TEMPUS ENERGY: UNBALANCE IN THREE PHASE GRIDS
1: Introduction
Consider a three phase grid as visualized in Figure 1. Suppose the three phase voltages 𝑢𝑓1 (𝑡) ,
𝑢𝑓2 (𝑡) and 𝑢𝑓3 (𝑡) are sinusoidal voltages having three times the same magnitude. The phase
differences between these voltages equal 120°. In each phase, there is a grid impedance and assume
three times the same grid impedance is obtained i.e. 𝑍1 = 𝑍2 = 𝑍3 . In case the load is symmetric,
the phase currents 𝑖𝑓1 (𝑡) , 𝑖𝑓2 (𝑡) and 𝑖𝑓3 (𝑡) have the same magnitude and the phase differences
equal 120°. This implies the line voltages 𝑢𝐿1𝐿2 (𝑡), 𝑢𝐿2𝐿3 (𝑡) and 𝑢𝐿3𝐿1 (𝑡) also have the same
magnitude and the phase differences also equal 120°.
When the load of the grid is not symmetric, the phase currents 𝑖𝑓1 (𝑡) , 𝑖𝑓2 (𝑡) and 𝑖𝑓3 (𝑡) do not have
the same magnitude and/or do not have phase differences equal to 120° (i.e. the loads have
different power factors implying the currents do not have the same phase differences with respect to
the phase voltages 𝑢𝑓1 (𝑡) , 𝑢𝑓2 (𝑡) and 𝑢𝑓3 (𝑡)). This implies the three line voltages 𝑢𝐿1𝐿2 (𝑡),
𝑢𝐿2𝐿3 (𝑡) and 𝑢𝐿3𝐿1 (𝑡) do not have the same magnitude and/or do not have phase differences equal
to 120°.
Figure 1: Three phase power system
2: The use of vectors or complex numbers
To simplify calculations, sines are very often respresented by complex numbers (vectors). The
sinusoidal voltage
𝑢(𝑡) = √2 𝑈 𝑠𝑖𝑛(𝜔𝑡 + 𝜑)
̅ with magnitude 𝑈 and phase φ. More precisely,
is represented by a complex number 𝑈
̅ = 𝑈𝑒 𝑗𝜑 = 𝑈(𝑐𝑜𝑠𝜑 + 𝑗𝑠𝑖𝑛𝜑).
𝑈
In a symmetric three phase power system, there are three phase voltages 𝑢𝑓1 (𝑡), 𝑢𝑓2 (𝑡) and 𝑢𝑓3 (𝑡)
having the same amplitude and having phase differences equal to 120°. More precisely, the voltages
𝑢𝑓1 (𝑡) = √2 𝑈𝑓 𝑠𝑖𝑛(𝜔𝑡 + 𝜑)
2𝜋
𝑢𝑓2 (𝑡) = √2 𝑈𝑓 𝑠𝑖𝑛 (𝜔𝑡 + 𝜑 − )
3
2𝜋
(𝑡) = √2 𝑈𝑓 𝑠𝑖𝑛 (𝜔𝑡 + 𝜑 + )
{𝑢𝑓3
3
̅𝑓1 = 𝑈𝑓 𝑒 𝑗𝜑
𝑈
̅𝑓2 = 𝑈𝑓 𝑒 𝑗(𝜑−2𝜋⁄3)
{𝑈
̅𝑓3 = 𝑈𝑓 𝑒 𝑗(𝜑+2𝜋⁄3)
𝑈
are visualized in Figure 2.
Figure 2: Vector diagram of a symmetric three phase power system
Figure 3 visualizes a vector diagram for a non symmetric three phase power system. In general, the
voltages have different amplitudes and the phases are arbitrary (i.e. there are no phase differences
equal to 120°) giving
𝑢𝑓1 (𝑡) = √2 𝑈𝑓1 𝑠𝑖𝑛(𝜔𝑡 + 𝜑𝑓1 )
{𝑢𝑓2 (𝑡) = √2 𝑈𝑓2 𝑠𝑖𝑛(𝜔𝑡 + 𝜑𝑓2 )
𝑢𝑓3 (𝑡) = √2 𝑈𝑓3 𝑠𝑖𝑛(𝜔𝑡 + 𝜑𝑓3 )
̅𝑓1 = 𝑈𝑓1 𝑒 𝑗𝜑𝑓1
𝑈
̅𝑓2 = 𝑈𝑓2 𝑒 𝑗𝜑𝑓2
{𝑈
̅𝑓3 = 𝑈𝑓3 𝑒 𝑗𝜑𝑓3
𝑈
Figure 3: Vector diagram of a non symmetric three phase power system
3: Symmetrical components
A three phase power system can always be considered as the sum of a direct component (positive
sequence), an inverse component (negative sequence) and a homopolar component (zero sequence).
