Hi,

Db have been invented to have a meaningful way to compare ratio level between different fields: electronic, acoustic, digital...

When used without anything with them ( eg: db ) this is what happen.

Once you have something added to them things change and they are now

**referenced **to something:

DbFS : used in digital and Full Scale refer to the absolute maximum level digital can produce ( =0dbfs, iow in digital you'll often see -xx dbfs ( minus xx dbfs)

DbSPL: acoustic level referenced to 1Pa which equal to 94dbSPL

DbU: in electronic, voltage related to 0.7746 Volts ( without relation to any given impedance)

Etc,etc,etc,...

When talking about Voltage the logarithmic scale is based on 20log.

Eg 20log(2)= 6.

Iow +6db equal doubling voltage.

When talking about power then its based on 10log.

Eg 10log(2)= 3.

Iow when you double power ( of an amp) you gain +3db.

So to answer your initial question: 2,83v into 4ohm load gives 86db ( @ 1meter)

How much power do you send to your loudspeaker to have 86db?

You need an Ohm's law wheel to the rescue if like myself you can't remember the variations around Ohm's law( i'll let you google it

).

We know to define power from known Voltage and Resistance ( impedance), Power= (V)2/R - V squared, divided by R

So (2,83)2 / 4 = 2

2w.

Now we add our 0,5v to 2,83v it gives ( 3,33)2 /4 = 2,77w

2,77/2= 1.35 ratio.

As we are talking power we'll use 10log:

10log(1,35)= 1.3db

86+1,3= 87,3db for 2,77 watts (3,33V into 4 ohms).