Special Edition, Part 3.
Welcome back to our special inSync feature on the dB and what it means. We are working our way to a discovery of the difference between -10 dBV and +4 dBu levels, but along the way we are learning the background that will give us the knowledge to truly understand the implications of the difference. By the time we are finished you will know more about the dB than you ever thought you wanted to.
If you’ve missed the last two editions of inSync you should go back and study them now or you will be lost. So far we’ve explored logarithms, Ohm’s Law, and begun to explain the nature of the dB and how it is used. These concepts have given us a great foundation to build from, but now it’s time to dig in to the specifics.
dB power, the dBm
We will start with the dBm because it most directly applies to what we’ve done so far. 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.
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.
That’s where we’re going to table it for today. Tomorrow we’ll see what all this means.