Coupled Circuits J. R. Lucas
Resonance Resonance in sound. At resonance maximum vibration (string) and hear the greatest sound (water column) Same idea is present in electrical engineering. Resonance basically occurs when a quantity, such as voltage or current, becomes a maximum. Maximum one quantity minimum with another
So that resonance could occur at a minimum value of a quantity as well. For example, for a given source voltage, current would would be be maximum when impedance of the circuit is a minimum minimum..
I
Z R jX R 2 X 2
V Z
In an a.c. circuit, would have a minimum value when X is zero (Possible because inductive and capacitive reactances have opposite signs).
So that resonance could occur at a minimum value of a quantity as well. For example, for a given source voltage, current would would be be maximum when impedance of the circuit is a minimum minimum..
I
Z R jX R 2 X 2
V Z
In an a.c. circuit, would have a minimum value when X is zero (Possible because inductive and capacitive reactances have opposite signs).
When X is zero, the circuit is purely resistive and the power factor of the circuit circuit becomes unity. unity. Thus resonance is also defined in terms of the power factor of a circuit becoming unity. Three main methods of defining resonance condition: (a) current through a circuit for a given source voltage becomes a maximum maximum:: voltage across a circuit for a given source current becomes a minimum, admittance of the circuit becomes a maximum, or impedance of the circuit becomes a minimum.
(b) voltage across a circuit for a given source current becomes a maximum: current through a circuit for a given source voltage becomes a minimum, impedance of the circuit becomes a maximum, or admittance of the circuit becomes a minimum. and (c) power factor of the circuit becomes Unity:
when the impedance of the circuit is purely real, when the admittance of the circuit is purely real, or
when the voltage and the current are in phase.
Condition (a) occurs in series circuits, and is usually referred to as Ser ies Resonance . Condition (b) occurs in parallel circuits, and is usually referred to as Parallel Resonance . While series resonance and parallel resonance are exclusive conditions, unity power factor condition could correspond to either series resonance or parallel resonance.
In complicated circuits, the unity power factor condition could also give displaced answers from the other two conditions.
When does oscillations occur ? What determines natural frequency of oscillations ? Consider the simple pendulum. Why does it oscillate ? Because of stored energy in the system – potential and kinetic energy. During oscillations, energy transfers between potential energy and kinetic energy. Natural frequency of oscillations, depends on the value of gravity, length and so on. Friction in the medium would reduce the energy and cause the pendulum to slow down and finally stop.
In an electric circuit, Energy is stored in the electrostatic field in the capacitance electromagnetic field in the inductance. Oscillations occur when the energy gets transferred between these two forms. Resistance would cause energy losses results decreasing magnitude of oscillations.
Ser ies Resonance
Occurs in a circuit where the energy storage elements are connected in series. L
R
At angular frequency of Z = R + j L +
1 j C
,
1
j C
= R+j
( L
1
) C
Consider value of when | I | becomes a maximum for a given V | Z | becoming a minimum, or | V| becoming a minimum for a given I.
2
2
magnitude of impedance = | Z |, | Z | = R + phase angle of Z,
L 1 / C = tan R
( L
1
)2 C
1
Condition for maximum or minimum | Z | can either be obtained by
differentiation of | Z | or 2 differentiation of even | Z | or inspection from physical considerations. 2
Since | Z | consists of the sum of two square terms, the minimum value for any of components would be zero.
Since only the second term is dependant on , the minimum value could occur when second term is zero. i.e. for minimum value of | Z | 1 1 L or o C LC
2
( L
1
) ,= 0 or C
This would also correspond to the minimum value of Z. If the current were considered, E
| I | R j ( L phase angle of I=
E
1
) C
R ( L 2
1
) C
2
=,
|I|
Q R 0 high Q low R low Q high R
phase /2 anglelead
lag /2
o low Q hi h R
Q R 0
high Q low R
e c n a t c a e R e v i t c u d n I
e c n a t c a e R e v i t i c a p a C
XL
XL+XC
XC
Now consider the frequency at which the power factor of the circuit becomes unity.
Occurs when imaginary-part(imp imaginary-part(impedance) edance) 0 or imaginary-part(current) 0. i.e. j ( L
1
) C
= 0, or at
= o. (same as before).
Shape of curve is dependant on value of r r of of the circuit. As r As r 0 0,, perfect resonance occurs occurs,, I at o As r 0 0,, angle of current with respect to voltage tends to either /2 or /2, changing at o. Series resistance r r defines defines quality of the resonance. If r is low, Ploss is low and hence the quality is high. At resonance, low r means r << Lo and r << 1/Co.
