Subject: BASE DB and Watts
From: "Saved by Windows Internet Explorer 8"
Date: Wed, 22 Sep 2010 16:06:32 -0700

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.

DECIBELS

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):

P1
10x log ------
P2

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):

V1
20x log ------
V2

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

2
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).

*******************************************************************

ADDING POWERS, SPL
LEVELS AND VOLTS IN DECIBELS

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)

equals

15
20x Log ------
.775

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:

-5
10
10x LOG -------------
-12
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:

-5
2 x 10
10x LOG -----------
-12
10

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:

-5
4 x 10
10x LOG ---------
-12
10

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

lets try 10 guitar amps:

-5
10 x 10
10x LOG ------------
-12
10

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!!!!

*****************************************************************

ACOUSTIC POWER and SOUND PRESSURE LEVEL
READING IN DBs

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...

P1
10x LOG --------
P2 (our reference power)

and for voltages:

V1
20x LOG --------
V2 (our ref voltage)

So when you measure ACOUSTIC POWER in WATTS PER SQUARE METER

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

P1
db = 10x log -------
P ref (our reference here will equal .001 watt)

-55 db = .000 000 003 watts

2 db noise =

P1
2 db = 10x log ---------
P2 (P2 equals the absolute minimum noise
level possible which is -128 dbm )

to find wattage of minimum noise

P1
-128 dbm = 10x log ----------
.001 watt

which equals .000 000 000 000 000 15 watts

then

P1
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!)

THEN:

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 ----------------------------
.001

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.

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