The Filter

Once we have rectified the AC output from the transformer we have a series of DC pulses as described in The Rectifier previously. In the U.S. and other places where the power grid operates at 60 Hz, a half-wave rectifier produces 60 pulses per second. In places where the power grid operates at 50 Hz this will be 50 pulses per second.

Unrectified AC. This is about 12 volts RMS. It swings between +16 volts and -16 volts.

Half-wave rectified AC. The negative half of each cycle has been removed, leaving a series of DC pulses. There are 60 pulses per second where the power grid operates at 60 Hz.

A full-wave rectifier fills in between these pulses by flipping the polarity of one half of the sine wave. This produces 120 pulses per second on a 60 Hz power grid and 100 pulses per second on a 50 Hz power grid.

Full-wave rectified AC. The negative half of each cycle has been flipped to positive. This fills in the gaps between pulses. There are now 120 pulses per second where the power grid operates at 60 Hz.

This stream of pulses will work fine for a DC motor or a light bulb, but most electronic circuits require a steady DC voltage. A power supply needs some kind of filter to smooth out these pulses. The power supply we are building will use a large capacitor for this filter.

A capacitor filter is like a dam on a river

The capacitor filter acts a lot like a dam used to control a river. In the western United States we have the Colorado River. During the snow melt, spring rains and summer monsoon the Colorado River used to be a raging torrent from Colorado to the Gulf of California. During the fall and winter it became little more than a creek. In 1936 the U.S. Bureau of Reclamation finished Boulder Dam (renamed Hoover Dam) across the Colorado River near the village of Las Vegas, Nevada. The Colorado River backed up behind the dam forming Lake Mead. This makes water available year round instead of only in the wet season.

Hover Dam with Lake Mead. The dam and lake act like a filter capacitor in a filtered unregulated power supply.

If no water were released from Lake Mead, it would rise to the top of the dam and remain there (with the excess running over the spillway). However, water is steadily released from the lake for irrigation and to produce electricity. During the dry season this causes the height of the lake to drop steadily, only to be refilled during the wet season. However, there is always some water in the lake to be released when needed.

A power supply's filter capacitor acts much like a dam. When a pulse of electricity comes from the rectifier the capacitor charges to the highest voltage received from the rectifier (this is near the peak voltage of the sine wave). If no electricity were taken from the capacitor it would remain charged to this voltage. Now you can attach a circuit across the capacitor just as you can connect a circuit to a battery. Electricity now flows from the positive side of the capacitor, through the circuit (aka, the load) to the negative side of the capacitor. This removes some electricity from the capacitor (partially discharges the capacitor) and the voltage stored in the capacitor will drop. However, when the next pulse arrives from the rectifier, the capacitor is refilled. This is just like Lake Mead getting refilled during the wet season.

Let's take a look at a practical filtered power supply. First let's look at one using a half-wave rectifier.

A filtered unregulated power supply using a half-wave rectifier.

Once connected to an AC power source (such as a power receptacle from the power grid) a train of DC pulses will come from the rectifier diode. The first pulse will charge the capacitor to the peak voltage of the sine wave coming from the receptacle. The voltage in the capacitor will remain at this voltage indefinitely if no load is connected to the power supply. If you could capture this event on an oscilloscope it would appear like the following illustration.

A filtered unregulated power supply using a half-wave rectifier. The first pulse from the rectifier charges the capacitor to the peak voltage of the sine wave and the voltage remains at this level. The voltage from the transformer dropps to 0 volts between pulses (gray lines) but the voltage in the capacitor remains at the peak voltage (+16.3 volts here).

Assumed and real voltages

We make a few assumptions to arrive at the 16.3 volts DC shown above. First we assume that the power from the receptacle is 120 volts RMS. We also assume that the transformer has a 10:1 turns ratio. This would give us 12 volts RMS on the secondary. Multiply this by 1.414 to get the peak voltage of about 17 volts. We lose approximately 0.7 volts across the diode leaving us with approximately 16.3 volts to charge the capacitor. In reality, let's say the transformer is specified with a 117 volt primary and a 12 volt secondary (transformers are not specified by turns ratio). The most common voltage in the U.S. is 117 volts so this matches your transformer. The secondary voltage will probably not be 12 volts. There is some variability in the manufacturing so the secondary voltage will likely be some percentage above or below the specified voltage. In addition, the voltage drop across the diode varies depending on the current flowing through it. With no load on the power supply this current is going to be tiny, so the voltage across the diode will be considerably less than 0.5 volts (remember that the 0.7 volts we expect across a silicon diode is typical, not an absolute). If the power supply is delivering a high current, the voltage across the diode could be 1.5 volts or more (you might find 3 volts across a large power diode with a large current) All this reminds us that the numbers we use in electronics are usually imprecise figures. With that in mind, let's simplify the numbers a bit. Let's start with 16 volts DC across the capacitor just to make the numbers even.

