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Notes 16: Voltage Regulation2
Single phase step voltage regulators
16.0 Configuration and basic operation
As a result of our conclusion in “Notes15”
that the pu impedance of an autotransformer
is quite small, we will neglect it in the
analysis contained in these notes.
Step voltage regulators have both single
phase and 3 phase applications. We focus in
these notes on the single phase applications.
There are two broad classes of single phase
step voltage regulators, referred to as Type
A and Type B. They differ in whether the
 (Type A) source voltage is to be applied
across one winding (and load voltage
taken across both windings) or
 (Type B) source voltage is to be applied
across both windings (and load voltage
taken across one winding).
1
Figure 1 shows a type A step-up voltage
regulator. The diagram on the right is a
simplification of the diagram on the left.
+
L
R
2
IL
L
L
I2
+
E2
Preventive
autoxfmr
+
N
IL
IS
S
+
VL
V
S
E
1
1
-
I
-
VS
VL
-
+
IS
N1
+
S
SL
-
-
SL
Fig. 1: Type A boost configuration
Figure 2 shows a type A step-down voltage
regulator. Note the position of the switch
inside the dotted circle in Figs. 1 and 2.
+
L
E
2
Preventive
autoxfmr
2
I2
IL
L
+
N
L
R
+
IL
IS
+
V
S
E
1
-
-
VS
VL
1
V
L
-
+
I
S
-
IS
N
1
+
-
S
SL
SL
Fig. 2: Type A buck configuration
2
Figure 3 shows a type B step-up voltage
regulator. The diagram on the right is a
simplification of the diagram on the left.
Preventive
autoxfmr
IS
N2
I2
+
S
R
E2
+
I
S
S
+
L
IS
IL
N1
+
VS
IL
L
+
E
1
L
V
S
+
VL
VL
I
1
-
-
SL
-
SL
Fig. 3: Type B boost configuration
Figure 4 shows a type B step-down voltage
regulator. Note the position of the switch
inside the dotted circle in Figs. 1 and 2.
Preventive
autoxfmr
IS
N2
+
S
R
S
E2
+
I
L
I2
S
+
IS
IL
N1
VS
+
+
I
IL
L
L
V
S
+
1
E
1
VL
VL
-
-
SL
-
SL
Fig. 4: Type B buck configuration
3
Some things to note about Fig. 1-4…
The notation S is “source,” L is “load,” and
SL is “source-load” which indicates to what
the respective terminal is connected.
The switch inside the dotted circle is called a
reversing switch. This switch toggles the
regulator between the boost and buck
configurations.
The upper winding in each figure has the
taps. The taps enable changing the number
of turns for the indicated winding and in
doing so provide adjustment to the turns
ratio and therefore the load voltage.
The preventive autotransformer, shown in
each figure between a terminal and the taps,
has the function of preventing the regulator
from being disconnected from the circuit
each time the tap is changed.
4
A more detailed diagram is shown for the
preventive transformer and associated
components in Fig. 5.
Fig. 5: [1]
In Fig. 5, one observes the preventive
autotransformer, the transfer switches, and
the selector switches.
Without the preventive autotransformer and
the transfer switches, when making a tap
change, if opening selector switch j occurs
before the closing of selector switch j+1, the
main line through the regulator will open
circuit.
5
We can see how this is avoided by
discussing the case of moving from tap
position 1 to tap position 2.
When the tap is in position 1, transfer switch
A is closed, B is open, and C is closed. All
selector switches are open except 1. Fig. 6.
When switching to position 2, the transfer
switches A and B close, C opens, and
selector switches 1 and 2 are closed. Fig. 7.
When the tap is in position 2, transfer switch
B is closed, A is open, and C is closed. All
selector switches are open except 2. Fig. 8.
Fig 6
Fig 7
6
Fig 8
Since Figs. 6 and 8 represent steady-state
configuration and Fig. 7 represents the
switching configuration, we see that at no
time is it possible to encounter an open
circuit.
The preventive autotransformer is required
in order to smoothen the transient
encountered when switching from the Fig. 6
configuration to the Fig. 7 configuration,
since there will be, initially, a voltage
difference between the tap 1 and tap 2
position.
The preventive transformer provides a low
impedance path to the load current since the
two sides are effectively in parallel. But it
provides a high impedance path to
circulating currents encountered in the Fig. 7
configuration, since the two sides are in
series. This is good because we desire
minimum impedance to load currents and
maximum impedance to circulating currents.
7
Table 1 provides the sequence of operation
of full operation from taps 1 to 9.
Table 1 [1]
Additional
explanation
of
operation can be found in [2].
regulator
16.1 Regulator rating
Step voltage regulators differ from standard
autotransformers in that they are rated on the
2-winding transformer basis rather than the
autotransformer basis.
8
This means step voltage regulators are rated
on the kVA transformed rather than the kVA
passed through.
Recall Fig. 3, repeated here for convenience:
Preventive
autoxfmr
IS
N2
I2
+
S
R
E2
+
I
S
S
+
L
IS
IL
N1
+
VS
+
E
1
IL
L
L
V
S
+
VL
VL
I
1
-
-
SL
-
SL
Fig. 3: Type B boost configuration
Example 1:
Assume we need 1000 kVA of flow-through
capacity. What should be the capacity of the
regulator if the turns ratio is nt=N2/N1=1/11?
The input voltage is
1
1
VS  E1  E2  E2  E2  E2 (  1)
nt
nt
9
The autotransformer will need to carry
1
1
Sauto  VS I S  E 2 I S (  1)  S xfmr (  1)
nt
nt
With Sauto=1000 kVA, then we need:
Sauto
1000
S xfmr 

