Table of Contents
Design of a SingleTuned Filter
The singletuned filters are designed to filter out (usually) characteristic harmonies of a single frequency. The equivalent circuit seen by the harmonic current I_{h} generated by the converter is shown in Fig. Z_{Fh }and Z_{Sh} are the filter and system impedances at the harmonies frequency (hf).
The harmonic current in the filter is given by
I_{Fh} = I_{h} Z_{Sh}  / Z_{Sh} + Z_{Fh}
The harmonic voltage at the converter bus is
V_{h} = I_{Fh} Z_{Fh} = I_{h} / Y_{Fh} + Y_{Sh} = I_{h} / Y_{h}
The basic objective in designing the filter is to select the filter admittance Y_{Fh} to minimize V_{h} or satisfy the constraints on V_{h} The problem of designing a filter is complicated by the uncertainty about the network admittance (Y_{Sh}) There are two possible representations of system impedance in the complex plane. These are (a) the impedance angle is limited (see Fig.(a) and (b) the impedance is limited both in angle and impedance. The impedance is assumed to lie in the region shown in Fig. (b) in which R_{1}, R_{2,} and θ_{m }obtained from the system characteristics.
It is to be noted that (a) or (b) reflect the broad assumptions about the nature of system impedance. The assumption (b) requires more knowledge of the system than (a). Having complete information on the network impedance characteristic will result in the economic design of filters and also a realistic assessment of the harmonic distortion of the converter bus voltage.
Although assumption (a) is rather simplistic, it allows a simplified computation of the optimum value of Q. In computing the optimum value of Q, we need to minimize the maximum value of V_{h.} Since we do not know the exact network impedance (except that its angle is limited), we assume the worstcase scenario for the network admittance that results in maximizing V_{h.} The optimum value of Q corresponds to the lowest value of the upper limit on V_{h.}
The value of Y_{h} (the net admittance seen by the converter) is reduced if the detuning parameter 5 is maximum = δm. For a specified δ_{m }and X_{0}, the locus of the filter impedance(as Q is varied) is a semicircle in the 4th quadrant of the GB plane as shown in Fig. The diameter of the circle is 1 / (2δ_{m} X_{0}). From the figure’s geometry, It can be seen that minimum Y_{h} occurs when θ = θ_{m.} The optimum value of Q can be obtained from gametheoretic analysis. Suppose one selects YFH (OP) arbitrarily (the tip of YFH lying along the semicircle).
In that case, the network can select Y_{SH} such that the vector Y_{h} (OD) is perpendicular to PD (the vector Y_{SH}) and ensure Y_{h} is a minimum. To maximize the minimum magnitude of Y_{h}, it is necessary to have PD tangential to the circle at point P. Thus, we select Y_{FH} to maximize Y_{h} when the network (viewed as an adversary, not under our control) tries to minimize it.
To compute optimum Q, we note that triangle OPC is isosceles with angle COP = angle CPO = θ_{m}/2. The angle of the impedance Z_{Fh} is (90 – θ_{m} / 2) and we have,
tan(90 – θ_{m} / 2 ) = Q_{opt} × (2δ_{m}) = cot(θ_{m} /2)
Hence,
Q_{opt }= cot(θ_{m} /2) / 2δ_{m}= 1+cosθ_{m} / 2δ_{m}sinθ_{m}
Since Y_{h} = Y_{Fh} cos(θ_{m}/2), Y_{Fh } = cos(θ_{m}/2)/(2δ_{m}X_{0}), we can derive the expression for V_{h} using Eq. We finally get,
V_{h} = I_{h} / Y_{Fh} +Y_{Sh} = 4 δ_{m} X_{0}I_{h} / 1+cosδ_{m}
Minimum cost tuned filters
The costs of the reactor and the capacitor which make up the tuned filter are dependent on their respective ratings. The rating of the capacitor is given by
S_{C} = I^{2}_{F1}X_{C1} + I^{2}_{Fh}X_{0}
where I_{F1} is the fundamental frequency current in the tuned filter given by
I_{F1} = V_{1} . (X_{C1} – X_{L1}) = V_{1} / X_{0} (h – 1/h) = (hV_{1} / (h^{2} – 1) . X_{0}
In deriving the above equation, the effect of the resistance, R is neglected. V_{1} is the fundamental frequency component of the voltage at the filter bus. (Actually, the harmonic components in the bus voltage in steady state, are neglected as filtering of harmonics in the source current supplied to the converter is assumed to be effective). X_{C1} = X_{0}h is the reactance of the capacitor and X_{L1} = X_{0}h is the reactance of the inductor at the fundamental frequency.
