Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

14.6.1 
Input Impedance function of a two-Winding transformer
Consider a two-winding transformer with zero winding resistance. Let the self-inductance of primary 
and secondary windings be L
p
and L
s
. Let n be the ratio of secondary turns to primary turns. Let the 
secondary be terminated in a load circuit that has an input impedance of Z(s). See Fig 14.6-1. 
M
V
p
(
s
)
V
s
(
s
)
V
s
(
s
)
Z
(
s
)
V
p
(
s
)
sM
Z
(
s
)
–
+
–
+
–
±
+
–
+
L
p
L
s
s
(
L
p 
M
)
s
(
L
s
M
)
Fig. 14.6-1 
A passively terminated two-winding transformer (winding resistance negligible) 
We do not specify the relative polarity. Instead, we cover both options by using proper signs for the 
inductance values in the equivalent circuit shown.
Now, the primary-side input impedance function Z
p
(s) can be derived as shown below.
Z s
s L
M
s L
M
Z s
sM
s L
M
Z s
sM
s L
M
p
p
s
s
p
( )
(
)
[ (
)
( )] [
]
(
)
( )
(
)
=
+
+
× ±
+
±
=
+
∓
∓
∓
∓
[[ (
)
( )] [
]
( )
(
)
(
)
( )
(
s L
M
Z s
sM
sL
Z s
s L
M
sL
s L
M
Z s
s L
s
s
p
s
p
∓
∓
∓
+
× ±
+
=
×
+
×
±
ss
s
p s
p
s
M
sM sMZ s
sL
Z s
s L L
M
sL Z s
sL
Z s
∓
)
( )
( )
(
)
( )
( )
×
±
+
=
−
+
+
2
2
We observe that this impedance function is independent of relative polarity of windings.
Now, assume that the coupling coefficient is unity. Thus, for a perfectly coupled two-winding 
transformer with zero winding resistance, the input impedance function is given by,
Z s
sL Z s
sL
Z s
sL Z s
s n L
Z s
p
p
s
p
p
( )
( )
( )
( )
(
)
( )
(
=
+
=
+
2
∵
self-inductance square of number of turns)
P
a
=
+
=
sL Z s n
sL
Z s n
p
p
[ ( ) /
]
[ ( ) /
]
2
2
aarallel combination of and 
L
Z s
1
( )
This impedance is shown in Fig. 14.6-2. 

---

Analysis of Coupled Coils Using Laplace Transforms 
14.21
k = 
1
Z
p
(
s
)
Z
(
s
)
L
p
L
s
Z
p
(
s
)
sL
p
Z
(
s
)
n
2
Fig. 14.6-2 
Reflection of secondary impedance function to primary with 
k
= 
1
Now, we can incorporate the primary winding resistance r
p
as extra impedance in series at the input 
and the secondary winding resistance r
s
as a part of the load.
L
p
L
s
r
p
r
s
k = 
1
Z
p
(
s
)
Z
(
s
)
sL
p
r
p
Z
p
(
s
)
r
s
n
2
Z
(
s
)
n
2
Fig. 14.6-3 
Primary-side input impedance in a perfectly coupled transformer
A perfectly coupled transformer is called an ideal transformer if L
p
, L
s
and M have infinitely high 
values. Practical iron-cored transformers can be modeled approximately by an ideal transformer. Such 
an ideal transformer will have input impedance that is a turns-ration ratio transformed version of the 
secondary load impedance. We have discussed the application of such transformers in impedance 
matching applications earlier in this chapter.
14.6.2 
transfer function of a two-Winding transformer
Now, we look at the voltage transfer function of a two-winding transformer. Let I
p
(s) and I
s
(s) be the 
Laplace transforms of the primary and secondary mesh currents in the s-domain equivalent circuit 
shown in Fig. 14.6-1. Then, the mesh equations in matrix form is
sL
sM
sM
sL
Z s
I s
I s
V s
p
s
p
s
p
∓
∓
+











 =






( )
( )
( )
( )
0
Solving for I
s
(s), I s
sMV s
s L L
M
sL Z s
s
p
p s
p
( )
( )
(
)
( )
=
±
−
+
2
2
.
Since V
s
(s) 
= 
Z(s) 
× 
I
s
(s), 
V s
V s
sMZ s
s L L
M
sL Z s
sMZ s
s L L
M
L
s
p
p s
p
p s
p
( )
( )
( )
(
)
( )
( )
(
)
=
±
−
+
=
±
−
+
2
2
2
Z
Z s
r
r
p
s
( )
for 
= =
0
(14.6-1)
L
p
L
s
= 
M
2
for a perfectly coupled transformer ( i.e., k 
= 
1). Therefore, for a perfectly coupled 
transformer with r
p
= 
r
s
= 
0, the voltage transfer function is given by 
V s
V s
sMZ s
sL Z s
M
L
L L
L
L
L
n
k
r
r
s
p
p
p
p s
p
s
p
p
( )
( )
( )
( )
,
=
±
= ±
= ±
= ±
±
=
=
for 
1
ss
=
0 
(14.6-2)

