Inductor Filter
The output of the rectifier passes through an inductor. The inductor blocks the ac component and allows only the dc component to reach the load.
Fig 9.7 Inductor Filter
Ripple factor
G = RL / 3Ö 2 wL (9.35)
The ripple factor is inversely proportional to inductance and directly proportional to the load resistance. However, it should be kept in mind that a larger value of the inductor can reduce the ripple but the inductor has a higher dc resistance.
To analyse this filter for a full-wave rectifier, the Fourier series can be written as follows:
VO = 2Vm /p - 4Vm /p (1/3 Cos 2wt + 1/15 Cos 4wt + 1/35 Cos 6wt + ....)
2Vm / p being the dc component.
Assuming the output of 3rd and higher terms to be negligible, the out-put voltage is
VO = 2Vm / p - 4Vm / 3p Cos 2wt (9.36)
The resistances of the diode, choke and transformer can be neglected as they are small compared to the load resistance, RL.
The impedance of series combination and RL at 2w is,
Z = Ö(RL2 + (2wL2 ) = Ö (RL2 + 4w2L2 ) (9.37)
D.C component of current
Im = Vm / RL(9.38)
A.C component of current,
Im = Vm / Ö (RL2 + w2L2) (9.39)
The resulting current, i, is given by,
i = 2Vm/p RL - 4Vm /3p Cos (2wt-f)/Ö (RL2 +4 w2L2) (9.40)
where
f = tan-1 (2wL)/RL
If the load resistance is infinity, i.e., the output is an open circuit, then the ripple factor is
¡ = 2/3Ö2
= 0.471
This is only slightly less than 0.482 (for a full-wave rectifier). The difference is attributable to omission of higher harmonics . The inductor filter is used only where RL is small.
EXAMPLE 20.0
Calculate the value of the inductance to be included in the filter circuit of a FW rectifier circuit operating at 60 Hz to provide a d. c. output with 4% ripple and load load resistance of 100 W .
Solution
For a FW rectifier circuit using an inductor filter,
G = RL / 3Ö 2 (w L)
0.04 = 100 / 3Ö 2(2p x 60 x L)
L = 0.0625 / 0.04
= 1.5625 H