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Microphone Month

Special +4/-10 Edition Summit

Over the past few years we’ve brought you the news and answered hundreds of technical questions about everything from burning CD’s to synchronizing MIDI to film, to tracking down ground loop hums, to buying gear on the Internet. During that time we’ve received hundreds of other questions that we answered privately via e-mail. But there’s been one burning question through all of this that we’ve never really been able to get to… until now.

What is the actual difference between -10 and +4 and what does it mean at a practical and technical level?

Before we could try to cover it we had to have a small library of words defined in our Word for the Day archives. I’ve been sneaking these words in here and there for literally two years all in anticipation of this glorious day where we finally hit this very confusing and popular topic. So, in honor of our 800th issue of inSync (why 800th? Why not?), we shall now get down to business on this subject. Rather than give a quick answer that doesn’t produce a real understanding of it I have decided to include additional background about this thing we call the dB. There’s a lot of general confusion about the dB and its applications we will be able to clear up along the way, and in the end most of you should have a good practical understanding of what’s going on.

The Set Up
If you’ve managed to get this far and are still reading we shall assume you are one of the many who’ve noticed that most audio equipment seems to come with either or both -10 dBV audio connections or +4 dBu audio connections and that these refer to different signal levels. You may be aware that +4 dBu gear is sometimes considered “pro” and -10 dBV gear is considered semi-pro or consumer. You may also know, possibly by learning it the hard way, that the two levels aren’t very compatible with one another. You can’t just plug a +4 dBu device into a -10 dBV device and expect ideal results. There is a significant difference between the two. But what is it? It’s not “14.”

Due Diligence
Before you try to tackle the upcoming sections make sure you have consulted our WFTD archives and checked out the definition for Ohm’s Law, Voltage, Current, Resistance, Impedance, dB, dBm, dBu, and dBV and followed the various linked terms in those records. Don’t worry if you don’t understand everything. Just get acquainted with the terms. You should now understand that the dB ratings we use represent some value compared to some “reference” value. The “reference” for 0 dBu = .775 volts so that means +4 dBu = 1.23 volts. Don’t worry too much if you don’t fully grasp all the math behind this (though it does really help if you do). As you read through this keep in mind that voltage is provided by a supply or source, and generally doesn’t change with different loads (resistance or impedance). The amount of current flow or amperage is what changes with different loads. Changing any of the three (voltage, current, resistance) will change the amount of power or work being done. This is a simplified generalization with exceptions, but it will help you to keep it in mind as you read.

Also of note: It is merely a matter of convention that +4 dBu gear often has balanced connections and -10 dBV gear is often unbalanced. The operating level of gear can be (for the most part) independent of the type of connections it has. For the sake of this discussion we will not concern ourselves with the implications of balancing or anything beyond the nominal operating levels of equipment.

A Word About Logs (no, not lumber)
Logarithmic equations seem to intimidate people fairly easily. But they don’t have to. Logs are a critical part of working with and understanding the dB. You will need to work with logs at a basic level to fully benefit from the explanation that is to follow. A key purpose of the logarithmic function is to make very unwieldy numbers easier to work with and visualize. Here’s a good example. Suppose you take two measurements of sound intensity in different parts of a studio. Without getting too deep into physics and all the background of how these measurements are really made just accept that you end up with numbers that represent sound power (Be careful, this is not the same as sound pressure (SPL)). So we get two numbers: 0.00012301269 and 0.000008909014. Now quickly, what is the difference between the two values? Sure, with calculators a dime a dozen, and even built into watches today, it’s pretty easy to punch in a few numbers and come up with 0.000114103676, assuming your calculator can deal with numbers this long. But for us humans these kinds of numbers are so abstract they really don’t mean anything. We can’t just look at them and draw any conclusions. But way back when scientists were wrestling with them they didn’t even have the benefit of calculators so they came up with a short cut – the log.

The logarithm is just a simple math trick. It lets us put these numbers into a form that makes them easier for us to handle and relates better to the way in which we actually perceive sound. A log is the exponent that shows the power to which a fixed number (the base) must be raised to produce a certain number. For example, 3 to the power of 2 is 9. The log of 9, with a base of 3 equals 2.
Log39 = 2
In science the base is always assumed to be 10 unless otherwise specified. The base 10 log of 100 equals 2. This is just another way of saying 10 to the power of 2 equals 100 and would normally be written as log100 = 2.