 100kVA
1
11  1
(  1)
nt
Thus we see that the iron and windings
required for the regulator need be able to
handle only 10% of the flow-through of the
circuit it is regulating.
But what will be the voltage transformation
in this case?
VL
E1


VS E1  E 2
1
1
1
11



 1.1
N2
1 10 / 11 10
1
1
N1
11
So we see that we obtain a 10% voltage
boost in this case.
10
16.2 Voltage and current equations
The defining voltage and current equations
for the Type B step-voltage regulator in the
raise (boost) position are developed in Table
1, resulting in eqs. (6) and (7).
Table 1: Type B – Boost Position
Voltage equations Current equations
(1)
E1 E 2
N I N I

1 1
2 2
N1 N 2
VS  E1  E2
I L  I S  I1
(2)
VL  E1
I2  IS
(3)
N2
N2
E2 
E1 
VL
N1
N1
I1 
(4)
N2
N
I2  2 IS
N1
N1


N2 
N 2  (5)
VL I L   1 
 I S
VS   1 
N1 
N1 


(6)
VS  aRVL
I L  aR I S
aR  1 
N2
N1
11
(7)
The defining voltage and current equations
for the Type B step-voltage regulator in the
lower (buck) position are are developed in
Table 2, resulting in eqs. (13) and (14).
Table 2: Type B – Buck Position
Voltage equations Current equations
E1 E 2

N1 N 2
VS  E1  E2
N1 I 1  N 2 I 2
(8)
I L  I S  I1
(9)
VL  E1
I2  IS
(10)
N2
N2
(11)
N
N
E2 
E1 
VL I 1  2 I 2  2 I S
N1
N1
N1
N1
(12)


N2 
N2 
VL I L   1 
 I S
VS   1 
N1 
N1 


(13)
VS  aRVL
I L  aR I S
N2
aR  1 
N1
12
(14)
In comparing the final equations for the
boost configuration of Table 1, eqs. (6) and
(7), with the final equations for the buck
configuration of Table 2, eqs. (13) and (14),
we see that the only difference between
them is the sign of the turns ratio per eqs. (7)
and (14).
16.3 Regulator ratio
The parameter aR in eqs. (7) and (14) is
called the regulator ratio. Typical maximum
boosts and bucks are ±10%. What will aR
and nt=N2/N1 be in the maximum boost and
buck positions?
Maximum boost position:
Consider eq. (6):
VS  aRVL
1 VL

 1.1 aR  1  .9091
 aR VS

1 .1
13
Then from eq. (7),
N2
aR  1 
 .9091
N1
N2
 1  .9091  .0909
 N1
This is a ratio of 1/11, which was the ratio
used in example 1.
Maximum buck position:
Consider eq. (13) (same as eq. (6)):
VS  aRVL
1 VL