The rating of the reactor (S_{L}) is given by S_{L} = I^{2}_{F1}X_{L1} + I^{2}_{Fh}X_{0}
The size of the filter (S) is defined as the reactive power supplied by the capacitor (at the fundamental frequency). Thus, we get
S_{C} = S + I^{2}_{Fh}X_{0}
S_{L} = S / h^{2} + I^{2}_{Fh}X_{0}
The total cost of the filter (K) is given by
K = S_{C}U_{C} + S_{L}U_{L}
where U_{C} and U_{L}, are the unit costs of the capacitor and inductor, respectively. Substituting Eqs. we get an expression for the total cost as a function of the size of the filter. Since our objective is to determine the optimum size of the filter that will minimize the total cost, we need to express X_{0} in terms of the size S. It can be easily shown that
X_{0} = N / S, N = V^{2}_{1}h^{3} / (h^{2} – 1)^{2}
Substituting Eqns. , we get,
K = AS + BS^{1}
where A = U_{C} + U_{L}, B = I^{2}_{Fh} N(U_{C} +U_{L})
It is to be noted that I_{Fh} is expressed in terms of I_{h }(the harmonic current injected by the converter)
Since A and B are constants, the cost of the filter is minimum when the size of the filter is minimum given by
S_{opt} = (B/A)^{1/2}
The filtering would improve with the increased size of the filter. Hence the choice of the filter size will not be below that given by the above expression.
Design of a High Pass Filter
This is a secondorder (damped) filter described earlier. For harmonic frequencies equal to or higher than 17, a secondorder highpass filter is usually provided. By defining the following parameters,
h_{0}ω=1/√ LC, X_{0} = √ L/C, Q = R/X_{0}
the following values can be chosen
0.5 < Q < 2
h_{0} ≤√2h_{min}
where h_{min} min is the smallest value of h to be handled by the filter. The choice of h_{0} given above implies that the filter impedance at the h_{min} has decreased approximately to the value of R. The filter impedance is given by Eqn. The reactive power supplied by the filter Q_{F} is
Q_{F }= (h_{0} / (h_{o}^{2} – 1))(V_{1}^{2} / X_{0})
In deriving the above, the voltage across the inductor at the fundamental frequency is neglected. It can be shown that the filtering is improved if Q_{F}, is increased and a higher value of h_{0} can be chosen. Hence, it is advantageous to design a high pass filter to exclude six pulse operations. There is no optimum quality factor, Q in the case of highpass filters. Q is chosen for providing the best performance of the filter over a band of frequencies higher than those covered by tuned filters. The performance of these filters is insensitive to deviations in the frequency or filter parameters.
Double and Triple Tuned Filters
A singletuned filter can be used to filter one characteristic harmonic (or a noncharacteristic harmonic when required). Multiple singletuned filters are required to filter several (discrete) harmonics. For example, at least two singletuned filters are required for twelve pulse converters (to filter 11^{th} and 13^{th} d harmonics). Since AC filters also provide the required reactive power which is a function of the power flow in the HVDC link, there is a need to arrange the filters into switchable sections (filter banks) which adds to the cost. An alternative is to provide switchable shunt reactors that are switched in at low DC power levels.
This also is a costly option with increased space requirements. Weak AC systems (with low SCR) also impose an upper limit on the size of the filter bank to limit the voltage fluctuations (flicker) caused by the switching of a filter bank. Thus, coordination of the filtering and supplying reactive power considerations dictate the adoption of double and tripletuned filters. The advantages also include (i) only one inductor (L_{1}) is subject to full line impulse voltage and (ii) power loss at the fundamental frequency is considerably reduced.
Electrically, neglecting losses, a doubletuned filter is identical to two singletuned filter branches shown in Fig. (b). Similarly, a lossless tripletuned filter is identical to three singletuned filter branches.
The double and tripletuned filters shown in Figs have a single resister R each. Additional damping resistors can be applied in parallel with reactor L, and/or Ly and L to produce a damped characteristic.
Analysis of a DoubleTuned Filter
The impedance of the filter shown in Fig. (a) (neglecting the effect of R_{1}) as a function of the complex frequency (s) is given by
where
The zeros of Z_{F}(s) lie on the imaginary axis corresponding to the two frequenciesω_{1} and ω_{2.} The poles are also on an imaginary axis corresponding to ω = 0 and ω = Ω. From the properties of the impedance of oneport LC networks, we can state
0 < ω_{1} < Ω_{1} < ω_{2}
Since the doubletuned filter is equivalent to the two singletuned filters shown in Fig.(a), we can compare the impedances of the two circuits and obtain the following results
L_{1} = L_{a}L_{b }/ (L_{a} + L_{b}), L_{2} = L_{a} + L_{b}, C_{1} = C_{a} + C_{b,} C_{2} = C_{a}C_{b }/ (C_{a} + C_{b})
From the above, it follows that
L_{1}L_{2} = L_{a}L_{b}, C_{1}C_{2} = C_{a}C_{b}
The impedance of the doubletuned filter as a function of the frequency is obtained by substituting s = jωh in the expression given in Eq. Note that ω is the fundamental (system) frequency and & is viewed as a continuous variable.
