Power and Decibel relationship.

Here is a link to a chart:

http://www.allis.com/charts/watts_dbm.html

Excellent Chart!

Also, as proved by the chart, if you are interested in the effect of an increase (or decrease) in power there are two (approximate but very close) formulaic relationships:

1. Multiplying (or dividing) power by 10 results in an addition (or subtraction) of 10 dB.

2. Doubling (or halving) your power results in an increase (or decrease) of 3 dB.

There is a (somewhat hairy!) mathematical formula (from which the table is derived), but I do not know it.

The formula of X dB + Y dB is :SPL sum = 10 log (10^x/10 + 10^y/10). Thus Sound Pressure Levels sum of 60dB + 61dB =10 log(10^6 + 10^6.1)

Power(dB)=10log(Watts) Watts=10^(dB/10) If using dBm then divide the power in Watts in the first equation by 1mW(1*10^-3)

English Translation: A decibel, dB, [10log(x)] is just a measure of something relative to a standard reference point (usually 1 milliwatt, dBm, 1 watt, dBW, the SPL at the standard threashold of human hearing, dB or antenna gain relative to omnidirectional). Usually the reference point is 1 unit so its logarithm is 0.

To say something is twice as loud, say a stereo at 2W compared to 1W is 10log(2) - 10log(1) = 10*.3 = +3dB. Its just a way of avoiding long division in lunchroom conversations on a napkin.

If you only have one value of power you cannot convert it to dB, because for this you will need a reference power. For example, if you have a device that produces 100 watts at the input and 1000 watts at the output, then you can determine the device power in db by taking the log-base-10 of the ration 1000/100 (the reference in this case is the input power) and then multiply the result by 10. In this case you will get 10dB.

On the other hand, if you know only one value of power, then you cannot determine the number of dB, but you can calculate the power in dBm. The dBm unit uses a reference power (always) of 1 mW. For example, if you know that a device produces 1000 watts, to calculate the power in dBm you must calculate the log-base-10 of the ration 1000/1mW or 1000/0.001 and then multiply the result by 10. In this case you will get 60dBm. I hope tyhis explanation will help you.

Abe

thank you very much now i understand it

RE: Power and Decibel relationship
It is not power but 'Power_Ratio' ie P[in] / P[out] .
Deci-Bell [db] = 1/10 of 'Bell' a a power-ratio unit to denote power gain [in amplifiers] or power loss [in Attenuators or impedance-matching networks, Filters etc.
Bell is too large a unit so deci of bell [decibell or db] is used in calculations.
Gain = 10x log[to the base 10] P[in]/ P[out] + value =Gain - value = loss
Note: You will be well aware of calculations that all units should be the same for + - / x
DBm is wrt 1mw
For voltage gain it is =20log V[in] / V[out].
Mind in logarithms x2 means square & divided by 2 is square-root

When I was at school in the forties I well understood decibels in relation to sound levels but could never find anyone who knew what the reference level was (0 db), it was not till I left school was I able to find it was 0.00002 dyne's/cm^2 at 2 KHz in old pre SI units

Is there some sort of conversion formula to go from dB10 (watts) to dB20 (voltage) or vise versa?

define and explain please

Decibel and its relationship to power