Two Port Networks
Y Parameter Example
I
1
y
11 V
1
I1 = y11V1 + y12V2
V
2
I
1
y
12 V
2
0
V 0
1
I2 = y21V1 + y22V2
I1
I2
1
+
V1
y
I
2
21 V
1
V
2
y
0
I
2
22 V
2
+
1
s
s
_
V2
_
short
1
To find y11
2
V1 I 1 (
2
s ) I
1
2 1s
2s 1
so
We use the above equations to
evaluate the parameters from the
network.
I
1
y
11 V
1
V
2
0
=
s + 0.5
V 0
1
Two Port Networks
Y Parameter Example
I1
y
I
2
21 V
1
V
2
0
+
V1
I2
1
+
1
s
s
_
1
We see
V1
2I 2
I
2
y
21 V
1
= 0.5 S
V2
_
Two Port Networks
Y Parameter Example
I1
To find y12 and y21 we reverse
things and short V1
I
1
y
12 V
2
short
+
V1
I2
1
+
1
s
s
_
V2
_
1
V 0
1
We have
y
I
2
22 V
2
V 0
1
We have
V2
2I1
V2
y
12
I
1
V2
= 0.5 S
2s
I2
( s 2)
y22
1
0 .5
s
Two Port Networks
Y Parameter Example
Summary:
Y
=
y11
y12
y21
y22
s 0.5
0.5
0.5 0.5 1 s
Now suppose you want the Z parameters for the same network.
Two Port Networks
Going From Y to Z Parameters
For the Y parameters we have:
For the Z parameters we have:
I YV
From above;
V
1
V Y I
ZI
ZI
Therefore
y
Z Y
1
z
11
z
21
z
12
z
22
22
y
12
Y
y
21
Y
y
11
Y
Y
where
Y
det Y
Two Port Parameter Conversions:
Two Port Parameter Conversions:
To go from one set of parameters to another, locate the set of parameters
you are in, move along the vertical until you are in the row that contains
the parameters you want to convert to – then compare element for element
z11
H
h22
Interconnection Of Two Port Networks
Three ways that two ports are interconnected:
Y parameters
ya
* Parallel
y
yb
yb
Z parameters
za
* Series
ya
z
zb
za
zb
ABCD parameters
* Cascade
Ta
Tb
T
Ta Tb
Interconnection Of Two Port Networks
Consider the following network:
I1
Find
V2
R1
I2
R1
+
+
V1
V1
T1
_
R2
T2
R2
V2
_
Referring to slide 13 we have;
R1
V1
I1
R2
R2
1
R2
R1
1
R1
R2
R2
1
R2
R1
1
V2
I2
Interconnection Of Two Port Networks
R1
V1
I1
R2
R2
1
R2
R1
1
R1
R2
R2
1
R2
R1
1
V2
I2
Multiply out the first row:
V1
R1
2
R2
R2
R1
R2
V2
R1
R2
R2
R1
R1 ( I 2 )
Set I2 = 0 ( as in the diagram)
R2 2
V2
V1
R1
2
3 R1 R2 R2 2
Can be verified directly
by solving the circuit
Basic Laws of Circuits
End of Lesson
Two-Port Networks
AC STEADY STATE ANALYSIS
In AC steady state analysis the frequency is assumed constant (e.g., 60Hz).
Here we consider the frequency as a variable and examine how the performanc
varies with the frequency.
