#### Transcript Document

```ECE 3336
Introduction to Circuits & Electronics
Set #16
Transformers
Fall 2012,
TUE&TH 4:00-5:30 pm
Dr. Wanda Wosik
1
Inductors
Inductors model the interaction
between magnetic fields and
voltage and current.
df (t)
v L (t) = N
dt
L=
LX= #[H]
iL
+
vL
-
diL
vL = LX
dt
f (t)[Wb]
i(t)[A]
 - Magnetic flux in webers [Wb]
2
Mutual Inductance
We model this effect with what
we call “mutual inductance”,
and we call it M [H]
di1
v2 = M
dt
We also rename L [H], as the
“self inductance” v1 = L
Magnetomotive
force
di1
dt
vp(t)
vs(t)
vp(t)
vs(t)
Voltage induced in
v(t)=N•d/dt the secondary coil
in response to
current changes
turns
in the primary coil.
Current direction =
polarity important
3
The Dot Convention
Phases of the voltages in the primary and secondary windings are
identified by dots.
The same phase is obtained
for both instantaneous voltages
v1(t) and v2(t)
180° phase shift between
instantaneous voltages
df (t)
v1(t) = N1
dt
time domain
df (t)
v 2 (t) = N 2
dt
n2
v2
i1
N=
=
=
.
n1
v1
i2
If i1 and i2 are both defined as entering (or
leaving) the dotted terminal, then
The same flux
n2 / n1 = -i1 / i2 = N;
n2 / n1 = i1 / i2 = N.
Otherwise,
4
From Time Domain to Phasor Domain
time domain
phasor domain
The same flux
Hambley
df (t)
v1(t) = N1
dt
df (t)
v 2 (t) = N 2
dt
Voltage and
current ratios
n2
v2
i1
N=
=
=
.
n1
v1
i2
n2 V2
I1
N= =
=
.
n1 V1
I2
5
Power in Ideal Transformers
Note that if the voltage increases going from one
side of a transformer to the other, the current
decreases by the same factor.
There is no power gain. The factor is the ratio of
the number of turns. We named this as the turns
ratio, N.
n2 V2
I1
N= =
=
.
n1 V1
I2
Center-tapped transformer
˜2
V
* ˜
˜
˜
˜
˜
S1 = I V = NI
= I2 V2 = S2
N
*
1 1
*
2
240 V
120V line
We can tap the secondary voltage at two
(or more) points
120V line
6
Examples of transformers
Other configurations of transformers
Center-tapped transformer
Figure
Power transformers
7.31, 7.32can be huge
or small
7
Impedance Reflection with Transformers
Transformers can be used to match
V2
V1 = , and I1 = I 2 N .
N
If we divide the first equation by the
second, we get
V1 V2
N
= , or
I1 I 2
2
Z2
N Z1 = Z 2 , or Z1 = 2 .
N
2
8
Impedance Reflection with Transformers
Transformers can be used
(impedances).
This means that if we look
at the apparent impedance
seen at the primary side of
a transformer (Z’) we will
see the impedance at the
secondary side divided by
the turns ratio squared.


Z2
N Z1 = Z 2 , or Z1 = 2 .
N
2
This can be very useful. It is often referred
to as the reflected impedance.
9
The maximum power transfer in AC circuits
Maximum power transfer to Rload (in DC)
In AC circuits we will need very similar
impedance matching with the source:
Zs=Rs+jXs
2) Now, from the complex power
1) The real power absorbed by Rload
Z LVS
VS*
VS2
SL = V I =
´
=
ZL
Z S + Z L (Z S + Z L )* | Z S + Z L ) |2
PL = V˜L I˜L cosq = Re(V˜L I˜L )
Where:
ZL
V˜L =
V˜S
ZS + ZL
˜
˜*
V
V
*
*
S
S
I˜L = (
) = Figure
*
ZS + ZL
(Z S + Z7.35
L)
Maximum power
transfer if:
ZL=ZS*
RL=RS and
XL=-XS
*
L L
We calculate the real power (again)
VS2
PL = Re(V I ) = Re(
)Re(Z L )
| Z S + Z L |2
*
L L
VS2
=
Re(Z L )
(RS + RL )2 + (X S + X L )2
=
V RL
(RS + RL )2 + (X S + X L )2
2
S
PL=PLMAX
0
When
will
PLMAX? 10
Impedance transformation improves
power delivery - example
a)
The heaters (62.5Ω and 15.625Ω) of
1,000W power rating operate in two
circuits a) and b).
b)
If we use the heater a) with 125 V
source the power will decrease
P=250 W i.e. [(125V/62.5Ω)x125V]
because I=2A
Now if we use a step-up
(N=2) transformer: the
again I=4A and the power is
restored to P=1,000 W
V2
V1 = , and I1 = I 2 N .
