Page 110 - Handbook of Electrical Engineering

AUTOMATIC VOLTAGE REGULATION 91

The output signals from the saturation block are therefore,

V a1 − V fd1 = Ae BVf d1 (4.9)

And

V a2 − V fd2 = Ae BVf d2 (4.10)
Taking natural logarithms of both sides of equations (4.9) and (4.10) gives,

log (V a1 − V fd1 ) = log A + BV fd1
e
e
And
log (V a2 − V fd2 ) = log A + BV fd2
e
e
Eliminating log e A by subtraction gives
log (V a1 − V fd1 ) − log (V a2 − V fd2 )
e
e
B =
V fd1 − V fd2

V a1 − V fd1
log
e
V a2 − V fd2
B = (4.11)
V fd1 − V fd2
And therefore from (4.9) and (4.10)
V a1 − V fd1 V a2 − V fd2
A = or (4.12)
e BVf d1 e BVf d2
It has become the custom to choose the two pairs of data at the 100% and 75% excitation
levels of the exciter. The purpose being to suit computer simulation programs that require these
specific data points.

The 100% pairs are those at the ceiling output voltage of the exciter whilst the 75% pair are at
75% of the ceiling output voltage. The saturation level can be described by dividing the difference in
V a that is needed above that required on the linear non-saturated line, by the non-saturated value of
V a . Hence at V fd100 the value of V a is V a100S from the saturated curve and V a100U from the straight
line. Similarly at the reduced output voltage V fd75 the two values of V a are V a75S and V a75U .The
two saturation levels S E100 and S E75 are given by,

V a100S − V a100U
S E100 = per unit
V a100U
And
V a75S − V a75U
S E75 = per unit
V a75U
From the data for the exciter V fd100 and V a100S should be available together with V a75S .The
manufacturer may also provide S E100 and S E75 . V fd75 is easily calculated from V fd100 .