If you have looked at specifications for audio gear over the years, you may have run across a term called dBV. If you have worked in broadcast, live sound, recording, or field audio, then you most certainly have seen this term, and probably already have a firm grasp of the concept and its implications.

However, for those unfamiliar with the nomenclature, what follows should help you better understand the concept. If you have read either part I or part II of my posts on decibels, you'll notice some striking similarities between what you see in the world of the dBV and what you see in the world of dB SPL.

Firstly...why the term dBV?

"dBV" simply means "decibels referenced to volts". Contrast this with "dB SPL" which simply means "decibels referenced to sound pressure" (even though the units are Pascal).

But wait... aren't decibels used to describe the intensity of a sound?

Yes, they are. However, you need to keep in mind that decibel can be used to describe any two like quantities. That is, if you wanted to express the day to day changes in gas prices, you could do that. Granted, it doesn't make a lot of sense, but the point is, as long as you stuck to the same currency unit, the math would be valid.

Anytime the decibel notation is cited, there must be a reference value. The purpose here is to define the value of 0 dB - and this is true whether for SPL or dBV. In the case of the former, that reference value is 20 micro Pascal, while in the latter, the reference is taken as 1.0 VRMS. Where these two differ can often confuse people; in SPL, 0 dB refers to the threshold of hearing (the point below which a healthy ear can no longer discern any sound), and thus, all of the dB SPL values we are used to seeing are all greater than zero, and most of us probably have a feel for the value that corresponds to a very loud sound (i.e. rock concert...approx 110 dB etc).

However, in dBV (and related other electrical concepts such as dBF, dBm, etc), it is quite common to see dB values greater than or less than zero.

Keep in mind this one thing: the only thing that defines 0 dB (regardless of what is being expressed) is the chosen reference value - nothing more. In some instances it may seem arbitrary (the reference) and in otehrs it may seem more logical.

The Math:

The general expression for calculating dBV is exactly the same as the math required to calculate dB SPL; they are of the exact same form. The only differences are that:

a) the units are different.
b) the reference values are different.

However, the equation is of the same exact form, and as such, once you are familiar with the math behind dB SPL, the math behind dBV (or the other way aorund) will seem logical to you. It will also (hopefully) then seem logical that when we discuss speakers, there is a 1:1 relationship between the SPL we get from a speaker, and the drive voltage applied to it.

One note of caution though...be sure that you are using like scaling. That is, if you are working in RMS, stick with RMS. If you are working in peak, stick with peak...if peak-to-peak...stick with peak to peak. However, in dB SPL and dBV, the scaling is RMS.

OK, so I said earlier that the reference for dBV was 1.0 VRMS. Let's revisit the general expression:

(1) dB = 20 * LOG10(quantity 1/ reference quantity)

So, let's check the math behind 0 dBV... The math specific to dBV is given by:

(2) dBV = 20 * LOG10 (V/Vref); where Vref = Reference voltage = 1.0 VRMS

Se what I mean? If you compare equation (1) to (2), you will see they are the same things, save for the nomenclature - the math, however, is identical.

OK, so let's verify how we get 0 dV for 1.0 VRMS. Suppose we have a DMM handy and we look at some point in a circuit, or the voltage going to a speaker, and we observe exactly 1.0 VRMS. This is what happens:

(3) dBV = 20 * LOG10 (1.0/Vref)

...but Vref = 1.0 VRMS and thus, we have

(4) dBV = 20 * LOG10 (1.0/1.0) ... and this yields...

(5) dBV = 20 * LOG10 (1) ... but the base-10 log of 1.0 is equal to zero. Thus zero multiplied by twenty is still zero. Thus, we see that for a value of 1.0 VRMS, we indeed get 0 dBV.

So what's the "take-away" here? If you know the reference...and the stated dB value is 0.0 (zero), then you know the quantity (in this case, voltage) being cited is equal to the reference.

Patterns and ratios...

