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2φ φ CT Configuration
The purpose of this Appendix is to illustrate how two
CT's may be used to sense three phase currents.
The proper configuration for the use of two CTs rather
than three to detect phase current is shown. Each of
the two CTs acts as a current source. The current that
comes out of the CT on phase 'A' flows into the inter-
posing CT on the relay marked 'A'. From there, the cur-
rent sums with the current that is flowing from the CT
on phase 'C' which has just passed through the inter-
posing CT on the relay marked 'C'. This 'summed' cur-
rent flows through the interposing CT marked 'B' and
from there, the current splits up to return to its respec-
tive source (CT). Polarity is very important since the
value of phase 'B' must be the negative equivalent
of 'A' + 'C' in order for the sum of all the vectors to
equate to zero. Note, there is only one ground con-
nection as shown. If two ground connections are made,
a parallel path for current has been created.
In the two CT configuration, the currents will sum vec-
torially at the common point of the two CTs. The dia-
gram illustrates the two possible configurations. If one
phase is reading high by a factor of 1.73 on a system
that is known to be balanced, simply reverse the polar-
ity of the leads at one of the two phase CTs (taking care
that the CTs are still tied to ground at some point). Po-
larity is important.
To illustrate the point further, the diagram here shows
how the current in phases 'A' and 'C' sum up to create
phase 'B'.
Once again, if the polarity of one of the phases is out
by 180°, the magnitude of the resulting vector on a bal-
anced system will be out by a factor of 1.73.
On a three wire supply, this configuration will always
work and unbalance will be detected properly. In the
event of a single phase, there will always be a large
unbalance present at the interposing CTs of the relay. If
for example phase 'A' was lost, phase 'A' would read
zero while phases 'B' and 'C' would both read the mag-
nitude of phase 'C'. If on the other hand, phase 'B' was
lost, at the supply, 'A' would be 180° out of phase with
phase 'C' and the vector addition would equal zero at
phase 'B'.
APPENDIX D
D-1
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