Quall i ty of a Cir cui t Qua
In simplest terms, we could defined quality of the circuit in terms of either ratio of Lo to r or r or 1/C 1/Co to r . Quality =
L o r
1
C o r
at resonance, for a series circuit
[For a parallel circuit, quality would actually increase when value of resistance R rather R 0, and hence inverse ratio would define the quality] A means of measuring quality of a resonant circuit has been defined, but which would appear to cause confusing confusing results if the series and parallel circuits are considered.
Quality Factor Q Necessary to have a unique definition In terms of quantities which do not vary with circuit. Quality of resonance is a relationship between maximum energy stored in energy storing elements (L or C)
and energy dissipation in resistive elements in the circuit. For consistency with original definitions, Q factor has a multiplying constant 2 , in addition to ratio of energies. Q
2
total stored energy energy dissipation per cycle
=
Determination of total stored energy Consider the simple pendulum. maximum kinetic energy occurs at the lowest point maximum potential energy occurs at the highest point total stored energy does not change if friction were absent.
It is easier to calculate either the maximum kinetic energy only at the lowest point, or the maximum potential energy only at the highest point, rather than calculate both at any other point.
Total stored energy in Electrical Circuit
maximum energy stored in the electromagnetic field occurs in the inductor when current is a maximum, maximum energy stored in the electrostatic field occurs in the capacitor when voltage is a maximum. Total energy does not change unless dissipated by the resistive elements in the circuit. For calculation purposes, energy stored at either the peak current or the peak voltage is considered. For a series circuit, it is easier to consider the current through the inductance rather than the voltage across the capacitor.
Verification of Energy based and Impedance based formulae
Necessary to verify whether energy based formula for Q is compatible with the impedance based formula for Q. For the series circuit, for a current of i(t) = Im sin t 2 Maximum energy stored in inductor = ½ L I m 2 Energy dissipated per cycle in series resistor = r Irms T, where T = 2 Q-factor for series circuit would be given by Q =
1 2
LI m2 2 rms
rI T
2 .
1 2
LI m2
r ( I m / 2 ) 2 / 2
L r
which is the expected result. Similar verification can be done for parallel circuit.
Relationship between Q-factor and bandwidth In a resonance curve, P max occurs at Imax (or Vmax) at I m resonance frequency (o). I m 2 2 Since P I (or V ), half 2 power occurs when I (or V) magnitude decreases by a factor of 2. 0
o
Points (½ and ½) on the resonance curve where half power occurs are known as half power points.
Frequency range between the half power points is defined as the bandwidth of the resonance curve. Fractional deviation of frequency from resonance =
o
For a series resonant circuit, we have the condition E
|I|
and
1
R j ( L
) C
L o
1
Q
R
E
C o R ,
E L o
R ( L 2
o
=,
1
) C
1
LC
1
.
R L
2
o
2
2
o
so that | I |
E L o
.
1 1
1 2 2 2 o Q o E 1
1 2 2 o o L o 2 o Q
near resonance, if the Q factor is high (as usually is), o o 2 , = o and
| I | =
o
E 1
2 1 2 L o 2 4 Q
E
R 1 4Q 2 2
| I | is a maximum when = 0,
1
2
giving |Imax| =
at the half power points,. P 12 P max , I i.e. i½ =
I max 2
=
1
E
2 R
=
E
R 1 4Q 2 12 2
1
2
1 2
I max
E R
,
i.e. 1 + 4 Q
½ = +
½ = 2, giving
2
2
1 2Q
½ =
,
4Q
2
½ = 1 2
1 2Q
Bandwidth = ( ½ ½) = ( ½ ½) =
½ ½
=
o
Q R
Q-factor =
resonance frequency bandwidth
[This same definition can be shown to be true for parallel resonance as well] Parallel Resonance
L 1
Parallel resonance occurs in a circuit where the different energy storage elements are connected in parallel. At an angular frequency of Y=
R
1
1
1
+
j L
,
+
j C
=
R
j ( C
1 L 2
)
magnitude of admittance = | Y |, | Y | =
1
R
2
( C
1
L
)
2
parallel resonance is seen to occur when imaginary part is zero. 1
i.e.
L
= C
Also corresponds to unity power factor.