OK, so in "Part Deux" of the decibel posts I created a table...and I'm going to do the same thing here...towards a purpose. Here we go ( I will start at +10 dBV and work my way downward):

[dBV] {V} /Change/
[+10] {3.16 V} /N/A/
[0] {1.0 V} /3.16/
[-10] {316 mV} /3.16/
[-20] {100 mV} /3.16/
[-30] {31.6 mV} /3.16/
[-40] {10 mV} /3.16/
[-50] {3.16 mV} /3.16/
[-60] {1.0 mV} /3.16/

Again we see a pattern emerging...note how at + 10 dBV we have 3.16 V, and at 0 dBV we have 1.0 V, and at - 10 dBV we have .316 V (316 mV). Now here's something that I previously glossed over - look at any 20 dB step. That is, look at +10 dBV to - 10 dBV, or from -20 dBV to - 40 dBV or whatever...if you look closely, you will see that each 10 dB step is a factor of 3.16 different from the first. But, 3.16 is actually the square root of 10. Hmmm... so, if we multiply 3.16 x 3.16 we get 10...but also, note the difference in magnitude of the voltage across a 20 dB span. As an example, look at -30 dBV, and now look at -50 dBV; at -30 dBV the voltage is 31.6 mV (millivolts) but at - 50 dBV the voltage is 3.16 mV - which is a factor of 10.

Thus, another 'keep this one in your head" ratio is this:

"any change of 20 dBV represents a factor of ten in the measured quantity"

(incidentally, I neglected to state this for the dB SPL relationshipos, but the same axiom holds true.)

Let me repeat that...

For any change of 20 dB (when working with dBV), the ratio of the measured values is always a factor of 10 *

* unless...you are working with power. That is a siliar but different relationship than that used for SPL, voltage, or current. But if you are working in dB notation with SPL, voltage, and current, this relationship is valid.


Thus, if you can remember that, and you can remember that the reference is 1.0, then you know that...wait for it... -20 dBV is equal to 0.1V (100 mV), and that -40 dBV is equal to .01V (10 mV), and then logically, - 60 dBV would be equal to 0.001 (1.0 mV). Just to reiterate something from before, look at the value associated with - 10 dBV (316 mV)... if you jump down the chart by 20 dB to - 30 dBV, what's the associated voltage? Yep, it's a factor of ten different, that is, it's 31.6 mV. Likewise, if you were to drop down to - 50 dBV, you would see the value is 3.16 mV. The same pattern is inherent here...over, and over, and over. Likewise, were you to mrach up the scale (to positive dBV numbers) this relationship would still hold.

Also, note the change as shown in the last column; 10 dB steps always follow the 3.16 scale factor - exactly as in dB SPL. Now, while you may not have a dBV value that lands precisely on one of these 10 dB steps, if you look at the pattern, you can at least get a ball-park approximation of the value in volts. That is, if you saw " - 17 dBV" you would know (from the table above) that the value is a bit more than 100 mV, but definitely less than 316 mV; remember, the more negative the dBV value, the smaller will be the number of volts associated with it.

Looking at SPEC SHEETS...

So, if you look at a spec sheet and see that some device (whatever it is) lists some test conditions (perhaps for distortion or some other parameter) and the values are given, and then in parenthesis you see something like @ -14 dBV. How many volts is that?

As with SPL, all you need to do is to re-arrange the expression shown in (3) so that rather than entering the unknown voltage to determine its dBV value, you enter the known dBV value and it return the unknown value in units (that is, volts RMS). Let's do that...starting with equation (2):

dBV = 20 * LOG10 (V/Vref) ... but we want to make this equation work for us (not the other way around), and by that I mean we have to isolate "V" (the numerator inside the parenthesis). So let's do that, step by step:

(6) dBV/20 = LOG10(V/Vref) - (we just divided both sides of the expression by 20, nothing more...)

(7) 10^(dBV/20) = (V/Vref) (this is by the properties of logarithms). Remember, the carat (^) means "raised to the power".

(8) 10^(dBV/20) * Vref = V (by algebra), but let's rewrite it in a form we're used to seeing:

(9) V = 10^(dBV/20) * Vref ...but we know that Vref = 1.0, and thus, equation (9) reduces to:

(10) V = 10^(dBV/20)

The 'spec sheet' defined the test condition as having been done at - 14 dBV. So, what's the voltage? We just 'plug and chug' and we get:

(11) V = 10^(-14/20)

and thus... - 14 dBV equals...

(12) V = 10^(-0.7) .. which equals...

(13) V = 0.199 V (or 199 mV).

Let's check our math... we'll use equation (2) and our value of 199 mV to see if we end up with - 14 dBV. Here goes...

(14) dBV = 20 * LOG10(0.199/1)

(15) dBV = 20 * - 0.7

(16) dBV = -14

So, there you have it: anytime you are given a dBV value and want to know its actual RMS voltage, you just have to invoke equation (10), plug in the known dBV value, and you'll get the (previously unknown) voltage corresponding to the (known) dBV value.

For those who read parts "One" and "Deux", you'll note this is the exact same thing we did to solve for the unknown RMS pressure when we knew the SPL.

Mark