Don’t worry if you don’t fully follow all this background about logs. We have calculators today that handle the details for us. So getting back to the long numbers above, the log of the first number is -3.91, and the log of the second number is -5.05. I did a little rounding there, but we’re close enough. You can verify this with any decent calculator that has a log function. Just type in one of the long numbers above and hit the log button. So the difference between -5.05 and -3.91 = 1.14. In some applications it is easier to find the difference between the two numbers and then just find the log of that difference, but to do that you need to divide (not subtract) them. That’s just the way it works. You can accept it or go get your old math books out and learn all the subtleties. In the example we are using that would look like this.

Log (


) =1.14

In this example that happens to be 1.14 “Bels,” a term coined by the folks at Bell Labs years ago when they were figuring all this stuff out. In practice they opted to use the decibel, which is 1/10th of a Bel. The decibel is easy to obtain by multiplying the log by 10. This means 10log100 = 20. Applying this to our example above leaves us with a difference of 11.4 decibels.

10Log (


) =11.4

The decibel made a lot of the numbers those Bel scientists were crunching even easier to handle. What they left us by doing this is a legacy we will likely never depart from. The decibel is a mainstay of the audio industry. It is abbreviated dB and nowadays there are several different forms of the dB that mean different things. We will cover a few of those.

Don’t worry if you don’t understand everything so far. Read through this and the linked words a couple of times, and if possible get your calculator out and punch in the numbers. Unless you are already familiar with this material do not expect to be able to skim over it and walk away with any understanding. It is more likely you will have to thoroughly read through this several times before the light begins to come on. This is normal with new and difficult material so do not feel stupid or slow. If you take your time and study this you will eventually be able to understand it. Then you can amaze your audio friends with your mastery.

We’ve laid some groundwork by explaining logarithms and how they relate mathematically to the decibel. One critical aspect of this is that Bels and Decibels always, always, always are a comparison between two power levels. When an audio engineer says the volume in a concert was 120 dB SPL it is easy to assume that dB SPL is an absolute value. It isn’t. It is simply 120 dB above a “reference level” of 0 dB SPL. 0 dB SPL is considered to be the minimum threshold of human hearing (at least based on the limited research available at the time the reference was defined). Further confusing matters is the fact that dB SPL is not really a ratio of power levels, but of pressure levels. Power can be obtained and is inferred once you know the pressure (as you will see in a minute with Ohm’s Law), but it isn’t strictly power and this means the math for calculating it is different. The important thing to remember for now is that the dB is always a ratio between two levels. If you go to an audiologist and they say your hearing is down 6 dB at 4 kHz that means you hear 6 dB less at that frequency compared to the rest of your hearing, or compared to some reference they may define as “normal” hearing. If you have a 100 watt power amp driving a speaker and you substitute a 200 watt amp you will get 3 dB more level (theoretically; it doesn’t exactly work in practice). But 200 watts is not 3 dB. The DIFFERENCE between 100 and 200 watts is 3 dB. The corollary to this is that the difference between 200 watts and 100 watts is -3 dB. We’ll expose the math behind this shortly. Just make sure you understand that all versions of the dB always are comparing two power values with each other. And even when power isn’t explicitly part of the equation, as in dB SPL, it is still inferred. More dB SPL means more sound power (if everything else stays the same), but the math is a little different.

What is watt?
Now that we know all dB references relate to power it is important to understand what power is. Power equates directly to doing work. More power means more of some work gets done per unit of time. In electronics a common measure of power is the watt. Look at a speaker for a good example. The more power (in wattage) applied to the speaker the more work it is going to do.

Ohm’s Law defines the relationship between voltage, current, and resistance in electrical circuits. Once you know that a specific voltage applied to a load of some resistance will produce a specific amount of current flow, it becomes possible to figure out how much work is being done.

I know; that was kind of confusing. Look at it this way. You have a light bulb. You plug it into an electrical outlet known to have 120 volts of electricity present. If you know your light bulb presents a resistance (to current flow) of 144 ohms you can use Ohm’s Law and a little basic math to figure out that it will draw .8333 amps of current. The formula for this is Current = Voltage/Resistance. Voltage, current, and resistance (or impedance) all relate to each other in this simple mathematical way. We’re going to spare you the background on this, but it turns out that wattage relates to current, voltage, and resistance according to Ohm’s Law as well. If you know two of the other three you can figure out wattage according to the following formulas.

