Page 104 - Handbook of Electrical Engineering

AUTOMATIC VOLTAGE REGULATION 85

is loaded from zero to full-load at rated power factor. In order to achieve this low level of regulation
the gain G a needs to be high.
The power amplifier has a practical lower limit of zero and an upper limit of typically 10.0
per unit. The upper limit should be high enough to ensure that the full output of the exciter can be
obtained during field forcing of the main generator, e.g. during short circuits that are at or near to
the generator.

4.1.3.1 Worked example

Find the value of the gain G a for an AVR fitted to a generator that has a synchronous reactance of
2.0 pu. Assume the full-load has a power factor of 0.8 lagging and a terminal voltage V of 0.995 pu
i.e. (0.5% regulation)
Step 1. Find the equivalent series impedance Z that can represent the load. The full volt-ampere load
on the generator is S,
S = P + jQ pu MVA

When the terminal voltage is V< 1.0, the load impedance Z is,
VV ∗
Z = R + jX =
S ∗
Where ∗ denotes a conjugate quantity.
Hence
0.995 2
Z = = 0.792 + j0.594 pu
0.8 − j0.6
Step 2. Find the emf in the generator
The emf feeds a series circuit consisting of the load plus the synchronous reactance X s .Itcan be
shown that the emf E is,

V 2
E = (Z + X.X s ) + jR.X s
Z 2
Hence
V
2
2
|E|= (Z + X.X s ) + (R.X s ) 2 (4.1)
Z 2
Now
2
2
2
Z = R + X = 0.9801 pu
0.995 2 2
|E|= (0.9801 + (0.594)(2.0)) × ((0.792)(2.0))
0.9801
= 2.7259
Hence the gain G g of the generator in its full-load steady state is,
|V | 0.995
G g = = = 0.365 pu (4.2)
|E| 2.7259