Variation in impedance of basic components
Resistor
ZR
R
R 0
Inductor
ZL
j L
L 90
Capacitor
Zc
1
j C
1
C
90
Frequency dependent behavior of series RLC network
Z eq
R
j L
1
j C
( j ) 2 LC j RC 1
j C
" Simplification in notation" j
Z eq ( s )
s 2 LC
j
j
RC
C
s
sRC 1
sC
2
| Z eq |
( RC )
2
(1
C
2
LC )
2
Z eq
tan
1
2
j(
LC 1
RC
LC 1)
Simplified notation for basic components
1
Z R ( s ) R, Z L ( s ) sL, Z C
sC
For all cases seen, and all cases to be studied, the impedance is of the form
Z ( s)
am s m am 1s m 1 ... a1s a0
bn s n bn 1s n 1 ... b1s b0
Moreover, if the circuit elements (L,R,C, dependent sources) are real then the
expression for any voltage or current will also be a rational function in s
LEARNING EXAMPLE
Vo ( s )
sL
1
sC
R
VS
R sL 1 / sC
s j
R
Vo
Vo
sRC
VS
s 2 LC sRC 1
j RC
VS
2
( j ) LC j RC 1
j (15 2.53 10 3 )
10 0
2
3
3
( j ) (0.1 2.53 10 ) j (15 2.53 10 ) 1
NETWORK FUNCTIONS
Some nomenclature
When voltages and currents are defined at different terminal pairs we
define the ratios as Transfer Functions
INPUT OUTPUT TRANSFER FUNCTION SYMBOL
Voltage
Voltage
Voltage Gain
Gv(s)
Current
Voltage
Transimpedance
Z(s)
Current
Current
Current Gain
Gi(s)
Voltage
Current
Transadmittance
Y(s)
If voltage and current are defined at the same terminals we define
Driving Point Impedance/Admittance
EXAMPLE
I 2 ( s ) Transadmit tance
V1 ( s) Transfer admittance
V2 ( s)
Gv ( s)
Voltage gain
V1 ( s)
YT ( s)
To compute the transfer functions one must solv
the circuit. Any valid technique is acceptable
LEARNING EXAMPLE
VOC ( s)
sL
V1 ( s)
sL R1
The textbook uses mesh analysis. We will
use Thevenin’s theorem
ZTH ( s )
I 2 ( s ) Transadmit tance
V1 ( s) Transfer admittance
V2 ( s)
Gv ( s)
Voltage gain
V1 ( s)
YT ( s)
I 2 ( s)
ZTH (s)
VOC (s)
I 2 ( s)
R2 V2 ( s )
ZTH ( s )
1
1
R1 || sL
sC
sC
s 2 LCR1 sL R1
sC ( sL R1 )
VOC ( s)
R2 ZTH ( s)
YT ( s)
s 2 ( R1
Gv ( s)
R2
sLR1
sL R1
sL
V1 ( s )
sL R1
s 2 LCR1 sL R1
sC ( sL R1 )
s 2 LC
R2 ) LC s( L R1R2C ) R1
V2 ( s)
V1 ( s)
R2 I 2 ( s)
V1 ( s)
R2YT ( s)
sC ( sL R1 )
sC ( sL R1 )
POLES AND ZEROS (More nomenclature)
H ( s)
am s m am 1s m 1 ... a1s a0
bn s n bn 1s n 1 ... b1s b0
Arbitrary network function
Using the roots, every (monic) polynomial can be expressed as a
product of first order terms
H ( s)
z1, z2 ,..., zm
K0
( s z1 )(s z2 )...(s zm )
( s p1 )(s p2 )...(s pn )
zeros of the network function
p1, p2 ,..., pn
poles of the network function
The network function is uniquely determined by its poles and zeros
and its value at some other value of s (to compute the gain)
EXAMPLE
zeros : z1
1,
poles : p1
H ( 0) 1
2
j 2, p2
2
j2
H ( s)
K0
H (0)
K0
(s 2
1
1
8
( s 1)
j 2)( s 2
H ( s) 8
j 2)
K0
s 1
s2 4s 8
s 1
s2 4s 8
LEARNING EXTENSION Find the pole and zero locations and the value of K o
Vo ( s)
for the voltage gain G ( s)
VS ( s )
H ( s)
K0
Zeros = roots of numerator
Poles = roots of denominator
( s z1 )(s z2 )...(s zm )
( s p1 )(s p2 )...(s pn )
For this case the gain was shown to be
sCin Rin
1
G ( s)
[1000 ]
1 sCin Rin
1 sCo Ro
zero : z1
poles : p1
K0
s
s 100
[1000 ]
0
50 Hz , p2
(4 10 7 )
20,000 Hz
40,000
s 40,000
Variable
Frequency
Response
SINUSOIDAL FREQUENCY ANALYSIS
A0e j (
t
B0 cos( t
A0 H ( j )e j (
)
H (s)
B0 | H ( j ) | cos
)
t
t
)
H( j )
Circuit represented by
network function
To study the behavior of a network as a function of the frequency we analyze
the network function H ( j ) as a function of .
Notation
M( ) | H( j )|
( )
H( j )
H( j )
M ( )e j
( )
Plots of M ( ), ( ), as function of
are generally called
magnitude and phase characteri stics.
20 log10 (M ( ))
BODE PLOTS
vs log10 ( )
( )
HISTORY OF THE DECIBEL
Originated as a measure of relative (radio) power
P2 |dB (over P1 ) 10 log
P
V2
I R
R
2
P2 |dB (over P1 )
P2
P1
V22
10 log 2
V1
I 22
10 log 2
I1
V |dB 20 log10 | V |
By extension I |dB 20 log10 | I |
G |dB 20 log10 | G |
Using log scales the frequency characteristics of network functions
have simple asymptotic behavior.
