Constant current transformer for
Base circuits
In nominal operation mode the inductance L1 is
in resonance operation mode with the capacitor C and supplying the maximum
constant current through load Z. This output current is
"impressed" and does not depend on the size of the load impedance Z. When the SCR controller is switched on, the Lscr inductance and the C capacity are switched on in parallel
and resonance mode is interrupted. This controls the
amount of output current Iout within a certain area.
At 3-times mains frequency, inductance L3 is in
resonance with capacity C and derives the third current harmonic from the
constant current transformer.
Transformer T is used to adjust the voltages. Figure 2 shows a
version of the constant current transformer illustrating the transformer's
combined primary winding and the L1 winding for the L1L2 choke. For design reasons, the laminations for transformer T and choke L1L2
must be identical which always results in higher material costs. To this must be added the extra costs for fitting a common winding. However, when supplying power at several input voltages, only one set of
connection terminals is required.
To optimise the size of the Lscr choke, it is
possible to provide 2-3 taps at the secondary winding.
The
taps on the primary side are used to connect the various primary voltages: 208V, 240V, 347V, 400V and 480V.
Figure 3 shows the auto-transformer connection
between the C winding and the load winding. This
design saves on materials and it would be an advantage for constant current
transformers with output voltages up to 2000-3000V.
Figure 4 shows the design for a 4kVA constant
current transformer (U2=U3=500V) and figure 5 shows a 1-2 k VA constant current
transformer (U2=U4=225-250V).
Basic circuits and basic equations
-
-
R'2 ohmic resistance of L2 winding, of load winding, of C winding, of Lscr winding and of L3 winding, converted to number of turns on primary winding.
-
L1 - inductance of L1 winding (when L2 winding is open) and the leakage inductance on primary winding
-
L'2 - inductance of L2 winding /when L1 winding is open) and the leakage inductance on secondary winding , converted into number of turns on primary winding
-
Mutual inductance of inductances L1 and L'2 : M = K*(L1*L'2)^0.5
-
C'e - capacity of capacitor, converted to number of turns on primary winding
-
Z' - load impedance, converted to number of turns on primary winding
The basic equations now read as follows:
·
Uin = R1*Iin
+ jωL1*Iin +jωM*I'out + jωL'2*I'out +
jωM*Iin + (R'2+Z')*I'out (2)
·
Iin = I'out + I'ce (3)
The underlined parameters are the complex
variables, referred to the input voltage Uin.
K => factor of the mutual inductance (0.90 -
0.95)
ω = 2*π"f
j = (-1)^0.5
f => mains frequency (50/60Hz)
(R'2+Z'+ j(ωL'2+1/ω/C'e))*(R1+j ωL1-j/
ω/C'e)] (4)
L1 = L'2 = L
M = K*(L1*L'2)^0.5 = K*L
ωL1 = ω*L = 1/ω/C'e
Q = ω*L/r
For Q(K+1)<< Q^2(K+1)^2 equation (5) can
be further simplified:
Constant current transformer parameters in resonance
operation mode
I'out.nom = U'out.nom/R'nom
and
L = L1 =
R'nom/ (ω*(K+1)) (10)
U'out.nom=Uin.min (11)
The nominal output current I'out.nom should be
supplied in resonance operation mode at the minimum input voltage Uin.min. At a different, greater input voltage the resonance operation mode is
controlled with the current by the Lscr choke, and the nominal output current
is kept constant.
Nominal current for
capacitor
First, equation (8) is simplified:
I'ce.nom = I'out.nom
- jI'out.nom
and
I'ce.nom
= I'out.nom (2) 1/2 = 1.41*I'out.nom (13)
Nominal voltage for C
winding
and
U'3.nom = 1.41*U'out.nom/(K+1)
or
U3,nom
= U'ce.nom = 1.41*Uout.nom/(K+1) (14)
Capacity Ce is a resultant value from the
serial circuit of the L3 choke and capacitor C (see figure 6).
