Handbook of Electrical Engineering

AUTOMATIC VOLTAGE REGULATION 91

The output signals from the saturation block are therefore,

Va 1 −Vfd 1 =AeBVf d^1 (4.9)

And

Va 2 −Vfd 2 =AeBVf d^2 (4.10)

Taking natural logarithms of both sides of equations (4.9) and (4.10) gives,

loge(Va 1 −Vfd 1 )=logeA+BVfd 1

And

loge(Va 2 −Vfd 2 )=logeA+BVfd 2

Eliminating logeA by subtraction gives

B=

loge(Va 1 −Vfd 1 )−loge(Va 2 −Vfd 2 )

Vfd 1 −Vfd 2

B=

loge

(

Va 1 −Vfd 1

Va 2 −Vfd 2

)

Vfd 1 −Vfd 2

(4.11)

And therefore from (4.9) and (4.10)

A=

Va 1 −Vfd 1

eBVf d^1

or

Va 2 −Vfd 2

eBVf d^2

( 4. 12 )

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
Vathat is needed above that required on the linear non-saturated line, by the non-saturated value of
Va. Hence atVfd 100 the value ofVaisVa 100 Sfrom the saturated curve andVa 100 Ufrom the straight
line. Similarly at the reduced output voltageVfd 75 the two values ofVaareVa 75 SandVa 75 U.The
two saturation levelsSE 100 andSE 75 are given by,

SE 100 =

Va 100 S−Va 100 U

Va 100 U

per unit

And

SE 75 =

Va 75 S−Va 75 U

Va 75 U

per unit

From the data for the exciterVfd 100 andVa 100 Sshould be available together withVa 75 S.The
manufacturer may also provideSE 100 andSE 75 .Vfd 75 is easily calculated fromVfd 100.