Coupling Element and Coupled circuits



Coupled inductor
Ideal transformer
Controlled sources
Coupling Element and Coupled circuits
Coupled elements have more that one branch and branch voltages or branch
currents depend on other branches. The characteristics and properties of
coupling element will be considered.
Coupled inductor
Two coils in a close proximity is shown in Fig.1
i1
+
v1
-
i2
+
v2
-
Fig.1 Coupled coil and reference directions
Coupled inductor
Magnetic flux is produced by each coil by the functions
1  f1 (i1 , i2 )
Where
f1
and
f2
2  f 2 (i1 , i2 )
are nonlinear function of
i1
and i2
By Faraday’s law
d 1 f1 di1 f1 di2
v1 


dt
i1 dt i2 dt
d 2 f 2 di1 f 2 di2
v2 


dt
i1 dt i2 dt
Coupled inductor
Linear time-invariant coupled inductor
If the flux is a linear function of currents
and
1 (t )  L11i1 (t )  Mi2 (t )
2 (t )  Mi1 (t )  L22i2 (t )
di1
di2
v1  L11
M
dt
dt
di1
di2
v2  M
 L22
dt
dt
In sinusoid steady-state
V1  j L11 I1  j M I 2
Note that the signs of
L11 and L22
V2  j MI1  j L22 I 2
are positive but the sign for M can be
 or 
Coupled inductor
Dots are often used in the circuit to indicate the sign of M
i1
H1
i2
+
+
v1
v2
-
-
H2
Fig. 2 Positive value of M
Coupled inductor
Coefficient of coupling
The coupling coefficient is
k
|M |
L11 L22
If the coils are distance away k is very small and close to zero and equal
to 1 for a very tight coupling such for a transformer.
Coupled inductor
Multi-winding Inductors and inductance Matrix
For more windings the flux in each coil are
1  L11 I1  L12 I 2  L13 I 3  ..
2  L21I1  L22 I 2  L23 I3  ..
3  L31I1  L32 I 2  L33 I3  ..
L11 , L22 , L33
are self inductances and
L12  L21 , L13  L31 , L23  L32
In matrix form
φ  Li
are mutual inductances
Coupled inductor
1 
   2 
3 
 i1 
i  i2 
 i3 
 L11

L   L21
 L31
L12
L22
L32
L13 

L23 
L33 
i2
i
d1 + 1
v1 
dt
+ v  d2
2
dt
-
-
Fig 3 Three-winding inductor
i3
+
-
d3
v3 
dt
Coupled inductor
Induced voltage
The induced voltage in term current vector and the inductance matrix is
Example 1
di
vL
dt
Fig. 4 shows 3 coils wound on a common core. The reference direction of
current and voltage are as shown in the figure. Since H1 and H 2 has the
same direction but H3 are not therefore L12 is positive while L13 and
i2
L23 are negative.
i1
-
v1
+
+
H1
-
H2
Fig. 4
H3
i3 + v 3
v2
Coupled inductor
It is useful to define a reciprocal inductance matrix
 L1
i  
which makes
i1  111  122
i2  211  222
where
11 
L22
L
L
, 22  11 and 12  21  12
det L
det L
det L
Thus the currents are
t
t


t0
0t
i1 (t )  11 v1 (t ')dt '  12 v2 (t ')dt '  i1 (0)


0
0
i2 (t )  21 v1 (t ')dt '  22 v2 (t ')dt '  i2 (0)
Coupled inductor
In sinusoid steady-state
11

V1  12 V2
j
j
 21
 22
I2 
V1 
V2
j
j
I1 
Series and parallel connections of coupled inductors
Equivalent inductance of series and parallel connections of coupled
inductors can be determined as shown in the example 2.
Coupled inductor
Example 2
Fig. 5 shows two coupled inductors connected in series. Determine the
Equivalent inductance between the input terminals.
i
+
Fig. 5
v
i1
+
v1
L1  5
i2
M 3
-
+
-
v2
-
L2  2
1  L11i1  Mi2  5i1  3i2
2  Mi1  L22i2  3i1  2i2
i  i1  i2 , v  v1  v2
d d1 d2  (0)  0



dt
dt
dt
   1  2  8i1  5i2  13i

H
L 
i
 13
Coupled inductor
Example 3
Fig. 6 shows two coupled inductors connected in series. Determine the
Equivalent inductance between the input terminals.
i
+
Fig. 6
i1
+
v
v1
L1  5
-
i2
M 3
-
+
v2
-
Note
L2  2
L  L11  L22  2 | M |
1  L11i1  Mi2  5i1  3i2
2  Mi1  L22i2  3i1  2i2
i  i1  i2 , v  v1  v2
d d1 d2
 ( 0)  0



dt
dt
dt
   1  2  2i1  i2  i

H
L 
i
1
for series inductors
Coupled inductor
Example 4
Two coupled inductors are connected in parallel in Fig 6. Determine the
Equivalent inductance.
i
+
i1
+
v1
L1  5
-
Fig 6
i2
v
-
M 3
+
v2
-
L2  2
L22

det L
2
5
det
3
L11
5
22 

detL
5
det 
3
L
3
12  12 
detL
5
det 
3
11 
2
3
2
3
2
3
2
5
 3
Coupled inductor
The currents are
i1  111  122  21  32
i2  211  222  31  52
KVL
v1 (t )  v2 (t ) and 1 (0)  1 (0)  0
By integration of voltage
Therefore
1 (t )  2 (t )
i  i1  i2  1  22  
L 
Note

i
1
H
  11   22  2 | 12 |
for parallel inductors
Ideal transformer
Ideal transformer is very useful for circuit calculation. Ideal transformer
Is a coupled inductor with the properties
 dissipate no energy
 No leakage flux and the coupling coefficient is unity
 Infinite self inductances
Two-winding ideal transformer
i1
Fig. 7
i2
+
+
v1
v2
-
-
Ideal transformer
Figure 7 shows an ideal two-winding transformer. Coils are wound on ideal
Magnetic core to produce flux. Voltages is Induced on each winding.
If

