How to Convert dBV and dBA to dBW?

[KyleGranger]

TL;DR Summary

I am trying to figure out how to get dBW from dBV and dBA

I'm starting to consider filter design and notice that everything is in dB. I remember seeing this at some point in some previous coursework, but its completely slipped my mind. I know that there are plenty of online calculators, but I would really like to understand how to do the conversions myself.

Here is a really generic example that I am trying to figure out.

If I have a certain dBV and a certain dBA, how do I get dBW? I know that I can't simply multiply dBV*dBV since they're logarithms. Here what I think and where I get stuck.

if I have (e dBV) and (f (dBA), I can convert those to Volts and Amps using
(e dBV)=20log10(X)
X=10^(e/20) Volts

and

(f dBA)=20log10(Y)
Y=10^(f/20) Amps

knowing that P=V*I
P=10^(e/20)*10^(f/20)
P=10^((e+f)/20) Watts

I think this is right, but I'm not sure where to go to get back to dBW.

 

[KyleGranger]

I know that dBVand dBA use 20log10(x) but power is 10log10(x). I'm wondering why the difference in 10 and 20 as a multiplier. I read this has to do with power being proportional to V^2, but I'm still not sure how to do this and actually work it out.

 

[Averagesupernova]

I've never heard of dBA. You need to realize you cannot convert from volts to watts without knowing impedance. If you are simply increasing or decreasing volts into a given impedance, then it simply scales up or down. Make a few examples and do the math. It will make sense.

 

[tech99]

I know that dBVand dBA use 20log10(x) but power is 10log10(x). I'm wondering why the difference in 10 and 20 as a multiplier. I read this has to do with power being proportional to V^2, but I'm still not sure how to do this and actually work it out.

Actually, dBA is not used as a unit in everyday electronics because it is confused with the acoustic unit dBA.
But your calculation is correct. Remember that decibels refer to ratios. For instance, dBV is decibels relaitve to 1 Volt. Here is an example.
Power is proportional to Voltage^2. So to express a voltage in decibels relative to 1 Volt we can use either dBV = 10 log V^2 or dBV = 20 log V.
To express a power ratio in decibels relative to 1 Watt we can use dBW = 10 log P.

 

[DaveE]

dBuA, however, is commonly used in EMC standards. 0 dBuA = 1uA, 20 dBuA = 10uA.

 

[sophiecentaur]

I've never heard of dBA.

That's a good thing (lol) - although Power is I²R, same as V²/R. So dB's can be used and the factor is 20 for both.

Summary:: I am trying to figure out how to get dBW from dBV and dBA

I'm starting to consider filter design and notice that everything is in dB

That's a bit of an overstatement, I think. The Maths of circuit design tends to use a linear scale because Voltages and Currents in circuits need to be Added and Subtracted. You can't do that with dBs. I'd suggest you go to another source of information for your filter design.

 

[Averagesupernova]

That's a good thing (lol) - although Power is I2R, same as V2/R. So dB's can be used and the factor is 20 for both.

Oh I realize increase or decrease in current means *20 the same as voltage. I've just never heard of spec'ing current in decibels. Somewhere I am sure it's done. Everything I think wouldn't have a reason to exist usually does in some form for a good reason.

 

[jsg2021]

I know that dBVand dBA use 20log10(x) but power is 10log10(x). I'm wondering why the difference in 10 and 20 as a multiplier. I read this has to do with power being proportional to V^2, but I'm still not sure how to do this and actually work it out.

Log10(x^N) = N * Log10(x)

High school log identity.

P = V^2 / R

If you standardize the resistance/impedance, the R falls away and it's P ~ V^2 vs. V which with log10 gives a factor of 2.