Defining ω_{r} = 1 / √ L_{1}C_{1} , X_{1} = ω_{1}L_{1} = 1 / ω_{1}C_{1} , h_{r} = ω_{r }/ ω, H_{1} = Ω _{1} / ω, X_{2} = L_{2 }= 1 / Ω _{1}C_{2}
We can express the impedance as a function of h as
Z_{F} (jωh)= jX_{1} = (h / h_{r} – h_{r }/ h) + X_{2 }/ j(h / H_{1} – H_{1 }/ h)
J[X_{1} . (h^{2 }– h^{2}_{r} ) / (hh_{r}) – (X_{2}H_{1}h) / (h^{2 }– H_{1}^{2}]
To compute ω_{1}, and ω_{2 }we equate the RHS of Eq to zero and get the following quadratic equation in h^{2} given below
h^{4 }– (h_{r}^{2 }+ H_{1}^{2 }+ X_{2 }/ X_{1} . H_{1}h_{r}). h^{2} + h_{r}^{2}H_{1}^{2} = 0
The reactive power (Q_{F}) supplied by the filter is given by
Q_{F} = V_{1}^{2} / Z_{F1}
where Z_{F1} = (h_{r}^{2} – 1) . X_{1 }/ h_{r }– (X_{2}H_{1}) / (H_{1}^{2} – 1)
Analysis of TripleTuned Filters
The impedance of a lossless tripletuned filter as a function of complex frequency s is given by
Z_{F}(s) = L_{1}(s^{2} + ω_{1}^{2})(s^{2 }+ ω_{2}^{2})(s^{2} + ω_{3}^{2}) / (s(s^{2} + Ω_{1}^{2})(s^{2} + Ω_{2}^{2}))
where ω_{1}, ω_{2 }and ω_{3 }are frequencies which have to be filtered Ω_{1} and Ω_{2} are defined by
Ω_{1} = 1 / √L_{2}C_{2} , Ω_{2} = 1 / √L_{2}C_{3}
The impedances of the filter as a function of ho (h is treated as a continuous variable) are given by
Z_{F }(jhω) = jX_{1 }(h / h_{r} – h_{r }/ h) + X_{2 }/ j (h / H_{2} – H_{1 }/ h + X_{3 }/ j(h / H_{2} – H_{2 }/ h
where
H_{1} = Ω_{1} /ω, H_{2} = Ω_{2} /ω, h_{r} = ω_{r }/ ω, ω_{r} = 1 / √L_{1}C_{1}
X_{1} = ω_{1}L_{1} = 1 / ω_{1}C_{1}, X_{2} = Ω_{1}L_{2} = 1 / Ω_{1}C_{2}, X_{3 }= Ω_{2}L_{3} = 1 / Ω_{2}C_{3}
To compute ω_{1}, ω_{2,} and (the zeroes Z_{F}) we equate the RHS of Eq. to zero. We finally get a cubic equation in h^{2 }given by
h^{6} – a_{1}h^{4} + a_{2}h^{2} – a_{3} = 0
a_{1} = h_{r}^{2} + H_{1}^{2} + H_{2}^{2} + h_{r }(X_{2 }/X_{1 .} H_{1} + X_{3 }/ X_{1}. H_{1})
a_{2} = H_{1}^{2}H_{2}^{2 }+ h_{r}^{2} (H_{1}^{2} + H_{2}^{2}) + h_{r}H_{1}H_{2} . (X_{2 }/X_{1} . H_{2} + X_{3 }/ X_{1} . H_{1})
a_{3} = H_{1}^{2}H_{2}^{2}h_{r}^{2}
h_{1} = ω_{1 }/ ω, h_{2} = ω_{2 }/ ω, h_{3} = ω_{3 }/ ω are the three real roots of Eq. It is easily observed that the product of the real roots is √a_{3. }This implies that
ω_{1}ω_{2}ω_{3} = ω_{r}
we have to select Ω_{1} and Ω_{2} such that
0 < ω_{1} < Ω_{1} < ω_{2} < Ω_{2} < ω_{2}
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