Figure 3-32. - Half-wave rectifier with an LC choke-input filter.

This means that the capacitor (C1) offers 265 ohms of opposition to the ripple current. Note, however, that the capacitor offers an infinite impedance to direct current. The inductive reactance of L1 is:

X_L = (2) (3.14) (60) (10)

X_L = 3.8 Kilohms

This shows that L1 offers a relatively high opposition (3.8 kilohms) to the ripple in comparison to the opposition offered by C1 (265 ohms). Thus, more ripple voltage will be dropped across L1 than across C1. In addition, the impedance of C1 (265 ohms) is 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.

Let's go a step further and redraw the filter circuit so that you can see the voltage divider action. (Refer to figure 3-33.) Remember, the 165 volts peak-to-peak 60 hertz provided by the rectifier consist of both an ac and a dc component. The first discussion will be about the ac component. Looking at figure 3-33, you see that the capacitor (C1) offers the least opposition (265 ohms) to the ac component; therefore, the greatest amount of ac will flow through C1. (The heavy line indicates current flow through the capacitor.) Thus the capacitor bypasses, or shunts, most of the ac around the load.

By combining the X_C of C1 and the resistance of R_L into an equivalent circuit, you will have an equivalent impedance of 265 ohms.

Figure 3-33. - AC component in an LC choke-input filter.

You now have a voltage divider as illustrated in figure 3-34. You should see that because of the impedance ratios, a large amount of ripple voltage is dropped across L1, and a substantially smaller amount is dropped across C1 and R_L. You can further increase the ripple voltage across L1 by increasing the inductance:

Figure 3-34. - Actual and equivalent circuits.

Now let's discuss the dc component of the applied voltage. Remember, a capacitor offers an infinite impedance to the flow of direct current. The dc component, therefore, must flow through R_L and L1. As far as the dc is concerned, the capacitor does not exist. The coil and the load are, therefore, in series with each other. The dc resistance of a filter choke is very low (50 ohms average). Therefore, most of the dc component is developed across the load and a very small amount of the dc voltage is dropped across the coil, as shown in figure 3-35.

Figure 3-35. - DC component in an LC choke-input filter.

As you may have noticed, both the ac and the dc components flow through L1, and because the coil is frequency sensitive, it provides a large resistance to ac and a small resistance to dc. In other words, the coil opposes any change in current. This property makes the coil a highly desirable filter component. Note that the filtering action of the LC capacitor input filter is improved when the filter is used in conjunction with a full-wave rectifier as shown in figure 3-36. This is due to the decrease in the X_C of the filter capacitor and the increase in the X_L of the choke. Remember, the ripple frequency of a full-wave rectifier is twice that of a half-wave rectifier. For a 60-hertz input, the ripple will be 120 Hertz. Let's briefly calculate the X_C of C1 and the X_L of L1:

Figure 3-36. - Full-wave rectifier with an LC choke-input filter.

It should be apparent that when the X_C of a filter capacitor is decreased, it provides less opposition to the flow of ac. The greater the ac flow through the capacitor, the lower the flow through the load. Conversely, the larger the X_L of the choke, the greater the amount of ac ripple developed across the choke; consequently, less ripple is developed across the load. This condition provides better filtering.

Q.24 In an LC choke-input filter, what prevents the rapid charging of the capacitor?
Q.25 What is the value usually chosen for a filter choke?
Q.26 If the inductance of a choke-input filter is increased, will the output ripple voltage amplitude (E_r) increase or decrease?