Speedskater
Major Contributor
It's not about 'damping factor' it's about 'source impedance' .
Actually, an amp with low DF should sound like less low frequency power regarding the frequency response measurement. But exactly the opposite happens!
Then even more power is applied at a higher DF. But exactly the opposite is audible compared to a lower DF.An amplifier with low DF provides more power at the woofers resonance frequency, so nothing opposite happening.
Then even more power is applied at a higher DF. But exactly the opposite is audible compared to a lower DF.
Due to the low output impedance of the amp, the voltage at the chassis and thus the raw SPL is higher, isn't it?With a higher DF the higher impedance of the woofer at resonance won't cause a rise in SPL at resonance, so no relative bass boost. Again, nothing opposite happening.
No, it's the other way round.Due to the low output impedance of the amp, the voltage at the chassis and thus the raw SPL is higher, isn't it?
But how does this fit with the measurements in post #237, which actually point to a lower SPL?With DF = 0 (current drive) you'll get the highest terminal voltage (and SPL) at resonance where impedance is highest.
But how does this fit with the measurements in post #237, which actually point to a lower SPL?
But I mean the overall volume difference at 2 ohms between the high and low DF, ie at a high DF the SPL is also higher!
Yes, we're aware that you don't understand the basic voltage divider. And apparently, Ohm's Law also eludes you.The two measurements at DF=700 are on top of each other and show no level deviation independent of the applied load (2/8ohms).
For better representation, the two measurements at DF=60 were placed below at -6dB. The lower level of 0.4dB at 2ohms can only be seen at the low DF (blue line) and in my opinion this tendency does not fit with the statement of KSTR: With DF = 0 (current drive) you'll get the highest terminal voltage (and SPL) at resonance where impedance is highest.
Uh huh.I don't see my error in thinking
Let's say the ideal amp (with Zout=0) outputs 1V, regardless of load.But how does this fit with the measurements in post #237, which actually point to a lower SPL?
All correct but I mean the loss of SPL (example of bass loss at 119Hz, as it is also described here: Link) and the loss is just higher the lower the df is. And this is also well demonstrated by the measurements on the T+A A200 at df=60/2ohms, meaning an SPL loss of 0.4dB compared to the high df.Obviously, as you lower the load resistance, a higher proportion of the amp's (unloaded) output voltage will drop across the output resistance.
All correct but I mean the loss of SPL (example of bass loss at 119Hz, as it is also described here: Link) and the loss is just higher the lower the df is. And this is also well demonstrated by the measurements on the T+A A200 at df=60/2ohms, meaning an SPL loss of 0.4dB compared to the high df.
That you still don't understand voltage dividers and Ohm's Law.What explanation do you have for this observation?