By Herb Friedman, W2ZLF
HOW TO UNDERSTAND THE DECIBEL WITHOUT HIGHER MATH:
The Decibel frightens many electronic hobbyists, who see it as something dark and mysterious. But if you can grasp Ohm's Law, you should have no trouble understanding the decibel.
In the early days of electronics the decibel was confined almost exclusively to the world of broadcast and telephone engineers. But today the concept is used just about everywhere and it's inescapable in the areas of audio, ham radio, and Citizens Band radio.
IT'S JUST A RATIO
Telephone engineers found that the human ear perceives sound-level changes on a proportaional, or logarithmic scale rather than linearly. It is on this characteristic of the ear that the concept of the decibel is based.
For example, let's say you are listening to a hi-fi system when its output power is 10 watts. Logically you'd think that boosting the output to 20 watts would make the sound twice as loud, but it wouldn't. You'd notice only a slight increase in volume. In fact, it would take nearly a ten-fold power increase for your ear to sense the volume! Going a step further, the power would have to be increased 100--from 10 to 1000 watts--for your ear to notice four times the volume. When a sound increases in intensity you receive an impression of increased loudness approximately equal to the (logarithm) of the ratio of the two electrical powers.
The (bel) is the name given the unit that expresses the logarithm of this ratio. In mathematical language, this is written as bel=log P1/P2, where P1 is the larger power and P2 is the smaller power. In working with the formula you first divide the larger power, P1, by the smaller power, P2. The number you get is the ratio of the two. Take this number, look up its log in a table of logarithms and you'll have a reading in bels, or effective increase in loudness.
Let's try an example. Assume P1 is 48 watts and P2 is 6 watts. Forty-eight divided by 6 is 8, of course, and the logarithm of 8 is 0.9030. Stated another way, the ratio in bells of 48 watts to 6 watts is 0.9030. However, the bel is a fairly large unit, so the (decibel), which is a tenth of a bel and is usually referred to by the abbreviation db, is used instead. This modifies the formula slightly, making it db=10 log P1/P2. Using this formula in the previous example would give an answer of 9.030 db, just ten times as great. Whether the answer represents a db loss (-db) or a db gain depends on the circuit you are working with. We'll have more about this later.
Though the db is a (ratio) of two powers, it also can be used to express absolute power, provided we have a reference or zero-db level. (Take a look at microphone specs in an electronics catalog and you'll see the output stated as just -55db, -48 db, etc.) In the early days of radio, many different reference levels were established, and there were almost as many different meters used to indicate these levels. By 1939, audio engineers had decided to do something about the lack of standardization. They came up with what is called the VU (volume unit) meter and a standard reference level of 1 millliwatt (.001 watts) in a 600-ohm circuit.
This standard reference is referred to as 0dbm, which is another way of saying 0db is synonymous with 1 milliwatt of power output. A statement that power output is 10dbm is talking about a power 10db more than 1 milliwatt. The VU meter is calibrated so when connected to a 600-ohm circuit and when measuring sine-wave power, 0 VU corresponds to 0dbm and 3 VU corresponds to 3dbm. (The mike referred to with a -55 db output, you see, gives you a fairly weak signal.) We'll show later how greatly a 1 or 2db difference in mike output can affect amplifier power output.
The db is also used to deal with the output power of transmitters, receivers and antennas, since this power is ultimately delivered to a speaker.
You can solve most db problems without bothering with logs if you use the table shown with this article. (At the end of the series) Let's work a problem to see how it's done. One day you decide your 10-watt hi-fi amplifier can be souped up a little. You change a few tubes, alter the circuit slightly and, with the same input as before, the amplifier now delivers 20 watts. The ratio of the power change (P1/P2) is 20/10 or 2. Look down the column in the table headed (Power Ratio) until you reach 1.99 (this is close enough to 2 for our purposes). In the left column you see 3db. Therefore, the increase in output power from 10 to 20 watts amounts to a 3 db improvement. If you had started with a 20-watt amplifier and reduced the output to 10 watts, the power ratio would still be 2, but the now the 3db change represents a loss. It would be expressed as -3db.