P = E x I Where P = Power in watts, I = Current in Amps,
P = I2 x R E = Voltage in volts, and R = Resistance in ohms.
P = E2/R

This means you can also reverse the logic and use power as a known quantity in figuring out voltage, current, or resistance. Knowing any two of these gives you a path to discovering the other two because of these relationships. Lets go back to your light bulb. Most light bulbs say how much wattage they draw right on them. You have a 100 watt bulb and you know your electricity is 120 volts. Plug those numbers into the appropriate formula above, apply a little algebra and you now know that your light bulb draws .83333 amps and that its resistance is 144 ohms.

P = E x I You can solve the equation before or after you plug in the numbers.
P = E x I
I = P/E 100 = 120 x I
I = 100/120 100/120 = I
I = .83333 .83333 = I

That formula solved for current in amps. We can apply the same technique to find the resistance in Ohms.

P = E2/R You can solve the equation before or after you plug in the numbers.
P = E2/R
R/P = E2 100 = 1202/R
R = E2/P R = 1202/P
R = 1202/100 R = 14400/100
R = 14400/100 R = 144
R = 144

Of course, once you know the voltage from the first solution, you can also find the resistance using the more basic R = E/I formula. The answer is 144 ohms.

Actually its “impedance” is 144 ohms. Impedance applies to AC (Alternating Current) circuits; resistance applies to DC (Direct Current) circuits. For our discussion we are going to use impedance and resistance interchangeably to simplify things, but understand that any time you are working with AC waveforms the impedance and resistance are NOT the same.

Don’t worry if you don’t fully comprehend all the math here. You can take our word for it and you’ll still be able to understand where we end up with all this. However, it will help if you take the time to plug in some numbers and do the math.

Pop Quiz – You’re playing a live gig and setting up your stage lights. How many 500 watt lighting fixtures can you run off of one 15 amp outlet (assume 120 volts are present)?

One path to the answer is as follows. You need to know how many amps of current each 500 watt fixture draws. The formula to use is I = P/E. Plug in the numbers.
I = P/E
I = 500/120
I = 4.1666

That’s a little over 4 amps per fixture. Four fixtures will add up to a little over 16.6 amps. You may or may not get away with this in practice, but you’re safest to limit it to three fixtures per 15 amp circuit (a 12.5 amp draw). See? This stuff is really useful. Don’t just gloss over the math. It can help you in many different ways.

Next we will begin to apply these concepts to the dB. We’re getting close guys, hang in there. Meanwhile feel free to practice your Ohm’s Law calculations. Here are a couple of questions to keep you busy.

  1. If a power amp delivers 100 watts into an 8 ohm speaker how many watts would it theoretically deliver (if able) into a 4 ohm speaker? You may know the answer is 200, but prove it with the math.
  2. A signal driven into a resistance (impedance) of 600 ohms produces .001 watts of power. What is the voltage? If the impedance goes up to 10,000 ohms (with everything else staying the same) how much power will now be transferred? This answer will be revealed as we work through the next few sections.

dB power, the dBm
As stated in the WFTD Archives, dBm is a measure of power or wattage compared to a fixed reference. The formula we used for sound power in the first installment of this special edition also applies to electrical power. It is 10log (P2/P1), where P1 & P2 are the two power levels in question and the resulting number is in dB. Looking back at the first homework question from yesterday we can see how this applies. A 100 watt amp at 8 ohms can become a 200 watt amp at 4 ohms. In terms of dB this represents a 3 dB increase. The math is pretty easy if you’ve followed along so far:
10log(200/100) = 3 dB
So what does dBm mean? The little “m” just specifies a reference point for us in milliwatts (an upper case “M” means it’s in watts). The zero reference for dBm is 1 milliwatt. That is the standard that has been set, so: 0 dBm = 1 milliwatt, or .001 watts. Now, if you know a piece of equipment has an output rated at 0 dBm then you can assume that means it can deliver 1 milliwatt of power into a reasonable load (ideally the load would be specified as it is in power amps, but sometimes loads are assumed based on common convention). If the device can deliver +20 dBm that means it can produce 100 milliwatts. Going back and forth between milliwatts and dBm is kind of a pain, but we’re just doing it to explain things. A lot of equipment has been rated in dBm over the years. Now you know that this specifies a power level. Converting the dBm rating into watts is generally not necessary in practice. If a device has a nominal input rating of 0 dBm then you probably want to drive it with a device whose output is a nominal 0 dBm to insure you are feeding it the proper amount of power. Quite often it’s that simple.