 0.9 aR  1  1.111
 aR VS

1 .1
Then from eq. (14),
N
aR  1  2  1.111
N1
N2
 1.111  1  .111
 N1
This is a ratio of 1/9.
14
16.4 Voltage regulation per step
Most regulators today will have 32 steps for
±10% regulation, with 16 steps in either
direction.
Since 16 steps provides 10% boost (or
buck), then a single step will result in a
voltage change of 1/16 of 10%, which is:
1
25
5
 10% 
% %
16
28
8
or 5/8% per step.
Since “% voltage” is “pu voltage” x 100, the
pu voltage change per step is
5
1

 0.00625
8 100
So we obtain 0.00625 pu voltage change per
step.
15
16.5 Regulator ratio from tap
We have previously used the term “tap
position” to denote the position on the
tapped winding of the tapping terminal.
It is common to use the term “Tap” to refer
to the numerical identity of the position on
the tapped winding if the tapping terminal.
Therefore, the range of Tap is from 1 to 16.
Note that information characterizing a
regulator will not include the effective ratio
(N2/N1)effective corresponding to each tap
position. Normally, you will have only the
numerical identity of the tap position, Tap.
The ratio (N2/N1)effective is then given by the
pu voltage change per step times the number
of steps (Tap):
 N2 


 0.00625(Tap)
 N 1  effective
16
(15)
Note that if Tap is 0, then the effective turns
ratio is 0. In this case, there is no boost, and
VL=VS.
For the boost configuration, we recall eq.
(7), repeated here for convenience:
N2
aR  1 
(7)
N1
Substituting eq. (15) yields:
aR  1  0.00625(Tap)
(16)
Likewise, for the buck configuration, we get
aR  1  0.00625(Tap)
(17)
16.6 abcd model
Repeating eqs. (6) and (13) here,
VS  aRVL
I L  aR I S
17
So the abcd model would be as follows:
VS  aVL  bI L
I S  cVL  dI L
(19)
(20)
where
a  aR
b0
c0
1
d
aR
(21)
(22)
(23)
(24)
16.6 Control
The tap changing is controlled by a
 line drop compensator actuating a
 voltage relay with time delay which drives
 motor operating circuit controling a motor
to push up and down the taps as needed
A conceptual (block) diagram of the control
process is shown in Fig. 9 where the LDC
receives inputs proportional to (a) voltage
18
across and (b) current through the feeder
being regulated.
Preventive
Autotransformer
R
S
+
L
Series
Winding
L
VSource
Motor operating
circuit
Reversing
Switch
Control
CT
Shunt
Winding
Control
PT
-
+
VLoad
LDC
SL
Timedelay
relay
Voltage
relay
Fig. 9
We will describe each of the main blocks
shown in Fig. 9.
16.4 Line drop compensation operation
Without line drop compensation (LDC), a
voltage regulator will control the voltage at
its own location. However, it is typical that
we desire to control the voltage at a location
19
downstream from the voltage regulator, i.e.,
at the far-end of the feeder.
Fig. 10 shows a detailed diagram of an LDC.
kVA rating
kVhi - kVlow
Vsend
Vload
I line CTp:CTs
Rline + jXline
I comp
R
X
Load
LDC side
1:1
+ V drop kVlow:120v
+
V reg
Relay side
+
Vrelay
-
-
Voltage
Relay
Fig. 10: Line drop compensator circuit
Some items to note in regards to Fig. 10:
 Choose the potential transformer (PT) to
give a 120 v nominal voltage on the relay
side. So when substation low side voltage
is kVlow (rated voltage), Vreg=120. Thus,
if we choose the line-to-neutral voltage
kVlow as base voltage on feeder side, then
120 is base voltage on relay side. In perunit, Vsend=Vreg.
20
 Because the isolation transformer between
relay side and LDC side is 1:1, the voltage
base on the LDC side is also 120.
 The voltage relay draws no current, it only
monitors voltage. Therefore, the current
through the isolation transformer is zero.
 Choose the current transformer (CT) to
have current rating of CTp:CTs, where CTp
is usually chosen as rated feeder current.
 The drop across the 1:1 isolation
transformer, Vdrop, equals the drop across
the LDC impedance R+jX.
 If R and X are chosen so that they are, in
per-unit, equal to Rline+jXline in per-unit,
then (Vsend-Vload)=Vdrop.
Solve the last relation for Vload, we have, in
per-unit,
Vload=Vsend-Vdrop
(25)
Note from the relay circuit that
Vrelay=Vreg-Vdrop
(26)
21
However, in per-unit, Vsend=Vreg.
From (25) and (26), we conclude that, in
per-unit, Vrelay=Vload !!!
Even better, since the relay circuit voltage
base is 120, Vrelay gives the actual service
voltage (at the meter) seen by the load.
Because we want to maintain load voltage
between 114 and 126, we could set the
voltage relay to actuate when it sees voltage
drop below 120 v.
Now how to design the circuit to make this
happen? By “design the circuit,” we mean
we need to choose appropriate values of
LDC resistance R and reactance X.
16.5 LDC Design
Key to LDC design is choosing a consistent
set of bases for the per-unitization. You
22
must do this correctly. Table 3 summarizes
this consistent set of bases.
Table 3: Base Quantities
Base
Feeder side
Relay &
quantity
LDC side
Voltage
VLN,rated=kVlow
120
Current
CTp
CTs
Impedance Zbase,L=VLN,rated/CTp 120/CTs
In what follows, we will assume that we
know the feeder impedance, and it is
denoted in ohms as RlineΩ+jXlineΩ.
The per unit value of the (feeder and LDC)
impedance is given by
Rpu  jX pu 
Rline  jX line
Z base, L
(27)
With Zbase,L as given in Table 3, eq. (27) is:
Rpu  jX pu 
Rline  jX line CTp
VLN , rated
23
(28)
Now we want the ohmic value of the
compensator impedance. This is obtained by
multiplying the per-unit value by the LDCside impedance base given in Table 3, as in
eq. (29).
R  jX  Rpu  jX pu Zbase, LDC