The direct component is visualized in Figure 4 and given by
𝑢𝑑1 (𝑡) = √2 𝑈𝑑 𝑠𝑖𝑛(𝜔𝑡 + 𝜑𝑑 )
2𝜋
𝑢𝑑2 (𝑡) = √2 𝑈𝑑 𝑠𝑖𝑛 (𝜔𝑡 + 𝜑𝑑 − )
3
2𝜋
𝑢 (𝑡) = √2 𝑈𝑑 𝑠𝑖𝑛 (𝜔𝑡 + 𝜑𝑑 + )
{ 𝑑3
3
̅𝑑1 = 𝑈𝑑 𝑒 𝑗𝜑𝑑
𝑈
̅𝑑2 = 𝑈𝑑 𝑒 𝑗(𝜑𝑑−2𝜋⁄3)
{𝑈
̅𝑑3 = 𝑈𝑑 𝑒 𝑗(𝜑𝑑+2𝜋⁄3)
𝑈
Figure 4: Direct component
The inverse component is visualized in Figure 5 and given by
𝑢𝑖1 (𝑡) = √2 𝑈𝑖 𝑠𝑖𝑛(𝜔𝑡 + 𝜑𝑖 )
2𝜋
𝑢𝑖2 (𝑡) = √2 𝑈𝑖 𝑠𝑖𝑛 (𝜔𝑡 + 𝜑𝑖 + )
3
2𝜋
(𝑡) = √2 𝑈𝑖 𝑠𝑖𝑛 (𝜔𝑡 + 𝜑𝑖 − )
{𝑢𝑖3
3
̅𝑖1 = 𝑈𝑖 𝑒 𝑗𝜑𝑖
𝑈
̅𝑖2 = 𝑈𝑖 𝑒 𝑗(𝜑𝑖+2𝜋⁄3)
{𝑈
̅𝑖3 = 𝑈𝑖 𝑒 𝑗(𝜑𝑖−2𝜋⁄3)
𝑈
Figure 5: Inverse component
The homopolar component is visualized in Figure 6 and given by
𝑢ℎ1 (𝑡) = √2 𝑈ℎ 𝑠𝑖𝑛(𝜔𝑡 + 𝜑ℎ )
{𝑢ℎ2 (𝑡) = √2 𝑈ℎ 𝑠𝑖𝑛(𝜔𝑡 + 𝜑ℎ )
𝑢ℎ3 (𝑡) = √2 𝑈ℎ 𝑠𝑖𝑛(𝜔𝑡 + 𝜑ℎ )
̅ℎ1 = 𝑈ℎ 𝑒 𝑗𝜑ℎ
𝑈
̅ℎ2 = 𝑈ℎ 𝑒 𝑗𝜑ℎ
{𝑈
̅ℎ3 = 𝑈ℎ 𝑒 𝑗𝜑ℎ
𝑈
Figure 6: Homopolar component
4: The use of symmetrical components
As already mentioned, an arbitrary non symmetric three phase power system is given by
𝑢𝑓1 (𝑡) = √2 𝑈𝑓1 𝑠𝑖𝑛(𝜔𝑡 + 𝜑𝑓1 )
{𝑢𝑓2 (𝑡) = √2 𝑈𝑓2 𝑠𝑖𝑛(𝜔𝑡 + 𝜑𝑓2 )
𝑢𝑓3 (𝑡) = √2 𝑈𝑓3 𝑠𝑖𝑛(𝜔𝑡 + 𝜑𝑓3 )
with arbitrary 𝑈𝑓1 , 𝑈𝑓2 , 𝑈𝑓3 , 𝜑𝑓1 , 𝜑𝑓2 and 𝜑𝑓3 . Starting from 𝑈𝑓1 , 𝑈𝑓2 , 𝑈𝑓3 , 𝜑𝑓1 , 𝜑𝑓2 , 𝜑𝑓3 it is
possible to calculate appropriate 𝑈𝑑 , 𝑈𝑖 , 𝑈ℎ , 𝜑𝑑 , 𝜑𝑖 and 𝜑ℎ in such a way the sum of the direct,
inverse and homopolar voltage sequences gives the original non symmetric power system (voltages).
