4-55. The choke (L1) is usually a large value (from 1 to 20 henries) and offers a large

inductive reactance to the 120-Hz ripple component produced by the rectifier. Therefore,

the effect that L1 has on the charging of the capacitor (C1) must be considered. Since L1 is

connected in series with the parallel branch consisting of C1 and RL, a division of the

ripple (AC) voltage and the output (DC) voltage occurs. The greater the impedance of the

choke, the less the ripple voltage that appears across C1 and the output. The DC output

voltage is fixed mainly by the DC resistance of the choke.

4-56. Now that you know how the LC choke-input filter functions, we will look at it

with actual component values applied. For simplicity, the input frequency at the primary of

the transformer will be 117 volts, 60 Hz. Half-wave and full-wave rectifier circuits will be

used to provide the input to the filter.

4-57. Figure 4-23 shows the half-wave configuration. The basic parameters are as

follows:

With 117 volts AC rms applied to the T1 primary, 165 volts AC peak-to-peak

is available at the secondary [(117 V) x (1.414) = 165 V]

Remember that the ripple frequency of this half-wave rectifier is 60 Hz. Therefore, using

the below formula, the capacitive reactance of C1 is computed as follows:

1

XC =

2π fC

1

XC =

-6

(2) (3 .14) (60) (10) (10

)

(1) (10 6 )

XC =

3768

XC = 265 ohms

This means that the capacitor (C1) offers 265 ohms of opposition to the ripple current.

However, the capacitor offers an infinite impedance to DC. Using the following formula,

the inductive reactance of L1 is computed as follows:

XL = 2π f L

XL = (2) (3.14) (60) (10)

XL = 3.8 kilohms

4-58. The above calculation shows that L1 offers a relatively high opposition

(3.8 kilohms) to the ripple in comparison to the opposition offered by C1 (265 ohms).

Therefore, more ripple voltage will be dropped across L1 than across C1. The impedance

of C1 (265 ohms) is also relatively low in respect to the resistance of the load

(10 kilohms). Therefore, more ripple current flows through C1 than the load. In other

words, C1 shunts most of the AC component around the load.