Introduction to Audio Electronics
Chapter 3 : The Decibel

Thanks to Mr. Ray Rayburn for his proofreading of and suggestions for this article.

Logarithms

Before getting into what a decibel (dB) is, we have to review some math:

Once upon a time you learned to do multiplication, after which someone explained that you can use division to do the reverse. For example:

if A = B * C then A / B = C and A / C = B

Well, we all know what an exponent is. Its equivalent backwards process is called a "logarithm." Logarithms (or "logs") work like this:

If 102 = 100 then log10 100 = 2

Actually, it's:

If AB = C then logA C = B

If the A in logA C = B is not specified then it's assumed to be 10.

Now we have to go through some properties of logarithms.

log 10 = 1 or log 101 = 1

log 100 = 2 or log 102 = 2

log 1000 = 3 or log 103 = 3

This should come as no great surprise - you can check them on your calculator if you don't believe me. Now, let's play with these three equations.

log 1000 = 3

log 103 = 3

3 * log 10 = 3

Therefore:

log AB = B * log A

Power and Bels

Sound in the air is a change in pressure. The greater the change, the louder the sound. The softest sound you can hear according to the books is 20 * 10-6 Pascals (abbreviated "Pa") (it doesn't really matter how big a Pa is - you just need to know the number for now...). The loudest sound you can tolerate without screaming in pain is about 200 000 000*10-6 Pa (or 200 Pa). This ratio of the loudest sound to the softest sound is therefore a 10,000,000:1 ratio (the loudest sound is 10,000,000 times louder than the softest). This range is simply too big. So a group of people at Bell Labs decided to represent the same scale with smaller numbers. They arrived at a unit of measurement called the Bel (named after Alexander Graham Bell - hence the capital B.) The Bel is a measurement of power difference. It's really just the logarithm of the ratio of two powers (Power1:Power2 or Power1/Power2). So to find out the difference in two power measurements measured in Bels (B). We use the following equation.

D Power (Bels) = log (Power1/Power2)

Let's leave the subject for a minute and talk about measurements. Our basic unit of length is the metre (m). If I were to talk about the distance between the wall and me, I would measure that distance in metres. If I were to talk about the distance between Vancouver and me, I would not use metres, I would use kilometres. Why? Because if I were to measure the distance between Vancouver and me in metres the number would be something like 5,000,000 m. This number is too big, so I say I'll measure it in kilometres. I know that 1 km = 1000 m therefore the distance between Vancouver and me is 5 000 000 m / 1 000 m/km = 5 000 km. The same would apply if I were measuring the length of a pencil. I would not use metres because the number would be something like 0.15 m. It's easier to think in centimetres or millimetres for small distances - all we're really doing is making the number look nicer.

The same applies to Bels. It turns out that if we use the above equation, we'll start getting small numbers. Too small for comfort; so instead of using Bels, we use decibels or dB. Now all we have to do is convert.

There are 10 dB in a Bel, so if we know the number of Bels, the number of decibels is just 10 times that. So:

1 dB = 1 Bel / 10

D Power (dB) = 10 log (Power1 / Power2)

So that's how you calculate dB when you have two different amounts of power and you want to find the difference between them. The point that I'm trying to overemphasize thus far is that we are dealing with power measurements. We know that power is measured in watts (Joules per second) so we use the above equation only when the ratio is comparing two measurements in watts.

What if we wanted to calculate the difference between two voltages (or electrical pressures)? Well, Watt's Law says that:

Power = Voltage2 / Resistance

or

P = V2 / R

Therefore, if we know our two voltages (V1 and V2) and we know the resistance stays the same:

D Power (dB) = 10 log (Power1 / Power2)

D Power (dB) = 10 log ((V12/R) / (V22/R))

D Power (dB) = 10 log ((V12/R) * (R/V22))

D Power (dB) = 10 log (V12 / V22)

D Power (dB) = 10 log (V1 / V2)2

D Power (dB) = 2 * 10 log (V1 / V2)

(because log AB = B * log A)

D Power (dB) = 20 log (V1 / V2)

That's it! (Finally!) So, the moral of the story is, if you want to compare two voltages and express the difference in dB, you have to go through that last equation.

Remember, voltage is analogous to pressure. So if you want to compare two pressures (like 20 * 10-6 Pa and 200 000 000* 10-6 Pa) you have to use the same equation, just substitute V1 and V2 with P1 and P2 like this:

D Power (dB) = 2 * 10 log (Pressure1 / Pressure2)

This is all well and good if you have two measurements (of power, voltage or pressure) to compare with each other, but what about all those books that say something like "a jet at takeoff is 140 dB loud." What does that mean? Well, what it really means is "the sound a jet makes when it's taking off is 140 dB louder than......" Doesn't make a great deal of sense..... Louder than what? The first measurement was the sound pressure of the jet taking off, but what was the second measurement with which it's compared?

This is where we get into variations on the dB. There are a number of different types of dB which have references (second measurements) already supplied for you. We'll do them one by one.

dBspl

The dBspl is a measurement of sound pressure (spl stand for Sound Pressure Level). What you do is take a measurement of, say, the sound pressure of a jet at takeoff (measured in Pa). This provides Power1. Our reference Power2 is given as the sound pressure of the softest sound you can hear, which we have already said is 20 * 10-6 Pa.