In practical circuits, series resonance and parallel resonance can occur in different parts of same circuit. The unity power factor resonance may correspond to one of these. Example 1
Find the types of resonance and resonance frequencies of the circuit shown. Solution Z = R + jL1 +
j L2 j L2
R
L1
L2 C
1
j C 1 j C
j L2
= R + jL1 + 1 2 L2C
Consider each type of resonance condition in turn (a) when the power factor is unity equivalent impedance is purely real. Therefore
L1 +
L2 1 L2C 2
i.e. L1 + L2 - L1L2C = 0, 2
where Leq =
L1 L2
=0 or
2
L1 L2 L1 L2C
1
LeqC
L1 L2
It is seen that the equivalent inductance is L 1// L1. Thus unity power factor resonance frequency corresponds to =
1 Leq C
,
(b) when the impedance of the circuit is a minimum Examination shows that this also corresponds to a minimum value of impedance or series resonance. This need not be the case for all examples. (c) when the impedance of the circuit is a maximum 2
L2 | Z | = R L1 + 2 1 L2C 2
2
Z will have a maximum value of
at (1 - L2 C) = 0
resonant frequency for parallel resonance =
1
L2C
L
E x amp ampll e 2
Find the unity power factor resonance frequency of circuit.
R C
Also determine the parallel resonance frequency if R = 20 , L = 10 mH, and C = 4 F. Solution ( R j L)
Z
1
R j L j C = = 2 1 (1 LC ) j CR R j L j C
for unity power factor f actor resonance, impedance impedance Z is purely real.
R
(1 LC ) 2
=
L CR
L CR ,
i.e. C R = L (1 - L C) 2
2
L CR 2
2
, or = L2 C Unity power factor resonance 2
=
L2 C
3
6
10 10 4 10 400 400 6
100 100 10 4 10
6
4582.6 rad / s 729.3 Hz
Now consider the magnitude of the impedance impedance 2 2 2 R L 2 |Z| = 2 2 2 2 2 (1 LC ) C R 2
for maximum value of | Z |, | Z | must be a maximum.
2
d | Z |
i.e.
= [(1-2LC)2 + 2C2R 2].2L2
d 2 2 2 2 2 2 – (R (R + L )[2(1- LC).( 2LC) + 2C R ] = 0
i.e. (1- LC) L + C L R + 2R LC(1- LC) 2 2 2 2 2 2 2 2 + 2 L LC(1- LC) LC) – – (R (R + L )C R =0 2
2
2
2
2
2
2
2
2
L C + [-2L C + C L R - 2R L C +2 L C 4
4
2
2
2
2
2
3
2
2
2
2
2
2
2
2
2
3
4
– L L C R ] +L + 2R LC - C R = 0
substituting values for the components -0.1610
-12 2 -4 1.2810 + 1.294410 = 0 4 – 1.28 15 0.809010 = 0 4 8106 2 – 0.8090 -18
can be solved as follows 2 = 4106 (161012 + 0.80901015) 6 6 6 = 410 28.72310 = 26.72310 rad/s = 791.4 Hz
This quadratic equation in
It can be seen that this resonance frequency is close but not the same as the earlier result of 729.3 Hz.
Loci Diagrams for RL and RC circuits
(a) Series RL circuit
L
R
I
For a constant voltage V at constant frequency applied,
V,
If resistance R of the circuit varies, VR + VL = V = a constant. V = (R + j L) .I = (R + j X) .I If V is taken as reference, the current I would lag the voltage by a phase angle .
VL
I VR
P
VR is in phase with I and V L would be quadrature o leading I (or current I lagging the voltage V L by 90 ). Since VR and VL must be mutually perpendicular, when R varies, point P must move along a semi-circle. This semi-circle is the locus of point P as R is varied. P Consider the variation of I as R varies VL and X is kept fixed. VR V = VL + VR . VL is first drawn perpendicular to I o I (such that current lags voltage by 90 ). The locus of the point P will again be a semi-circle.
Since X is fixed, there will be a definite proportion between length of phasor I and length of phase V L. o Also, I is always lagging the voltage V L by 90 . Thus the locus of I must also be a semi-circle lagging o the semi-circle for voltage V L by 90 as shown. (b) Series RL circuit with inductance L having a finite resistance r L r R I In practical inductances winding resistance r may not be negligible. VX VX would have two components V, corresponding to inductive and resistive components respectively.
The locus of the node between VX VL the resistor and the inductor is VR no longer a semi-circle. I However, for total resistance PV r R + r the locus of that point with the pure inductance part remains a semi-circle. If X is variable, then ratio of V R to Vr remains a constant and value of the internal resistance of the coil can be determined.
(c) Series RC circuit Series RC circuit analysed in a similar manner to series RL circuit. However current would be leading I instead of lagging the voltage. Unlike in the practical inductor, the practical capacitor does not have VC significant losses. Thus the practical locus diagram is same as the theoretical diagram.