4.1: Calculation of symmetrical components
To calculate the symmetrical components, the voltages 𝑢𝑓1 (𝑡), 𝑢𝑓2 (𝑡) and 𝑢𝑓3 (𝑡) will be represented
̅𝑓1 = 𝑈𝑓1 𝑒 𝑗𝜑𝑓1 , 𝑈
̅𝑓2 = 𝑈𝑓2 𝑒 𝑗𝜑𝑓2 and 𝑈
̅𝑓3 = 𝑈𝑓3 𝑒 𝑗𝜑𝑓3 . In case 𝑈𝑓1 =
by the complex numbers 𝑈
𝑈𝑓2 = 𝑈𝑓3 , 𝜑𝑓2 = 𝜑𝑓1 − 2𝜋⁄3 and 𝜑𝑓3 = 𝜑𝑓1 + 2𝜋⁄3, a symmetrical three phase system is
obtained which equals
̅𝑓1 = 𝑈
̅𝑓1
𝑈
̅𝑓2 = 𝑎2 𝑈
̅𝑓1
{𝑈
̅𝑓3 = 𝑎 𝑈
̅𝑓1 .
𝑈
Here, 𝑎 = 𝑒 𝑗2𝜋⁄3 and notice that 1 + 𝑎 + 𝑎2 = 0 and 𝑎3 = 1.
̅𝑑 for the first phase, 𝑎2 𝑈
̅𝑑 for the second phase and 𝑎 𝑈
̅𝑑 for the
The direct sequence contains 𝑈
̅𝑖 for the first phase, 𝑎 𝑈
̅𝑖 for the second phase and
third phase. The inverse sequence contains 𝑈
2 ̅
̅ℎ for the first phase, 𝑈
̅ℎ for the
𝑎 𝑈𝑖 for the third phase. The homopolar sequence contains 𝑈
̅
second phase and 𝑈ℎ for the third phase. This implies
̅𝑓1 = 𝑈
̅ℎ + 𝑈
̅𝑑 + 𝑈
̅𝑖
𝑈
̅𝑓2 = 𝑈
̅ℎ + 𝑎2 𝑈
̅𝑑 + 𝑎 𝑈
̅𝑖
{𝑈
̅𝑓3 = 𝑈
̅ℎ + 𝑎 𝑈
̅𝑑 + 𝑎2 𝑈
̅𝑖
𝑈
or equivalently in matrix notation
̅𝑓1
𝑈
1 1
̅𝑓2 ] = [1 𝑎2
[𝑈
1 𝑎
̅𝑓3
𝑈
̅ℎ
1 𝑈
̅𝑑 ]
𝑎 ] [𝑈
2
̅𝑖
𝑎
𝑈
̅𝑓1 , 𝑈
̅𝑓2 and 𝑈
̅𝑓3 are known, it is possible to calculate the symmetrical
In case the voltages 𝑈
̅ℎ , 𝑈
̅𝑑 and 𝑈
̅𝑖 as
components 𝑈
̅ℎ
𝑈
1 1 1
̅𝑑 ] = [1 𝑎
[𝑈
3
̅𝑖
1 𝑎2
𝑈
̅
1 𝑈𝑓1
̅𝑓2 ]
𝑎 2 ] [𝑈
𝑎 𝑈
̅𝑓3
This implies in case 𝑈𝑓1 = 𝑈𝑓2 = 𝑈𝑓3 , 𝜑𝑓2 = 𝜑𝑓1 − 2𝜋⁄3 and 𝜑𝑓3 = 𝜑𝑓1 + 2𝜋⁄3, there is only a
̅ℎ = 0, 𝑈
̅𝑑 =
direct component and there is no inverse nor homopolar component. More precisely, 𝑈
̅𝑖 = 0. In case 𝑈𝑓1 = 𝑈𝑓2 = 𝑈𝑓3 , 𝜑𝑓2 = 𝜑𝑓1 + 2𝜋⁄3 and 𝜑𝑓3 = 𝜑𝑓1 − 2𝜋⁄3, there
𝑈𝑓1 𝑒 𝑗𝜑𝑓1 and 𝑈
̅ℎ =
is only an inverse component and there is no direct nor homopolar component. More precisely, 𝑈
̅𝑑 = 0 and 𝑈
̅𝑖 = 𝑈𝑓1 𝑒 𝑗𝜑𝑓1 . The direct and the inverse components are actually symmetrical
0, 𝑈
three phase systems having another phase sequence (positive or negative).
By adding direct, inverse and homopolar voltage sequences, it is not only possible to obtain an
arbitrary non symmetric power system of phase voltages. By adding direct, inverse and homopolar
sequences, it is also possible to obtain line voltages, phase currents and line currents.