The asymptotes can be used as reasonable and efficient approximations
Constant Term
the x - axis is log10
this is a straight line
Poles/Zeros at the origin
(j )
N
|(j )
N
|dB
(j )
N 20 log10 ( )
N
N 90
Simple zero
Simple pole
LEARNING EXAMPLE
Draw asymptotes
for each term
Draw composites
Generate magnitude and phase plots
10(0.1 j
1)
Gv ( j )
Breaks/cor ners : 1,10,50
(j
1)(0.02 j
1)
dB
40
20
10 |dB
20 dB / dec
0
20dB / dec
20
90
45 / dec
45 / dec
0.1
1
10
100
1000
90
Generate magnitude and phase plots
25( j
1)
Draw asymptotes for each Gv ( j )
Breaks (corners) : 1, 10
( j )2 (0.1 j
1)
Form composites
LEARNING EXAMPLE
dB
40
28dB
20
0
40dB / dec
20
90
45 / dec
45
90
180
0.1
1
10
100
270
Sketch the magnitude characteristic
breaks : 2, 10, 100
10 4 ( j
2)
G( j )
But the function is NOT in standard form
(j
10)( j
100 )
20( j / 2 1)
We need to show about
Put in standard form G ( j )
( j / 10 1)( j / 100 1)
4 decades
LEARNING EXTENSION
dB
40
26 |dB
20
0
20
90
1
10
100
1000
90
Sketch the magnitude characteristic
10 j
not in standard form
Put in standard form
G( j )
(j
1)( j
10)
zero at the origin
j
G( j )
breaks : 1, 10
(j
1)( j / 10 1)
LEARNING EXTENSION
dB
40
20
0
20
20dB / dec
20 dB / dec
90
90
0.1
1
10
Once each term is drawn we form the composites
100
270
LEARNING EXTENSION
Determine a transfer function from the composite
magnitude asymptotes plot
C
E
A
B
D
A. Pole at the origin.
Crosses 0dB line at 5
5
j
B. Zero at 5
C. Pole at 20
D. Zero at 50
E. Pole at 100
G( j )
5( j / 5 1)( j / 50 1)
j ( j / 20 1)( j / 100 1)
Sinusoidal
Properties of resonant circuits
At resonance the impedance/admittance is minimal
Z( j )
R
j L
| Z |2 R 2 ( L
1
j C
1 2
)
C
Y(j ) G
1
j L
| Y |2 G 2 ( C
j C
1 2
)
L
Current through the serial circuit/
voltage across the parallel circuit can
become very large
1
0L
Quality Factor : Q
R
0CR
Given the similarities between series and parallel resonant circuits,
we will focus on serial circuits
LEARNING EXAMPLE Determine the resonant frequency, the voltage across each
element at resonance and the value of the quality factor
1
0C
I
VC
0L
1
j
0C
Q
0
1
LC
1
3
6
(25 10 H )(10 10 F )
At resonance Z 2
VS 10 0
I
5A
Z
2
3
3
L
(
2
10
)(
25
10
) 50
0
VL j 0 LI j 50 5 250 90 (V )
2000 rad / sec
50
I
j 50 5 250
0L
R
50
2
90
25
At resonance
VS
R
| VC | Q | VS |
| VL |
0L
Q | VS |
Resonance for the series circuit
Z( j )
R
1
M( )
1/ 2
1 Q2 (
1
j C
j L
)
0
1 2
)
C
| Z |2 R 2 ( L
0 2
0
BW
Q
Claim : The voltage gain is
VR
1
Gv
V1 1 jQ (
0
)
0
R
Gv
R
1
j C
j L
At resonance :
0L
QR,
0C
R
Z( j )
Half power frequencie s
( )
tan 1 Q (
0
1
QR
Z( j )
R
j
QR
j
0
QR
0
)
0
Gv
R
Z
0
R1
jQ (
0
M ( ) | Gv |, ( ) |
Gv
LO
0
1
2Q
1
2Q
2
1
)
LEARNING EXTENSION
A series RLC circuit as the following properties:
R 4 , 0 4000 rad / sec, BW 100 rad / sec
Determine the values of L,C.