This is the output of the same power supply. The voltage is assumed to be an even 16 volts just for convenience.

Now let's put a load across the output of the power supply (across the capacitor) and see what happens.

Adding a load to the power supply partially discharges the capacitor between DC pulses. This gives us the characteristic sawtooth-shaped ripple as the capacitor quickly charges through the rectifier and discharges more slowly through the load.

The load draws off some of the capacitor's charge and the capacitor partially discharges. When a pulse of voltages comes from the rectifier the capacitor is topped off. This fast charge and slow discharge cycle causes the DC voltage to go up and down. This is called ripple. With a half-wave rectifier this ripple occurs 60 times every second where the power grid frequency is 60 Hz and 50 times every second where the grid frequency is 50 Hz. Therefore, in the U.S., a filtered unregulated power supply with a half-wave rectifier will have a ripple frequency of 60 Hz.

If we increase the load (by decreasing the load resistance or impedance) the capacitor will discharge to a lower voltage during the ripple cycle.

A heaver load causes the capacitor to discharge more between pulses. The capacitor discharges to a lower voltage before it is topped off.

If the load is heavy enough the capacitor may not be able to charge to the peak of the input voltage. If electricity is being removed from the capacitor at a high rate, the rectifier may not be able to deliver enough electricity to make up for it.

If the load is very light the ripple will by small. In this case we can probably use the output of the power supply as is. In that case we have a finished power supply. Battery chargers, for example, are usually quite tolerant of ripple. However, some circuits can't tolerate ripple. Particularly, audio circuits will produce an audible hum if there is ripple coming from the power supply (troubleshooting hint: hum in your speakers could indicate a bad filter or bad regulator in your power supply). Video circuits will produce light and dark horizontal bands in the image if there is ripple in the power supply. These circuits usually need regulated power supplies. This will be discussed later. For now let's look at a filtered unregulated power supply with a full-wave rectifier.

Power supply with a full-wave rectifier

A filtered unregulated power supply using a full-wave rectifier.

A full-wave rectifier reverses the polarity of half of the sine wave coming from the transformer. This fills the gaps between the pulses that come from a half-wave rectifier, giving us twice as many pulses per second. A full-wave rectifier on a 60 Hz power grid will produce 120 pulses per second.

DC pulses from a full-wave rectifier. There are twice as many pulses as from a half-wave rectifier.

The main advantage of a full-wave rectifier is that it reduces the ripple caused by loading the power supply. Let's compare the ripple when the two types of power supplies are lightly loaded.

Ripple of a filtered unregulated power supply with a half-wave rectifier (top) and a full-wave rectifier (bottom).

Notice that the ripple in the lower diagram (full-wave rectifier) is noticeably less than with the upper diagram (half-wave rectifier). This is even more dramatic with a heavy load.

Same power supplies as above, but with a heavy load. There is less ripple with a full-wave rectifier.

Notice that the voltage doesn't drop nearly as low with the full-wave rectifier (lower illustration). It is boosted back to the peak voltage before it has a chance to drop as low.

Disadvantages of a power supply with a full-wave rectifier

A full-wave bridge rectifier requires three more diodes than a half-wave rectifier. This cost is trivial compared to the advantages. However, if you are producing a million power supplies, using a half-wave rectifier would save anywhere from $30,000 to $120,000 over the production run compared to a full-wave bridge rectifier. In such a case, where every penny can count big, you may use a half-wave rectifier if it is adequate for the product being produced.

Another disadvantage is that there are always two diodes in series with the load. This doubles the voltage loss between the transformer and the load. This increases the typical loss from 0.7 volts to 1.4 volts. It can actually be anywhere from 0.6 volts to 6 volts (each diode may drop 0.3 to 3 volts), depending on how much current the load is using. This may require a transformer with a slightly higher secondary voltage than a half-wave rectifier would require.