You may have noticed in the WFTD archives for dBm that a voltage of .775 volts into a load impedance of 600 ohms will produce 1 milliwatt of power, or 0 dBm. This is significant. There are references to 600 ohms in a lot of old audio books.

Why the heck does 600 Ohms keep coming up?
This is old Ma Bell at work. We’re not going to get too far into the history right now (maybe another time). Back in the early days of audio, Bell had to define many audio standards and ways of working. Many of these methods are still used today (even though some of them are brutally difficult to deal with). Much of their equipment was set up to have 600 ohm input and output impedances in order to drive really long lines (like phone lines). Back then audio was much more of a brute force kind of science and involved transferring much higher power (wattage) levels between equipment than we typically use in modern studio and stage equipment. Because they were concerned mostly with power transfer, the impedances of the circuits were critical. Input and output impedances had to be matched, and 600 ohms was the major standard for what we think of as line level gear. Nowadays input impedances are very high (thousands of ohms), and output impedances are very low. The result is almost no power transfer between equipment. In fact, we usually only look at the voltage levels (ignoring current) between devices these days. The obvious exception being power amps and speakers.

This brings us to our second homework question from yesterday. A signal driven into 600 ohms of impedance produces .001 watts of power. You can find the voltage with Ohm’s Law. Looking at the formulas unveiled yesterday and applying some algebra you can find that voltage is equal to the square root of the power (watts) times the resistance.

E = √PR
E = √600 x .001
E = √0.6
E = .775

Our voltage in that scenario is 0.775 volts. Of course we just said this two paragraphs above, but who’s paying attention? Now, if we raise the impedance to 10,000 ohms, how much power transfer do we have? If you’re good at Algebra there are several ways to find this, but now that we know the voltage an easy one is:
P = E2/R
P = 0.7752/10000
P = 0.6/10000
P = 0.00006 watts

The voltage squared is .775 x .775, which equals 0.6. Divide 0.6 by the resistance of 10,000 ohms and you get: 0.6/10000 = 0.00006 watts, or 0.06 milliwatts. That’s not much power. What is that in dBm?
dBm = 10log (P2/P1)
10log (0.06/1) = -12.22 dBm.

This means you are 12.22 dB below the zero (1 milliwatt) reference for dBm.

Now that may not sound like a big difference, but it is. In terms of current flow (back to Ohm’s Law again) you go from 1.3 milliamps down to .078 milliamps into the 10,000 ohm load. At a voltage of .775 volts that’s not enough current to worry about. In fact, many engineers ignore it when it gets that small.

We’ve shown that 0 dBm only specifies a level of 1 milliwatt. It doesn’t really specify a voltage or a current. However, if you know the impedance (resistance) you can figure out voltage and current. That’s where the dBu comes in. At 600 ohms there is one and only one value of voltage that will give you 1 milliwatt, and it is 0.775 volts (we can prove this with Ohm’s Law).

E = √P x R
E = √0.001 x 600
E = √0.6
E = 0.775

Years ago it was determined that this would also equal 0 dBu. Now that we know the significance of 600 ohms, we can see why it was handy to say that 0 dBm is equal to 0 dBu so long as the impedance is 600 ohms. This isn’t really the whole story, but we’ll get there. Run with us for now…

Who cares about power?

The significant thing about the dBu standard is that it doesn’t explicitly relate to power. “But I thought you said, ‘all dB values are a ratio of two power levels?'” I did, and they are. However, power can be looked at as a product of voltage and/or current. Go back and look at your Ohm’s Law again. One of the formulas you will see is P (power) = E (voltage) squared and divided by R (resistance or impedance).
P = E2/R
If you know that the impedance or resistance is the same among two voltages then you can assume voltage (or the signal level) is the only remaining variable that can change the power level. You can prove this with the math we’ve outlined or take our word for it, but this means that we can now use voltage as the sole determining factor to describe level differences of equipment (assuming impedance doesn’t change). The math looks a bit different for this because we have to account for the fact that power is the result of a product. It’s either voltage times current, or in the case we’ve shown here, voltage squared divided by resistance. The formula for calculating dB differences in voltage is thus:
20log (V2/V1) = dB
Don’t worry if you don’t fully follow all of the logic for how we got here. Read over it a few times and get what you can. Just keep in mind that calculating dB differences for voltage (or pressure) are different than for power. You use 20log instead of 10log.