VR
P
(d) Parallel RL and RC Circuits Loci diagrams of parallel RL and RC circuits is obtained in a manner similar to obtaining that for the series circuits.
Mutual Inductance Mutual coupling between coils exist, when one coil is in the magnetic field created by another coil. L p
Ls
p m i p e p
M
es
is
When a varying current i p(t) flows in the primary winding, then a varying flux p is produced in the same coil and produces a back emf e p(t).
Part of the flux produced m can link with a second coil. Since this flux will also be varying, an induced emf e s will be produced in the second coil. d p
e p = N p
d t
d m
, es = Ns d t ,
m = k . p
k – coefficient of coupling mutual flux m
primary flux p
k < 1, k 1 (slightly less but very close to unity). In the linear region of the magnetisation characteristic,
p current i p producing the flux. Also, mutual flux m current i p. flux produced
Thus, induced emf es rate of change of current i p. The constant of proportionality is defined as the Mutual inductance Msp. N s m kN s p d i p d i p i.e.
e s
d t
M sp
d t , where
M sp
i p
i p
Thus the (coefficient of) Mutual inductance is defined as the flux linkage produced in a secondary winding per unit current in the primary winding. The sign associated with the mutual inductance can be positive or negative dependent on the relative directions of the two windings.
Direction of voltage induced in second coil will depend on the relative direction of winding of the two coils. Flux p in a coil is related to the l current i p producing the flux N p N s through the self inductance of coil L p and may be expressed in terms of the dimensions of the magnetic path (length l and cross-section A) as follows. N p p
i.e. L p =
i p
N p B p A i p
N p N p i p A
L p =
i p l
N p H p A i p
2
N p A l
, but N p i p = H p l
The mutual inductance may be similarly derived in terms of the dimensions. k sp N s p k sp N s B p A k sp N p H p A Msp =
i p
k sp N s N p i p A
=
i p l
i p
i p
k sp N p N s A l
k ps N s N p A
similarly, M ps = l Coupling between the primary and the secondary, for all practical purposes, will be identical to the coupling between the secondary and the primary, so that k ps = k sp.
Thus it can be seen that for all practical purposes, M ps is identical to M sp, and would usually be denoted by a single quantity M and a single coefficient k. M =
kN s N p A l 2 p
,
N A L p =
l 2
, Ls = 2
2 s
N A
giving M = k L p Ls or
l M=k
L p L s
Energy stored in a pair of mutually coupled coils
Inductor stores energy in electromagnetic field =
1 2
LI 2 .
When mutual coupling is present, total energy stored by two coils is different from addition of 1 LI 2 terms. 2
i p Difference is the effective energy stored in the mutual inductance. V p L p Total energy stored
M
= v p i p dt + vs is dt ( L p =
d i p dt
M
d i s dt
).i p dt ( L s
d i s dt
M
d i p dt
).i s dt
= L p i p di p M i p dis + Ls is dis M is di p
is Ls
Vs
This may be grouped to give Total energy stored = L p i p di p + Ls is dis M i p dis M is di p = L p i p L s i s M i p i s Therefore the effective energy stored in the mutual inductance corresponds to M i p is . 1 2
2
1 2
2
Equivalent inductance of 2 mutually coupled coils in series
L1
i
L2
i
L1
L2
Coils can either be connected in series so that fluxes aid each other, or fluxes oppose each other. Since each current is i, Total energy stored =
1 2
L1i L2 i M i.i or
=
1 2
L1i L2i M i.i
2
2
1 2
1 2
2
2
If a single equivalent inductor L eq is considered, 1 2
2
L i eq total energy stored = . Thus equating the energies 1 2
2
Leqi =
1 2
L1i 2 12 L2i 2 M i.i or
i.e. Leq = L1 + L2 + 2 M
1 2
L1i 2 12 L2i 2 M i.i
or L1 + L2 2 M
The effective inductance can either increase or decrease due to mutual coupling dependant on whether the coils are wound in the same direction or not.
Example 3
Consider a couple of coils connected in series as shown. Let each coil have N turns, I I2 1 then total series connected L2 L1 coil has 2N turns. If the dimensions of the common magnetic circuit on which these are wound have area A and length l 2
N A L1 = L2 =
,
l 2
(2 N ) A
2
4 N A
and total coil L = . l l L1 + L2 is obviously not equal to this total.
