EE 205 Dr. A. Zidouri
Electric Circuits II
The Ideal Transformer
Lecture #24
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EE 205 Dr. A. Zidouri
The material to be covered in this lecture is as follows:
o
o
o
o
The Ideal Transformer
Determining the Voltage and Current Ratios
Rules For Assigning Proper Algebraic Sign For Relating The Voltage And Current
Impedance Matching
After finishing this lecture you should be able to:
¾
¾
¾
¾
Understand the Behavior of Ideal Transformers
Determine the polarity of the Voltage and Current Ratios
Analyze Circuits Containing Ideal Transformers
Use The Ideal Transformer For Impedance Matching
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EE 205 Dr. A. Zidouri
The Ideal Transformer
An ideal transformer consists of two magnetically coupled coils having N1 and N2 turns
respectively, and exhibiting these three properties:
o The coefficient of coupling is unity k = 1,
o The self-inductance of each coil is infinite L1 = L2 = ∞
o The coil losses, due to parasitic resistance, are negligible.
Understanding the behavior of ideal transformers begins with equation (33-11) which describes
the impedance Zab we repeat this below
ω2M2
ω2M2
Z ab = Z11 +
− Z S = R1 + jωL1 +
(34-1)
Z 22
R2 + jωL 2 + ZL
Let us consider the same circuit as in Fig.33-1 of previous lecture
Fig.34-1 Frequency-domain model of a Linear Transformer
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EE 205 Dr. A. Zidouri
Exploring Limiting Values
To show how Zab changes when k = 1, L1 and L 2 approach infinity we use the notation:
Z 22 = R2 +RL + j ( ωL 2 + XL ) = R22 + j X 22
(34-2)
Then rearrange
Xab
Rab
47444
8
644744
8 644
ω2M2R22 ⎛
ω2M2 X22 ⎞
Z ab = R1 + 2
+ j ⎜ ωL1 - 2
2
2 ⎟
R22 + X22
R22 + X22
⎝
⎠
Before we let L1 and L2 increase, we write the coefficient as:
⎛
ωL ωL X
ωL X ⎞
Xab = ωL1 - 21 2 2 22 = ωL1 ⎜1 − 2 2 222 ⎟
R22 + X22
⎝ R22 + X22 ⎠
(34-3)
(34-4)
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EE 205 Dr. A. Zidouri
2
Using the fact that when, k = 1 then M = L1L 2
Putting the term multiplying ωL1 , over a common denominator gives
⎛ R222 + ωL 2 XL + XL2 ⎞
Xab = ωL1 ⎜
⎟
2
R222 + X22
⎝
⎠
Factoring ωL 2 out from numerator and denominator yields
⎛
⎞
⎜
⎟
R222 + XL2
X
+
L
ω2L1L 2 ⎜
ωL 2 ⎟
Xab = 2 2 ⎜
⎟
ω L 2 ⎜ R2
⎡ ⎛ X ⎞2 ⎤ ⎟
L
⎜ 22 ω2L2 + ⎢1+ ⎜
⎟ ⎥⎟
⎜
⎟
ωL
2
⎢⎣ ⎝
2 ⎠ ⎥
⎦⎠
⎝
(
(34-5)
)
(34-6)
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EE 205 Dr. A. Zidouri
⎛
⎞
⎜
⎟
R222 + XL2
XL +
⎜
L
ωL 2 ⎟
Xab = 1 ⎜
⎟
L2 ⎜ ⎛ R ⎞2 ⎡ ⎛ X ⎞2 ⎤ ⎟
22
L
⎜⎜
⎟ + ⎢1+ ⎜
⎟ ⎥⎟
⎜ ⎝ ωL 2 ⎠ ⎢ ⎝ ωL 2 ⎠ ⎥ ⎟
⎣
⎦⎠
⎝
(
)
(34-7)
2
⎛N ⎞
L
As k → 1 ⇒ 1 → constant value of ⎜ 1 ⎟
L2
⎝ N2 ⎠
The reason is that; as the coupling becomes extremely tight, the two permeances p1 and p2 become
equal.
Therefore as
k → 1, L1 → ∞, L2 → ∞
2
⎛N ⎞
Xab = ⎜ 1 ⎟ XL
⎝ N2 ⎠
The same reasoning leads to simplification of the reflected resistance:
(34-8)
2
⎛ N1 ⎞
ω2M2R22 L1
R
=
=
⎜ ⎟ R22
22
2
R222 + X22
L2
⎝ N2 ⎠
Applying the results given by the two previous equations yields
2
2
⎛ N1 ⎞
⎛ N1 ⎞
Zab = R1 + ⎜ ⎟ R2 + ⎜ ⎟ (RL + jXL )
⎝ N2 ⎠
⎝ N2 ⎠
(34-9)
(34-10)
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EE 205 Dr. A. Zidouri
Comparing the two results we see that when k → 1, L1 → ∞, L2 → ∞
¾ The transformer reflects the resistance of the secondary and load impedance to the primary side
2
⎛ N1 ⎞
by a scaling factor equal to ⎜ ⎟ (the turns ratio squared).