N
11
What happens with power? How to play it loud?
Maximum power would
of 500 Ω. But we have
8Ω.
AC Thevenin circuit
We want to supply
power from a high
impedance (V high I
low) amplifier to a low
impedance (low V
high I) speaker.
N 2 Z1 = Z 2 , or Z1 =
Z2
.
N2
A transformer will give
impedance
transformation ratio
500:8 so that the
delivered power will
reach its maximum
12
Electric power transmission
(a) direct power transmission is affected by the line
resistance
h=
=
Psource V˜source I˜source
=
Figure 7.37a,
b
(b) power transmission with transformers
1
N
1
2
R
=
N
M
What will be =?
We will use
impedance
transformation for

Figure 7.37a, b, Rizzoni
13
Electric power transmission - reduction of Rline by 1/N2
(c) equivalent circuit seen by generator
(d) equivalent circuit seen by load
R'source = N Rsource
2
Figure 7.37c, d
h=
1
2
(R'
+R
)
=
R
+
N
Rline
line
2
N
=
Psource Rsource + (1/N 2 )Rline + Rload 14
Figure 7.37c, d
Balanced three-phase Power (AC circuit)
Positive, or abc, sequence for balanced three-phase voltages (“-” acb)
Wye-wye (Y-Y) connection
neutral
Line voltages
ab, bc, ca
Figure 7.40,
7.41
V˜ab = V˜an + V˜nb = V˜an - V˜bn
V˜an = V˜bn = V˜cn = V˜
All line voltages
Phase voltages
V˜an = V˜anÐ0°
V˜bn = V˜bnÐ(-120°)
V˜cn = V˜cnÐ(-240°) = V˜cnÐ120°
V˜ab = V˜Ð0° - V˜Ð(-120°) = 3V˜Ð30°
V˜bc = V˜Ð(-120°) - V˜Ð(120°) = 3V˜Ð(-90°)
V˜ca = V˜Ð120° - V˜Ð0° = 3V˜Ð150°
Figure 7.41 Rizzoni
V˜an + V˜bn + V˜cn = 0
15
Balanced three-phase AC circuit (redrawn)
Three circuits are in parallel.
I˜n = I˜a + I˜b + I˜c
1 ˜
(Van + V˜bn + V˜cn ) = 0
Z
Can be eliminated
V˜
pa (t) = (1+ cos2wt )
R
V˜
pb (t) = [1+ cos(2wt - 120°)]
R
=
Figure
7.42
V˜
pc (t) = [1+ cos(2wt + 120°)]
R
Advantage of the 3 phase also in less wiring (3)
Compared to single phase (6 wires).
Figure 7.42
p(t) = p a (t) + p b (t) + pc (t)
3V˜ 2 V˜ 2
=
+
[cos2wt + cos(2wt - 120°)
R
R
3V˜ 2
+cos(2wt + 120°)] =
R
Constant! power
16
a delta connection
Delta-connected generators
Line (-to-line) voltage
Figure
7.43
V=0
I=0
Currents drawn by wyeand by delta connected
V˜
V˜
˜
(Ian ) y =
=
Ð(-q )
Zy | Zy |
For both currents to be the same we have to have
Z=3Zy
Rizzoni, Figure 7.43
Delta draws 3 times more current than a wye load does. 17
Line voltage convention for residential circuits
A 3-wire AC system supplied by the power company
red
Higher line loss will be
from the 120V source.
To reduce power loss
(I2R) thick wires are
used.
white (earth ground)
black
83.3A
Figure
7.50
Power loss=69.4 W
Rs=0.02Ω
41.7.A
V˜B - V˜R = V˜BR = V˜B - (-V˜B ) = 2V˜B = 240Ð0°
Power loss=34.7 W
This is the voltage (rms)
between the hot wires
18
A typical residential wiring arrangement
•Limit power
dissipation by
appropriate
connections fro
Figure
7.52
Figure 7.52
•Avoid heat
generation (safety
aspects)
19
Structure of an AC power distribution network (just for your curiosity)
That reduces power losses in transmission lines
Step-up transformer
Figure
Substations 7.58
An electric power
network
=the Power grid
allows for
redistribution of
power to various
substations
(various V levels
obtained after
stepping-down).