What went wrong ? If the two coils are wound on the same magnetic circuit, very closely, then there would be mutual coupling with the coefficient of coupling almost unity. Then, M = k L p Ls
1
2 2 N A N A
l
l
2 N A
l
Thus the total inductance would be 4 N 2 A
L1 + L2 + 2 M =
l
as expected.
Magnetic circuit Analysis When a coil is wound round a magnetic core, the core becomes magnetised
one side becomes a north pole, and other side becomes a south pole. There are different methods of remembering which side is which. One of the simplest methods to remember is
When we look at a coil from one side,
if current direction is anti-clockwise, nearer side is a north pole (also seen from arrow direction of “N”);
if current direction is clockwise, nearer side is a south pole (also seen from arrow direction of “S”). left hand side
right hand side
A
B
Consider coil A and coil B, we can easily visualise that they are both wound in the same direction.
i.e. if we examine the coil from the left hand side when a current is entering the left hand side end of the coil, each coil would produce a north pole at the left hand side and a south pole at the right hand side. left hand side
right hand side
A
B
A B
With a slightly differently drawn diagram it would be less obvious. Further if the magnetic circuit was not a straight line, confusion could tend to enter the decision.
A method would be to open out the bent path to make it a straight line and then compare directions. Each time we analyse a magnetic circuit, we would need to look at the relative directions of the two coils. However, once wound, the relative directions of the coils would not change, independent of how we look at the coils. Thus we can use a simpler method to know the relative directions of the coils. For this purpose we use dots to denote similar ends of different coils.
Dot Notation
Relative directions of coils is important in determining direction of induced voltage with mutual coupling. Consider the following magnetic circuits, and the corresponding electrical circuit with polarities defined by the dotted ends. A A
B
A
B B
Note that there are two possible ways of drawing the dotted ends, but both give the same relationship to each other.
The dots do not indicate that one end is a north pole or a south pole, as this would depend on which direction the current passes in the coil. Dots indicate similar ends, in that if a changing flux passes in magnetic core, then induced voltages would either all correspond to direction of dot or all to opposite direction. Thus they indicate similar ends of windings only. Consider the same core as before, but with the coil B wound in the opposite direction. P P
A
Q R
B
S
A A
Q
R
B B
S
The position of the dots again indicate the different winding directions. Once the dots are drawn to indicate similar ends, unlike in the magnetic circuit, there is no real necessity to physically place the winding diagram in the same physical position. Thus the following diagrams would be identical. R P
Q
R
S
S
P
Q
Thus it is seen that once the dots have been marked to identify similar ends, the physical positions have no meaning and we can draw them where it is convenient.
Let us now see how, once the dots have been marked, in an electrical circuit the correct directions of the induced voltage scan be obtained. V1
I1
A
V2
I2
B
V1 I1
A
V2 I2
B
If a single coil is considered, the induced voltage would always be opposite to the direction of current flow in that winding. The voltages V 1 and V2 have been drawn as such in the above diagrams.
Let us now see what would happen due to each individual current. Since each winding is wound in the same direction, the current I1 would induce voltages in A and B in the same sense. Also if the current I 2 is marked as shown, then it too would induce voltages in A and B in the same sense, and also in the same sense as due to the current I1. The fluxes and the corresponding induced emfs are thus additive as the coils are wound in the same direction. This is also shown by the non magnetic equivalent circuit.
In other words, the voltage drop due to the self inductance term and the voltage drop due to the mutual inductance term have the same sign. Consider what would happen if direction of current I 2 is changed without changing any physical considerations. V1
V2
V1 I1
A I1
B
A
V2 I2
B
I2
Obviously, the direction of the voltage V 2 due to this current I2 on its own winding will change in direction.
However, the induced voltage due to the current I 1 will have the same direction for both coils, as the physical directions of the two coils are still the same as before. Thus it is seen that the voltage drop due to the self inductance term and the voltage drop due to the mutual inductance term have opposite signs. Let us see what would happen if the direction of one of the coils were changed but with the currents entering each winding at the left hand end of the winding. V1
I1
I2
V2
V1 I1
A
B
A
V2 I2
B
If the direction of one of the windings is changed, the mutual induced voltage will change in direction relative to the self inductance term. Thus again the voltage drop due to the self inductance term and the voltage drop due to the mutual inductance term have opposite signs. The final possibility is if both the direction of the winding and the current direction are changed. V1
I1
I2
V2
V1 I1
A
B
A
V2 I2
B
It will easily be seen that the voltage drop due to the self inductance terms and that due to the mutual inductance term must have the same sign. The above derivations were done primarily based on the magnetic circuits appearing in the above 4 cases. Let us now see what has happens with the electrical circuits. The summary of the 4 cases are shown.