How the dBu works in practice.

All of these dB references compare some value to a reference value. We’ve shown that 0 dBu is referenced to .775 volts. While that reference is derived from a relationship between dBu and dBm at 600 ohms the reference itself is not specifically concerned with resistance or impedance. It simply means that .775 volts equals 0 dBu. We are concerned only with voltage levels here. More voltage coming in or out of a piece of gear means more volume. That part is easy. If you learn to plug the numbers into the formulas we’ve provided here and in the WFTD archives, you can learn some interesting things. For example, you can now see that 1.23 volts is +4 dBu. This is just a matter of relating the 1.23 volt level to the reference level for 0 dBu, which is .775. The math:
20log (V2/V1) = dB
20log (1.23/.775) = +4 dB

Since .775 volts is the reference for 0 dBu, we can say that 1.23 volts is +4 dBu.

+4 dBu is a common zero reference for pro audio equipment. Now you know it specifies a voltage of 1.23 volts. We can do a lot with this knowledge. Here’s a simple one:

If the maximum voltage you can get out of your mixer is 15.5 volts, how many dBu is that? Assuming impedance is fixed, that is about +26 dBu, which is a typical maximum for a lot of audio gear. The math looks like this:
20log(15.5/.775) = +26 dBu
Next we will look at dBV, solve the +4/-10 question, and begin to tighten up the loose ends. Please take the time to go over what we’ve done so far. If you spend some time with the math, you’ll be surprised how easy it gets and how much more thoroughly you will understand these principles. Hang in there.

The dBV is very similar to dBu. In fact the only important difference is just that is has a different reference voltage level. The two references aren’t really designed to be intermingled, which is the crux of the problem we so often run in to. Because we buy gear that references each type of level we do intermingle them in practice. We’ll get back to all this. For now we shall just look at how the dBV works.

Finally we have an easy one. It works exactly the same way as the dBu, but we’ll quickly go over the math to illustrate (hold the groans please). 0 dBV was conveniently established with a reference level of 1 volt. That means 1 volt equals 0 dBV. If you’ve been following the math you will see that voltages of less than 1 volt will give us dBV levels in negative numbers, while voltages greater than 1 volt will yield dBV values greater than 0. It works out that our -10 dBV equipment produces a voltage of .316 volts. Here’s the math:
20log (.316/1) = -10 dBV
Look back at the mixing board example from yesterday. If your -10 gear could somehow muster up that same 15.5 volts that your +4 gear can (which it typically cannot), the corresponding dBV value would be +23.8 dBV. The math looks like this:
20log (15.5/1) = +23.8 dBV
Got it? In dBu land that same 15.5 volts is +26 dBu. Strange, but that’s what happens when you have two different references.

Be careful jumping to conclusions
So that means there’s a 2.2 dB difference between +4 and -10 (26 – 23.8 = 2.2)? No. All we’ve shown is that there is a 2.2 dB difference between a voltage that is compared to two different references (in this case .775 & 1 volt) into the same load. Put more simply, there is a 2.2 dB difference between the “reference” voltages for dBu and dBV. Using the 20 log formula it’s easy to show the 2.2 difference between the voltage referenced in dBV versus dBu:
20log(1/0.775) = 2.2 dB
So 2.2 is not the answer for the difference between +4 dBu and -10 dBV. It’s the answer for the difference between their reference voltages. +4 and -10 clearly are not zero, so you can see why we need to go a bit further. If you plugged the actual voltage levels for +4 dBu and -10 dBV into our formula what would you get?
20log (1.23/0.316) = 11.8044 dB
Let’s call it 11.8 for short. Voila! That’s it! The difference between +4 dBu and -10 dBV is 11.8 dB. You can now amaze your friends with your command of our audio language.

It can’t be that easy
But it is that easy. And hopefully it actually makes some sense now that we’ve provided a lot of the background. But there is more to the story. What does this 11.8 dB difference really mean? In short, it means there is an 11.8 dB difference between the zero reference or nominal signal level of equipment rated at +4 dBu versus -10 dBV. There are many ways this affects us, but before we go too much further we need to close a few holes we left behind along the way.