4.2: Exercise
Consider a non symmetrical three phase grid having (𝜔 = 314 𝑟𝑎𝑑⁄𝑠) line voltages
𝑢𝐿1𝐿2 (𝑡) = √2 240 𝑠𝑖𝑛(𝜔𝑡)
{𝑢𝐿2𝐿3 (𝑡) = √2 210 𝑠𝑖𝑛(𝜔𝑡 − 2.178)
𝑢𝐿3𝐿1 (𝑡) = √2 210 𝑠𝑖𝑛(𝜔𝑡 + 2.178)
Calculate the symmetrical components and study the impact of this non symmetrical voltage on an
induction motor (suppose the stator windings are delta connected). Suppose the slip 𝑠 of the motor
equals 𝑠𝑘 ⁄2 (𝑠𝑘 is the slip at the breakdown torque 𝑇𝑘 ) and 𝑠𝑘 = 0.2. Calculate the decrease of the
motor torque due to the inverse component in the grid voltage.
Solution:
̅𝐿1𝐿2 = 240, 𝑈
̅𝐿2𝐿3 = 210 𝑒 −𝑗2.178 and 𝑈
̅𝐿3𝐿1 = 210 𝑒 𝑗2.178, one obtains that 𝑈
̅ℎ = 0, 𝑈
̅𝑑 =
Since 𝑈
̅𝑖 = 20. Using the Kloss equation, the torque in the positive direction equals
220 and 𝑈
2 𝑇𝑘𝑑
4
𝑇𝑑 = 𝑠
𝑠𝑘 = 5 𝑇𝑘𝑑
𝑑
𝑠𝑘 + 𝑠𝑑
due to the direct component (𝑠𝑑 = 𝑠𝑘 ⁄2) (𝑇𝑘𝑑 is the breakdown torque due to the direct voltage
component). With respect to the inverse rotating magnetic field due to the inverse component in the
grid voltage, the slip equals
𝑠𝑖 = 1 + (1 −
𝑠𝑘
𝑠𝑘
)=2−
2
2
The inverse voltage component gives a breakdown torque
2
𝑈𝑖 2
20
𝑇𝑘𝑖 = ( ) 𝑇𝑘𝑑 = (
) 𝑇𝑘𝑑
𝑈𝑑
220
implying a torque in the negative direction which equals
2 𝑇𝑘𝑖
𝑇𝑖 = 𝑠
𝑠 .
𝑖
+ 𝑘
𝑠𝑘
𝑠𝑖
In case 𝑠𝑘 = 0.2, 𝑠𝑑 = 0.1 and 𝑠𝑖 = 1.9 which implies the total torque
20 2
𝑇 = 𝑇𝑑 − 𝑇𝑖 = 0.8 𝑇𝑘𝑑 − 0.208 𝑇𝑘𝑖 = (0.8 − 0.208 (
) ) 𝑇𝑘𝑑 = 0.798 𝑇𝑘𝑑 .
220
In the present example, the impact of the inverse voltage component on the torque is limited. In case
𝑠𝑘 is larger and/or the inverse voltage component is larger, the impact of 𝑇𝑖 will be larger. Consider
e.g. the situation when 𝑠𝑘 = 0.5.
5: The relationship between voltages and currents
Consider a three phase power system as visualized in Figure 1. The generators generate a symmetric
direct sequence 𝑢𝑓1 (𝑡) , 𝑢𝑓2 (𝑡) and 𝑢𝑓3 (𝑡) (with sinusoidal voltages). In case the load is symmetric
(with sinusoidal currents), also the currents 𝑖𝑓1 (𝑡), 𝑖𝑓2 (𝑡) and 𝑖𝑓3 (𝑡) only contain a direct component.
This implies the line voltages 𝑢𝐿1𝐿2 (𝑡), 𝑢𝐿2𝐿3 (𝑡) and 𝑢𝐿3𝐿1 (𝑡) only contain a direct component and
are symmetric.
In case the load is non symmetric, the currents 𝑖𝑓1 (𝑡), 𝑖𝑓2 (𝑡) and 𝑖𝑓3 (𝑡) contain a direct component,
an inverse component and a homopolar component. The voltage drops across the grid impedances
𝑍1 , 𝑍2 and 𝑍3 will also contain a direct component, an inverse component and a homopolar
component. Relying on the voltage law of Kirchoff, the line voltages 𝑢𝐿1𝐿2 (𝑡), 𝑢𝐿2𝐿3 (𝑡) and 𝑢𝐿3𝐿1 (𝑡)
will also contain a direct component, an inverse component and a homopolar component. This
implies the voltages 𝑢𝐿1𝐿2 (𝑡), 𝑢𝐿2𝐿3 (𝑡) and 𝑢𝐿3𝐿1 (𝑡) will not be symmetric anymore.
6: Consequences of non symmetric grid voltages
6.1: Induction motors
Consider an induction motor fed by a non symmetric three phase voltage and this voltage is the sum
of a direct component and an inverse component. This implies the stator current of the motor will
contain a direct component and an inverse component (assume the direct component is larger than
the inverse component).
