ECE 202: Ideal Transformers
Charity Pettis
March 29, 2004
Objectives
• Establish relationships between the
primary and secondary sides of an ideal
transformer
• Define the turns ratio of an ideal
transformer
• Work an example problem using circuit
analysis and the relationships between the
primary and secondary sides of an ideal
transformer
The Ideal Transformer
a=
v1 − i2
=
v2
i1
1
Idealizations
1. The coupling coefficient is one.
k=
M
=1
L1 L2
2. The mutual and self-inductances are very
large.
M , L1 , and L2 → ∞
a=
v1 − i2
=
v2
i1
Development
V1 = sL1 I1 + sMI 2
V2 = sMI1 + sL2 I 2
V1
sL1 I1 + sMI 2
=
MV2 M ( sMI1 + sL2 I 2 )
V1
sL1 I1 + sMI 2
1
=
=
MV2 L2 ( sL1 I1 + sMI 2 ) L2
V1 M
=
=
V2 L2
L1 L2
=
L2
L1
L2
2
Recall that
a=
V1 ( s )
V2 ( s )
From the previous slide
V1
=
V2
L1
L2
The self - inductance of a coil is
L = N 2P
where N is the numbers of turns in the coil,
and P is the permeance.
V1
=
V2
N12 P N1
=
=a
N 22 P N 2
Development
V1 = sL1 I1 + sMI 2
V1
sL I + sMI 2
= 11
=0
sM
sM
L1
I1 + I 2 = 0
M
L1
I1 = − I 2
M
L1
L1
=
=
M
L1 L2
a=−
L1
=a
L2
I2
I1
The Impedance Transformation
Property
The impedance transforma tion property states
Z in ( s ) = a 2 Z ( s )
We can verify this by
Z in ( s ) =
 V (s) 
V1 ( s ) aV2 ( s )
 = a 2 Z ( s )
=
= a 2  2
I1 ( s ) − I 2 ( s )
 − I 2 (s) 
a
3
Example 1
 1 
Z ( s ) = 10Ω // 

 s (0.1F ) 
10
10
s = 10
Z ( s) =
10 s + 1
10 +
s
Z in ( s) = a 2 Z ( s)
 10  1000
Z in ( s ) = (10) 2 
=
 s +1 s +1
Example 2
Step 1: Find Z in ( s ).
Step 2 : Use voltage division to find V1.
Z in ( s ) = a 2 Z ( s )
Z ( s ) = Rload = 10Ω
Zin( s ) = (10) 2 (10Ω) = 1kΩ
V1 =
Z in ( s)Vs
Rs + Z in ( s)
V1 =
(1kΩ)Vs
1
= Vs
(1kΩ + 1kΩ) 2
Example 3
Step 1: Find the turns ratios a1 and a2 .
a1 =
N1 100
=
= 10
Na1 10
a2 =
Na2 2
= =2
N2 1
4
Step 2: Break the Circuit into manageable pieces.
From the impedance
transforma tion property
Z a ( s ) = a22 Z ( s ).
Z a ( s ) = a22 Rload
Z a ( s ) = ( 2) 2 (5Ω)
Z a ( s ) = 20Ω
Step 3: Find the equivalent impedance
Z a ' ( s ) = Ra // Z a ( s )
Z a ' (s) =
Ra Z a
Ra + Z a
Z a ( s ) = 20Ω
Za ' ( s ) =
(20Ω)(20Ω)
= 10Ω
(20Ω + 20Ω)
Step 4 : Find Z in (s ).
Z in ( s ) = a12 Z a ' ( s )
Z in ( s ) = (10) 2 (10Ω) = 1kΩ
5
Step 5 : Find V1 and I1.
Zin( s ) = 1kΩ
Zin( s )Vs ( s )
V1 =
Zin( s ) + Rs
V1 =
V1
Zin( s)
1
Vs ( s)
I1 = 2
= 0.0005Vs ( s)
(1kΩ)
I1 =
(1kΩ)Vs ( s ) 1
= Vs ( s )
(1kΩ + 1kΩ) 2
Step 6 : Find Va and I a 2 .
V1
10
1 1

Va =  Vs ( s ) 
10  2

Va = 0.05Vs ( s )
Va =
I a1 = aI1
I a1 = 10(0.0005Vs ( s ))
I a1 = 0.005Vs ( s )
Ia2 =
20Ω
(0.005Vs ( s ))
(20Ω + 20Ω)
= 0.0025Vs ( s )
Ia2 =
Ia2
Ra
I a1
Ra + Z a ( s )
Step 7 : Find V2 and I 2 .
Va
a
0.05Vs ( s )
V2 =
= 0.025Vs ( s )
2
V2 =
I 2 = − aI a 2
I 2 = −2(0.0025Vs ( s ))
I 2 = −0.005Vs ( s )
6
Example 4
100
= 10
10
Z in = (10) 2 (10Ω)
a=
Z in = 1kΩ
V1 =
(10Ω)(1kΩ)
I s ≈ 10 I s
1.01kΩ
7