Since
is the flux of a one-turn coil then
d1
v1 
dt
1  n1 , and 2  n2
d2
and
v2 
dt
v1 (t ) n1

v2 (t ) n2
we have
(1)
In terms of magnetomotive force (mmf) and magnetic reluctance
mmf  
n1i1  n2i2  

Ideal transformer
If the permeability  is infinite

becomes zero then
n1i1  n1i1  0
i1 (t )
n2

i2 (t )
n1
and
(2)
From (1) and (2)
v1 (t )i1 (t )  v2 (t )i2 (t )  0
(3)
The voltage v1 does not depend on i1 or i2 but it depends only on v 2
Ideal transformer
For multiple windings
n2 i2
n1i1  n2i2  n3i3  0
+
v2
v1 v2 v3


n1 n2 n3
-
i1
Ideal i3
+
+
v1
v3
-
n1
n3
Fig. 8
(equal volt/ turn)
Ideal transformer
Impedance transformation
Ideal i
2
i1
+
+
v1
v2
-
n1
Rin
n2
(n1 n2 )v2
-
v1
Rin   n

i1  ( 2 n )i2
1
Rin 
 
n1
n2
2
RL
RL

n1 2 

n2 
v2 

  i2 
v2   RL i2
Impedance transformation
In sinusoid stead state
Ideal i
2
i1
+
+
v1
v2
-
Z in
n1 : n2
V1
Zin ( j ) 

I1
 
n1
n2
2
ZL
Fig. 9
-
 V2

 I2



 
n1
n2
2
Z L ( j )
Controlled sources
Controlled sources are used in electronic device modeling. There four kinds
of controlled source .
i1
+
v1
-
0
 Current controlled current source
 Voltage controlled current source
 Voltage controlled voltage source
 Current controlled voltage source
i2
i1  0
+
 i1
v2
-
+
v1
-
+
 v1
+
-
+
v1
-
i2
i1  0
v2
g m v1
-
i2
i1  0
+
v2
-
i2
+
v1  0
-
rm i1
+
-
v2
Fig. 10
Controlled sources
Current controlled current source : Current ratio
i2

i1
Voltage controlled current source :
i2
gm 
v1
Voltage controlled voltage source :
Current controlled voltage source :
Transconductance
Voltage ratio
Transresistance
v2

v1
v2
rm 
i1
Controlled sources
Example1
Determine the output voltage from the circuit of Fig.11
1
Mesh 1
( Rs  R1 )i1  vs
R1
v1  i1 R1 
vs
Rs  R1
Mesh 2
Rs
R2
i1
+
v_s
+
R1
+
v1
-
v2   v1
1'
RL
RL
vL  i2 RL 
v2 
 v1
R2  RL
R2  RL
RL
R1


vs
R2  RL Rs  R1
2
+
i2 RL
vL
-
-
2'
Fig.11
Controlled sources
Example 2
Determine the node voltage from the circuit of Fig.12
1
2
+
G1
is
C1
C2
v1
+
G2
v2
-
1'
Fig.12
i2  g m v1
2'
KCL
dv1
d (v1  v2 )
G1v1  C1
 C2
 is
dt
dt
d (v2  v1 )
C2
 G2v2  i2
dt
(1)
Controlled sources
(1)  (2)
Diff. (3)
d (v2  v1 )
 G2v2  gmv1  0
dt
dv1
(G1  gm )v1  C1
 is  G2v2
dt
C2
dv1
d 2 v1 dis
dv
(G1  g m )
 C1 2 
 G2 2
dt
dt
dt
dt
from (1) dv2
then
d 2 v1
dt 2
1

dt
C2
(2)
(3)
(4)
dv1


(
C

C
)

G
v

i
2
1 1
s
 1
dt


 G1  g m  G2 G1 G2  dv1 G1G2
G2
1 dis




v


is (5)

1
C1
C1 C2  dt C1C2
C1 dt C1C2

Controlled sources
The initial conditions
v1 (0)  V1 , v2 (0)  V2
From (3)
dv1
1
(0) 
is (0)  G2V2  ( g m  G1 )V1 

dt
C1
From (5) and (6)
v1 (t ) and
v2 (t ) can be solved
(6)
Controlled sources
Other properties
The instantaneous power entering the two port is
p(t )  v1 (t )i1 (t )  v2 (t )i2 (t )
Since either v1 (t ) or
i1 (t ) is zero thus
p(t )  v2 (t )i2 (t )
If
R2 is connected at port 2
Therefore
v2  i2 R2
p(t )  i22 R2
Power entering a two port is always negative
Controlled sources
Example 3
Consider the circuit of Fig. 13 in sinusoid steady-state. Find the input
impedance of the circuit.
1
I1
1'
IL
2
+
Is
V
I 2   I1
2'
Z in
Fig. 13
ZL
Controlled sources
I s  I1
I 1  I 1  I L
IL
I1 
1
ZLIL
V
Z in 

 (1   ) Z L
Is
I1
Note if   1 the input impedance can be negative and this two port
Network becomes a negative impedance converter.