Rline  jX line CTp 120
VLN , rated
CTs
(29)
One interesting issue is that LDC impedance
settings are typically dialed in as a voltage,
and the dialed in voltage is the voltage, in
volts, seen across each impedance element
when rated current is flowing.
To obtain the voltage settings corresponding
to the ohmic values of R and X given in
equation (29), we need to multiply by CTs.
We denote the corresponding LDC settings
using primed notation, i.e.,
24
R  jX    R  jX CTs


 R pu  jX pu Z base, LDC CTs


Rlne  jX line CTp
VLN ,rated
120
CTs
CTs
120 Rln e  jX line CTp
(30)
VLN ,rated
Example 2:
A substation transformer is rated 5000 kVA,
115kV:4.16kV, and the equivalent line
impedance from the regulator to the load
center is 0.3+j0.9 ohms.
1.Determine the voltage transformer and
current transformer ratings for the
compensator circuit.
2.Determine the R and X settings of the
compensator in ohms and volts.
Assume that the phases are balanced, and
you are designing the regulator for a single
one of the phases.
25
Solution:
1.PT and CT settings:
PT setting:
The rated line-to-ground voltage of the
substation
transformer
is
4160/sqrt(3)=2401.8 volts.
To obtain 120 volts in the relay circuit, we
need a PT with a turns ratio of
2401.8:120=20.015.
CT setting:
The rated current in the feeder is obtained
based
on
|S3|=sqrt(3)|V||I|
|I|=|S3|/(sqrt(3)|V|), i.e.,
I rated 
5000 E 3
3 (4160)
26
 693.9
The primary rating of the CT on the feeder
side is selected to be 700 A. To work with
reasonably small current levels in the LDC
circuit, we will choose a CT ratio to give us
a rated LDC-side current of CTs=5 A.
Therefore the CT ratio is
CTp/CTs=700/5=140
2. R and X settings:
Applying eq. (29), we obtain the ohmic
value of the compensator impedance as:
R  jX 
Rline  jX line CTp 120
VLN , rated
CTs