Where to ground a power supply with a full-wave rectifier

Review from What is Ground (part 1) and What is Ground (part 2). Ground is the point in a circuit that you decide is zero volts and you measure all other voltages compared to that point. One of the most common points to designate as ground is the lowest possible voltage in the circuit. This not true for power supplies with a full-wave rectifier. Here you want ground (assuming the system has a negative ground) to be connected directly to the negative side of the load. This should make sense because you want to control the voltage across the load. A power supply with a full-wave rectifier has voltages that are lower than the voltage at the negative side of the load. Let's assume we are correctly using the negative side of the load as ground and see what the voltages are around the power supply circuit.


Here we are assuming that the transformer secondary voltage peaks at 17.4 volts and we are at that peak voltage now. We are in the "positive" half of a cycle, meaning the conventional current is flowing in a clockwise direction in the transformer secondary. Therefore the current is flowing to the right out of the top of the transformer. The upper-right diode is forward biased and we are losing 0.7 volts as the current passes through this diode. This leavs +16 volts at the top of the load. The bottom of the load is 0 volts because that is where the black lead of the voltmeter is. We still have a diode to go through before we reach the bottom connection of the transformer secondary. We expect to lose another 0.7 volts across this diode. That means the voltage at the bottom of the secondary will be -0.7 volts. How do we end up with less than 0 volts? If this is unclear to you please review What is Voltage, What is Ground (part 1) and What is Ground (part 2). Also review the video lecture Measuring Voltage. The bottom of the load is 0 volts because that is where the black lead of the voltmeter is. This is not the lowest voltage in the circuit. Any lower voltages are negative voltages. Therefore, the voltage at the bottom of the secondary is 0.7 volts lower than the negative side of the load and appears as -0.7 volts on the voltmeter. Again, we want to control the voltage across the load, so we measure our voltage with the negative side of the load as ground. In practical application, any point in the circuit that is directly connected to the negative side of the load can be used as a ground connection.

Ground can be anywhere that is directly connected to the negative side of the load. That is anywhere along the darkened line above.

What would happen if you grounded the bottom side of the transformer secondary? Recall that ground is a reference voltage that must not change when the current changes. The voltage at the bottom of the secondary will change compared to the voltage at the negative side of the load as the current changes. When there is very little current flowing, the voltage across the diode will be less than 0.5 volts. However, increase the load and the voltage across the diode will be around 0.7 volts. With a heavy load it could be 1 volt or more (for high current diodes). Therefore, the voltage at the bottom of the transformer secondary can be anywhere from 0.3 volts to 1 volt or more lower than the negative side of the load. If you make the bottom of the secondary your ground, the negative side of the load will vary from +0.3 volts to +1 volt or more, when it should always be 0 volts. I am at a loss to explain when this would be a good thing. Therefore, do not do the following:

Bad ground. DON'T DO THIS..

That's better.

Choosing the capacitor

Choosing the correct capacitor for a filter is a process of trial and error. The capacitor must be large enough that it doesn't discharge too much between pulses, but not so large that it doesn't charge fully with each pulse. Actually, if smoothing the ripple is more important than maximum voltage output, an oversized capacitor may be desireable. Tens of thousands of microfarads, or even hundreds of thousands are not uncommon. However, small power supplies may have 1,000 microfarads or less. Some large power supplies may have many capacitors connected in parallel to get enough capacitance to do the job.

Inductive filtering

The power supply we are building for our project uses a capacitor filter. Filtered unregulated power supplies may also use inductive filtering. An inductor in series with the load will tend to block changes in current (see  DC Circuits, Part 5, Lesson 2). This tends to smooth the inherent ripple of the unregulated power supply. You may hear that the inductor cuts off the peaks of the ripple where the capacitor fills in the valleys. This is, at best, an over simplification. The following circuit is a filtered unregulated power supply with a typical capacitor-inductor filter.

A filtered unregulatged power supply using capacitors and an inductor for filtering.

The above filter is called a 'Pi' filter because the components are arranged in the shape of the Greek letter Pi.

A filtered unregulated power supply is a finished circuit. It is perfectly suitable to power circuits that can handle a supply voltage that drops with current demand (when the load increases) or can tolerate ripple. Next we will add a regulator that will completely remove the ripple and greatly reduce the voltage drop when the load increases.



Building a Linear Power Supply - Part 2 - The Rectifier and Filter