POWER SUPPLIES

A power supply (PS) is an electronic circuit that converts alternating current (AC) to direct current (DC). Most types of electronic equipment require direct current. In some cases, small transistor TV’s & radios, MP3 players, etc., the direct current is supplied by a battery. More often however, electronic devices are operated from the AC power line. In these cases, a special circuit must convert the alternating current supplied by the power company to the direct current required by the electronic device. We will study several different circuits which are used for this purpose.

The heart of the power supply is the rectifier circuit, which converts the AC sine wave to a pulsating DC voltage. This is the first step in producing the smooth DC voltage required by electronic circuits. We will consider three different types of rectifier circuits. These can be classified as Half-wave rectifiers, Full-wave rectifiers, and Bridge rectifiers. Because it is the simplest, we will consider the Half-wave rectifier first.

Most types of electronic equipment get their power from the 115 V AC, 50 Hz power line. The most convenient way for power companies to mass produce and distribute electricity is by AC.

The waveform produced by the power companies is a sine wave. The average value of a sine wave, as shown in Figure 4-1A, is zero volts because it has equal positive and negative alternations. To produce a net positive or negative voltage, it is necessary to distort the sine wave. For example, if we clip off the negative alternation as shown in Figure 4-1B, the resulting waveform will have a net positive value. This is the principle behind the Half-wave rectifier.

BASIC CIRCUIT

The most basic form of the Half-wave rectifier is shown in Figure 4-2A. The load which requires the direct current is represented by the resistor. A silicon diode is placed in series with the load so that current can flow in one direction but not in the other.

Figure 4-2B illustrates the operation of the circuit during the positive half cycle of the sine wave from the power line. The anode of the diode is positive. Consequently, the diode conducts, allowing current to flow through the load as shown. Since the diode acts as a closed switch during this time, the positive half cycle is developed across the load.

Figure 4-2C shows the next half cycle of the input sine wave. Here, the anode is negative and the diode cannot conduct. It acts as an open switch. Consequently, no current flows through the load and no voltage is developed across the load. As you can see, the AC sine wave is changed to a pulsating DC voltage.

Of course, the pulsating DC voltage is not suitable for most loads. Remember, this is only the first stage of the power supply. Later stages smooth out the ripples and change the pulsating DC to a steady DC.

EFFECTIVE, PEAK, AND AVERAGE VALUES

Figure 4-3 compares the effective, peak, and average values of the waveforms associated with the Half-wave rectifier. As explained in an earlier unit, AC voltages are normally specified in terms of their effective RMS values. Thus, when we speak of the 115 V AC power line, we are specifying an effective or RMS value of 115 VAC. In terms of peak values:

Vrms = Vpeak x 0.707 ( 1 / Ö2 = 1 / 1.414 = 0.707)

The peak value is always higher than the RMS value. In fact:

Vpeak = Vrms x 1.414

Therefore, if the RMS value is 115 VAC, then the peak value must be:

Vpeak = Vrms x 1.414 = 115 x 1.414 = 162.6 V (using 110 VAC è 155.5 V)

The average value of the sine wave is 0 volts. Figure 4-3B shows how the average value changes when the negative half cycle is clipped off. Since the waveform swings positive but never negative, the average voltage is positive. For this type of waveform, the average voltage (Vave) is determined by the equation :

Vavg = Vpeak / p = Vpeak x (1 / 3.14) = Vpeak x 0.318

Thus, in our example:

Vavg = 162.6 x 0.318 = 51.7 V

If we ignore the small voltage drop across the conducting diode, the output voltage across the load will appear as shown in Figure 4-3B. The voltage drop across the silicon diode is about 0.6 volts. Compared to the other voltages, this tiny voltage is negligible.

The average voltages specified in this section pertain to the rectifier itself. When a filter is added at the output, these values change. Since the rectifier is normally operated with a filter, the average voltages here will not be the average voltages from the overall power supply. As used in this section, the average voltage simply gives us a convenient way of comparing rectifier circuits.

RECTIFIER WITH TRANSFORMER

The basic rectifier circuit always produces the same peak and average voltages when connected across the 115 V AC line. Obviously, this is a disadvantage since some devices require higher or lower voltages.

This problem can be overcome by connecting a transformer between the AC line and the rectifier. If higher voltages are required, a step-up transformer can be used. When lower voltages are required, as with most transistorized equipment, a step-down transformer can be used.