0.3  j 0.9 700 120

 2.1  j 6.3
2401.8
5
And we obtain the voltage settings by
multiplying by CTs:
R  jX   2.1  j6.35  10.5  j 31.5v
27
16.6 Obtaining equivalent line impedance
In Example 2, the ohmic value of the line
impedance was given. If the feeder radially
feeds a single load at the end of the feeder,
this impedance is the actual impedance of
the single phase line between the point of
regulation and the load.
However, if there are multiple taps along the
line, then the current at the sending end does
not flow all the way to the load.
In this case, we use an equivalent impedance
computed as if the sending end current did
flow all the way to the load. Define the
source voltage Vsend, the feeder-end voltage
Vload, and the line current at the source Iline.
Then the so-called equivalent line
impedance is given by
Rline  jX line 
Vsend  Vload
I line
28
(31)
This impedance is NOT the same as the
actual line impedance. The only way to
obtain it is to run a power flow without the
regulator and extract the three quantities of
eq. (31) from the power flow solution.
16.7 Regulator settings
In addition to the load drop compensator
impedance values, there are three other
important regulator settings (Fig. 9 may be
useful to view while reading the below):
 Set voltage: The desired output of the
regulator. It is the voltage level that the
regulator tries to hold, and is a setting of
the voltage relay. (The voltage relay is the
receiver of the intelligence from the circuit
in which the regulator is located, and it
initiates operation of the tap-changer).
29
 Bandwidth: In order to prevent continuous
tap-changing, (and therefore increase the
regulator life), the voltage relay does not
actuate until the difference between the set
and measured voltage exceeds half the
bandwidth. Settings of 1.5, 2, and 2.5 volts
are common for ±10%, 32 step regulators.
If set voltage is 120 volts and bandwidth is
2 volts, the regulator will change taps until
load voltage lies between 119-121 volts.
 Time delay: In order to prevent tap
changing during a transient or short-time
change in current, and therefore reduce the
frequency of tap changing, a time delay
relay is inserted between the voltage relay
and the motor operating circuit. This is the
waiting time between the time when the
voltage goes out of bandwidth and when
the motor operating circuit actually begins
to move the tap, i.e., the length of time that
a raise or lower operation is called for
before actual execution of the command.
Typical time delays are 30 to 60 seconds.
30
Fig. 11 illustrates the above 3 settings [3].
Fig. 11
Example 3:
The substation transformer of Example 2 is
supplying 2500 kVA at 4.16 kV (L-L) and
0.9 pf lag. The line-drop compensator
impedance was set as computed in Ex 2, i.e.,
R  jX  2.1  j 6.3 ohms. The set voltage is
120 volts and the bandwidth is 2 volts.
Determine the regulator tap position that
will hold the load center voltage at the
desired voltage level and within the
bandwidth.
31
So the tap on the regulator needs to be set so
that the voltage at the load center lies
between 119 and 121 volts.
Solution:
Again we assume the phases are balanced
and we are designing a regulator for one of
the phases.
Our approach is to compute the voltage seen
by the voltage relay assuming no voltage
regulator action, and then to compute the
minimum number of tap changes necessary
to move the voltage to either 119 (if it is
below) or 121 (if it is above).
The first step is to compute the actual line
current. The magnitude is given by
I line
S3
2500 E 3


 346.97
3VLL
3 (4160)
Since pf=0.9 lag, the current angle
(assuming the L-N voltage is reference) is:
32
   cos 1 (0.9)  25.84
So the current phasor is:
I line  346.97  25.84
Now this can be used to obtain the
compensator current via use of the CT ratio,
which is CTp/CTs=700/5=140. Therefore
5
I comp 
346.97  25.84  2.4783  25.84
700
The L-N voltage at the transformer low side
is given by 4160/sqrt(3)=2401.8. Since we
are assuming this is the reference, then it has
angle of 0 degrees. Then the regulator
voltage is transformer by the PT ratio, which
is 2401.8:120, therefore the compensator
voltage is 120 volts.
The relay voltage in the compensator circuit
equals 120 volts less the drop across the
compensator impedance. The drop across
the compensator impedance is given by the
compensator current times the compensator
impedance.
33
The drop across the compensator impedance
is:
Vdrop  I comp R  jX 
 2.4783  25.842.1  j 6.3  16.45845.72
The relay voltage is:
Vrelay  Vreg  Vdrop
 120  16.45845.72  109.24  6.19
Now we need at least 119 volts.
Recalling from Section 16.4 that we obtain
0.00625 pu voltage per step, this will be
0.00625*120=0.75 volts/step.
Therefore we will need to move an amount
of steps given by
119  109.24
 13.02
Tap= 0.75
Now we can only take an integer number of
steps, so let’s assume we take 13 steps.
From eq. (16), we can get the regulator ratio:
aR  1  0.00625(Tap)  1  .00625(13)  0.9188
34
Example 4:
Assume that in Example 3, the 2500 kVA at
4.16 kV (L-L) and 0.9 pf lag is measured at
the low side of the substation transformer
terminals.
Compute the actual voltage at the load.
Solution:
Recall that we know, from Example 2, the
equivalent impedance between the regulator
and the load is 0.3+j0.9 ohms. Therefore if
we obtain the voltage and current on the
load side of the regulator, we can compute
the load voltage.
To obtain the voltage and current on the load
side of the regulator, we make use of the
generalized constants for the regulator (see
eqs. 21-24):
35
a  aR  0.9188
b0
c0
1
1
d

