Time for one of those long, boring semi-technical posts that no one here reads...
1. The decibel is a relative rather than an absolute measurement, i.e. it is used to measure the ratio of one signal to another. I am sure everyone is most familiar with it's use in representing the signal to noise ratio of equipment.
- Richard B.
I know this is a confusing topic for we who don't work with these kinds of calculations. I've looked at this thread and have tried to walk through some figures for myself...
My example is my Sansui AU-6500
+25db Power Amp
Did I make any mistakes?
Pauln
This would infer that your power amp makes 320 watts. I'm kind of doubting that.
How are you getting the 320W figure?
Here is what I have tried for the power amp stage:
If I use voltages I have the input voltage of 0.8V to get rated output of 28W,
The main power amp has a sensitivity of 800mv at which it provides rated power of 28W, so I need to figure how much volts are being output from those watts.
Anyone has an explanation why many manufacturers continue to insist on 500mV or so input sensitivity? Seems modern digital sources tend do send 2Vrms out which significantly compresses the useful range of volume controls, attenuators can be used in between but may lead to degraded SQ.
How are you getting the 320W figure?
Here is what I have tried for the power amp stage:
If I use voltages I have the input voltage of 0.8V to get rated output of 28W, so I'm looking to comprare the output voltage at rated power.
28W=V^2/8ohm so V=15 at rated power, then 20*log(15V/.8V)=+25.5dB
I'm still comparing volts, but is that still considered "potential or pressure" at this stage?
Mistreating volts as power or intensity, 10*log(15V/.8V)=+12.7dB
Looks better?
If I convert volts to power, I have 0.8V^2/8ohm=0.8W as if the pre-amp output was treated as "power" and applied to the speaker load, so then 10*log(28W/0.8W)=+15.5dB
This also looks better, but I'm not sure I can use the pre-amp output like that since it really sees the power amp's input impedance... hmm...
If I use the input impedance of the power stage I have 0.8V^2/40Kohms=0.000016W so 10*log(28W/0.000016W)=+62.4dB
That's not right, but maybe it is unfair to use two different loads... 15V^2/40Kohms=0.005625W so 10*log(0.005625W/0.000016W)=+25.5dB
I'm bouncing around and running out of things to compare...