Nominal voltage for
capacitor with L3 choke
The L3 choke and capacitor C (see figure 6) are
in resonance at 3-times mains frequency.
Zc = j[ωoL3-1/(ωoC)] =
j[1/(9ωoC)- 1/(ωoC)]= -j8/(9ωoC)
and
Uc.nom
= 1.13*Uce.nom (16)
At the same time the equivalent capacity Ce of
the L3-C serial circuit must be in resonance with inductance L1.
Z'c = 1/(ωo*C'e)]= ωoL1
and
C'e=1/
ωo^2/L1 bzw.
C'e=1/ ωo2/L'2 (17)
C =
8*Ce/9 (18)
Nominal power for
capacitor
Pce.nom
= U'ce.nom*I'ce.nom =2*Pout.nom/(K+1) (19)
Pc.nom
= U'c.nom*I'ce.nom =(9/8)*Pce.nom (19a)
Vector diagram
Figure 7c shows short circuit operation. In Figure 7d K=0.9 and the power factor of the load is 0.7 (inductive). In this operating mode the capacitor voltage can be considerably greater
than in nominal operation mode, which may result in an increase of the
capacitor current and can cause an induction in the core of transformer T. At this point, it should also be mentioned that the power factor of the
input current is capacitive with an inductive load.
f=60Hz
ωo = 2*π*f = 377
Output power=15kVA
Output current = 6.6 A
Capacitor's nominal operating voltage
Uc.nom=550V
Factor of mutual inductance K = 0.9
Calculated
data:
·
Rnom = Uout.nom/Iout.nom =
2272/6.6 = 344 Ohm
·
L2 = Rnom/ ωo/(K+1) =
344/377/1.9 = 480mH
·
U2.nom =
1.41*Uout.nom/(K+1)= 1686V
·
U1.nom = 1.41*Uin.min/(K+1)
= 1.41*228/1.9 = 170V
·
L1 = L2*(U1.nom/U2.nom) 2 = 480*(170/1686) 2 = 4.83mH
·
Pce.nom = 2*Pout.nom/(K+1) =
15789 (for Pout - pure ohmic load)
·
Ice.nom = Pce.nom/Uc.nom = 15789/550
= 28.7A
·
Uc3.nom = Uce.nom =
Uc.nom/1.13 = 486V
·
Ce = (1/L2/ ωo2)*(U2.nom/U3.nom) 2 = (1/0.480/3772)*(1686/486) 2 = 176μF
·
C=8*Ce/9 = 156
·
L3 = 1/(9ωoCe)=1/(9*3772*156e-6)
=5mH
·
Pce.nom = U3.nom^2*ωCe
= 15670Var (OK)
Power factor of
constant current transformer in resonance operation mode
The power factor of the input current is
calculated using equation (7): This equation can be
simplified as follows:
I'in.nom =
Uin.min* Z'/(( ω L**(K+1)) 2 = Uin.min* (R'+jX')/((
ω L**(K+1)) 2
Thus :
Cos(φ)
= R'/Z' = R'/(R'^2+X'2) 1/2 (20)
where:
R' => Ohmic value of load impedance Z'
X' => Inductive value of load impedance Z'
With an inductive load, the power factor is
lower than 1 and is of a capacitive nature! (see
figure 7d)
The operating efficiency of the constant
current transformer is first calculated without taking into account magnetic
losses of the magnetic components.
R'nom = ω L*(K+1)
Z' = R'nom >>r
it is easy to establish the following from
equations (6) and (7):
Iin.nom/I'out.nom ˜ 1
and
η
= Q*(K+1)/[Q*(K+1)+2] (22)
In power range from 1 to 70kVA, the Q factor of
choke L1L2 is normally between 50 and 200. Taking
into account Fe and Cu magnetic losses in the other magnetic components, we can
talk about an equivalent Q factor for the constant current transformer of
10-80.