Let's say we go to the end of an airport runway with a sound pressure meter and measure a jet as it flies overhead. Let's also say that, hypothetically, the sound pressure turns out to be 200 Pa. Let's also say we want to calculate this into dBspl. So, the sound of a jet at takeoff is :

Pressure (dBspl) = 20 log (Pressure1 / Reference)

Pressure (dBspl) = 20 log (Pressure1 / 20 * 10-6 Pa)

Pressure (dBspl) = 20 log (200 Pa / 20 * 10-6 Pa)

Pressure (dBspl) = 20 log 10 000 000

Pressure (dBspl) = 20 log 107

Pressure (dBspl) = 20 * 7

Pressure = 140 dBspl

So what we're saying is that a jet taking off is 140 dBspl which means "the sound pressure of a jet taking off is 140 dB louder than the softest sound I can hear."

dBm

When you're measuring sound pressure levels, you use a reference based on the threshold of hearing (20 * 10-6 Pa) which is fine, but what if you want to measure the electrical power output of a piece of audio equipment? What is the reference that you use to compare your measurement? Well, in 1939, a bunch of people sat down at a table and decided that when the needles on their equipment read 0 VU, then the power output of the device in question should be 0.001 W or 1 milliwatt (mW). Now, remember that the power in watts is dependent on two things - the voltage and the resistance (Watt's law again). Back in 1939, the impedance of the input of every piece of audio gear was 600 ohms. If you were Sony in 1939 and you wanted to build a tape deck or an amplifier or anything else with an input, the impedance across the input wire and the ground in the connector would have to be 600 ohms.

As a result, people today (including me until my error was spotted by Ray Rayburn) believe that the dBm measurement uses two standard references - 1 mW across a 600 ohms impedance. This is only partially the case. We use the 1 mW, but not the 600 ohms. To quote John Woram's Sound Recording Handbook "...the dBm may be correctly used with any convenient resistance or impedance."

By the way, the m stands for milliwatt.

Now this is important: since your reference is in mW we're dealing with power. Decibels are a measurement of a power difference, therefore you use the following equation:

Power (dBm) = 10 log (Power1 / 1 mWrms)

Where Power1 is measured in mWrms.

What's so important? There's a 10 in there and not a 20. It would be 20 if we were measuring pressure, either sound or electrical, but we're not. We're measuring power.

dBV

Nowadays, the 600 ohms specification doesn't apply anymore. The input impedance of a tape deck you pick up off the shelf tomorrow could be anything - but it's likely to be pretty high, somewhere around 10 kohms. When the impedance is high, the dissipated power is low, because power is inversely proportional to the resistance. Therefore, there may be times when your power measurement is quite low, even though your voltage is pretty high. In this case, it makes more sense to measure the voltage rather than the power. Now we need a new reference, one in volts rather than watts. Well, there's actually two references..... The first one is 1Vrms. When you use this reference, your measurement is in dBV.

So, you measure the voltage output of your piece of gear - let's say a mixer, for example, and compare that measurement with the 1Vrms reference, using the following equation.

Voltage (dBV) = 20 log (Voltage1 / 1 Vrms)

Where Voltage1 is measured in Vrms.

Now this time it's a 20 instead of a 10 because we're measuring pressure and not power. Also note that the dBV does not imply a measurement across a specific impedance.

dBu

Let's think back to the 1mW into 600 ohms situation. What will be the voltage required to generate 1mW in a 600 ohms resistor?

P = V2 / R

V 2 = P * R

V = sqrt (P * R)

V = sqrt (1 mWrms * 600 ohms)

V = sqrt (0.001 Wrms * 600 ohms)

V = sqrt 0.6

V = 0.774596669 Vrms

Therefore, the voltage required to generate the reference power was about 0.775 Vrms. Nowadays, we don't use the 600 ohms impedance anymore, but therounded-off value of 0.775Vrms was kept as a standard reference. So, if you use 0.775 Vrms as your reference voltage in the equation like this:

Voltage (dBu) = 20 log (Voltage1 / 0.775 V rms)

your unit of measure is called dBu. Where Voltage1 is measured in Vrms.

(It used to be called dBv, but people kept mixing up dBv with dBV and that couldn't continue, so they changed the dBv to dBu instead. You'll still see dBv occasionally - it is exactly the same as dBu.... just different names for the same thing.)

Remember - we're still measuring pressure so it's a 20 instead of a 10, and, like the dBV measurement, there is no specified impedance.

Addendum: "Professional" vs. "Consumer" Levels

Once upon a time you probably learned that "professional" gear ran at a nominal operating level of +4 dB compared to "consumer" gear at only -10 dB. (Nowadays, this seems to be the only distinction between the two.....) What few people ever notice is that this is not a 14 dB difference in level. If you take a piece of consumer gear outputting what it thinks is 0 dB VU (0 dB on the VU meter), and you plug it into a piece of pro gear, you'll find that the level is not -14 dB but -11.79 dB VU....... The reason for this is that the professional level is +4 dBu and the consumer level -10 dBV. Therefore we have two separate reference voltages for each measurement.

0 dB VU on a piece of pro gear is +4 dBu which in turn translates to an actual voltage level of 1.228 Vrms. In comparison, 0 dB VU on a piece of consumer gear is -10 dBV, or 0.316 Vrms. If we compare these two voltages in terms of decibels, the result is a difference of 11.79 dB.

The Summary

dBspl

Pressure (dBspl) = 20 log (Pressure1 / 20 * 10 -6)

where Pressure1 is measured in Pa.

dBm

Power (dBm) = 10 log (Power1 / 1 mWrms)

where Power1 is measured in mWrms.

dBV

Voltage (dBV) = 20 log (Voltage1 / 1 Vrms)

where Voltage1 is measured in Vrms.

dBu

Voltage (dBu) = 20 log (Voltage1 / 0.775 Vrms)

where Voltage1 is measured in Vrms.


If you have any suggestions or comments, please mail martin@music.mcgill.ca.

Return to the index