Yup. Sensitivity matters dB/mW, the power is where it matters. So, even if its high impedance, it can have high sensitivity. Vice Versa. Depending on how high the output impedance is, would limit the voltage to the load, not to mention the current is effected.
Look up Bode plot. Real gain is
Logarithmic units and decibels
[edit]Power gain
Power gain, in
decibels (dB), is defined by the
10 log rule as follows:
where
Pin and
Pout are the input and output powers respectively.
A similar calculation can be done using a
natural logarithm instead of a decimal logarithm, and without the factor of 10, resulting in
nepers instead of decibels:
[edit]Voltage gain
When power gain is calculated using voltage instead of power, making the substitution
(P=V 2/R), the formula is:
In many cases, the input and output impedances are equal, so the above equation can be simplified to:
and then the
20 log rule:
This simplified formula is used to calculate a
voltage gain in decibels, and is equivalent to a power gain only if the
impedances at input and output are equal.
[edit]Current gain
In the same way, when power gain is calculated using current instead of power, making the substitution
(P = I 2R), the formula is:
In many cases, the input and output impedances are equal, so the above equation can be simplified to:
and then:
Quote:
EDIT: Although.. I am thinking that maybe the output impedance matters in all this.
The gain is a voltage gain.. but I think the loudness works with the power gain.
So, I think this is wrong.
Ok, so, it's 20* instead of 10* for voltage gain. I've fixed above.
The conclusion is the same.. 90% of the pot to get to the same voltage.
http://www.diyaudio.com/forums/tubes-valves/32334-db-into-voltage-gain.html
I am still thinking output impedance matters in all this....