Thus equation (10) results in an operating efficiency
of 90.5% to 98.7%.
Non-resonance operation mode of constant current
transformer
Lscr-Inductor
This choke is wired via the C winding parallel to
capacitor C (see figure 1). Choke Lscr generates an inductive current by means of the
two anti-parallel wired thyristors (SCR). This
compensates the capacitive current of capacitor C, the resonance operation of
the constant current transformer is disturbed and the output current is
reduced.
The circuit shown in figure 8 is used to
calculate the output current in non-resonance operation mode.
Starting with equation (4) and using the
following relations:
r = 0
L1 = L =
Kl*Lo
L'2 = Ka*L
M = K*Kl*Lo
C'e = Kc*C'eo
ω = Kω*ωo
K=1
1/(C'eo*Lo) = ωo2
(Resonance operation mode)
the output current in non-resonance operation
mode can be calculated as follows:
I'out=Uin*(Kω^2*Kc*Kl*Ka1/2+1)/[Z'*(Kω2*Kc*Kl-1) - jωoLo*(1+Ka1/2) 2] (23)
Calculating the inductance of the Lscr choke
This investigation calculates the factor Kc. The other factors are set to 1.
I'out = Uin*(Kc+1)/[Z'*(Kc-1)
- 4*jωoLo] (24)
The required control range for the output
current can be described as follows:
1.
The output current is I'out.nom, in resonance operation mode (Kc=1) at
the minimum input voltage Uin.min.
2.
The output current is I'out.min, at maximum input voltage Uin.max and
load Z'=R' = 0.
and
Uin.max/Uin,min=1.1/0.95=1.157
we can calculate::
Kc=
- 0.5
This means that the impedance of the Lscr choke
must be smaller than the impedance of the capacitor by a factor of 1.5.
ωoL'scr
= 0.67/(ωoC'eo) (25)
It follows that the core power on the Lscr
choke is around 50% greater than for capacitor Ceo.
Pl.nom
= 1.5*Pce.nom = 3*Pout.nom/(K+1) (25a)
Optimum core power
for for Lscr choke
and
Pl.nom
= 0.5*Pce.nom = Pout.nom/(K+1)
When taps are provided on the output winding,
only parameters Z' and Ka, or L'2 change in equation (23). In resonance
operation mode of inductance L1 and capacitor Ce, equation (23) reads as
follows:
I'out=Uin/(- jωoLo*(1+Ka1/2)) = Uin/(-
jωoLo*(1+(W2.1/W2.2)))
where:
W2.1
= Turns of the load winding
W2.2 = Turns of the tap on the load winding
and:
Ka = (W2.1/W2.2) 2
For a change in output current between 7A and
2A, the number of turns on the tap must be the smallest output voltage.
W2.2 = W2.1/6
The main advantage of this version can be
described as follows:
1.
The power factor is always approx. equal to 1
2.
With some taps the core power of the Lscr choke can be reduced by a
factor of 3-5.