⎝ N2 ⎠
¾ Hence we may describe the ideal transformer by two characteristics:
v1
v2
v1 N1
=
⇒
=
9
N1
N2
v 2 N2
(34-11)
i1 N2
i
N
i
N
=
⇒
=
2 2
9 1 1
i2 N1
(34-12)
In the following section we show how to determine the voltage and current ratios.
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EE 205 Dr. A. Zidouri
Determining the Voltage and Current Ratios
2
L
¾ For unity coupling, M = L1L 2 or M = L1L 2 therefore (34-15) becomes: V2 = 2 V1 (34-16)
L1
¾ For unity coupling, the flux linking coil 1 is the same as the flux linking coil 2, so we need only one
permeance to describe the self inductance of each coil, thus (34-16) becomes
o V2 =
N22 p
N2
V
V1 (34-17)
=
1
N12 p
N1
or
V1 V2
=
(34-18)
N1 N2
Ampere-turn Ratio: Short Circuit Coil
Summing the voltages around the shorted coil of Fig.34-3 (b) yields
o
0 = -jωMI1 + jωL 2I2
(34-19)
From which for k=1,
I1 L 2
L2
L
N
=
= 2 = 2
=
I2 M
L1 N1
L1L 2
(34-20)
This is equivalent to I1N1 = I2N2
(34-21)
In practice, coils wound on a ferromagnetic material behave very much like an ideal
transformer. Fig.34-4 shows the graphic symbol for an ideal transformer
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EE 205 Dr. A. Zidouri
Determining the Polarity of the Voltage and Current Ratios
In the this section we show how to establish reference polarities for currents and voltages and
remove the magnitude signs from equations (34-11) and (34-12)
Rules for assigning proper algebraic sign for relating the voltage and current
o If the coil voltages v1 and v2 are both positive or negative at the dot-marked terminal, use a
plus sign, otherwise, use a minus sign
o If the coil currents i1 and i2 are both directed into or out of the dot-marked terminal, use a
minus sign, otherwise, use a plus sign
The following four circuits illustrate these rules:
Fig.34-6 shows three ways to represent the same turn ratio of an ideal transformer (8 in this case)
V1 V2
=
N1 N2
I1N1 = −I2N2
V1
V
=- 2
N1
N2
I1N1 = I2N2
V1 V2
=
N1 N2
I1N1 = I2N2
V1
V
=- 2
N1
N2
I1N1 = −I2N2
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EE 205 Dr. A. Zidouri
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EE 205 Dr. A. Zidouri
Example 34-1
The load impedance connected in Fig.34-7 consists of ZL = 0.2375 + j0.05 Ω and ZS = 0.25 + j2 Ω . If
v g = 2500cos400t find the steady state expressions for a) i1, b) v1, c)i2, d)v2
Fig.34-7 Circuit for Example 34-1
a)
and
Because
we have
Vg = 2500 ∠0o = ( 0.25 + j2) I1 + V1
V1 = 10 V2 = 10 ⎡⎣( 0.2375 + j0.05 ) I2 ⎤⎦
I2 = 10I1
V1 = 10V2 = 100 ( 0.2375 + j0.05 ) I1
= ( 23.75 + j5 ) I1
Therefore
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EE 205 Dr. A.
Thus h e steady state expression for i l k h . m I # ) E
Hence
,
-
--:T,
...
<
EE 205 Dr. A. Zidouri
The Use of Ideal Transformer for Impedance Matching
Fig.34-8 Coupling a Load to a Source by an Ideal Transformer
V1 =
V2
a
I1 = aI2
and
Therefore
Zin =
and ZL =
V2
I2
⇒
Zin =
V1 V2 a 1 V2
=
= 2
I1
aI2
a I2
ZL
a2
Thus the ideal transformer’s secondary coil reflects the load impedance back to the primary coil, with
1
the scaling factor 2
a
Zin is greater or less than ZL depends on the turns ratio
a.
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EE 205 Dr. A. Zidouri
Self Test:
o
1) The source voltage in the phasor domain circuit in the Fig.34-9 below is 25∠0 kV . Find the
amplitude and phase angle of V2 and I2.
V2 =1868.15∠142.39o V
I2 =125∠216.87o A
Fig.34-9 Circuit for Self Test
2) The output impedance of the amplifier
is 192 Ω and the internal impedance
of the speaker is 12 Ω. Determine the
required turn ratio n to achieve
impedance matching for maximum
power transfer.
Answer: n=0.25
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