0
1
LC
Q
0L
R
1
0CR
BW
0
Q
1. Given resonant frequency and bandwidth determine Q.
2. Given R, resonant frequency and Q determine L, C.
4000
0
Q
40
BW 100
L
QR
0
C
1
L
2
0
40 4
4000
1
0 RQ
0.040 H
1
4 10
2
16 10 6
1.56 10 6 F
LEARNING EXAMPLE Determine
0,
max
when R 50
and R 1
50 mH
0
5 F
1
LC
umax
0
Q
1
LC
max
0
1
2
Q
6
0L
R
1
1
0CR
1
2Q 2
2000 rad / s
(5 10 )(5 10 )
2000 0.050
R
max
2000
1
1
R
50
1
2Q 2
Q Wmax
2
1871
100 2000
Evaluated with EXCEL and rounded to zero decimals
Using MATLAB one can display the frequency response
FILTER NETWORKS
Networks designed to have frequency selective behavior
COMMON FILTERS
High-pass filter
Low-pass filter
We focus first on
PASSIVE filters
Band-reject filter
Band-pass filter
Simple low-pass filter
1
j C
1
R
j C
V0
V1
Gv
1
1 j
Gv
M ( ) | Gv |
Gv
( )
M max
1, M
1
;
1
j RC
1
RC
1
2
1
tan
1
1
1
2
half power frequency
BW
1
Simple high-pass filter
Gv
Gv
V0
V1
R
1
j
M ( ) | Gv |
Gv
( )
M max
1
1
j C
R
j
1, M
j CR
1 j CR
;
RC
2
1
2
tan
1
1
1
LO
1
2
half power frequency
Simple band-pass filter
Band-pass
( R / L)
Gv
V0
V1
R
R
j
LO
1
C
L
( R / L)
R/ L
2
HI
RC
M( )
RC
M
R/ L
2
1
LC
2
2
1 M(
LO )
0)
2
BW
M(
1
LC
0
M(
LC 1
1
2
M(
HI
2
)
) 0
HI
LO
2
2
0
4
4
R
L
2
0
Simple band-reject filter
0
1
LC
j
0L
1
0C
0
at
0 the capacitor acts as open circuit
V0 V1
at
the inductor acts as open circuit
V0 V1
LO ,
HI
are determined as in the
band - pass filter
LEARNING EXAMPLE
Depending on where the output is taken, this circuit
can produce low-pass, high-pass or band-pass or bandreject filters
Band-reject filter
Band-pass
Bode plot for R 10 , L 159 H , C 159 F
VL
VS
VC
VS
j L
R
j
L
1
C
1
j C
R
j
L
1
C
VL
VS
VC
VS
0
0
0,
1,
VL
(
VS
VC
(
VS
) 1
) 0
High-pass
Low-pass
ACTIVE FILTERS
Passive filters have several limitations
1. Cannot generate gains greater than one
2. Loading effect makes them difficult to interconnect
3. Use of inductance makes them difficult to handle
Using operational amplifiers one can design all basic filters, and more,
with only resistors and capacitors
The linear models developed for operational amplifiers circuits are valid, in a
more general framework, if one replaces the resistors by impedances
These currents are
zero
Ideal Op-Amp
DC Machines
Introduction:
The electrical machines deals with the energy
transfer either from mechanical to electrical
form or from electrical to mechanical form, this
process is called electromechanical energy
conversion.
DC GENERATOR
• An electrical machine which converts
mechanical energy into electrical energy is
called an electric generator and a generators
works on the principle of dynamically induced
emf according to Faraday’s Law of electro
Magnetic Induction
Cont…
A D. C. machine consists of two main parts
a) Stationary part: It is designed mainly for
producing a magnetic flux.
b) Rotating part: It is called the armature, where
mechanical energy is converted into electrical
Energy
Construction of a DC Generator
Parts of a Dc Generator:
• Yoke
• Magnetic Poles
a)Pole core
b)Pole Shoe
• Field Winding
• Armature Core
• Armature winding
• Commutator
• Brushes and Bearings
Yoke
1)It serves the purpose of outermost cover of
the dc machine so that the Insulating
materials get protected from harmful
atmospheric elements like moisture, dust
and various gases like SO2, acidic fumes etc.
2)It provides mechanical support to the poles.
3)It forms a part of the magnetic circuit. It
provides a path of low reluctance for magnetic
flux.
Poles
Each pole is divided into two
parts
a) pole core
b) pole shoe
1) Pole core basically carries a field
winding which is necessary to
produce the flux.
2) It directs the flux produced
through air gap to armature core to
the next pole.
3)Pole shoe enlarges the area of
armature core to come across the
flux, which is necessary to produce
larger induced emf.
Armature
It is further divided into two parts namely,
(1) Armature core (2) Armature winding
Armature core:
Armature core is cylindrical
in shape mounted on the shaft. It consists of
slots on its periphery and the air ducts to
permit the air flow through armature which
serves cooling purpose.
Cont…
Armature winding
Armature winding is nothing but the inter
connection of the armature conductors, placed
in the slots provided on the armature core.
When the armature is rotated, in case of
generator magnetic flux gets cut by armature
conductors and emf gets induced in them.
Field winding
Cont..
a)The field winding is wound on the pole core
with a definite direction
b)To carry current due to which pole core on
which the winding is placed behaves as an
electromagnet, producing necessary flux.
3)As it helps in producing the magnetic field i.e.
exciting the pole as electromagnet it is called
‘Field winding’ or ‘Exciting winding’.
Commutator
The rectification in case of dc generator is done
by device called as commutator.
1.To facilitate the collection of current from the
armature conductors.