A TV receiver may require DC voltages as low as + 5 volts and as high as +50,000 volts. Transformers are used with rectifiers to produce this wide range of voltages.

Figure 4-4 shows a rectifier used with a step-down transformer. The step-down ratio is five to one (115:23). Therefore, the effective voltage across the secondary is :

115 / 5 = 23 VAC

When rectified, this gives a peak value of:

Vpeak = Vrms x 1.414 = 23 x 1.414 = 32.5 V

The average value of the output is:

Vavg = Vpeak x 0.318 = 32.5 x 0.318 = 10.3 V

By using the proper step-up or step-down ratio, we can make the peak or average voltage of the rectified output any value we like.

ISOLATION

Another purpose of the transformer is to isolate the electronic device from the power line. To see why this is important, let's consider a situation in which a transformer is not used.

In some devices, the rectifier is connected directly to the power line. For reasons of economy, a power transformer is not used. In devices of this type, one side of the metal chassis is connected directly to one side of the power line. Figure 4-5A illustrates an oscilloscope and a signal generator which are constructed in this manner. Unless the metal chassis is completely insulated from the user, a dangerous shock hazard exists. Since the two-prong plug can be connected two ways, we never know which side of the AC line is connected to the chassis. Anyone touching the chassis and ground simultaneously could receive a dangerous shock.

An even worse situation is shown in Figure 4-5B. Here, the two pieces of test equipment are connected to the AC line. If one of the two-prong plugs is reversed, the chassis of the oscilloscope will be connected to one side of the line while the chassis of the generator will be connected to the other side of the line. The voltmeter shows that a 115 volt difference in potential exists between the two devices. Anyone who touches both chassis simultaneously will find himself connected directly across the 115 V AC line.

If transformers are used in both power supplies, no hazard will exist. The power lines connect only to the primary of the transformer and not to the chassis. Even if one side of the secondary connects to the chassis, the chassis is isolated from the line by the transformer.

Transformerless power supplies are not uncommon, especially in low cost consumer products. They are perfectly safe if the "hot" chassis is completely insulated from the user by plastic cabinets, plastic knobs, etc. But while these devices present no hazard to the user, they can still be dangerous to anyone who must service them. To repair such a device, the protective plastic cabinet must be removed. If you ever work on a device of this type, protect yourself by using an isolation transformer between the AC line and the chassis. Recall that an isolation transformer has a turns ratio of 1 to 1. Thus, it takes 115 VAC from the line and delivers 115 VAC to the chassis. However, it isolates the chassis from the AC line and helps protect against accidental shock.

RIPPLE FREQUENCY

The Half-wave rectifier gets its name from the fact that it operates only during one half of each input cycle. Its output consists of a series of pulses. The frequency of these pulses is the same as the input frequency. Thus, when operated from the 50 Hz line, the frequency of the pulses will be 50 Hz. This is called the ripple frequency.

OUTPUT POLARITY

A rectifier can produce either a negative or a positive output voltage. The output polarity depends on what point is connected to ground and which way the diode is connected.

In Figure 4-6A, current can flow only in the direction indicated. This produces a positive voltage at the top of the load with respect to ground. Thus, this rectifier produces a positive output voltage.

If we need a negative output voltage, the diode can be turned around as shown in Figure 4-6B. Or, ground can be connected to a different point in the circuit as shown in Figure 4-6C. Either of these arrangements will produce a negative output voltage. However, if both changes are incorporated, the output voltage is positive again as shown in Figure 4-6D.

The Half-wave rectifier has a serious disadvantage. It allows current to flow through the load for only one-half of each cycle. As you will see later, this makes filtering more difficult. A circuit which overcomes this problem is the Full-wave rectifier.

The Full-wave rectifier circuit is shown in Figure 4-7. It uses two diodes and a center-tapped transformer. When the center-tap is grounded, the voltages at the opposite ends of the secondary are 180º out of phase with each other. Thus, when the voltage at point A swings positive with respect to ground, the voltage at point B swings negative. Let's examine the operation of the circuit during this half cycle.

Figure 4-7A shows that the anode of D1 is positive while the anode of D2 is negative. Obviously, only D1 can conduct. As shown, current flows from the center tap up through the load and D1 to the positive potential at the top of the secondary. When D1 conducts, it acts like a closed switch so that the positive half cycle is felt across the load.