This is not your father’s impedance
We have shown how these references fit into a historical context, specifically the 600 ohm world of the telecommunications industry. In modern equipment designs, there is no concern over 600 ohms because the actual input impedances are much higher. As stated before, they are so high that for all practical purposes the power transfer can be ignored given the typical voltages involved. Further, when the load impedance stays the same, two voltages can be accurately compared without even knowing what the impedance is. So in modern audio equipment we just look at voltage differences. There are exceptions, but in terms of the line level gear we deal with, almost everything is referenced to a voltage; it’s just that it may be a dBu reference or a dBV reference. In fact the dBV came into existence as a standard way of looking at only voltage into very high impedances (1000 ohms or higher, typically more like 10,000 ohms). This is where our second homework question from two days ago is relevant. You remember, the one about raising the impedance of a load from 600 ohms to 10,000 ohms and what that does to power transfer. Coincidentally, putting 1 volt of electricity across a load of 1000 ohms will cause enough current flow to produce 1 milliwatt of power (0dBm).
P = E2/R
P = 12/1000
P = 0.001 watts

How handy, and actually it isn’t a coincidence, however it’s not that important because we aren’t too concerned with power transfer these days as input impedances are generally thousands of ohms. But…

He said “but.” Heh, Heh…
A complication arises when you connect your gear to something that does have a 600 ohm input impedance. It’s not too common to find such a device these days, but 20 or 30 years ago it wasn’t hard at all, and this made things quite confusing. If you were to plug a device rated at 0 dBV (1 volt) nominal output into a device rated at 0 dBm (1 milliwatt @ 600 ohms, or .775 volts) less current will be required by the lower .775 volt device to produce the same power (1 milliwatt) at it’s input. Most devices these days can’t easily produce this much output power (remember they are designed to be driven into loads much higher than even 1000 ohms), but if one could what would happen?

Hang on
The device with the 600 ohm input is designed to register “zero” on its dBm metering when it is hit with an input of .775 volts because at 600 ohms this is 0 dBm (the all important power transfer level “required” by the device to properly drive it). If we had a device with a 0 dBV nominal output it would reach “zero” on its meter at 1 volt. Hitting the receiving device with a 1 volt signal is going to cause its meter to show +2.2 dB. This is the quandary caused by the two different standards. There really is no easy way to reconcile them. 2.2 dB is close enough that it’s not that big of a deal and you can probably just live with it, assuming the upstream device can really deliver this much current into the 600 ohm load. If it can’t, then it is going to go into distortion before it even gets to its 1 volt output. Meters aren’t always going to tell you this so you have to listen and be aware of what you are doing.

The discrepancy described above between equipment rated at 0 dBV and 0 dBm is actually what caused the dBu standard to come into existence. The Bell Labs guys wanted a way to differentiate between equipment that was true 600 ohm 0 dBm equipment and equipment with higher input impedances that might still be interfaced with equipment rated in dBm. The dBu was adopted and used the .775 volt reference that arises from a 600 ohm load, but it really ignored impedance. Actually, when it first came out it was known as the dBv (little “v”), but people began to use the little “v” and the upper case “V” interchangeably. This obviously caused a lot of confusion and is probably one of the many reasons why this whole topic is so misunderstood to this day. Anyway, they changed the little “v” to a “u” and that was that. So the dBv and dBu are actually the same thing, but dBu is the proper term. It is widely believed that the “u” stands for “unterminated,” because the dBu reference isn’t concerned with the termination impedance.

2.2 dB and you
Because of these two “standards” you know right away that there is a built in 2.2 dB discrepancy between anything rated dBV versus dBu. At a practical level, this means that a 1 volt signal applied to a device rated in dBV will register 2.2 dB lower on its meters than a device calibrated to dBu. We rarely observe this phenomenon, however, because most of the time we are interfacing equipment rated at -10 dBV zero reference with equipment rated at +4 dBu, in which case the levels are much further off than a mere 2.2 dB. How much? 11.8 dB.