The direct component generates a stator rotating magnetic field. Due to this rotating field, the rotor
rotates in the same direction as this rotating field (positive direction). The inverse component
generates a second rotating field which rotates in the opposite direction. This second rotating field
also affects the rotor.
Due to each rotating magnetic field, there is a torque speed characteristic as visualized in Figure 7.
The first rotating field (due to the direct component) has a synchronous speed 𝑁𝑆 and the second
rotating field (due to the inverse component) has a synchronous speed −𝑁𝑆 . The actual speed 𝑁 of
the rotor is somewhat smaller than 𝑁𝑆 .
Figure 7: Torques due to the direct and inverse rotating fields
Due to the direct rotating magnetic field there is a large driving torque in the positive direction. The
inverse rotating magnetic field generates a smaller torque in the opposite direction (negative
direction). The total torque is the direct torque reduced with the inverse torque. Moreover, due to
the large slip of the rotor with respect to the inverse rotating magnetic field additional voltages and
currents are induced in the rotor causing a significant additional heat dissipation. This reduces the
energy efficiency of the motor.
When comparing with 𝑁𝑆 , the rotor has a slip
𝑠𝑑 =
𝑁𝑆 − 𝑁
𝑁𝑆
and when comparing with −𝑁𝑆 , the rotor has a slip
𝑠𝑖 =
−𝑁𝑆 − 𝑁
.
− 𝑁𝑆
There are two stator rotating magnetic fields, the first one has the synchronous speed 𝑁𝑆 and the
second one has the synchronous speed −𝑁𝑆 . This first stator rotating magnetic field generates
voltages and currents having a frequency 𝑠𝑑 50 𝐻𝑧. This generates a rotor rotating magnetic field
having a speed 𝑠𝑑 𝑁𝑆 with respect to the rotor. Since the rotor has a speed (1 − 𝑠𝑑 ) 𝑁𝑆 , the rotor
rotating field has an absolute speed 𝑁𝑆 .
The second stator rotating magnetic field generates voltages and currents having a frequency
𝑠𝑖 50 𝐻𝑧. This generates a rotor rotating magnetic field having a speed 𝑠𝑖 (−𝑁𝑆 ) with respect to the
rotor. Since the rotor has a speed (1 − 𝑠𝑖 )(−𝑁𝑆 ) = (1 − 𝑠𝑑 )𝑁𝑆 , the rotor rotating field has an
absolute speed −𝑁𝑆 .
This means there are four rotating magnetic fields:
-
Rotating field 1: stator field with speed 𝑁𝑆
Rotating field 2: stator field with speed −𝑁𝑆
Rotating field 3: rotor field with speed 𝑁𝑆
Rotating field 4: rotor field with speed −𝑁𝑆 .
The interaction of the rotating fields 1 and 3 gives a constant torque. This is the desired useful torque
(coming from the direct torque speed characteristic having a positive direction). The interaction of
the rotating fields 2 and 4 also gives a constant torque. This is the torque coming from the inverse
torque speed characteristic having a negative direction.
The interaction between the rotating fields 1 and 4 gives a pulsating torque (which is on average
zero). The interaction between the rotating fields 2 and 3 gives a pulsating torque (which is on
average zero). Due to these pulsating torques, the torque of the induction motor is no longer
constant.
6.2: The neutral conductor
Consider a three phase load (wye connected with neutral conductor) as visualized in Figure 8. The
impedances 𝑍1 , 𝑍2 and 𝑍3 are different. In general, this implies the currents 𝑖𝑓1 (𝑡), 𝑖𝑓2 (𝑡) and 𝑖𝑓3 (𝑡)
have different magnitudes and have phase differences equal to 120°.
Figure 8: Non symmetric three phase load
Using Kirchoff’s circuit law,
𝑖𝑁 (𝑡) = 𝑖𝑓1 (𝑡) + 𝑖𝑓2 (𝑡) + 𝑖𝑓3 (𝑡).
The currents 𝑖𝑓1 (𝑡), 𝑖𝑓2 (𝑡) and 𝑖𝑓3 (𝑡) are the sum of a direct component, an inverse component and
a homopolar component. The direct component contains three currents which have the same
magnitude and which have phase differences equal to 120°. The sum of these three currents is zero
implying there is no current in the neutral conductor due to the direct current component. The
inverse component contains three currents which have the same magnitude and which have phase
differences equal to 120°. The sum of these three currents is zero implying there is no current in the
neutral conductor due to the inverse current component.
The homopolar component contains three currents which have the same magnitude and which have
the same phase implying their sum is not zero. Due to the homopolar phase current component, a
current is flowing in the neutral conductor. This current in the neutral conductor is three times as
large as the homopolar component in the phase current.