Figure 4-7B shows what happens during the next half cycle when the polarity of the voltage reverses. The anode of D2 swings positive while the anode of D1 swings negative. Therefore, D1 cuts off and D2 conducts. As shown, current flows from the center tap, through the load and O2 to the positive voltage at the bottom of the secondary.

Notice that current flows through the load during both half cycles. This is the prime advantage of the Full-wave rectifier over the Half-wave rectifier.

For simplicity, assume that the transformer has a 1 to 1 turns ratio. That is, it neither steps-up nor steps-down the applied voltage. Therefore, the voltage across the entire secondary is 115 VAC. However, the voltage at each end of the secondary with respect to the center tap is only one-half this value, or 57.5 VAC. This is shown in Figure 4-8.

Figure 4-9A further illustrates the voltage present in one-half of the Figure 4-8 secondary. The specified value of 57.5 VAC is the RMS, or effective, value. Consequently, the peak value is :

Vpeak = Vrms x 1.414 = 57.5 x 1.414 = 81.3 V

The voltage in the other half of the secondary has exactly the same amplitude but is 180º out of phase.

Ignoring the tiny voltage drops across the diodes, the output waveform of the Full-wave rectifier appears as shown in

Figure 4-9B. Compare this with the output of the Half-wave rectifier shown earlier in Figure 4-3B. In this example, the peak voltage of the Full-wave rectifier is one-half the value of the peak voltage of the Half-wave rectifier. Nevertheless, the average voltages of the two are the same. Although the peak amplitude is only one-half as high with the Full-wave rectifier, the pulses occur twice as often. Thus, the formula for finding the average voltage from a Full-wave rectifier is :

Vavg = 2 x ( Vpeak / p ) = Vpeak x 0.636 = 81.3 x 0.636 = 51.7 V

As the name implies, the Full-wave rectifier works during both half cycles of the input sine wave. This gives a ripple frequency twice as high as the Half-wave rectifier. When operating from the 50 Hz line, the ripple frequency is 100 Hz. This higher ripple frequency is easier to filter than the 50 Hz ripple produced by the Half-wave rectifier.

Figure 4-10 illustrates a disadvantage of the Full-wave rectifier. For a given transformer, the Full-wave rectifier produces a peak output voltage which is half that of the Half-wave rectifier. In some cases this can be a disadvantage. Remember that a filter is added to the output of the rectifier circuit. As you will see later, when a filter is added, the average output voltage approaches the peak voltage. Thus, with a filter, the Half-wave rectifier can produce a higher output voltage than the Full-wave rectifier.

Fortunately, there is a type of rectifier which produces the same peak voltage as the Half-wave rectifier and the same ripple frequency as the Full-wave rectifier. Let's take a look at this circuit.

BRIDGE RECTIFIER

The Full-wave rectifier provides a direct current through the load on both half-cycles of the sine wave. This is a definite advantage over the Half-wave rectifier. However, for a given transformer, the average DC voltage output is no greater in the Full-wave rectifier than it was in the Half-wave rectifier. The bridge rectifier overcomes this disadvantage.

The bridge rectifier circuit is shown in Figure 4-11. It consists of four diodes arranged so that current can flow in only one direction through the load. This circuit does not require a center-tapped secondary as the Full-wave rectifier did. In fact, it does not require a transformer at all except for isolation and to provide a voltage other than that available from the line. Leads A and B could connect directly to the two-prong plug.

Figure 4-12A shows how current flows on the positive half cycle of the sine wave. Current flows from the bottom of the secondary through D1, the load, and D2 to the positive voltage at the top of the secondary. With D1 and D2 acting as closed switches, the entire secondary voltage is developed across the load.

On the next half cycle, the polarities reverse as shown in Figure 4-12B. The top of the secondary is now negative and the bottom is positive. Current flows from the top of the secondary through D3, the load, and D4 to the positive voltage at the bottom of the secondary. Notice that the current flow through the load is always in the same direction. With D3 and D4 acting as closed switches, the entire secondary voltage is again developed across the load.

Figure 4-13 shows the input and output waveforms of the bridge rectifier circuit. For the sake of explanation, again assume that the transformer has a turns ratio of 1 to 1. That is, assume that the secondary voltage is the same as the primary voltage, 115 VAC.