2.To convert internally developed alternating
emf to in directional (dc) emf
3.To produce unidirectional torque in case of
motor.
Brushes and brush gear
1) To collect current from commutator
and make it available to the
stationary external circuit.
2)Brushes are stationary and rest on
the surface of the commutator.
Brushes are rectangular in shape.
They are housed in brush holders,
which are usually of box type.
Cont..
• The brushes are made to press on the
commutator surface by means of a spring,
whose tension can be adjusted with the help
of lever. A flexible copper conductor called
pigtail is used to connect the brush to the
external circuit
Cont…
Bearings
Ball-bearings are usually used as they are more
reliable. For heavy duty machines, roller
bearings are preferred
Types of generators
1)Separately excited generators
2) Self-excited generators
i) Series Generator
ii)Shunt Generator
iii)Compound Generator
a) Long Shunt
b) Short Shunt
Separately excited generators:
These are generators whole field magnets are energized
from an independent external source of dc current
Ia
V
Field
Winding
IL
Ra
Eg
Load
Self-excited generators
• These are the generators whose field magnets
are energized by the current produced by the
generators themselves.
• Due to residual magnetism, there is always
present some flux in the poles when the
armature is rotated, some emf and hence some
induced current is produced which is partly or
fully passed through the field coils thereby
strengthening the residual pole flux.
Series Generator:
It is a generator where the field winding is
connected in series with the armature
Ia
Rse
Ra
Eg
IL
Load
Shunt Generator
It is a generator where the field winding is
connected in series with the armature
Ia
IL
RSh
Ish
Ra
Eg
Load
Compound Generator
It is generator which has both Series and Shunt
Windings. Compound generators are of two types
Long Shunt Compound Generator
IL
Ise
RSh
Rse
Ish
Ia
Ra
Load
Eg
Short Shunt Compound Generator
IL
Ise
Rse
Load
Ia
RSh
Ish
Ra
Eg
Types of armature winding
1)Lap winding:In this case, if connection is
started from conductor in slot 1 then the
connections overlap each other as winding
proceeds, till starting point is reached again.
2) There is overlapping of coils while
proceeding. Due to such connection, the total
number of conductors get divided into ‘P’
number of parallel paths, where
P = number of poles in the machine.
Wave winding
• In this type, winding always travels ahead avoiding
over lapping. It travels like a progressive wave hence
called wave winding
• Both coils starting from slot 1 and slot 2 are
progressing in wave fashion.
• Due to this type of connection, the total number of
conductors get divided into two number of parallel
paths always, irrespective of number of poles of
machine.
• As number of parallel paths are less, it is preferable
for low current, high voltage capacity generators.
EMF equation of a generator
Let
P = number of poles
= flux/pole in webers
Z = total number of armature conductors.
= number of slots x number of conductors/slot
N = armature rotation in revolutions (speed for
armature) per minute (rpm)
A = No. of parallel paths into which the ‘z’ no. of
conductors are divided.
E = emf induced in any parallel path
Eg = emf generated in any one parallel path in the
armature.
Cont..
Average emf generated/conductor = d /dt volt
In one revolution, the conductor will cut total
flux produced by all poles = x p
No. of revolutions/second = N/60
Therefore, Time for one revolution, dt = 60/N
second
According to Faraday’s laws of Electromagnetic
Induction, emf generated/conductor
= d /dt= x p x N / 60 volts
This is emf induced in one conductor
Cont..
For a simplex wave-wound generator
No. of parallel paths = 2
No. of conductors in (series)in one path = Z/2
EMF generated/path = PN/60 x Z/2
= ZPN/120 volt
Cont..
For a simple lap-wound generator
Number of parallel paths = P
Number of conductors in one path = Z/P
EMF generated/path = PN/60 (Z/P) = ZN/60
A = 2 for simplex – wave winding
A =P for simplex-- lap winding
Characteristics of dc generators
Magnetization characteristics
• It is also known as no-load saturation
characteristics or open-circuit characteristics
(OCC)
• This characteristics is the graph of the no-load
generated emf Eo and the field current If, at a
given speed
Cont..
Losses in a Dc Generator
Variable losses :
The armature current varies with load . The
copper losses that occur in the armature also
vary w.r.t to load Armature Cu loss = Ia2 Ra
Magnetic losses
• Hysteresis : The property of a magnetic material to
retain a part of the magnetism achieved by it, one
removal of the driving force is called hysteresis.
• The power required to overcome this hysteresis
effect is called hysteresis loss. It is denoted by Wh.
Wh = η Bm1.6 f V
• Where , η=Steinmetz Constant
•
B=Flux Density
•
F=Frequency
•
V=Volume Of the core
DC Motor:
• A dc motor is similar in construction to a dc
generator. As a matter of fact a dc generator
will run as a motor when its field & armature
windings are connected to a source of direct
current.
