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Old 10-25-2009, 10:47 PM
ninjackn ninjackn is offline
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Originally Posted by saratoga View Post
No thats not right. And if P_out is in units of watts, then its actually 10*log10().
No, P is units of pressure.

Originally Posted by saratoga View Post
Originally Posted by ninjackn
20*log10((P_out/P_ref) / (V_out/V_ref)) = output dB
Thats not right either. You need to convert V_out to dB and subtract, or else not use dB at all.

Let me put it this way, if you divide a logarithm by a number then you haven't done your math right. They add or subtract. Division isn't something thats very meaningful on them.
Oh, but it IS correct:
20*log10((P_out/P_ref) / (V_out/V_ref)) = output dB
First off we agree that if you have a sensitivity of X dB @ 1V (or "X dB/V" using AKG spec. notation) and if you have a voltage of Y dB applied then the output is = X dB - Y dB.

According properties of logarithm, division is equivalent to subtraction. We see this because dividing two exponential of the same base is equal to subtracting the exponents, that is (C^A)/(C^B)=C^(A-B).

Thus my math IS correct:
20*log10((P_out/P_ref) / (V_out/V_ref)) =
20*log10(P_out/P_ref) - 20*log10(V_out/V_ref) = X dB - Y dB = output dB.

This is also why I said i was off by a "factor of 2" because i'm used to working with power as milliwatts and converting it to dBm. To convert X milliwatts to dBm it's 10*log10(mW_power/1mW).

Often when dealing with voltages it's assumed that the reference has the same impedance and the measurement that was taken so it's converted to power:

P = V^2/Z (As it's been said on this thread may times)

10*log((V_out^2/Z) / (V_ref^2/Z))
we see that the impedance cancels out
10*log((V_out^2) / (V_ref^2))
the squares can be pushed out:
exponential in logs becomes multiplications and ta-dum:

So to clarify my point of view on "dB/V" further, I did not know if dB/V was a relationship of sound pressure in dB to volts or if it was a relationship of sound pressure in pascals to volts. I assumed that it was sound pressure in dB to volts which because I read the units "dB/V" as "dB per Volt" which is incorrect because it's actually describing dB of Pascals per 1V.

Last edited by ninjackn; 10-26-2009 at 02:30 AM. Reason: Clarifying that P is pressure and not power.
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