Figure 4-13A shows that the input sine wave has a peak voltage of 162.6 V. Ignoring the voltage drops across the conducting diodes, the voltage pulses developed across the load will have this same peak value as shown in Figure 4-13B. Compare this output with the outputs of the Half-wave rectifier (Figure 4-3B), and the Full-wave rectifier (Figure 4-9B). Notice that it has the same shape as the output of the Full-wave rectifier. However, the amplitude of the pulse is the same as with the Half-wave

rectifier.

As with the Full-wave rectifier, the average voltage is determined by the formula :

Vavg = Vpeak x 0.636

Also, the ripple frequency is the same as with the Full-wave rectifier or 100 Hz.

Strictly speaking, the bridge rectifier is a type of Full-wave rectifier since it operates on both half cycles of the input sine wave. However, in this unit, the name Full-wave rectifier will refer to the circuit shown in Figure 4-7. It uses two diodes and requires a center-tapped transformer.

Now let's compare the three basic rectifier circuits. Each has its own advantages, disadvantages, and characteristics.

HALF-WAVE RECTIFIER

Here, the advantage is simplicity and low cost. It requires only one diode and may be used with or without a transformer. It has several disadvantages. First, it is not very efficient since only half of the input wave is used. The average output voltage is low and the 50 Hz ripple frequency is hard to filter. When used with a transformer, the unidirectional secondary current can cause problems. Since current always flows in the same direction in the secondary, the core of the transformer can become magnetized. This is called DC core saturation. It tends to reduce the inductance and, therefore, the efficiency of the transformer. The Half-wave rectifier is generally restricted to low current applications, or to transformerless supplies, where economy is a prime factor.

FULL-WAVE RECTIFIER

The Full-wave rectifier has several advantages over the Half-wave rectifier. It is more efficient since it operates on both half cycles of the sine wave. Also, the 100 Hz ripple frequency is easier to filter. Since the currents in the two halves of the secondary are opposite, there is no problem with DC core saturation in the transformer.

A disadvantage of the Full-wave rectifier is that it requires a center-tapped transformer. Furthermore, the transformer must be somewhat larger than that required by a bridge rectifier. Another problem which we will discuss later involves the peak inverse voltage of the diodes. And finally, for a given transformer, the peak voltage is lower in the Full-wave rectifier than in the Half-wave rectifier.

BRIDGE RECTIFIER

The bridge rectifier has several advantages over the Full-wave rectifier. For one thing, it can be operated without a transformer. More often though, a step-up or step-down transformer is used to provide a voltage other than that available from the AC line. Even then, the design of the transformer is simplified because no center tap is required. Also, for a given transformer, the output voltage from the bridge is higher. Of course, the bridge rectifier does require four diodes, but diodes are relatively inexpensive and this is not a serious disadvantage. For these reasons, the bridge rectifier is very popular.

POWER SUPPLY FILTERS

As we have seen, the output of the rectifier is a pulsating DC voltage. Such a DC voltage is unsuitable for nearly all electronic applications. Most electronic circuits require a very smooth; constant supply voltage. For this reason, in virtually all power supplies, the rectifier circuit is followed immediately by a Filter. The purpose of the filter is to convert the pulsating DC provided by the rectifier into the smooth DC voltage required by electronic circuits.

THE CAPACITOR AS A FILTER

In its simplest form, the power supply filter may be nothing more than a capacitor connected across the output of the rectifier. This arrangement is shown in Figure 4-15A. The addition of the capacitor modifies the operation of the circuit.

The input voltage waveform is shown in Figure 4-15B. Notice that the peak value is about 162 volts. Let's examine the action starting with the point at which the sine wave initially swings positive.

When the anode of D1swings positive, D1 conducts, allowing current to flow through the load. Simultaneously, the capacitor charges to the polarity shown. Since there is negligible resistance in the charge path, C1 charges immediately in response to the rising input voltage. C1 charges for the first one-quarter cycle. After 90° of the input sine wave, C1 is charged to the peak value, or about 162 volts.

When the input sine wave begins to drop off, C1 tries to discharge. However, the only discharge path for C1 is through the load. Because the load has a certain resistance, the discharge of the capacitor is controlled by the RC time constant. The discharge time constant is long compared to the time for one cycle of the AC input. Consequently, after passing its peak at 90°, the AC line voltage drops off faster than the voltage across the capacitor. This means that after the first one-quarter of a cycle, the cathode of D1 will be more positive than the anode. Therefore, the diode cuts off, just after the sine wave passes 90°.

