## Decibels (dB) for Part 15 Use

Posted on October 20, 2012

Much of this information is posted around this site but here it is presented in one place.

**A Quick Reference:**

**For Voltage**

Much of this information is posted around this site but here it is presented in one place.

**A Quick Reference:**

**For Voltage**

**dB = 20log(V1/V2)**

**V1/v2 = 10^(dB/20)**

**dBuV = 20log(V/1uV)**

**Voltage = 1uV(10^(dBuV/20))**

**For Power**

**dB = 10log(P1/P2)**

**P1/P2 = 10^(dB/10)**

**dBm = 10log(P/1mW)**

**Power = 1mW(10^(dBm/10))**

**3dB is double the power and 10 dB is 10 times the power.**

**6dB is double the voltage and 20 dB is 10 times the voltage.**

*Detailed Information and Examples*

**The best way to understand what follows is to use a scientific calculator (one which does Logs and exponents of 10) and work the examples presented. As with many things, this can be hard at first but it gets easier with practice.**

**A fundamental to remember is that the decibel is always a ratio which makes dB convenient when comparing two measurements. For example if a voltage increases from 2 to 8 volts this increase is 20log(v1/v2) = 20log(8/2) = 12dB. If the voltage decreases then the dB is negative.**

**Decibels are also used to compare voltages at different parts of a circuit. For example, if the input to an amplifier is 0.7 Volts and the output is 15 Volts then the gain (recall that dB is a ratio) G = 20Log(15/0.7) = 26.6 dB.**

**There is a “trick” which can be used to express a voltage in dB as an absolute but in fact it is still a ratio. For example, dBuV is often used to express the strength of an electric field with the understanding that the reference voltage is 1 microvolt (uV). A field strength of 250 uV/m can be expressed as FS = 20log(250uV/1uV) = 47.96 dBuV.**

**A dBuV reading can be converted to Volts by using V = (1uV)10^(dBuV/20) which is for this example V = (1uV)10^(47.96dBuV/20) = 250 uV. (Try this by rounding 47.96dBuV up to 50dBuV and you will see how the log scale is not intuitive).**

**Of interest ot part 15 hobbyists is range so how are dBuV measurements used for this? Suppose we use a receiver which displays the received signal strength in dBuV and make a measurement of signal strength 200 feet from the transmit antenna and it is 55 dBuV. With tuning the signal increases to 60 dBuV. This doesn’t seem like much improvement but since dBs are logarithmic it actually is. Calculating: V = (1uV)10^(55dBuV/20) = 562 uV. After tuning V = (1uV)10^(60dBuV/20) = 1000 uV and since range is approximately proportional to field strength in the far field the increase in range here is 1000/562 = 1.78 or 78%. It is not necessary to calculate both voltages to get to the range change. Here is a shortcut: take the difference in dBuV (60dBuV – 55 dBuV = 5 dB) and calculate range increase = 10^(5dB/20) = 1.78 or 78%. You might wonder what happened to dBuV when the two readings were subtracted. Subtracting dB is the same as dividing volts so the 1 uV reference cancels leaving only a ratio expressed in dB.**

**At the transmitter, dBm can be used in a similar manner if the transmitter power delivered to the antenna can be measured. Lacking this, the output power ratios can be approximated by using the DC input power to the final stage. Take an example where the input power is increased from 75 to 100 mW and calculate the expected range increase. This increase is 10Log(100mW/75mW) = 1.25 dB. The increase in range using the equation from above is = 10^(1.25/20) = 1.15 or 15%. (This can also be done by calculating the range increase = SQRT(P2/P1) = SQRT(100mW/75mW) = 1.15 or 15%). You might wonder why the factor 10 was used to calculate the increase in dB yet the factor 20 was used to calculate the range increase. This is because 10 is used with dB calculations of power and 20 is used with dB calculations of voltage and the range change uses voltage.**

**There is a bit more about dB calculations which has been omitted here to focus on what is usable for Part15 calculations but in general it is good to understand that the dB is a logarithmic unit and a seemingly small change in dB represents a larger change in voltage, power, or range. For example a signal strength of 96 dBu is double that of 90 dBu yet they appear to be about the same.**

**Some points to remember:**

**dB is always a ratio.**

**dBuV and dBm are ratios to 1 uV and 1mW respectively.**

**Use the factor 10 when calculating with power and 20 when calculating with voltage.**

**Negative dB means a ratio less than 1**

**3dB is double the power and 6 dB is double the voltage.**

**Using dB can be daunting at first but if you are willing to work through the examples given using a scientific calculator and follow the meaning of the calculations then the quick reference given at the beginning of this article can become a “pocket reference”.**

**Neil**