Indeed, since
̅
𝐼𝑓ℎ
1 1 1
̅ ] = [1 𝑎
[𝐼𝑓𝑑
3
1 𝑎2
̅𝑖
𝐼𝑓
̅
1 𝐼𝑓1
2 ] [𝐼 ̅ ] ,
𝑎
𝑓2
𝑎 𝐼𝑓3
̅
̅ , 𝐼𝑓𝑑
̅ and 𝐼𝑓𝑖̅ . Since 𝐼𝑁̅ = 𝐼𝑓1
̅ + 𝐼𝑓2
̅ + 𝐼𝑓3
̅ and
it is possible to calculate 𝐼𝑓ℎ
̅
𝐼𝑓1
1
̅ ] = [1
[𝐼𝑓2
1
̅
𝐼𝑓3
1
𝑎2
𝑎
̅
1 𝐼𝑓ℎ
̅ ]
𝑎 ] [𝐼𝑓𝑑
2
𝑎
𝐼𝑓𝑖̅
one indeed obtains that
̅ .
𝐼𝑁̅ = 3 𝐼𝑓ℎ
In case the homopolar current component is large, this has to be taken into consideration when
choosing the neutral conductor. The cross section of the neutral conductor has to be sufficiently
large in order to reduce the heat dissipation.
Traditionally, the neutral conductor has a smaller cross section than the phase conductors. This
remains a good choice when there is no or only a limited non symmetry in the grid. The current in
the neutral conductor is zero or small. However, when there is an important non symmetry in the
grid, especially when there is an important homopolar component, it can be necessary to take the
cross section of the neutral conductor the same as the cross section of the phase conductors. In
extreme cases, it can be necessary to give the neutral conductor a larger cross section than the phase
conductors.
7: Reducing non symmetries in a grid
7.1: Compensating a non symmetrical load
It is important to reduce or (if possible) avoid non symmetries in a three phase grid. Using symmetric
three phase loads instead of single phase loads is useful. When using single phase loads, it is
important to spread them over the three different phases.
In case there is only one non symmetrical load as visualized in Figure 9 which can not be changed, it
is possible to compensate the non symmetry by using capacitors and inductors (resistors are not
used since resistors consume active power).
Figure 9: Non symmetrical load
As visualized in Figure 9, the current in phase 1 equals zero. In phase 2 and phase 3, the currents
have the same magnitude but there is a phase difference of 180°. These are important non
symmetries. Although there is no homopolar current component (there is no neutral conductor), the
direct current component and the inverse current component have the same magnitude.
Figure 10: Compensation of a non symmetrical load
Suppose in Figure 10 the admittance
𝑌̅23 =
1
𝑍̅23
is the useful non symmetric load (e.g. a resistor 𝑅 as visualized in Figure 9 giving 𝑌̅23 = 1⁄𝑅) and the
admittances 𝑌̅12 and 𝑌̅31 are used to obtain a symmetrical load behavior. In order to avoid heat