With the diode no longer conducting, the capacitor begins to discharge through the load. Thus, after the first one-quarter cycle, the current through the load is being supplied by the discharging capacitor. As C1 discharges, the voltage across it gradually decreases. However, before the capacitor can completely discharge, the next cycle of the sine wave begins. At some point in the first one-quarter cycle of the second sine wave, the anode of D1 again swings more positive than the cathode. D1 conducts once more, recharging C1 to the peak value.

Figure 4-15C shows the voltage across C1. Notice that the voltage initially rises to the peak value of the input sine wave. At 90°, D1 cuts off and C1 begins its discharge. Because the discharge time is longer than one cycle of the AC sine wave, the voltage across C1 never drops to zero volts. Instead, the capacitor charges again to the peak of the next pulse. This not only helps to smooth out the pulses, but it also raises the average voltage. Figure 4-16 illustrates this point more clearly.

Figure 4-16A shows the output of the Half-wave rectifier when no filter capacitor is used. As you saw earlier, the average voltage is only 31.8% of the peak voltage.

If a filter capacitor is added, the output voltage might appear as shown in Figure 4-16B. Here, the capacitor prevents the output from dropping to 0 volts. Therefore, the average output, voltage is higher.

If a larger capacitor is added, the RC time constant is increased. Consequently, the capacitor discharges more slowly. This raises the average voltage even higher as shown in Figure 4-16C. As the value of the capacitor is made larger, the average output voltage approaches the peak voltage.

In Figure 4-16, the shaded area represents the time that the diode conducts. It indicates the time that the diode is forcing current through the load. When a capacitor is added across the output, the diode conducts for a shorter period of time. Between the times that the diode is conducting, the capacitor supplies the current through the load. For the Half-wave rectifier, the capacitor must supply the current to the load for more than three-quarters of each cycle. If the current required by the load is high, a very large value of capacitance is required.

FULL-WAVE SUPPLY WITH CAPACITOR FILTER

So far we have considered filtering the output of a Half-wave rectifier. Now let's see how the capacitor behaves with the Full-wave and bridge rectifiers.

Figure 4-17A shows the unfiltered output of a Full-wave (or bridge) rectifier. The shaded area shows that one diode or the other is conducting at all times. Thus, the ripple frequency is twice as high as that of the Half-wave rectifier. When a capacitor is added, it charges to each peak. Since the peaks occur twice as often, the capacitor does not discharge very far before the next pulse occurs. As shown in Figure 4-17B, the average voltage is quite high. If a larger capacitor is added, the average voltage can be made almost equal to the peak voltage. Figure 4-17C illustrates this.

The shaded areas show when the diodes conduct. Making the capacitor larger reduces the time that the diodes conduct. Thus, the capacitor must still provide the current to the load for much of each cycle. However, with the Full-wave rectifier, the capacitor is charged twice during each cycle. Therefore, a given capacitor does a much better job of filtering in a Full-wave supply than it does in a Half-wave supply.

PERCENT RIPPLE

The purpose of the power supply filter is to smooth out the pulsating DC produced by the rectifier. There is a generally accepted figure of merit which tells how good the filter does its job. This figure of merit is called the Percent ripple.

% ripple = ( RMS of ripple / Vavg ) x 100%

When using this equation, the ripple in the output is assumed to be a sine wave. While it is not really a sine wave, the ripple does have a nearly symmetrical wave-shape, so our equation gives fairly accurate results.

Figure 4-18 shows the output of a rectifier-filter network. To compute the percent ripple of this waveform, first measure the peak-to-peak value of the ripple. As you can see, the ripple voltage has a peak-to-peak value of :

80 V – 60 V = 20 V

Therefore, the peak value must be one-half of this value, or 10 V. You can compute the RMS value of the ripple voltage by multiplying the peak by 0.707.

Vrms = 10 x 0.707 = 7.07 V

You can obtain the approximate average DC voltage by taking the value midway between the upper and lower values of the ripple.

In the example shown, the average DC voltage is 70 volts. Therefore, the percent ripple is :

% ripple = ( RMS of ripple / Vavg ) x 100 = (7.07 / 70 ) x 100 = 10.1%

The percent ripple can be made lower if a larger capacitor is used or if the value of the load resistance is increased. Since the load resistance is normally dictated by the circuit design, the amount of ripple is controlled by the value of the capacitor.