---

14.22
Magnetically Coupled Circuits
Eqn. 14.6-2 indicates that there is a polarity inversion in secondary voltage with respect to primary 
voltage for one relative polarity in windings.
Various factors prevent a practical transformer from meeting this ideal. Winding resistance and 
imperfect coupling are two such factors. If the mesh equations are written again with r
p
and r
s
included, 
the transfer function can be shown as 
H s
V s
V s
sMZ s
s L L
M
s L Z s
r L
r L
r
s
p
p s
p
s p
p s
p
( )
( )
( )
( )
(
)
(
( )
)
(
=
=
±
−
+
+
+
+
2
2
rr
Z s
s
+
( ))
(14.6-3) 
for a special load – resistive load. Let Z(s) 
= 
R. Then,
H s
V s
V s
sMR
s L L
M
s L R r L
r L
r r
R
s
p
p s
p
s p
p s
p
s
( )
( )
( )
(
)
(
)
(
)
=
=
±
−
+
+
+
+
+
2
2
(14.6-4)
This is a second order band-pass function (note the s term in the numerator polynomial).
We express this transfer function as the product of the ideally expected value (
= ±
n) and a factor 
involving s as follows.
H s
V s
V s
M
L
s
s
k L
R
s
r
R
r
R
L
L
s
p
p
s
s
p
s
p
( )
( )
( )
(
)
=
=
±
×
−




+
+ +




2
2
1
1

 +
+




r
L
r
R
p
p
s
1
(14.6-5)
M
k L L
p s
=
. Therefore, 
M
L
k
L
L
nk
p
s
p
=
=
. Therefore,
H s
n k
s
s
k L
R
s
r
R
r
R
L
L
r
L
r
R
s
s
p
s
p
p
p
s
( )
(
)
= ± × ×
−




+
+ +





 +
+
2
2
1
1
11




(14.6-6)
We want to determine the low-frequency cut-off and high-frequency cut-off for the sinusoidal 
steady-state frequency response function of this transfer function. In general, these two frequencies 
will be some complex functions of k, n, r
s
, r
p
, R and L
p
. However, we can use some approximations in 
the case of a practically important iron-cored transformer.
In an iron-cored transformer, the coupling coefficient is nearly unity. The load resistance is usually 
much larger than the transformer’s winding resistance. Hence, the effect of leakage flux and the effect 
of winding resistance are only second-order effects. The aggregate result of two second-order effects 
can be taken as the product of effects when each is acting alone. Therefore, when we consider the 
effect of winding resistance we assume that k
= 
1 and when we consider the effect of imperfect 
coupling we assume that r
p
= 
r
s
= 
0.
H s
k
n
s
s
r
R
r
R
L
L
r
L
r
R
s
p
s
p
p
p
s
( )(
)
with 
= = ±
+ +





 +
+




1
1
1

---

Analysis of Coupled Coils Using Laplace Transforms 
14.23
Further algebraic manipulation of the above expression leads to
H s
n
r
R
n
r
R
s
s
r
r
R
L
r
R
n
r
R
n
s
p
p
s
p
s
p
( )
= ×
+ +
×
+
+




+ +




= ×
1
1
1
1
2
2
bb
a
×
+
s
s
(14.6-7)
b
is a voltage reduction factor due to part of the applied voltage getting dropped across the winding 
resistance. 
-
a
is the pole of the first order transfer function. The transfer function is seen to be a first 
order high-pass function. The DC gain (evaluated by substituting s 
= 
j0) – i.e., the ratio between 
the steady-state response under DC conditions to the DC input value – is zero. The upper cut-off 
frequency is 
a
rad/s where
a
=
+




+ +




≈
r
r
R
L
r
R
n
r
R
r
L
p
s
p
s
p
p
p
1
1
2
for practical valuess of R  
We observe that both the voltage reduction factor 
b
and the cut-off frequency 
a
can be obtained 
from the equivalent circuit shown in Fig. 14.6-3 with Z(s) replaced with R.
The winding resistances prevent a two-winding transformer from passing DC voltage 
and low frequency AC voltage from primary to secondary. Hence, a signal containing low-
frequency content will suffer waveform distortion when it goes through a transformer.
The high-frequency behaviour of a tightly coupled transformer is studied after neglecting the 
winding resistances. Then, Eqn. 14.6-6 simplifies to 
H s
n k
s
k L
R
n k
R
k L
s
R
k L
n
s
s
s
( )
(
)
(
)
(
)
= ± × ×
−




+
= ± × ×
−
+
−
= ±
1
1
1
1
1
2
2
2
×× ×
+
k
s
l
l
(14.6-8)
This is a low-pass transfer function with a cut-off frequency of 
l
=
−
=
−
R
k L
R n
k L
s
p
(
)
(
)
1
1
2
2
2
rad/s.
Imperfect magnetic coupling prevents a two-winding transformer from passing high 
frequency AC voltage from primary to secondary. The two-winding transformer exhibits 
low-pass behaviour in the high-frequency end due to imperfect magnetic coupling. 
Therefore, a signal containing high-frequency content will suffer waveform distortion 
when it goes through a transformer.
The product of right sides of Eqn. 14.6-7 and Eqn. 14.6-8 provides an approximation for Eqn. 
14.6-6 in the case of a tightly coupled two-winding transformer with winding resistances very small 
compared to load resistance. Thus,

---

14.24
Magnetically Coupled Circuits
H s
n
k
( )
=
±
×
ideal gain 
expected
voltage-dropfactor 
due to imperrfect 
magnetic coupling
voltage-dropfactor 
due to windin
×
b
gg 
resistances
high-pass 
factor
low-pass 
factor
×
+
×
+
s
s
s
a
l
l
(14.6-9)
This transfer function has a band-pass frequency response function with lower cut-off frequency 
of 
a
and upper cut-off frequency of 
l
. Primary winding resistance and primary winding inductance 
decide the lower cut-off frequency. The upper cut-off frequency is decided by coupling coefficient, 
primary winding inductance and load resistance.
The bandwidth of a transformer can be increased by reducing the winding resistances and increasing 
the coupling coefficient. Bandwidth of a transformer is load dependent.