DB's and Watts

the material below is copyright 1993 by M. Gore.  used by permission. 
 Use of this material for  personal use only, is granted.
      Any other use is prohibited.  And is also very BAD Karma.

The decibel (which is one tenth of a BEL), is used to compare two 
levels of electric power in terms of the logarithm of their ratio.  In 
somewhat simpler terms, a decibel is a RATIO between two powers, 
voltages or Sound Pressure Levels (SPLs).

The advantage of using decibels (abbreviated as db) is that decibel 
units compress a huge range of RATIOS into numbers that are easy to 
understand. As an example, 10,000 watts vs. 1 watt is equal to a 40db 
difference between them, a bit easier to understand and talk about than 
to have said "a 10,000 to 1 ratio".  Plus a couple of db is about the 
minimum change the normal person can detect in audio signals, and 
thus if we can discuss specifications and measurements in detectable 
units, these units then mean something to us.  A 3 db change in noise 
is detectable.  A difference between .001 volt  and .00111 volt isn't.

A good thing to remember is that a full BEL (ten deci-bels) is 
"TWICE AS LOUD" to most people. 

Since the central figure in dbs is POWER, the readings must be the 
same unit, you cannot compare watts and voltages unless you first 
convert the readings into the SAME UNIT. 

Also note that for correct comparisons, wattage readings (and voltage 
readings) must be made using the same load (resistance in most cases).  
It does little to compare an amplifier that can drive an 8 ohm speaker 
to one that can only drive a 400 ohm device.  

There are two formulas used to find a db ratio between units:

The first is used ONLY FOR POWER (watts):

 10x  log   ------  

In words this is "10 times the log of the product of Power #1 Divided by Power #2"

First you find the RATIO by dividing the Power of one reading into the Power 
of the other reading, then find the Log of this result then multiply that 
figure by 10.

You must use this formula when comparing wattages, or Sound 
Intensity which is also measured in watts.

The second is used ONLY FOR VOLTAGES (and SPL levels):

             20x  log  ------

First find the ratio by dividing the two voltages, then find the Log of 
that figure, then multiply by 20.

This formula is for voltage relationships and Sound Pressure Levels. 
If you know math very well, you will see that since voltage squared 
divided by the resistance equals watts, when you have two voltages 
that you are comparing across the SAME RESISTANCE, the 
resistances cancel each other out and 20 log V1 / V2 equals 
10 log V1 squared divided by V2 squared.  

Thus the two formula really do give equal results when transformed to equal units!!!  

In other words, the two formulas are equal to one another, 
the voltage formula just cancels out the resistances and 
takes into account the squared voltages required in 
the power formula 

Power  = I  R 

Often we use a already known standard and compare a power or 
voltage to this standard.  Some of the standards are listed below:

           0 dbm  equals 1 milli Watt dissipated in a 600 ohm                              
		resistor.  Any reading with dbm in it, MUST have a 
		600 ohm resistor in which the power is measured.
             NOTE: 1 mW is equal to .775 volts rms across 600 ohms.

           -128dbm equals the noise contained in a resistor
                             without any external connections, 
                             at room temperature, due to the 
                             collisions of electrons. 

   Sorry, you can't have less noise than this in the "real" world".

           +4dbm  equals the standard for output at "0VU" from                             
		  professional equipment. This is 1.228 Vrms
		  across 600 ohms.
                  This is "NORMAL Professional OUTPUT LEVEL"

           -10 db equals different readings when made across                            
		different resistance loads.  Usually it is .3 Vrms 
		across 10,000 ohms. This is the norm
	 	      for SEMI-PRO gear. 

           dbV is used for references of 1 Volt RMS across a                            
		"given" load, usually a high impedance more
                   than 10,000 ohms

NOTE WELL that a 3 db increase in POWER is twice the POWER, 
while a 6 db increase in VOLTAGE is twice the VOLTAGE, 
and a 10 db increase in POWER is about what the "normal" 
person considers "twice as loud".

