ECE 3336
Introduction to Circuits & Electronics
Set #16
Transformers
Fall 2011,
TUE&TH 4-5:30 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
Coupled Inductors
• A current flowing in one inductor induces magnetic field that is felt by the second coil
• If this field changes due to changes in the original current, the second coil will try to respond
to eliminate these changes (the second coil wants to maintain the original magnetic field).
• This occurs by producing a voltage in the second coil that would result in a current opposing
these changes (through the magnetic field). Such interaction of both coils regarding their
currents and voltages is referred to as mutual inductance. Such voltage generation that
opposes the change in magnetic field will be a basis for transformers.
Produces
Constant
Magnetic
Field
Produces
Decreasing
Magnetic
Field
Produces
Increasing
Magnetic
Field
Lenz’s law
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indmut.html#c1
3
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
4
http://www.allaboutcircuits.com/vol_2/chpt_9/7.html
Mutual Inductance
“Mutual inductance”, the influence of one coil on the other, is
assumed in this course to be the same as the second on the first.
That is,
M12 = M21 = M.
Sometimes we will use L11 and L22 for the self inductance and L12 and L21 for
mutual inductance.
di1
v1 = L
dt
We have i1 current in the primary coil
di1
v2 = M
dt
We have i2 current in the secondary coil
Remember that transformers will operate in AC conditions. The DC input
voltage will not induce any voltage in the secondary winding (e.i. electrical
isolation will give an isolation transformer).
http://www.allaboutcircuits.com/vol_2/chpt_9/7.html
5
Ideal Transformers
We can derive equations for voltage and
current ratios in the ideal transformers
both in the time domain and phasor
domain
time 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
6
Dot Conventions for Ideal
Transformers
are both defined as the
If v1 and v2
dotted terminal with respect to the
undotted terminal, then
n2 / n1 = v2 / v1 = N.
If v1 and v2 are both defined as the
undotted terminal with respect to the
dotted terminal, then
n2 v2
i1
N= =
= .
n1 v1
i2
n2 / n1 = v2 / v1 = N.
Otherwise,
n2 / n1 = -v2 / v1 = N.
n2 V2
I1
N= =
=
.
n1 V1
I2
7
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
8
Dot Conventions for Ideal
Transformers
Analogous relations are for currents:
If i1 and i2 are both defined as entering
the dotted terminal, then
n2 / n1 = -i1 / i2 = N.
n2 v2
i1
N= =
= .
n1 v1
i2
If i1 and i2 are both defined as leaving
the dotted terminal, then
n2 / n1 = -i1 / i2 = N.
Otherwise,
n2 / n1 = i1 / i2 = N.
n2 V2
I1
N= =
=
.
n1 V1
I2
9
Power in Ideal Transformers
Note that if the voltage increases going
from one side of a transformer to the
other, the current decreases, and 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.
˜2
V
* ˜
˜
˜
˜
˜
S1 = I V = NI
= I2 V2 = S2
N
*
1 1
*
2
We can tap the secondary voltage at two
(or more) points
n2 V2
I1
N= =
=
.
n1 V1
I2
Center-tapped transformer
240 V
120V line
120V line
10
Examples of transformers
Other configurations of transformers
Center-tapped transformer
Figure
Power transformers
7.31, 7.32can be huge
or small
http://www.allaboutcircuits.com/vol_2/chpt_9/7.html
11
Impedance Reflection with Transformers
Transformers can be used to
match loads (impedances).
Note that
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
12
Impedance Reflection with Transformers
Transformers can be used
to match loads
(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.
13
The maximum power transfer in AC circuits
Maximum power transfer to Rload (in DC)
occurs when Rload=Rsource.
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? 14
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 e.i. [(125V/62.5Ω)x125V]
because I=2A
Now if we use a step-up
(N=2) transformer: the
current delivered to Rload is
again I=4A and the power is
restored to P=1,000 W
http://www.allaboutcircuits.com/vol_2/chpt_9/7.html
V2
V1 = , and I1 = I 2 N .
N
15
What happens with power? How to play it loud?
Maximum power would
be delivered to the load
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 =
http://www.allaboutcircuits.com/vol_2/chpt_9/7.html
Z2
.
N2
A transformer will give
impedance
transformation ratio
500:8 so that the
delivered power will
reach its maximum
16
17
Electric power transmission
(a) direct power transmission is affected by the line
resistance
Pload
V˜load I˜load
h=
=
Psource V˜source I˜source
Rload
=
Rsource + Rline + Rload
Figure 7.37a,
b
(b) power transmission with transformers
1
R''load = 2 (R'load +Rline ) = Rload + N 2 Rline
N
Reflected load here M=1/N
R'load =
1
2
R
=
N
Rload
2 load
M
What will be =?
We will use
impedance
transformation for

Figure 7.37a, b, Rizzoni
18
Electric power transmission - reduction of Rline by 1/N2
(c) equivalent circuit seen by generator
R''load =
(d) equivalent circuit seen by load
R'source = N Rsource
2
Figure 7.37c, d
h=
1
2
(R'
+R
)
=
R
+
N
Rline
load
line
load
2
N
Pload
Rload
=
Psource Rsource + (1/N 2 )Rline + Rload 19
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
20
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
21
Loads can be also in
a delta connection
Delta-connected generators
Line (-to-line) voltage
Figure
7.43
V=0
I=0
Currents drawn by wyeand by delta connected
loads
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. 22
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
23
A typical residential wiring arrangement
•Limit power
dissipation by
appropriate
connections fro
various loads.
Figure
7.52
Figure 7.52
•Avoid heat
generation (safety
aspects)
24
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).
Your house is carefully wired!
Figure 7.58
25