losses, the admittances 𝑌̅12 and 𝑌̅31 are capacitors or inductors (containing no resistors). Suppose the
̅𝐿1𝐿2, 𝑈
̅𝐿2𝐿3 and 𝑈
̅𝐿3𝐿1 supplied by the grid are symmetrical i.e. they only contain a direct
voltages 𝑈
component. By an appropriate choice of the admittances 𝑌̅12 and 𝑌̅31 , the currents supplied by the
grid must will only contain a direct component. The grid currents can be written as
̅
𝐼𝑓1
𝑌̅12
̅
[𝐼𝑓2 ] = [−𝑌̅12
̅
0
𝐼𝑓3
0
𝑌̅23
−𝑌̅23
−𝑌̅31
0 ]
̅
𝑌31
̅𝐿1𝐿2
𝑈
𝑌̅12 + 𝑌̅31
̅𝐿2𝐿3 ] = [ −𝑌̅12
[𝑈
̅𝐿3𝐿1
𝑈
−𝑌̅31
−𝑌̅12
𝑌̅23 + 𝑌̅12
−𝑌̅23
−𝑌̅31
−𝑌̅23 ]
̅
𝑌31 + 𝑌̅23
̅𝑓1
𝑈
̅𝑓2 ]
[𝑈
̅𝑓3
𝑈
̅𝐿1𝐿2 = 𝑈
̅𝑓1 − 𝑈
̅𝑓2 , 𝑈
̅𝐿2𝐿3 = 𝑈
̅𝑓2 − 𝑈
̅𝑓3 and 𝑈
̅𝐿3𝐿1 = 𝑈
̅𝑓3 − 𝑈
̅𝑓1 . Since
with 𝑈
̅𝑓1
𝑈
1 1
̅𝑓2 ] = [1 𝑎2
[𝑈
1 𝑎
̅𝑓3
𝑈
̅
𝐼𝑓1
̅ℎ
1 𝑈
1 1
̅ ] = [1 𝑎 2
̅𝑑 ], [𝐼𝑓2
𝑎 ] [𝑈
̅𝑖
𝑎2 𝑈
1 𝑎
̅
𝐼𝑓3
1 𝐼ℎ̅
𝑎 ] [𝐼𝑑̅ ]
𝑎2 𝐼𝑖̅
One obtains
𝐼ℎ̅
1 1
[𝐼𝑑̅ ] = [1
3
1
𝐼𝑖̅
giving
1
𝑎
𝑎2
𝑌̅12 + 𝑌̅31
1
2 ] [ −𝑌
̅12
𝑎
𝑎
−𝑌̅31
−𝑌̅12
̅
𝑌23 + 𝑌̅12
−𝑌̅23
−𝑌̅31
1
−𝑌̅23 ] [1
𝑌̅31 + 𝑌̅23 1
1
𝑎2
𝑎
1
𝑎]
𝑎2
̅ℎ
𝑈
̅
[𝑈𝑑 ]
̅𝑖
𝑈
𝐼ℎ̅
0
̅
[𝐼𝑑 ] = [0
0
𝐼𝑖̅
0
0
𝑌̅12 + 𝑌̅23 + 𝑌̅31
−(𝑎𝑌̅12 + 𝑌̅23 + 𝑎2 𝑌̅31 )]
−(𝑎2 𝑌̅12 + 𝑌̅23 + 𝑎 𝑌̅31 )
𝑌̅12 + 𝑌̅23 + 𝑌̅31
̅ℎ
𝑈
̅𝑑 ]
[𝑈
̅𝑖
𝑈
̅𝑑 , the
In order to avoid an inverse current component 𝐼𝑖̅ due to the direct voltage component 𝑈
condition
𝑎2 𝑌̅12 + 𝑌̅23 + 𝑎 𝑌̅31 = 𝑎2 𝑌̅12 +
1
+ 𝑎 𝑌̅31 = 0
𝑅
Since the real parts of 𝑌̅12 and 𝑌̅31 are zero (no resistive part in order to avoid heat losses), there are
two unknowns 𝐼𝑚(𝑌̅12 ) and 𝐼𝑚(𝑌̅31 ) which satisfy
2𝜋
2𝜋
1
2𝜋
2𝜋
(𝑐𝑜𝑠 ( ) − 𝑗𝑠𝑖𝑛 ( )) 𝑗 𝐼𝑚(𝑌̅12 ) + + (𝑐𝑜𝑠 ( ) + 𝑗𝑠𝑖𝑛 ( )) 𝑗 𝐼𝑚(𝑌̅31 ) = 0
3
3
𝑅
3
3
or equivalently
2𝜋
1
2𝜋
𝑠𝑖𝑛 ( ) 𝐼𝑚(𝑌̅12 ) + − 𝑠𝑖𝑛 ( ) 𝐼𝑚(𝑌̅31 ) = 0
3
𝑅
3
{
2𝜋
2𝜋
𝑐𝑜𝑠 ( ) 𝐼𝑚(𝑌̅12 ) + 𝑐𝑜𝑠 ( ) 𝐼𝑚(𝑌̅31 ) = 0
3
3
This implies
𝐼𝑚(𝑌̅12 ) = −
1
1
1
= −
𝑅 2 𝑠𝑖𝑛 (2𝜋)
𝑅 √3
3
𝐼𝑚(𝑌̅31 ) = +
1
1
1
= +
𝑅 2 𝑠𝑖𝑛 (2𝜋)
𝑅 √3
3
Therefore, 𝑌̅12 is an inductor having a value
𝐿=
𝑅 √3
𝜔
and 𝑌̅31 is a capacitor having a value
𝐶=
1
𝜔 𝑅 √3
.
Using these inductor 𝐿 and capacitor 𝐶 values as visualized in Figure 11, the grid current will only
contain a direct component in case the applied grid voltage only contains a direct voltage
component.
Figure 11: Compensation of a non symmetric load
In case the applied grid voltage has another phase sequence (it only contains an inverse voltage
component), a direct current component 𝐼𝑑̅ will be avoided when
𝑎 𝑌̅12 + 𝑌̅23 + 𝑎2 𝑌̅31 = 0
̅𝑖 , only an inverse current component 𝐼𝑖̅ will flow. Verify
Due to this inverse voltage component 𝑈
what values for 𝑌̅12 and 𝑌̅31 are needed.