Note that for every 10 db increase in POWER you get a decade 
increase in Wattage:

             0 db increase = same power
             3 db increase = 2 x power
            10 db increase = 10 x power
            20 db increase = 100 x power
            30 db increase = 1000 x power
            40 db increase = 10,000 x power

Note that for every 20 db increase in VOLTAGE you get a decade:

             0 db increase = same voltage
             6 db increase = 2 x voltage
            10 db increase = 3.16 x voltage
            20 db increase = 10 x voltage
            40 db increase = 100 x voltage
            60 db increase = 1000 x voltage
            80 db increase = 10,000 x voltage

Some "normal" Real-World db readings might be:

            noise of a tape recorder, in repro, 
             but having no tape running:            -58 dbm

    Overall noise level of a good Analog console:   -85 dbm

       Max output level of a older Neve Console:    +28 dbm
                 Tape recorder S/N :                 70 db

            Equivalent Mic Preamp noise:           -122 dbm

               Analog Console crosstalk:            -80 db

Note some readings are in dbm (referenced with a 600 ohm resistor 
and related to .775 Volts rms) and others are just db, meaning the ratio 
is between two measurements made on the same piece of gear under 
different conditions (ie: no signal vs. signal).                             



        Unfortunately YOU CANNOT JUST ADD DECIBELS!!!

Lets' say that we have a voltage of 7.75 volts rms (+20 dbm) across a 
600 ohm resistor and add to this another voltage with a dbm rating of 
+20 dbm.  Is the result +40 dbm?  NO WAY !!!!!

Since the two voltages are the same voltage (both are 7.75 volts) and 
we all know that doubling the voltage gains only a 6 db increase, the 
correct answer would be +26 dbm.

What we have to do in order to add dbs together (and SPL readings in 
db's) is to reduce the db figures BACK to voltages (or SPL 
readings or watts), then add these figures together then apply the 
correct db formula.....

so for 7.75 volts @ 600 ohms:

                     7.75 + 7.75
         20x Log    -------------
                        .775  (our reference voltage for 0 dbm)


         20x Log  ------

equals 25.73 dbm !!!  close enough to +26 dbm for me!!!

another way to state this is that if we have 1 guitar amp putting out 
70 db Sound Intensity (Sound INTENSITY is measured in WATTS, thus 
is power)... how many more amps to we need to sound twice as loud? 
(remember that we need a 10 db increase for the average listener to 
hear something "twice as loud")

1 amp = 70 db Sound intensity  (in Watts per square meter)

use the formula : 

   (since we've got the sound in Watts per square meter)

                  Intensity 1
         10x LOG ------------
                  Intensity 2  (the threshold of hearing will be our reference here)

  for 1 guitar amp:  

  10x LOG  -------------
              10   (threshold of hearing)

which equals 70 db Acoustic Intensity
   note this is the level ABOVE the threshold of hearing
   which is what "Acoustic Intensity" is defined as... 

So for 2 guitar amps:

                 2 x 10                      
      10x LOG  -----------           

which equals:

        73 db !!!

   ( Remember that if we add two equal wattages we get an increase 
                    of only 3 db!!!! )

now for 4 guitar amps:

                  4 x 10      
       10x LOG   ---------    

equals  76 db, not yet the required 10 db increase!!!!

lets try 10 guitar amps:

                10 x 10   
     10x LOG  ------------

which now equals 80 db !!!

  SO.... in order for the guitar player to sound TWOCE AS LOUD (a 10 db 
increase) she would need to have a total of 10 guitar amps (all the same) on stage!  

And quite a large stage if all the players needed 10 amplifiers each.....

  Then think if they wanted it "twice as loud" once again......

     100 amplifiers for each musician.... whew...

So how do we get around this over crowed stage yet 'sound twice as loud" ???

