We’ll cover the difference between dB, dBm, dBW, and dBc here and settle this confusion for good!

These are very important terms in RF and you should take whatever time you need to understand them thoroughly.

Although many of these units might seem confusing, the decibel and all of its variations are actually straightforward.

dB (Decibel)
dB is a ratio of two quantities. Since it is a ratio, it does not have any units. We typically talk about the ratio of two power levels, though occassionally we also use the ratio of voltage levels.

Power level ratio equation:

dB=10log(P2/P1)

If P1=Pin and P2=Pout

Then dB=10log(Pout/Pin)

And it’s a typical power gain of an electrical network, and it is only a relative value between two power levels.

Therefore,

*Gain (dB)=10log(Pout/Pin)

*If (Pout/Pin)=1, then Gain = 0 dB

*If (Pout/Pin)=10, then Gain = 10 dB

*If (Pout/Pin)=100, then Gain = 20 dB

*If (Pout/Pin)=1,000, then Gain = 30 dB

and

*If (Pout/Pin)=0.1, then Gain = -10 dB

*If (Pout/Pin)=0.01, then Gain = -20 dB

*If (Pout/Pin)=0.001, then Gain = -30 dB

Questions:

1. Input 10 mW to an RF amplifier and the measured output is 150mW, what is the gain in dB of this amplifier?

Ans.

Since Pin=10 mW and Pout=150 mW

Gain (dB)=10log(150/10)=11.8 dB

2. The gain of an RF amplifier is 18 dB, if the measured output power is 230 mW, then what is the input power?

Ans.

*Since Pout=230 mW and Gain=18 dB
*10log(230/Pin)=18 dB
*log(230/Pin)=(18/10)=1.8
*101.8=230/Pin
*Therefore, Pin=230/(101.8)=230/63.1=3.65 mW

dBm (Decibel relative to 1mW power level)
dBm is simply power measured relative to 1 milliwatt. So if we substitute P1 with 1 mW in the previous equation, the result is a measurement in dBm.

It is measured with respect to a fixed reference (1mW), therefore, it is an absolute value.

‘1 mW is 0 dBm as a reference.’ 0 dBm = 1 mW

Power in RF measurements is most commonly represented in dBm. Unlike dB, dBm is an absolute value and has its unit.

x dBm = 10 log(P/1 mW)

A power P of 10 mW is 10 times of 1 mW,

So 10log(10 mW/1 mW) = 10 dBm

I hope the following tests will help you clearly understand dB and dBm:

Questions:
1. How much is 12 dB over 30 dBm?

Ans.
12 dB over 30 dBm is 42 dBm.

And we can conveniently say

30 dBm (1W) + 12 dB (x 16) = 42 dBm (16W)

If we input 30 dBm (1000mW=1W) of power to an amplifier with 12 dB (x 16 of linear value) of gain, then we will get 42 dBm (16000mW = 16W) of output power.

2. How much is 30 dBm + 20 dBm?
Ans.

Isn’t it 50 dBm? Of course not, the correct answer is 30.4 dBm. Let’s find out how to get this answer.

2 random power sources, 30 dBm (1000 mW) and 20 dBm (100 mW), are combined together in a circuit. The total power is 1100 mW.

Therefore, 10log(1100 mW/1 mW) = 10log1100 = 30.4 dBm

And it’s only 0.4 dB more than 30 dBm.

Interesting, right?

30 dBm + 12 dB = 42 dBm, but

30 dBm + 20 dBm≠50 dBm

You need to know logarithm (dB, dBm) and linear value (times, mW) clearly and can convert them back and forth, so you won’t be confused.

Additional questions:
1. Input 200 mW to an amplifier and the measured output power is 35 dBm, what is the gain of this amplier? Ans. 12 dB

2. Apply 2 power source to a lossless combiner, power #1 is 15 dBm and power #2 is 8 dBm, what is the combined power in dBm? Ans. 15.8 dBm

dBW (Decibel relative to 1W power level)
dBW is not very different from dBm, the only difference being that we simply use 1 W as a reference in place of 1 mW used in dBm.

‘1 W is 0 dBW as a reference.’ 0 dBW = 1 W

x dBW = 10 log(P/1 W)

How many dBW is 3 watts? It is

x dBW = 10 log(3 W/1 W) = 4.8 dBW
dBW is an absolute value and has its unit.

Question:

How many dBm is 0 dBW?

Ans.

30 dBm

dBc (Decibel relative to carrier power level)
dBc is the power measured relative to the carrier power level.

This is typically used to specify how much are those spurs, harmonics, etc. below the main carrier power level.

If the carrier power level is 30 dBm and a spur is -20 dBm,