V1 I1
A
V2 I2
V1 I1
I1
A V1
A
V2 I2
I2
V1 I1
A
B
B V2
B
mutual term has same sign self inductance term mutual term has opposite sign to self inductance term mutual term has opposite sign to self inductance term
V2 I2
B
mutual term has same sign self inductance term
It is seen that in cases 1 and 4, the mutual inductance term has the same sign, while in cases 2 and 3 the mutual inductance term has opposite sign to the self inductance term. Let us see what properties actually cause the above situation. It is seen that in both cases 1 and 4, the currents marked enter the coil at the dotted end, while in both cases 2 and 3, one current enters at a dotted end, while the other current leaves at the dotted end. The above results can be stated in the following manner.
If both currents either enter at dotted end, or both currents leave at dotted end, the sign associated with mutual inductance is positive, and mutual inductance term in voltage drop would have the same sign as the self inductance term.
If one current enters at a dotted end and the other current leaves at dotted end, sign associated with mutual inductance is negative, and the mutual inductance term in the voltage drop would have the opposite sign to the self inductance term.
Example 4
For circuit shown, write down voltage drop across AB. e(t) L1 R i1 A B M L2
i2
VAB = R. i1 – e(t) + (L p i1 + M p i2) where p = d/dt VBA =
R. i1 + e(t) (L p i1 + M p i2)
Since both currents enter at respective dotted ends, mutual inductance term and self inductance term have same sign .
True even if voltage is measured in opposite direction. Usually advisable to treat the self inductance terms and the mutual inductance terms together within parenthesis to avoid wrong negation. With steady state a.c. analysis, the input quantities are sinusoidal with frequency , so the differential would give a multiplication of and a phase shift of Thus p = j would give the required equations. The impedance due to mutual inductance for a.c. would in general be j
Non-coupled Equivalent circuit of simple coupled circuits
When mutual coupling is present, voltage drops in a particular branch depends on
currents in that branch currents in other branch with which coupling exists Series and parallel equivalents of branch circuits containing mutual coupling cannot be easily implemented.
can sometimes be avoided by obtaining a noncoupled equivalent circuit.
(a) coupled coil s being on two ar ms of a T-juncti on Consider pair of mutually M L1 L2 i1 i2 coupled coils on two arms i1 i2 of a T-junction. P Q R
By marking currents i1 and i2 flowing in them, and a current (i 1 – i2) flowing in the common branch, Kirchoff‟s current law has been applied. Applying Kirchoff‟s voltage law between PR, and RQ,
VPR =
L1
d i1 d t
M
d i2 d t
= L1 p i1 – M p i2
or VPR = jL1 i1 – jM i2 with sinusoidal current
Current i1 leaves dotted end of L 1, while the current i 2 enters dotted end of L 2,
sign associated with the mutual inductance is negative, voltage drop term due to mutual inductance has opposite sign to that due to self inductance term
and VRQ = L2 p i2 – M p i1 If a non-coupled equivalent i1 circuit is to be obtained, P voltage drops in a branch should only correspond to currents in it own branch.
LA
LB i1 i2 Lm
R
i2
Q
Re-write the 2 equations as follows to achieve this. VPR = L1 p i1 - M p i2 – M p i1 + M p i1, i.e. VPR = (L1 – M) p i1 + M p (i1 – i2) similarly VRQ = (L2 – M) p i2 + M p (i1 – i2) These two equations would be satisfied with LA = L1 – M, LB = L2 – M, and Lm = M This transformation will be valid, independent of what directions marked for currents in the diagrams. [Compare with: Expression for parallel equivalent of two resistors is independent of original markings of the current directions when proving.]
L1
M
L1- M
L2
L2 - M
M
Note that L – M appears when the two coils are opposing each other. In this case the common branch has an added +M appearing on it. In like manner, if the two dots were at the further ends, the equations would be unchanged and the equivalent circuit would also be unchanged.
L1- M
M L2
L1
L2 - M
M
If position of one of the dots is changed, then the two coils would be aiding each other, and the terms can be shown to correspond to L+M with a common term of – M. M L1
L1+ M L2
L2 + M
M
Example 5
Write down the non-coupled equivalent circuit for the coupled circuit shown. R 1
L1
P
R 2 M
E
C
L2
Solution
R 1
L1
P R 2
To get correct sign in nonM C L2 coupled equivalent circuit, E consider imaginary current to flow through coil 1 and coil 2, forgetting other elements.