DC Motor Working principle
The principle of operation of a dc motor can be stated as when a
current carrying conductor is placed in a magnetic field; it
experiences a mechanical force. In a practical dc motor, the field
winding produces the required magnetic held while armature
conductor play the role of current carrying conductor and hence
the armature conductors experience a force.
As conductors are placed in the slots which are on the periphery,
the individual force experienced by the conductive acts as a
twisting or turning force on the armature which is called a torque.
The torque is the product of force and the radius at which this
force acts, so overall armature experiences a torque and starts
rotating.
Construction of a DC Motor
Parts of a Dc Generator:
• Yoke
• Magnetic Poles
a)Pole core
b)Pole Shoe
• Field Winding
• Armature Core
• Armature winding
• Commutator
• Brushes and Bearings
Yoke
1)It serves the purpose of outermost cover of
the dc machine so that the
Insulating
materials get protected from harmful
atmospheric elements like
moisture, dust
and various gases like SO2, acidic fumes etc.
2)It provides mechanical support to the poles.
3)It forms a part of the magnetic circuit. It
provides a path of low reluctance for magnetic
flux.
Poles
Each pole is divided into two
parts
a) pole core
b) pole shoe
1) Pole core basically carries a field
winding which is necessary to
produce the flux.
2) It directs the flux produced
through air gap to armature core to
the next pole.
3)Pole shoe enlarges the area of
armature core to come across the
flux, which is necessary to produce
larger induced emf.
Armature
It is further divided into two parts namely,
(1) Armature core (2) Armature winding
Armature core:
Armature core is cylindrical
in shape mounted on the shaft. It consists of
slots on its periphery and the air ducts to
permit the air flow through armature which
serves cooling purpose.
Cont…
Armature winding
Armature winding is nothing but the inter
connection of the armature conductors,
placed in the slots provided on the armature
core. When the armature is rotated, in case
of generator magnetic flux gets cut by
armature conductors and emf gets induced
in them.
Field winding
Cont..
a)The field winding is wound on the pole core
with a definite direction
b)To carry current due to which pole core on
which the winding is placed behaves as an
electromagnet, producing necessary flux.
3)As it helps in producing the magnetic field i.e.
exciting the pole as electromagnet it is called
‘Field winding’ or ‘Exciting winding’.
Commutator
The rectification in case of dc generator is done
by device called as commutator.
1.To facilitate the collection of current from the
armature conductors.
2.To convert internally developed alternating
emf to in directional (dc) emf
3.To produce unidirectional torque in case of
motor.
Brushes and brush gear
1) To collect current from commutator
and make it available to the
stationary external circuit.
2)Brushes are stationary and rest on
the surface of the commutator.
Brushes are rectangular in shape.
They are housed in brush holders,
which are usually of box type.
Cont..
• The brushes are made to press on the
commutator surface by means of a spring,
whose tension can be adjusted with the help
of lever. A flexible copper conductor called
pigtail is used to connect the brush to the
external circuit
Cont…
Bearings
Ball-bearings are usually used as they are more
reliable. For heavy duty machines, roller
bearings are preferred
Types of motors
1) Shunt motor
2) Series motor
3) Compound motor
i) Cumulative – compound motors
ii) Differential compound:
Type of Motor
Characteristics
Approximately constant
Shunt
speed. Adjustable speed
medium starting torque
Variable speed adjustable
Series
varying speed high
starting torque
Cumulative
compound
Variable speed adjustable
varying speed high
starting torque
Applications
For during constant speed
some centrifugal pumps
machine tools blows and fans
reciprocating pumps
For traction work i.e. electric
locomotives repaid transit
system trolley cars etc.
Crimes and hoists conveyors.
For intermittent high torque
load for shears and punches
elevators converge clearly
planer
Types of armature winding
1)Lap winding: In this case, if connection is
started from conductor in slot 1 then the
connections overlap each other as winding
proceeds, till starting point is reached again.
2) There is overlapping of coils while
proceeding. Due to such connection, the total
number of conductors get divided into ‘P’
number of parallel paths, where
P = number of poles in the machine.
Voltage equation of a Motor
•
The voltage v applied across the motor armature has to (1) over
core the back emf Eb and supply the armature holmic drop Ia Ro
v = Eb + Ia Rg
This is known as voltage equation of a motor
Multiplying both sides by Ia, we get
Via =Eb Ia + Ea2 Ra
VIa = electrical input to the armature
EbIa = electrical equivalent of mech. Power
developed in the armature
Ia2 Ra = un loss in the armature
•
•
Motor efficiency is given by the ratio of power developed by the
armature to its input i.e. Eb Ia / vIa = Eb/v.
