Application and Operation
6.14 Cell Bypass
Product User Manual
Operating Instructions, Version AE 12/2009, A5E01454341C
147
in Figure "Drive Output after Loss of 5 Cells", only the faulted cells are bypassed. The phase
angles of the cell voltages have been adjusted so that phase A is displaced from phase B by
61.1° and from phase C by 61.6°. The star point of the cells is far removed from the neutral
of the motor voltages, but the motor voltage is still balanced. The neutral-shift keeps 67% of
the original cells in use, and 50% of the original voltage is available.
C1
C2
B3
B2
B1
A1
61.6
61.1
C
B
A
VBAVAC
VCB
A3
A5
A2
A4
15 Module Drive after
Bypass of 2 Modules
in Phase B, and 3
Modules in
Phase C.
67% of the Modules
are in use. 50% of
full voltage is
available.
Figure 6-22 Drive Output after Loss of 5 Cells
Figure "Available Voltage after One Cell Bypass" compares the available voltage after one
cell is bypassed with and without using neutral-shift. In many cases, the extra voltage
available with neutral-shift will determine whether or not a cell fault can be tolerated. The
voltage capability of a drive after cell bypass can be calculated by using the following
procedure.
If X is the largest number of cells in bypass in two of the phases, then the maximum voltage
at the drive output will be:
Vout_bypass = Vout * (2*N - X) / (2*N)
where: Vout is maximum output voltage that the drive can deliver (Vout =
1.78*N*Vcell)
N is the number of ranks (i.e., number of installed cells per phase)
Vcell is the cell voltage rating.
Exemplary calculation of a cell bypass for a drive with 18 cells
Consider a drive with 18 cells, each rated for 750V.
The maximum output voltage that this drive can deliver is (with N = 6 and Vcell = 750, and
without O-M):
Vout = 1.78 * 6 * 750 = 8.01 kV
If after cell bypass, the drive has 6 cells operational in phase A, 5 cells in phase B, and 4
cells in phase C, then the maximum voltage that the drive can produce with neutral shift from
the above formula is (with X = 1 + 2 = 3, because 2 cells in phase C and 1 cell in phase B
are bypassed):
Vout_bypass = 8010 * (2 * 6 - 3) / (2 * 6) = 6 kV
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