Let's try a more difficult problem. Assume you are using a public address amplifier that is capable of delivering 25 watts of power. Your microphone, which has a -57db output level reference to 1 mw, drives the amplifier to only 10 watts output when the gain control is set to maximum. You decide to purchase a new microphone. What should its output be to get the amplifier to deliver a full 25 watts? First, we find the ratio of the power we WANT to the power WERE GETTING, which is 25 watts/10 watts, or 2.5. From the table we find that a power ratio of 2.5 equals 4db. Therefore, we need a microphone with an output 4db greater than -57db. The final answer, of course, is a microphone with an output of -53db (subtacting 4db from -57db.)
At this point you may wonder about intermediate values above 10db. In the table the values after 10db are 20, 30, 40, 50, etc. How do you determine the power ratio of 16db? It's simple. Split 16db into 10db and 6db, find the corresponding power ratios, and then MULTIPLY.
Now let's determine the DB GAIN of (two) amplifiers. But first, another important rule is that logarithms are added, not multiplied. A power GAIN of 4 equals a power RATIO of 4 which, from the table equals 6db. A power GAIN of 10 equals a power RATIO of 10, which equals 10db. By adding the logs, the total db gain is 6db+10db, or 16db. To do it another way, try splitting 16db into 8db and 8db. The power ratio corresponding to 8db is 6.31. So 6.31 X 6.31=40.
To work the other way, that is, to go from power ratio (gain) to db, divide the ratio into convenient units. A power gain of 40 equals 8 X 5. Under the column headed POWER RATIO in the table, note that a power ratio of 7.94 (sufficiently close to 8) equals 9db. A ratio of 5.01 (close to 5) equals 7db. Adding 7db and 9db we get 16db.
THE VOLTAGE RATIO
Instead of power, you can use voltages and the VOLTAGE RATIO column when determining ratios. However, the input and output impedances of the equipment you're checking MUST BE THE SAME. Now, notice in the table that 3db represents a POWER ratio of 1.99 but that 6db represents a VOLTAGE ratio of 1.99. We won't go into the math required to explain this but, when dealing with voltage or current, our early formula becomes db=20log E1/E2, where E1 is the larger voltage and E2 is the smaller voltage.
If everything in electronics were rated in db with the same reference level, db problems would not be difficult. As a general rule, most equipment IS rated in terms of the dbm reference level. The problem-children, however, are microphones, since their outputs may refer to different reference levels.
Some microphone output ratings are given in reference to 1 mw (0dbm). Some are rated in dbv (1 volt), and some are in reference to 1 millivolt. Some, judging from performance, have heaven-knows-what for a reference level. But, regardless of the reference level, the important thing to bear in mind is that since the db is just a ratio, it doesn't matter what reference is used as long as you consistently use the same one. If a microphone with a -57dbm rating driving an amplifier's output to 1 watt is replaced with another micropphone with a -54dbm output level (a 3db difference or a power ratio of 2), the amplifier will now deliver 2 watts.
If a microphone having a -60dbv rating driving an amplifier to 10 watts is replaced with a microphone having a -57dbv rating (again a 3db difference), other words, the power ratio always will be the same, regardless of the reference used, so long as both microphones have the SAME reference.
Most newcomers to amateur radio say the more power the better--if 100 watts output is good, 1 kilowatt is likely to be ten times better. This is not always true. For example, let's say the station your're listening to is running a "full gallon" (1 kilowatt) and you read him at 20db over S-9. (The S-meter reflects power, and each S unit nominally equals 6 db.) In the table you find that 20db is equivalent to a power ratio of 100. Now, by dividing the original 1,000 watts by 100 you get 10 watts. Tell the other station to cut back his power to 10 watts. You willl now receive him at S-9, still a powerful signal, with a considerable saving in power and tube life. That extra 20db was being wasted.
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