7.2: Avoiding and reducing non symmetries due to photovoltaic inverters
Avoiding non symmetries is also important to reduce the heat losses in the grid. Consider a three
phase grid without a neutral conductor and suppose a photovoltaic inverter injects power into the
grid. There exist single phase inverters and there exist three phase inverters. Suppose the grid has a
line voltage 𝑈 and the inverter injects an apparent power 𝑆. When a single phase inverter is used, a
current 𝐼1 will flow in two phase conductors. When a three phase inverter is used, a current 𝐼3 will
flow in all three phase conductors. This means
𝑆 = 𝑈 𝐼1 = √3 𝑈 𝐼3
implying 𝐼1 = √3 𝐼3. When 𝑅 is the resistance of a phase conductor, then the grid losses in case of a
single phase inverter equal
𝑃𝑙𝑜𝑠𝑠,1𝑝ℎ = 2𝑅 𝐼12
and the losses in case of a three phase inverter equal
𝑃𝑙𝑜𝑠𝑠,3𝑝ℎ = 3𝑅 𝐼32 =
𝑃𝑙𝑜𝑠𝑠,1𝑝ℎ
.
2
When a single phase photovoltaic inverter is used, the heat losses in the grid are doubled in
comparison with a three phase inverter. A similar situation occurs when another renewable and/or
distributed energy source generates power or when the three phase grid is loaded with a single
phase load (in parallel with a line voltage).
When using three identical photovoltaic panels with identical single phase inverters, it is also possible
to obtain a symmetrical situation (in case the panels produce the same electrical power). Figure 12
visualizes a three phase grid (having a line voltage of 230 𝑉) where three single phase inverters inject
their power. The currents 𝐼𝑎 , 𝐼𝑏 and 𝐼𝑐 have a magnitude which equals √3 times the magnitude of
the inverter currents 𝐼𝑎−𝑏 , 𝐼𝑏−𝑐 and 𝐼𝑐−𝑎 .
Figuur 12: Single phase inverters injecting power in a three phase grid
For instance the Sunny Tower of SMA Solar Technology allows to mount a number of single phase
inverters (Sunny Boy inverters) which allows them to inject their power symmetrically into a three
phase grid. In case the grid has a line voltage of 230 𝑉, the single phase inverters are placed in
parallel with the line voltage. A Sunny Tower with six Sunny Mini Central 8000TL inverters (each
having a nominal AC power of 8 𝑘𝑊) is able to inject a total power of 48 𝑘𝑊 into the grid.
Figuur 13: The use of the Sunny Tower
When a three phase grid with a line voltage of 400 𝑉 is used, then a neutral conductor is needed to
inject power using a 230 𝑉 single phase inverter. The inverter is mounted between a phase
conductor and the neutral conductor. A symmetrical injection of the power is obtained by injecting
the same power using a first inverter between 𝐿1 and 𝑁, a second inverter between 𝐿2 and 𝑁 and
finally a third inverter between 𝐿3 and 𝑁.
Figure 14: Single phase inverter between phase conductor and neutral conductor
When a three phase grid has a symmetrical load, no current is flowing in the neutral conductor which
implies this neutral conductor often has a smaller cross section than the phase conductors. In order
to limit the heat dissipation in the grid conductors, a symmetrical power injection is really useful
when the neutral conductor has a smaller cross section. Indeed, when the three single phase
inverters inject their power using 𝐿1 − 𝑁, 𝐿2 − 𝑁 and 𝐿3 − 𝑁 as already mentioned and they all
inject the same power, than they inject currents with the same magnitude which have phase
differences of 120°. This implies no current is flowing in the neutral conductor and this avoids heat
losses in this neutral conductor.
Notice Figure 14 not only visualizes the single phase inverter. Between the photovoltaic panels and
the inverter, an EMC/EMI filter has been mounted. This filter reduces the high frequent currents in
the photovoltaic cells due to the inverter. This avoids an accelerated aging of the photovoltaic cells.
There is also an EMC/EMI filter mounted between the inverter and the grid. This filter limits the
emission of high frequent disturbances into the grid in order to comply with national and/or
international EMC normalizations (e.g. see http://www.schaffner.com).
The principle that it is useful to spread single phase inverters of photovoltaic installations over the
phases is a very general principle. This principle is valid for all kind of distributed resources like the
Hydro Boy (an inverter fed by a fuel cell) and the Windy Boy (which injects the power of a micro wind
turbine) of SMA Solar Technologies.
References
J.D. Glover, M.S. Sarma, Power System: Analysis and Design, Brooks/Cole, Pacific Groove, California
USA, 2002.
J. Peuteman, Lokale productie: enkelfasig of driefasig?, Vlaams Elektro Innovatiecentrum VEI,
Technologiewacht, 2010.
J. Willems, Elektrische netten, Wetenschappelijke uitgeverij en boekhandel, Gent, 1978.