Generator and Electrical System - Small Wind Turbines: Analysis, Design, and Application - page 235

bus current (i.e. if diodes 1 and 2 are on, this is followed by diodes 2 and 3 on,
followed by diodes 3 and 4 on, and so forth).
Neglecting the voltage drop across a diode (typically 0.7-0.9 V when the diode
is on), and placing a pure resistance across the output (V dc in Fig. 11.6 ) the average
rectifier output voltage is

¼ 1 : 35V LL

p 2

V dc ¼ 3 = p V LL ð 11 : 2 Þ

where V LL is the generator line-to-line rms voltage (for example the rms voltage
that would be measured between terminals A and B in Fig. 11.6 ). Note that V dc is
not constant as rectification only makes it non-negative. Equation 11.2 is well
known by power converter designers for resistive and inductive type loads. In low
power rectifier design, however, the DC bus does not have inductive filtering but
rather a simple capacitor filter placed across the DC bus. In this case, the capacitor
charges up to a voltage very near the peak of the sinusoidal generator line-to-line
voltage, and the DC bus voltage becomes
V dc ¼ 2

p

V LL ¼ 1 : 41V LL ð 11 : 3 Þ

Whatever the output filtering, a generator will have a wide range of voltages as
the wind speed varies, as demonstrated in Fig. 11.7 . This has at least two conse-
quences. One is that at low wind speeds, the generator voltage may be so low that
the voltage drop across the diodes becomes significant compared to the generator
voltage. Rectification will then have a (possibly much) lower efficiency than at
rated generator voltage. If the loss of efficiency is too great, the diodes can be
replaced with transistor-type switching devices with a much lower voltage drop.
An electronic controller, however, is now necessary to turn the transistor switching
devices on and off. A second consequence is that at low wind speeds, the DC bus
voltage may not be adequate for the inverter to produce the needed AC output
voltage. One solution to this is to follow the rectifier with a boost converter.
For small generators in particular, a boost converter is required even for battery
charging. From Eq. 11.3 the output voltage of the Ginlong 500-A shown in

Fig. 11.7 Typical voltage
and current output of the
Ginlong 500-A PMG

35

30

Current
Voltage

25

20

15

10

5

0

0 100 200 300 400 500 600

Generator speed (r.p.m.)