It is seen that they are opposing. [You will notice that the two coils do not exactly meet at a common node, but that the node P would have that property if the positions of R 2 and L2 were interchanged.]
Since they are series elements, the voltage drop equation would not change even if the order were changed. [However, intermediate point of connection between these elements would have a different voltage from earlier, as the drops are occurring in a different order.]
The non-coupled equivalent circuit may thus be drawn R 1
L1-M
R 2 M
E
L2-M
C This circuit no longer has mutual coupled elements, but the elements have taken into account the affects of mutual inductance. Thus the problem may be solved as for any alternating current problems.
Transformer as a pair of mutually coupled coils Transformer is a pair of mutually coupled coils. is i p M From Kirchoff‟s voltage law V p = L p p i p – M p is V p L p Ls and Vs = – (Ls p is – M p i p) For near ideal transformers
|V p| : |Vs| a
a:1
|i p| : |is| 1/a Thus V p and aV s are comparable quantities i p and i s /a are comparable quantities.
Vs
i p
is/a
is Using these comparable variables i p-is/a V p = L p p i p – aM p is/a V p Vs aVs 2 and aVs = – (a Ls p is/a – aM p i p) /a If i p and i s are unequal, a:1 /a Leakage current = i p – i s re-formulated in terms of this leakage current as V p = L p p i p – aM p is/a + aM p i p – aM p i p = (L p – aM) p i p + aM p .(i p – is/a) 2 and aVs = – (a Ls p is/a – aM p i p) + aM p is/a – aM p is/a 2
M
= – a (Ls – a ) p is/a + aM p.(i p – is/a)
Note that the equations are expressed either in terms of i p and i p – i s /a or in terms of i s /a and i p – i s /a. Allows a non-coupled equivalent circuit to be formed L p- aM V p
i p
2
a (Ls – M/a) is/a
i p
aVs
aM
l p
2
a l s is/a
V p Lm
is
aVs
Vs
i p-is/a
i p-is/a
a:1
May also be drawn with ideal transformer added so that secondary quantities would remain unmodified on that side.
Also, M = k so that
L p Ls
M = k
, L p/Ls = a
L p
L p a2
2
giving aM = k L p, L p – aM = (1 – k) L p, 2 2 a (Ls – M/a) = a (1 – k) Ls Since k is the coefficient of coupling, (1 – k) corresponds to the leakage. (1 – k)Lp and (1 – k)Ls correspond to leakage inductances l p and l s of primary and secondary windings. Shunt inductance aM corresponds to the magnetisation inductance L m of the transformer. The transformer equivalent circuit is normally drawn with these variables rather than the self and mutual inductances.
Practical Transformer In addition has resistances r p and r s in primary and secondary windings has losses (eddy current loss and hysteresis) in the transformer core based on magnetic flux in core (or on corresponding voltage across windings). 2 r p , r s are included in primary and secondary sides ( r s drawn as a r s ). Core loss is represented by a shunt resistance R c across the magnetisation inductance.
i p
l p
r p
2
a l s
V p Lm
R c
a2r s
is/a
is
aVs
Vs
i p-is/a a:1
Dependent Sources Passive Circuit Elements Most basic elements – R, L and C. Do not generate any electricity. Either consume energy
in Resistive elements convert from electrical form to non-electrical form produce heat, light etc or store energy
in Capacitive or Inductive elements in electrostatic and electromagnetic fields.
Active Circuit Elements (Sources)
Circuit component capable of producing energy. Categorised into
voltage sources and current sources. independent sources ( generated voltage or current does not depend on any other circuit voltage or current)
dependent sources (generated voltage or current depends on another circuit voltage or current ).
Independent source
Terminal voltage (or current) depends only on the loading and internal source quantity, but not on any other circuit variable.
Symbol for Independent source
I ndependent Voltage and Cur r ent Sour ces e(t)
e(t) Z(p)
i(t)
i(t) v(t)
v(t)
Figure 5(a) – Ideal voltage source Figure 5(b) – Practical voltage source
v(t) = e(t) for all i(t) I(t) i(t) vt Figure 6(a) – Ideal current source
i(t) = I(t) for all v(t)
v(t) = e(t) – Z(p).i(t) I(t) i(t) Y(p)
v(t) Figure 6(b) – Practical current source
i(t) = I(t) – Y(p).v(t)
Dependent Sour ce Symbol for Dependent source
A dependent voltage source (or current source) would have its terminal voltage (or current) depend on another circuit quantity such as a voltage or current.
Four possibilities exist.