 1.0884
a R 0.9188
Applying eqs. (19-20):
VS  aVL  bI L  0.9188(VL )
I S  cVL  dI L  1.0884( I L )
But what is VS and IS? These quantities are
on the source side of the regulator, which
means the low side of the substation
transformer. Therefore:
VS  2401.8 volts and
I S  I line  346.97  25.84 (see ex. 3)
36
Then
1
1
VL 
VS 
2401.8  2614.2
0.9188
0.9188
1
IS
1.0884
1

( 346.97  25.84 )  318.77  25.84
1.0884
IL 
From this we may obtain the actual line-toground voltage at the load as:
VLoad  VL  I L (0.3  j 0.9)
 2614.2  318.77  25.84 (0.3  j 0.9)
 2412.8  5.15
On a 120 volt base, the load voltage is
 120 
2412.8  5.15
  120.55volts
 2401.8 
16.8 Three-phase voltage regulators
Three-phase regulators may be obtained by
interconnecting 3 single-phase regulators.
37
There are four different ways in use for
connecting single phase regulators for threephase circuit regulation. They are:
 Three regulators connected in grounded-Y.
 Three regulators connected in closed delta.
 Two regulators connected in “open-wye”
(sometimes also called “V” phase)
 Two regulators connected in open delta
In all of these cases, each regulator has its
own compensator circuit controlling the
voltage corresponding to its corresponding
voltage only. Therefore, the taps on each
regulator are changed separately.
We will only discuss the most common of
these, the grounded-Y configuration, which
is shown in Fig. 12.
38
B
IB
Ia
a
Ib
b
IA
A
+
+
V
An
Van
-
-
Ic
c
C
IC
Fig. 12
The following equations may be developed
for the three-phase grounded-Y regulator:
V An  a Ra
V    0
 Bn  
VCn   0
0
a Rb
0
0  Van 
0  Vbn 
a Rc  Vcn 
39
(32)
And
 1

 I A   a Ra
I    0
 B 
 I C  
 0

0
1
a Rb
0

0 
  Ia 
0   I b 

 
1   Ic 

a Rc 
Equations (32-33) are in the form of
(33)
VLN ABC   [a][VLNabc ]  [b][ Iabc ]
I ABC   [c][VLNabc ]  [d ][ I abc ]
where
a Ra
[a ]   0
 0
0
a Rb
0
0 
0 
a Rc 
40
 0 0 0
[b]  0 0 0
0 0 0
 0 0 0
[c ]  0 0 0
0 0 0
 1

0
0 

 a Ra

1

[d ]  0
0 


a Rb

1 
 0

0
a Rc 

The above model provides that the regulator
ratio for each phase may be different.
It is also possible to have all three regulators
using the same phase voltages and current
for input control signals in which case the 3
regulators will always be changed by the
same number of taps.
41
Finally, there is also available and in use the
so-called three-phase regulator (in contrast
to three interconnected single phase
regulators). The three phase regulator is
gang-operated so that the taps on all
windings change the same and, as a result,
only one compensator circuit is required.
In either of the previous two cases, it is up to
the engineer to decide which phase current
and voltage will be sampled by the
compensator circuit.
References
[1] A. Pansini, “Electrical distribution
engineering,” McGraw-Hill, 1983.
[2] B. Lloyd, “Distribution transformers,”
Chapter 6 in “Distribution Systems, Electric
Utility Engineering Reference Book,”
Westinghouse Corporation, 1965.
[3] T. Short, “Electric power distribution
handbook,” CRC press, 2004.
42