Nezer
Antibacterial soap... kills bacteria... bad karma?
- Joined
- Dec 6, 2001
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Now, Tangent you have my ass all confused on the dB to G relationship. I thought it was :
dB=20 log(G)
Where G=5 is about 14dB and where G=10 is 20? This gives a differenc of 6dB.
The formula that matchs your values (in the other thread I decided to keep this discussion out of) is
dB=10 log(G)
which is only useful when G is a watts value.
Uggh! Now I'm so confused again!!
So, in our case, calculating *voltage* gain, which of these formulas is accurate:
dB=10 log(Vout/Vin)
or
dB=20 log(Vout/Vin)
*when dealing with amps the Vin is assumed as 1 becasue we are more intrested in the gain ratio. So a gain of 5 would mean 5 volts out to every 1 volt in.
Would one of you creatures that is infinity more intelligent than I please clear this mess up for me!?
And while your at it, could you also invert the equation so that I can calculate voltage gain based on a dB?
dB=20 log(G)
Where G=5 is about 14dB and where G=10 is 20? This gives a differenc of 6dB.
The formula that matchs your values (in the other thread I decided to keep this discussion out of) is
dB=10 log(G)
which is only useful when G is a watts value.
Uggh! Now I'm so confused again!!
So, in our case, calculating *voltage* gain, which of these formulas is accurate:
dB=10 log(Vout/Vin)
or
dB=20 log(Vout/Vin)
*when dealing with amps the Vin is assumed as 1 becasue we are more intrested in the gain ratio. So a gain of 5 would mean 5 volts out to every 1 volt in.
Would one of you creatures that is infinity more intelligent than I please clear this mess up for me!?
And while your at it, could you also invert the equation so that I can calculate voltage gain based on a dB?