W4NEQ Christopher Scott |
How to cheat with dBs ... an intuitive approach |
Decibel relationships are relatively simple, yet many engineers forget that approximate calculations can be done mentally. We're so quick to hit the calculator LOG button that we sometimes fail to get a conceptual feel for the unit. When the calculator is left back at the office and these dB units come up, close approximations can still be made. Decibels, or dBs are very useful units, - one-tenth of a bel, (nobody uses bels anymore) allowing both small and large ratios to be expressed in a meaningful way with manageable numbers. In audio and RF work dBs are de rigeur. A simple method is presented that allows mental calculation. The key point to remember is that, logarithmic numbers (dBs) add and subtract from each other in the same way that standard numbers multiply and divide. This relationship is the premise underlying the functionality of the slide rule (analog computer for newbies), shown above. . |
While not quite as accurate as the calculated logarithmic machinery: 10 or 20 times the log(10) of the ratio, this approach does well enough for most work, and promotes a clearer understanding of the basic relationships. We will start by memorizing three pairs of numbers which apply to voltage (and current). Later on, with practice, these will be seen to relate via ohms law to power as well: For VOLTAGE 6 dB = multiply or divide the ratio by 2 10 db = multiply or divide the ratio by 3.33 (3 is close) 20 db = multiply
or divide the ratio by 10 Try to memorize these. Let's now prove that the dBs add and subtract, and the ratios multiply and divide: |
Mental example 1: An amplifier is said to have 26 dB of gain. If the input is 1 volt, what is the output? Put down the calculator - this is easy: Since 26 dB is at least 20 dB, we know that the output will be at least multiplied by 10 - or at least ten volts. So we subtract the 20 dB from the 26 dB and are left with 6 dB to "get rid of." We then multiply the ten volts by 2 (6 dB) to "get rid of" the remaining six. The amplifier output is 20 volts. |
Mental example 2: A microphone is said to produce - 60 dBV output. How many volts is this? First, let's review that for a dB term to be meaningful, it must represent either a CHANGE, as in example 1, or it must be RELATIVE to an identifiable quantity, as in this example. - 60 dBV means less than (because of the negative sign) one VOLT (the identifiable quantity is V) by 60 dB. So we simply divide ONE VOLT by the calculated ratio: 20... 20... 20... = 60 (dB) = 10 times 10 times 10 = 1000 One volt divided
by 1000 = .001 Volt or one millivolt (mV). |
Mental example 3: What is the Voltage
ratio represented by 16 dB? 6 dB equals a 2 multiplier. 3 x 2 = 6... a ratio of 6. (If we want to be more accurate, use 3.33 * 2 equals a 6.66 ratio) Our eyes have now glazed over, and a yawn is beginning to form... but if you cannot see a pattern developing, you have no soul. That's it. The rest is simply practicing these relationships. If we need to reverse the calculation, from a known ratio to dB, we just divide or multiply the ratio quantity at each reduction step: |
Last example: What is the Decibel expression for a Voltage ratio of 10,000 ? Hmm... looks like ten (20 dB) will divide into it well... 10,000 / 10 = 1,000 (20 dB) 1,000 / 10 = 100 (40 dB) 100 / 10 = 10 (60 dB) 10 / 10 = (end) (80 dB) The answer is 80 dB. |
POWER If the ratios are
POWER ratios like WATTS, the following chart applies: 3 dB = multiply or divide by 2 6 dB = multiply or divide by 4 10 db = multiply or divide by 10 20 db = multiply or divide by 100 Why the different constants for power? Because power, instead of changing linearly with voltage, changes with the square of voltage. Ohms law proves it: [P = V^2 / R]. Take some time to play with the numbers and you will see how the tables are consistent with Ohm's formula. |
Confusion often results from not just the dB calculation, but instead what UNIT the expression relates to. 0 dB(whatever) = (whatever) = no change. Here are some common Decibel unit expressions. dBV = dB change relative to a VOLT dBW = dB change relative to a WATT dBmV = dB change relative to a MILLIVOLT - common in CATV at 75 ohms dBm = dB change relative to a MILLIWATT - common in RF at 50 ohms, audio at 600 dBK = dB change relative to a KILOWATT - common in broadcast at 50 ohms dBu = dB change relative to a MICROVOLT - for RF receiver usage dBu = dB change relative to .775 volts rms at 600 ohms impedance for audio volume unit work. dBc = dB change relative to CARRIER (voltage or power) |
Now your friends will honestly believe you can do logarithmic calculations in your head.GoTo W4NEQ Main Page |