Higher the value of Eb as compared to v, higher the motor
efficiency.
Condition for maximum power
• The gross mechanical developed by a motor
pm = vIa – Ia2 Ra
Ia Ra = v/2
As v = Eb + Ia Ra and Ia Ra = v/2, Eb = v/2
• Thus gross mechanical power developed by a motor is
maximum when back emf is equal to half the applied
voltage. This conduction’s how ever at realized in
practice, because in that case current will be much
beyond the normal current of the motor.
• More over, half the input would be wasted in the form
of heat and taking other losses into consideration the
motor efficiency will be well below 50 %.
Problems
1. A 220v – dc machine has an armature resistance of 0.5
. If the full road armature current is 20A, find the
induced emf when the machine acts
(1)generator (2) motor.
The dc motor is assumed to be shunt connected in cash
case, short current in considered negligible because its
value is not given.
• (a) As generator : Eg = v + Ia Ra = 220 + 0.5 x 20 = 230 v
• (b) As motor
: Eb = v – Ia Ra = 220 – 0.5 x 20 =210 v
Problems
• 2. A 440 v, shunt motor has armature
resistance of 0.8 a and field resistance of 200
. Determine the back emf when giving an
output at 7.46 kw at 85% efficiency.
• Motor input power =
• Motor input current =
7.46 x 103
w
0.85
7460
19.95 A
0.85 x 440
Problems
•
A 4-pole dc motor has lap connected armature winding. The number of
armature conductors is 250. When connected to 230 v dc supply it draws
an armature current It 4 cm calculate the back emf and the speed with
which motor is running. Assume armature is 0.6 .
•
•
P = 4 A = P = 4 as lap connected
= 30 m wb = 30 x10-3 , V = 230v, z = 250, Ia = 40 A
From voltage equation V = Eb + Ia Ra
230 = Eb + 40 x 0.6
Eb = Pnz / 60A
206 = (30 x 10-3 x 4 x N x 250) / (60 x 4)
N = 1648 rpm.
Power Equation
• Power developed by the machine
= O/p + losses =
VI2 + losses
• Variable losses : The armature current varies
with load . The copper losses that occur in the
armature also vary w.r.t to load
Ar Cu loss = Ia2 Ra
Magnetic losses
•
Hysteresis losses : The property of a magnetic material to retain a part of
the magnetizing achieved by it, one removal of the driving force is called
hysteresis.
•
•
•
The power required to overcome this hysteresis effect is called hysteresis
loss. It is denoted by Wn.
Wh f
Bm1.6
Wh = 7 Bm1.6 f
•
Eddy current losses : When the iron part (rotor) rotates in the stationary
field, there is an induced emf in the iron part also apart from the emf
induced in the copper windings. This emf is called eddy emf. The iron part
provides a closed path for the eddy emf to circulate a current. This current
is called eddy current.
Mechanical losses
• Friction losses : The rotation of the armature
inside the bearings causes friction to be
developed. In order to over come this, the
prime more has to supply more power.
• Windage losses : An extra mounting on the
shaft of the armature such as fan to provide
air for cooling purposes causes an extra
burden on the prime motor. The extra power
required to drive the fan is termed as windage
loss. (constant)
Characteristics of motors
TA/IA CHARACTERISTICS.
•
•
•
•
For series motor
Ia
Ia
Ta
Ia2
Thus, torque in case of series motor is proportional to the square of the armature
current
Ta
Ta
Tsh
Ia
•
•
•
•
As the load increases, armature current increases and torque produced
increases proportional to the square of the armature current upto a certain
limit.
Hence Ta/Ia curve is a parabola.
After saturation, is almost independent Ia, hence Ta
Ia only.
So, the characteristic be comes a straight line.
Contd..
N/IA CHARACTERISTIC
•
N
(Eb/ )
[v – Ia (Ra + Rse)/Ia]
•
The values Ra and Rse are so shall as
Ia in motor that the effect of change in Ia on
speed avoids the effect of change in V – Ia Ra – Ia Rse on speed change in Eb for
various load currents is small and hence may be neglected.
N
Ia
•
With increased Ia,
also increases Hence, speed varies inversely as armature
current. When load is heavy, Ia is large. Hence speed is low (this increases Eb and
allows more armature current to flow). But when load current and hence Ia falls to a
small value speed becomes dangerously high.
Objectives
On completion of this period, you should be able to know
• Importance of a transformer
• Features of a transformer
• Working principle of a transformer
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Can you point where transformer is used ?
33,000
33,000
EE403.1-2
183
Transformer Is 120 Year Old
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184
How It All Began
• 1831 – Faraday’s law of Electromagnetic induction
discovered
• 1864 – Maxwell’s equations for mathematical models of
electromagnetic apparatus formulated
• 1885 – First real Transformer (single phase) Patented.
