Special Edition, Part 4.
We now continue with our dB saga. If you haven’t been following along for the past few days you need to back up and review before going on or this material will not make much sense.
dBu
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).
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
That’s going to do it for today. Tomorrow 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.