When you know the characteristics needed in the power supply, you can determine the value of filter capacitor required. The general equation for a Full-wave or bridge rectifier is :

Cmin = 1 / (2.828 x K x R_L x f )

K = percent ripple expressed as a decimal (ie. 12% = 0.12)

R_L = the resistance of the load in ohms

f = the frequency of the ripple in hertz

Cmin = the minimum capacitance value required in farads

During the next half cycle, the voltage across the secondary reverses. At the peak of the negative cycle, the anode of the diode is at -200 volts with respect to ground. At this point, the difference of potential across the diode is twice the peak value of the secondary, or 400 volts. Obviously, a diode must be selected which can withstand this voltage.

The maximum voltage that a diode can withstand when reverse biased is called the peak inverse voltage or PIV. The diode used in this example must have a PIV of 400 volts. Actually, since we do not want to push the diode to its limit, we should choose a diode with a somewhat higher PIV rating. A good rule of thumb is to operate the diode at no more than 80%, or 0.8, of its rated value. Thus, the PIV rating should be no lower than :

PIV = 400 / 0.8 = 500 V

Figure 4-19B shows that a similar situation exists in the Full-wave rectifier. With C1 charged to the positive peak and the top of the secondary at the negative peak, D1 experiences twice the peak voltage.

Interestingly enough, this situation does not hold true for the bridge rectifier. Here the diodes are never exposed to more than the peak of the secondary voltage. Figure 4-19C illustrates why. At first it might appear that D2 is being subjected to twice the peak value. However, closer examination will reveal that D3 is conducting. Thus, point A is effectively at ground potential and the voltage across D2 is equal to Vpeak. If you examine the circuit in various conditions, you will find that no diode is ever exposed to more than Vpeak. Therefore, another advantage of the bridge rectifier is that diodes with lower PIV ratings can be used.

RC FILTERS

The single capacitor filter is suitable for many, non-critical, low-current applications. However, when the load resistance is very low, or the percent of ripple must be held to an absolute minimum, the capacitor value required may be extremely large. While electrolytic capacitors are readily available in sizes up to 10,000 µF or larger, the larger sizes are quite expensive. A more practical approach is to use a more sophisticated filter that can do the same job but with lower capacitor values.

Figure 4-20 shows a Full-wave rectifier with a resistor-capacitor (RC) filter connected across its output. An RC filter of this type does a much better job than the single capacitor filter. Let's see how this filter works.

C1 performs exactly the same function that it did in the single capacitor filter. It is used to reduce the percent ripple to a relatively low value. Thus, the voltage across C1 might consist of an average DC value of + 100 volts with a ripple voltage of 10 volts peak-to-peak. This voltage is passed on to the Rl-C2 network, which reduces the ripple even further.

C2 offers an infinite impedance to the DC component of the output voltage. Thus, the DC voltage is passed to the load, but reduced in value by the amount of the voltage drop across R1. However, Rl is generally small compared to the load resistance. Therefore, the drop in the DC voltage caused by R1 is not objectionable.

C2 offers a very low impedance to the AC ripple frequency. Thus, the AC ripple sees a voltage divider consisting of Rl and C2 between the output of the rectifier and ground. Component values are chosen so that the resistance of Rl is much greater than the reactance of C2 at the ripple frequency. Therefore, most of the ripple voltage is dropped across R1. Only a trace of the ripple voltage can be seen across C2 and the load.

In extreme cases, where the ripple must be held to an absolute minimum, a second stage of RC filtering can be added. In practice, the second stage is rarely required. The RC filter is extremely popular because smaller capacitors can be used with good results.

The RC filter has some disadvantages. First, the voltage drop across R1 takes voltage away from the load. Second, power is wasted in R1 and is dissipated in the form of unwanted heat. Finally, if the load resistance changes, the voltage across the load will change. Even so, the advantages of the RC filter overshadow these disadvantages in many cases.

LC FILTERS

The next step in filters is the inductor-capacitor (LC) filter. A common type of LC filter is shown in Figure 4-21. C1 performs the same functions as discussed earlier. It reduces the ripple to a relatively low level. L1 and C2 form an LC filter which reduces the ripple even further.