# What is dBm and why is it used rather than watts?

Reading through an RF power amplifier datasheet, I found the sheet only referenced dBm output power, rather than watts.

• What is dBm?
• How do I convert it to watts?
• Why, or when, would you use dBm to specify power output rather than watts?
• Are there specific problems or equations that are easier to deal with in dBm vs watts?
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What is dBm?

dBm stands for decibels relative to one milliwatt. Decibels represent multiplicative factors, or ratios; by establishing a specific reference level they can instead be used as absolute values: 0 dBm is 1 milliwatt, 3 dBm is approximately 2 milliwatts, etc.

How do I convert it to watts?

Convert the decibel value to a scale factor and multiply by one milliwatt. That is,

$$x_{\mathrm{mW}} = 10^{x_{\mathrm{dBm}}/10} \cdot 1 \,\mathrm{mW}$$

$$x_{\mathrm{W}} = 10^{x_{\mathrm{dBm}}/10} \cdot 0.001 \,\mathrm{W}$$

For example, the datasheet you link mentions a value of

$$17 \,\mathrm{dBm} = 10^{17/10} \,\mathrm{mW} \approx 50.1 \,\mathrm{mW}$$

Why, or when, would you use dBm to specify power output rather than watts? Are there specific problems or equations that are easier to deal with in dBm vs watts?

Gain and loss in all stages of an RF system (feed line, filters, amplifiers) is multiplicative (if it were not, that would be nonlinearity), and therefore is typically written in dB so that the total gain or loss may be computed by adding, rather than multiplying, all the individual values together.

If you add a value in dB to a value in dBm, the result is in dBm. (Adding two dBm values is not usually meaningful since it would correspond to power squared.)

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Are there specific problems or equations that are easier to deal with in dBm vs watts?

Decibel units, dBm being an example of such, provide a more intuitive measure of some property that responds logarithmically, like power frequently does.

Consider, if you are transmitting now with 1W, and you add 1W more, you have doubled your transmit power. That's a big difference.

If you are transmitting with 100W, and you add 1W more, your transmit power is 101W. This is only 1% more power: hardly a relevant change.

Decibels account for this. From 1W to 2W is a +3dB change. From 100W to 101W is a +0.043dB change. If you were to increase +3dB from 100W, the result would be 200W, which is the same degree of improvement as 1W to 2W is.

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dBm is an incomplete term

You also need the missing parameter which the datasheet writers assume you know: the impedance. For audio / telegraphy this is 600 ohms but for RF this is usually 50 ohms.

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How does the impedance affect the dBm to watts conversion? What other dBm equations does it affect? – Adam Davis Feb 14 '14 at 1:15
@AdamDavisKD8OAS the impedance must be known if you want to express a voltage or a current in terms of dBm. This is a convenience when you know a voltage will end up eventually into a 50 ohm load, which is usually does, for RF applications. Essentially you are saying "this voltage here will become some power, after it goes through some current buffers". – Phil Frost Feb 14 '14 at 2:25
That said, it's a convenience, not a definition. dBm certainly is a complete term when properly used, and anyone specifying a voltage in dBm without making whatever conveniences are in effect obvious probably shouldn't be writing datasheets, or anything technical for that matter. – Phil Frost Feb 14 '14 at 2:28
I'm more familar (extremely) with audio where 0dBm is 0.775V regardless of the actual impedence. and equipment mostly refers to dB. Radio work (very rusty) varies more in what is used but impedances are usually critical, so yes you had better understand. – timc Feb 14 '14 at 3:43
@timc the impedance is only critical if you are measuring a voltage or a current in dBm. What's 1V relative to 1mW? I haven't a clue. But if you know the load has resistance R, then you can calculate P=V^2/R, now you have the power, and now you can use dBm. While there's a vestigial use of dBm as a measure of voltage into 600 ohms for audio, this isn't so in radio, where the load impedance very likely is 50 ohms, or we are measuring a power, not a voltage, so the impedance is irrelevant. – Phil Frost Feb 14 '14 at 12:29