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Voltage to decibel conversion

Hello, how do I convert voltage to decibels? E.g. if a speaker (with unity gain and no power losses) outputted a 1Vrms sine wave, how many decibels would this be?
Thanks. I'm just doing some maths of potential dividers as filters
 
A decibel is a measurement of a ratio of power, not a measurement of voltage.

dB = 10*log(P2/P1)

However, power can be calculated from voltage, P=(V^2)/R. So the decibel calculation for voltage is (for the same resistance):

dB = 10*log((V2^2)/(V1^2)) = 20*log(V2/V1)

http://www.hottconsultants.com/techtips/decibel.html

That is not a conversion from voltage to decibel, that is simply showing that a power ratios can be expressed as a decibels... Decibels express a ratio of power not an amount of power like voltage, thus as I said you don't convert between the two (voltage to decibel)...
 
True, to this point we are still not doing voltage to dB conversions. However, we have illustrated the math necessary to understand dBu and dBV, which are direct conversions from voltage to dB. In this case V1 is understood to be a defined reference voltage, 0.775V for dBu, and 1.0V for dBV.
 
The funny thing with dB is that f.i 3dB will always be 3dB. It doesn't matter if we're talking power or voltage, 3dB will always be 3dB. The strange thing is that people (including me) seams not to understand this. An increase of 3dB means double the power which is equivalent to +41% increase of voltage (keeping the resistance unchanged).
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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decibels represent a change in something.

3db means a doubling (actually something more like 3.010dB is a doubling, but 3dB close enough for government work.

The issue is that we're most often talking about an increase in power. Electrically, if that power is dissipated across a fixed resistance the change in voltage required is related to the square root of the change in power.

Looking at it mathematically:

P = V * I and I = V/R, so P = V^2 * R

2 * P = 2 * V^2 * R

Since R stays the same, What is the change in V required so that 2 * V^2 = (V * d) ^2?

2 * V^2 = V^2 * d^2

Cancelling out the V^2 terms we get

2 = d^2

therefore d = sqrt(2).

More generally, the change in voltage is the square root of the change in power.

Now we know that dB is 10 times the log of the change in power.

If the ratio of changes of power is (P1/P2) is p, then the change in decibels is 10Log(p).

We already know that the change in voltage is sqrt(p), so what is a conversion from dB to change in voltage?

(V1/V2)^2 = (P1/P2)

Taking logs of both sides we see that

log( (V1/V2) ^ 2) = log(P1/P2)

Applying a little log magic to get rid of the squared term in the brackets

2 Log (V1/V2) = log(P1/P2)

multiplying both sides by 10 we get

20 log (V1/V2) = 10 log(P1/P2)
 
ah, I realise I've made a terrible mistake. I forgot about PIV!
anyway, thanks for your help!
thanks for all the maths, Steve, that'll help me immeasurably.
 
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