• 1893 – First three phase Transformer was used in
Hellsjon 9.6 kv transmission system in Sweden
(dry type 3 phase transformers manufactured by
ASEA)
• 1900 – Oil Immersed Transformer was born.
9EE402.1 to 2
Transformer
• A transformer is a stationary electric machine which transfers
electrical energy (power) from one circuit to another circuit with
the same frequency but with different voltages and currents.
• A motor converts electrical power to mechanical power
• A generator converts mechanical power to electrical power
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186
Transformer
• Unlike in rotating machines,
there is no electrical to
mechanical energy conversion
• A transformer is a static device
and all currents and voltages are
AC
• The transfer of energy takes
place through the magnetic field
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187
Introduction to Transformers
Definition
• A transformer is a static
(device) AC Machine
which transfers electrical
energy from one electrical
circuit
to
another
electrical circuit without
change
of
frequency
through
a
common
magnetic (field) flux
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188
Introduction to Transformers contd…
• Electrical
energy
is
generated at places where
it is easier to get water
head and coal for hydro
and thermal power stations
respectively.
• Electrical energy generated
is to be transmitted to
considerable distances for
use in towns, cities and
villages
• Transmission of electrical
energy at high voltages is
economical
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189
Introduction to Transformers contd…
• Electrical Machines are
required for stepping up
the voltage at generating
stations
• Electrical Machines are
required for stepping
down the voltage at
places where it is to be
used
• Electrical machine used
for this purpose is known
as TRANSFORMER
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190
Introduction to Transformers contd…
• No electrical connection
is present between the
two winding
• Magnetic linkage exists
between the two windings
Primary
winding
• The magnetic linkage is
provided through a path
of low reluctance
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191
Introduction to Transformer contd…
Primary
winding
• A transformer basically
consists of two windings
which are wound on a
soft iron or silicon steel
core
• The winding which is
connected to the supply
mains is known as
primary winding
• The winding which is
connected to the load is
known as secondary
winding
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Working Principle of Transformers contd…
50Hz
50H
z
• The transformer works on
the principle of Mutual
induction between two
magnetically
coupled
coils
• Is based on Faraday’s
laws of Electromagnetic
Induction
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193
Working Principle of Transformer contd…
50
Hz
• When primary winding is
connected to AC supply
mains a current flows
through it
• This current produces an
alternating flux in the core
• This flux links with
primary
winding
and
produces self induced
e.m.f in the primary
winding which opposes
the applied voltage
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194
Working Principle of Transformer contd…
• This flux passes through
the core and links with
the secondary windings
to induce an E.M.F called
mutually induced E.M.F in
the secondary winding
Primary
winding
• The frequency of the emf
induced in the secondary
is same as that of the flux
or that of the supply
voltage
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195
Working Principle of Transformer contd…
50 Hz
• The induced emf in the
secondary winding will be
able to circulate the
current in the external
load connected to it
• Energy is transferred
from primary winding to
the secondary winding by
electro-magnetic
induction principle without
change of frequency
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Step Up Transformer
• When the transformer
rises the voltage it is
called step-up transformer
• In step-up transformer the
output voltage is higher
than input voltage
• In step-up transformer
secondary winding turns
are more than the primary
winding turns.
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Step Down Transformer
• When
the
transformer
reduces the voltage it is
called
step-down
transformer
• In step-down transformer
the output voltage is less
than input voltage
• In step-down transformer
secondary winding turns
are less than the primary
winding turns.
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Summary
In this session we have learnt about
•
Importance of a transformer
•
Features of a Transformer
•
Working principle of a transformer
•
Step-up transformer
•
Step-down transformer
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Quiz
1) Transformer is which type of Electrical Machine
a) Static
a.c. machine
b) Static
d.c. machine
c) Rotating a.c. machine
d) Rotating d.c. machine
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200
Quiz
2)Transformer works on the principle of
a) Mutual induction between two coils
b) Mutual conduction between two coils
c) Self induction between two coils
d) None of the above
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Quiz
3) Transformer converts electrical energy from one
electrical circuit to another electrical circuit without
change in
a) Frequency
b) Voltage
c) Current
d) None of the above
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Quiz
5) The winding connected to the supply mains is
called
a) Secondary winding
b) Primary winding
c) Auxiliary winding
d) None of the above
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Quiz
6) The winding connected to the load is known as
a) Secondary winding
b) Primary winding
c) Territory winding
d) None of the above
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Quiz
7) In step-up transformer the secondary winding voltage
is
a) More than the primary voltage
b) Less than the primary voltage
c) Equal than the primary voltage
d) Zero
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Frequently Asked Questions
1) Define Transformer
2) Explain the working principle of transformer
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