Decibels (dB) are a mysterious, mathematically complex concept for many people. 

Because the math behind decibels is slightly more advanced than the basic stuff, many folks unfortunately shy away from using decibels to their disadvantage. Cheat sheets, such as our handy decibel chart, can make it easier to employ this helpful tool, but even with such aids, folks without a basic understanding of the concept can still get lost.

My goal is to help you understand how decibels work and why they are important without drowning you in a sea of math. Yes, I will share a few formulas, but I promise you won't have to work out anything by hand. Keep your trusty calculator at the ready, and you'll be all set! Let's get started…

Decibels are a relative measurement.

This is an important concept to grasp. Decibels express a ratio between two levels of power (or, in some cases, voltage), meaning that a figure of "10 decibels" is meaningless. 10 decibels with respect to what? That's like saying "ten times." Ten times what? So, decibels represent a ratio of two levels. For example, a 10-watt transmitter is 10 dB more powerful than a 1-watt transmitter. There is a relative relationship between the two things. We'll define that relationship in a bit.

If decibels are a relative measurement, then what are all these terms like dBm, dBW, dBV, and so on?

These terms represent decibels related to some standard of measurement. dBm means decibels related to a milliwatt of power, dBW is decibels related to one watt, dBV is decibels related to one volt. Once decibels are referenced to a certain standard, they become a measurement with an absolute meaning. For example, 10 dBm is 10 dB above one milliwatt or 10 milliwatts.

Decibels can also be negative.

Wait, what? Yes, decibels can be negative. For example, the one-watt transmitter mentioned above is -10 dB compared to the 10-watt transmitter. More power equals a positive number; less power equals a negative number.

So, how are decibels calculated?

The formula for calculating decibels is relatively simple. It's not simple if you cannot handle logarithms. However, with modern calculators, you don't need to.

10 log (P1/P2)

P1 is the power in question, while P2 is the reference power. Allow me to explain… If you have a power source that is so many decibels above or below the power you are comparing it to, that is P1 in the equation above, while P2 is the power you are using to compare it to. Let me unpack that with an example: If I want an absolute measurement of power, such as dBm (power referenced to a milliwatt, as mentioned above), then P2 is one milliwatt. P1 is the measured power that I am trying to compare to the milliwatt. Let's say, for example, that I am measuring a power source, and the power meter shows that it is producing 10 watts.

Let's run the numbers to show how it works…

A milliwatt is 1/1000 of a watt. So, 10 watts would be 10,000 milliwatts. Plugging that into our handy-dandy dB formula, we get:

10 log (10,000/1) or

10 log 10,000

The log of 10,000, according to my convenient HP calculator, is 4. 4 times 10 is 40, so that means that my power source is producing 40 dB more power than a milliwatt, or +40 dBm.

Let's work that for a negative dB value…

So, lightning hit my power source, and now it's very weak. Instead of producing 10 watts, it now produces one-hundredth of a milliwatt. Running the numbers now gives a very different result.

10 log (0.01/1) or,

10 log 0.01

The log of 0.01 is -2. -2 times 10 is -20, so now my damaged power source is producing 20 dB less than a milliwatt, or -20 dBm. From +40 to -20 is a 60 dB difference. Time to get out the soldering iron!

If I know the decibels, how do I calculate the power ratio?

Basically, it's the same logarithmic process but in reverse. The formula looks like this:

Let's take the example of 30 dB. First, we divide the dB number by 10. That gives us 3 (thank you, Captain Obvious). Ten raised to the third power is 1000. So, depending on whether it is plus or minus 30 dB, the power is either 1000 times the original, or 1/1000th of the original. Obviously, numbers that are not multiples of 10 will be more complex to figure out, but that's what calculators are for.

Why are decibel figures different for voltages than power?

When I first learned about decibels, this confused me terribly. Why would the formula change from 10 log to 20 log? It turns out there is another equation that comes into play. Remember that decibels are a ratio of power. Well, for a given resistance or impedance, as the voltage changes, the power changes differently. The formula for power, with respect to voltage, is:

R (resistance or impedance) doesn't change, so the power changes by the square of the voltage.

In other words, if the voltage doubles, the power increases four times. Why? Because as the voltage increases, the current increases as well. Since power includes both voltage and current, the increase in each causes the power to go up as the square of the voltage.

The 20-log term takes that extra power increase into account. When working backward from dB to voltage change, you divide the dB figure by 20, not 10. Again, you're just reversing the process that got you to dB in the first place.

So there you have it! Everything you needed to know about decibels but were afraid to ask. Still have questions? Feel free to reach out to us to learn more or get a free consultation on your RF project.