V + –
Vo
I + –
Io
Fi ure 7 – De endent sources
Voltage dependent (controlled) voltage source o Current dependent (controlled) voltage source o Voltage dependent (controlled) current source o Current dependent (controlled) current source o
Example 1 Determine the current I and V o. Answer + – I Kirchoff‟s voltage law, gives 4V 4 7.5 = 4 I + 4 Vo + 5 – Vo + From Ohm‟s law 5V – + 7.5V 1.I = – Vo – 1 By substitution, – + I = 2.5 A Vo Figure 8 – Circuit for example 1 Vo = – 2.5 V
Example 2 Determine I and Vo. 5 Answer Kirchoff‟s voltage law, gives 6 = 5 (I – 0.8 Vo) – 4 – Vo I 0.8Vo – From Ohm‟s law + 6V 4V + – 1.I = – Vo 1 By substitution, – + Vo I = 1.0 A Figure 9 – Circuit for example 2 Vo = – 1 V
Example 3 Determine the current I. Answer Kirchoff‟s voltage law, gives
E = R 1 I + R 2 (1+) I From Ohm‟s law
R o I = – Vo By solution of equations E I R1 R2 (1 ) V o
Ro E R1 R2 (1 )
I
I
R 1
+ + E –
R 2
R o
Vo –
Figure 10 – Circuit for example 3
Operational Amplifier (Op Amp)
an active circuit element behaves as a voltage-controlled voltage source, very versatile can be used to add, subtract, multiply, divide, amplify, integrate and differentiate signals. A practical Op Amp, available in Integrated circuit (IC) packages would have inputs and outputs as shown in figure 11.
+
V Inverting Input
Output Non-inverting Input
+
V Null Figure 11 – Circuit connections of Op Amp +
–
Output voltage is limited to the linear range V to V . Ouside these limits, Op Amp is said to be saturated. „Null‟ determines the offset.
–
Vd
Rin
Rout
Vout
+ – A Vd
+ Figure 12 – Equivalent Circuit of Op Amp
Vd – Voltage difference of input terminals + and – A – Gain of Op Amp R in – Input resistance of Op Amp R out – Output resistance of Op Amp
Vout linearly proportional to V d with open-loop gain A. Op Amp has a dependent voltage source AV d. Parameter
Symbol Ideal
Open-loop gain
A
Input Resistance
R in
Output Resistance
R out
0
Supply Voltage
V ,V
+
–
Typical 5
10 to 10
8
10 to 100 6
10 to 10
13
5 to 24 V
Inverting Amplifier An inverting amplifier circuit is shown in figure 13.
R2 +
V
R1 –
Vin
+
V
Vout
Figure 13 – Circuit of Inverting Amplifier In an inverting amplifier output voltage decreases when input voltage increases and vice versa.
i
V in V d
R1
V d V out R2
V d Rin
V d V out R2
R2
V out AV d Rout
Figure 14 shows equivalent circuit of inverting amplifier.
i
R1
–
Vd
Rout
Rin +
+ – – A Vd
If Rout = 0, and Rin = , V out = – A V d AV in AV d R1
Figure14 – Equivalent Circuit
AV d A V d 2
R2
Vout
giving
AV in
V out
Thus V in
R1
V ou t AV ou t R2
V ou t R1
AR2 (1 A) R1 R2
If A as for an ideal Op Amp, V out
Gain = V in
R2 R1 ,
V d
V out A
0
Thus Vd is taken as virtual earth, when analyzing.
Non-inverting Amplifier Input voltage is applied directly to non-inverting (+) Vin input and a small part of output voltage is applied to the inverting ( – ) input from the R 1 R 2 potential divider.
+
V
– + V
For an ideal Op Amp, with Rout = 0, Rin = and A =
R1 R2 Gain =
R1
Vout
R
R2
Figure 15 – Non-inverting Amplifier
Summing Amplifier With 2 inputs, Summing amplifier is obtained . With R 1A = R 1B R2 V out (V inA V inB ) R1
If R 1A
R 1B V inA
R2 R1A V
VinA R1B
–
+
VinB
+ V
V inB
) V out R2 ( R1 A R1 B
Figure 16 – Summing Amplifier
Vout
Differential Amplifier
R2
Inputs are connected to both inverting or nonI A R1A VinA inverting terminals
Resultant output based on difference between V1A and V1B For ideal Op Amp,
VinB
IB R1B Vd R3
+
V – +
V
Vout
Figure 17 – Differential Amplifier
Rin = , Rout = 0, and A = . V out V ou t A 0 V d V d giving
