Special Edition, Part 2.
Welcome to part two of the Great Sweetwater -10, +4 special inSync dB feature. Last Friday we began an explanation that will lead us down the path to discovering the actual difference between -10 dBV and +4 dBu levels and what they mean to us in practice. If you are coming into today’s segment without reading the last issue we highly recommend you go back and read it before continuing.
Finally, I will issue a warning here. 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.
In Friday’s inSync we laid some groundwork by explaining logarithms and how they relate mathematically to the decibel. One critical aspect of this we saved for today 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.
Tomorrow 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.
- 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.
- 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.