This selection is normally made in the
following conditions:
-
For maximum input voltage Uin.max = 1.1 Uin.nom
-
For controlled output current Iout.nom
-
For power factor for load cosφ.min = 0.82 (operation with 30% open current transformer) or Z'=0.7*R'nom+j0.5*R'nom = 2ωoLo*(0.7+j0.5)
First the factor Kc is calculated with the aid
of equation (24) in resonance operation mode with Uin.min and from
non-resonance operation mode at Uin.max, as follows:
Iout.Kc=1 = Iout.Kc<1 =
Iout.nom
and
Uin.max*(Kc+1)/[(0.7+j0.5)*(Kc-1)
- j2] = Uin.min/(- j]
and
Uin.min/Uin.max) 2 = 0.75 =(Kc+1) 2/[(0.7*(Kc+1)) 2+(2-0.5*(Kc+1)) 2]
Thus :
Kc= 0.82
Now, using figures 6 and 8, we can calculated
the capacitor current:
I'ce = jIout*[Z'+jω(L'2+M)]/( ωM+1/(ωC'e)) = jIout*Kc*ωoCo
[Z'+j2ωoLo)]/(Kc+1)
With Kc = 0.82 and Iout = Iout.nom, we form the
following relation:
I'ce.max = jI'out.nom*Kc*[0.7+j1.5)]/(Kc+1)
and
I'ce.nom = jI'out.nom*[1+j]/2
Thus :
I'ce.max/I'ce.nom = 0.75/0.71=1.05
and
Uce.max/Uce.nom
= I'ce.max/I'ce.nom/Kc = 1.28 (26)
L3 choke and current harmonics
-
The two thyristors (SCR) wired anti-parallel.H
-
eavily saturated current transformers for load Z
-
Magnetizing current of the transformer
-
In a professional design for a constant current
transformer, the transformer's magnetizing current is not a source of
harmonics. Above all, with an induction of over 1.7T
(17000 Gaus), the third harmonic must be taken into account.
and
IL.5 = 0.045*IL.max (5th harmonic)
where:
IL.max = U4/ (ωoLscr)
When designing the Lscr throttle for:
Pl = 1.5*Pce.nom = 1.5*Pout.nom
Iout.3 = 0.1*IL.max*(U4/U2)
and
Iout.5
= 0.045 IL.max*(U4/U2)
Heavily saturated
current transformers for load Z
Figure 11 shows the primary voltage and the
primary current of the current transformer in nominal operation and in
"no-load operation" In both cases the same
"impressed" current Iout flows through the primary winding. At a ratio between the induction in nominal operation and the saturation
induction of the transformer of approx. 0.7, the mean value of voltage Uout in
"no-load operation" is approximately 50% greater than in nominal
operation. For a constant current transformer operating at
30% current transformers in "no-load operation", the mean value of
the voltage drop is:
Assuming that the no-load current of the
transformer in nominal operation is approximately 10%, width X of the pulse is
approximately 0.15π. We can now estimate the size
of the harmonics as follows:
Erms.3 = 0.33*Eav*sin(3*X/2)*( π /X) =
0.53*Uout.nom
Erms.5 = 0.2*Eav*sin(5*X/2)*( π /X) =
0.47*Uout.nom
and
Iout.3
= Erms.3/(0.7*Rnom^2+(3ωL2) 2) 1/ = 0.53*Uout.nom/(1.66*Rnom)
= 0.32*Iout.nom
Iout.5
= Erms.5/(0.7*Rnom2+(5ωL2) 2) 1/2 = 0.47*Uout.nom/(2.6*Rnom)
= 0.18*Iout.nom
IL3.1
= Ic.nom
IL3.3 =
0.32*Iout.nom*(U2/U3) (27)
IL3.5 = 0.18*Iout.nom *(U2/U3)
No-load operation
W1 =>Turns of the calculated tap
Uin.min =>
95% of the following values: 208V, 240V, 347V, 400V, 480V
B =>Induction in T (1T = 10000Gaus)
f => Frequency:
Kfe => stacking factor
K => factor of mutual inductance of L1 and L2,
taking into account the leakage inductance of transformer T.
In this design of primary winding and
inductance, L1 resonance operation can be calculated and installed only with a
tap.
The different leakage inductances of the taps and the
different factors of mutual inductance of the two taps result in slightly
differing resonance frequencies, which can be easily compensated with the
current of the Lscr choke.
Resonance frequency of tap 1:
ωo.12 = 1/(L1.1*C'e.1)
Resonance frequency of tap 2:
L1.2 ˜ L1.1*(W1.2/W1.1) 2
C'e.2 =
C'e.1*(W1.1/W1.2) 2
ωo.22= 1/(L1.2*C'e.2) ˜ 1/(L1.1*C'e.1) ˜
ωo.12