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  • (Power) Db to Watt

    This is so simple, ...but I can't nail it!!

    What is the conversion from Db to Watt?

  • #2
    Re: (Power) Db to Watt

    0 decibel is equal to 1 picowatt per square meter. This corresponds to the faintest sound that can be detected by a person with good hearing.

    A sound of 10 decibels means it is perceived as 10 times louder than 0 decibels.

    Since decibels is a logarithmic unit of perceptive sound, I do not think there is a direct conversion between it and power.

    I have seen conversions between decibel and power/area and between decibel and pressure, but I am fairly sure it would depend on the specific circumstances.

    Perhaps someone in the audio field could shed more light on the subject.

    Comment


    • #3
      Re: (Power) Db to Watt

      This would be power conversion in RF terms. ie., 20Db @ 850Mhz. There is a power value in Watts. As in, so many Db equals so many watts.

      I'm not clear on the conversion???

      Comment


      • #4
        Re: (Power) Db to Watt

        This is the closest thing I could find on the topic:

        http://www.mogami.com/e/cad/db.html

        Comment


        • #5
          Re: (Power) Db to Watt

          Originally posted by Scitnor
          This is so simple, ...but I can't nail it!!

          What is the conversion from Db to Watt?
          108 DB = WATT?

          Comment


          • #6
            Re: (Power) Db to Watt

            dbm = 10^((WATTS - 30)/10)

            Comment


            • #7
              Re: (Power) Db to Watt

              30db = 1 watt (1000 mw)
              27db = .5 watts (500mw)

              for each 3 db gain or loss it doubles or halves.

              33db = 2 watts (2000mw)

              36db = 4 watts (4000mw)

              24db = .25watts (250mw)

              Comment


              • #8
                Re: (Power) Db to Watt

                Forgot the formula!

                Converting db to mw and mw to db

                db=10*log(mw)
                mw=anti-log(db/10)

                Comment


                • #9
                  Re: (Power) Db to Watt

                  Originally posted by Scitnor
                  This is so simple, ...but I can't nail it!!

                  What is the conversion from Db to Watt?
                  DBM - 30 divide by 10 Inv log


                  60 dbm -30= 30 Divide by 10 = 3 log of 3 = 1000
                  50 dbm -30 = 20 Divide by 10 = 2 log of 2 = 100

                  +60 dbm =1000 watts
                  +50 dbm = 100 watts

                  each 3 db is twice or 1/2 of starting power

                  Comment


                  • #10
                    Re: (Power) Db to Watt

                    I am trying to convert dBm to Watts and none of the formula described so far work.

                    I have a reliable conversion from watts to dbm, but I cant reverse the situation, the formulas here don't reverse the result.

                    I am trying to do this in Excel, it is being used in a calculator for a Wifi Link budget.

                    This is the current Watts -> dBm (works) =10*LOG10(D8/ 0.001)

                    and in its current form, dBm -> Watts (doesn't work) =LOG((D9-30)/10)

                    'D8' (watts) and 'D9' (dBm) accept input from a user =)

                    If anyone can help, please email me [..guest link removed..]

                    Comment


                    • #11
                      Re: (Power) Db to Watt

                      Originally posted by ZachFlem
                      I am trying to convert dBm to Watts and none of the formula described so far work.

                      I have a reliable conversion from watts to dbm, but I cant reverse the situation, the formulas here don't reverse the result.

                      I am trying to do this in Excel, it is being used in a calculator for a Wifi Link budget.

                      This is the current Watts -> dBm (works) =10*LOG10(D8/ 0.001)

                      and in its current form, dBm -> Watts (doesn't work) =LOG((D9-30)/10)

                      'D8' (watts) and 'D9' (dBm) accept input from a user =)

                      If anyone can help, please email me [..guest link removed..]
                      You didn't reverse it correctly. If D9 is correct value in dN, then
                      Watts = (0.001 W)* 10^(d9/10)

                      ^ represents the funtion "raise to the power of" and ten raised to a power is the antilog function.

                      Comment


                      • #12
                        Re: (Power) Db to Watt

                        =(10^-3)*10^(db/10)

                        Comment


                        • #13
                          Re: (Power) Db to Watt

                          dBm = log10 (mW)*10
                          mW =10^(dBm/10)

                          Comment


                          • #14
                            Re: (Power) Db to Watt

                            The correct conversion is W=(1*10^-12)*10^(db/10)

                            W=power output in watts
                            db=sound intesity emitted
                            (1*10^-12) is the sound intensity threshhold
                            10^(db/10)= antilog function

                            Hope this helps. Most people forget to input the threshhold and thats what screws them up.

                            Comment


                            • #15
                              Re: (Power) Db to Watt

                              Originally posted by Physics Student View Post
                              The correct conversion is W=(1*10^-12)*10^(db/10)

                              W=power output in watts
                              db=sound intesity emitted
                              (1*10^-12) is the sound intensity threshhold
                              10^(db/10)= antilog function

                              Hope this helps. Most people forget to input the threshhold and thats what screws them up.
                              Different fields use different reference levels. Your formula is correct if the reference level is 1 pW. Audio frequently uses dB re 1 mW or even 1 W as the reference level. The reference level always needs to be clearly specified at least once to get everybody on the same page.

                              (Sound level measurement uses a reference of 20 µPa as a standard pressure.)

                              Comment

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