We use more EFFICENT speakers... expensive ones.  But we still have to have
a lot of them... just look at any large performance space and you'll see
lots of power amps and lots of speakers.  Usually these are built into the 
venue's own PA system or rented as needed.

             THE POWER OF SPEECH

The normal conversational speaker's voice is about 20 micro watts of 
SPL.  ( talkers during the classes I used to teach always seemed to 
have a lot more power... even whispers sounded loud to me!)

20 million people all talking at once (and saying the same thing, in 
exact phase with one another) would generate just enough power to 
light up one single 400 watt light bulb!!!!



There is confusion as to which of the two formulas for finding decibels 
one should use when dealing in SPL, and Acoustic Powers. 

as you remember:

for Powers (wattages)  P1 is Power measured for #1 unit
		       P2 is Power measured for #2 unit...

         10x LOG  --------
                     P2   (our reference power) 

and for voltages:

         20x LOG  --------
                    V2   (our ref voltage)


                    use the Power db formula

But when you measure SPL in dbs:

                   use the voltage db formula

In truth one should square the SPL readings and use the Power db 
formula, but using the easier Voltage formula works since:

            SPL1                                  SPL1 squared
   20x LOG -------     which equals    10x LOG   --------------
            SPL2                                  SPL2 squared


CONSOLE NOISE figures and Noise measurements in general

We cannot simply add  up db readings to get an overall noise level. 
To do correctly this we have to re- convert the db levels into watts, 
then add them, then find out the correct db ratios.
   (see note at the bottom of this page about noise phase !!)

Since the first stage of a mic pre-amp amplifies the noise level (as well 
as the signal) from the mic by say about 65 db, it creates a noise level 
at the output of the pre-amp of let's say -55 db, plus the noise inherent 
in the mic-pre amp itself.  We'll say the mic pre-amp has 2 db noise.

Then the total would be

         db = 10x log  -------
                        P ref    (our reference here will equal .001 watt)

    -55 db =  .000 000 003 watts


     2 db noise =

                      2 db =   10x log ---------
                                         P2  (P2 equals the absolute minimum noise
                                              level possible which is -128 dbm )

         to find wattage of minimum noise
             -128 dbm =  10x log  ----------
                                  .001 watt

           which equals .000 000 000 000 000 15 watts


           2 db noise = 10x log   ---------------------------
                                     .000 000 000 000 000 15

             2db noise = .000 000 000 000 000 24 watts

.000 000 000 000 000 24  watts plus  .000 000 003 watts

equals .000 000 003 000 000 24 watts  (whew!)


noise in db is...

                    .000 000 003 000 000 24 watts
        10x log    --------------------------
                         .001 watts

      which equals   10x log  -.000 003 000 000 24

     which then equals    -55.0001 db

next find out the power of the 5 db added in noise of the EQ amp:

                                  P eq
           5 db =  10x log  ---------------------------------
                               .000 000 000 000 000 15

      5 db noise = .000 000 000 000 000 475 watts

add this to the total noise before the EQ amp:

.000 000 003 000 000 000 24  +  .000 000 000 000 000 475 =

     .000 000 003 000 000 000 715 watts

then find the result:

                              .000 000 003 000 000 000 715
          noise =   10x log  ----------------------------
which  equals    10x log  -.000 003 000 000 000 715

which equals -55.0002 db

and so on......

          it isn't easy folks................

PLUS: Noise is not phase coherent, and thus will NOT add precisely the
same as two in phase signals.   While two in phase powers will add by
3 db, and two in phase voltages will add by 6 db, two noise voltages
will add by about 1.5 db to 2.5db.

This is due to the fact that each noise source has a different spectrum
of noise, and that noise spectrum is constantly changing, thus the phases of the
noise frequencies are always changing.

To find out the way two non-coherent signals add you can use this formula:

Total noise = square root of (signal 1 squared plus signal two squared)

This only approximates the 'total noise' of 2 noise
signals combined together, but generally is close enough for use.