What does this discrepancy really mean to my levels?
For these conclusions you can follow along proving each statement using the math unveiled in the discussion so far, or you can just take our word for it. A -10 dBV device outputting a signal at its “zero” metering level is outputting -10 dBV, which we have shown to be .316 volts (a high impedance is assumed, but not specified). This is its zero reference and that level is equal to -7.8 dBu, but shows up on the meters of a +4 dBu device as -11.8 dB. Of course this assumes the input device is of sufficiently high input impedance not to “load down” the output device. In order to get the +4 dBu device up to its “zero” level on its meter you must drive the -10 dBV device until its meters register +11.8 dB, at which point it is outputting 1.78 dBV, or 1.23 volts (which equals +4 dBu). Most -10 dBV devices can deliver this much voltage, but it puts them awfully close to clipping. At this point you are about out of headroom on the output device (at nearly its maximum level) while you are just beginning to drive the +4 dBu device. When the -10 dBV device peaks or clips there is still tons of headroom (probably at least 11.8 dB) available in the +4 dBu device. There is no way to drive the +4 dBu device to maximum level unless it has some sort of amplifier on the front end you can turn up to raise the voltage of the signal. Of course the tables get turned the opposite way if you are driving a +4 dBu device into a -10 dBV device. In most situations a reasonably high signal in the +4 dBu device will overload -10 dBV equipment. Maybe you can turn the level down at one end or the other, but you may be compromising your signal to noise ratio. For these and other reasons it is generally not advisable to interface the two different standards together without some sort of gain stage in between that properly changes from one level to the other. There are many such devices made in various price ranges.

Here is a handy chart that summarizes much of the above paragraph (some values have been rounded off for clarity).

-10 dBV Voltage +4 dBu
Meter Reading dBV dBu Meter Reading
+ 16
+ 6
+ 8.2
+ 4.2
+ 11.8
+ 1.8
1 .228
+ 4
+ 10
+ 2.2
+ 7.8
0 .775
+ 4
0 .5
+ 1.8
0 .338
0 .316
0 .250
0 .245
0 .1
0 .01
0 .001

Yeah, but does it “really” matter that much?
It depends on what you are doing. In critical situations where you want the lowest noise floor and maximum dynamic range, yes, it really is important to have the levels properly matched. In other applications you can often get by “fudging” it, especially on equipment with input and output level controls. As long as you understand what you are doing and what you are asking of the equipment you can make it work okay in many situations. Your ears will be the final judge.

Why is -10 dBV equipment sometimes considered consumer or semi-pro?
The -10 dBV standard arose because manufacturers were looking for less expensive ways to build equipment. The componentry required to drive .775 volts (with headroom to spare) into a 600 ohm load is relatively expensive. When input impedances got up in the 10,000 ohm range designers realized they could use much lower voltages and save money. The -10 dBV standard took off in the hi-fi industry while the pro audio and broadcast industries were still building and using gear referenced to dBu and dBm. Once -10 dBV gear began to crossover into professional use the trouble started and the rest is history, as they say.

It’s important to understand that there really isn’t much of an inherent disadvantage to a .316 volt (-10 dBV) nominal level compared to a 1.23 volt (+4 dBu) signal in the modern equipment we use today. However, since (as a matter of convention) the +4 dBu standard is usually applied to high quality balanced connections on high quality “pro” gear, while the -10 dBV standard is often used with unbalanced connections – sometimes of unknown quality on gear of sometimes questionable quality – the stigma that -10 dBV is “not as good” as +4 dBu has lingered in our industry. It really isn’t always true, and in fact it is sometimes preferable to use the -10 dBV unbalanced connections on gear rather than routing your signal through another gain stage to balance it and raise the level up to +4 dBu. It all just depends on the specific equipment and your specific circumstances.

Is all equipment rated at -10 dBV or +4 dBu?
Of course not. Why have two standards when you can confuse the public with more? 0 dBu, +8 dBu, 0 dBm, 0 dBV, and 0 dB (whatever that means) are a few of the operating levels you may find on equipment. And don’t forget about the old dBv (little v). The cool thing is that you now have the background to figure out what they actually mean in terms of voltage and real world interfacing with your existing gear. Sometimes impedances will be specified, but usually not. If no impedance is specified you should assume a very high input impedance and a very low output impedance, which means you only need to worry about voltage levels, not power transfer.

That’s all for now. We’ve laid a great foundation we can build on. There are many other related issues we can now delve more deeply into. We will create a direct link to this segment of inSync out in our Summits link area so you can always refer back to this discussion when you get confused. And you will get confused from time to time about some of this stuff.

Now go connect your equipment and make music.

By the way, did anyone notice that +11.8 + 2.2 = 14?

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