• WANTED: Happy members who like to discuss audio and other topics related to our interest. Desire to learn and share knowledge of science required. There are many reviews of audio hardware and expert members to help answer your questions. Click here to have your audio equipment measured for free!

Amplifier Output Impedance (Damping Factor) and Speakers

Geert

Major Contributor
Joined
Mar 20, 2020
Messages
1,938
Likes
3,526
Actually, an amp with low DF should sound like less low frequency power regarding the frequency response measurement. But exactly the opposite happens!

An amplifier with low DF provides more power at the woofers resonance frequency, so nothing opposite happening.
 

pogo

Major Contributor
Joined
Sep 4, 2020
Messages
1,242
Likes
382
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.
And this is probably due to the fact that with a music signal, proper braking cannot be applied here when a low DF is present.
 
Last edited:

Geert

Major Contributor
Joined
Mar 20, 2020
Messages
1,938
Likes
3,526
Then even more power is applied at a higher DF. But exactly the opposite is audible compared to a lower DF.

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.
 

pogo

Major Contributor
Joined
Sep 4, 2020
Messages
1,242
Likes
382
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.
Due to the low output impedance of the amp, the voltage at the chassis and thus the raw SPL is higher, isn't it?
 

KSTR

Major Contributor
Joined
Sep 6, 2018
Messages
2,690
Likes
6,013
Location
Berlin, Germany
Due to the low output impedance of the amp, the voltage at the chassis and thus the raw SPL is higher, isn't it?
No, it's the other way round.
With DF = 0 (current drive) you'll get the highest terminal voltage (and SPL) at resonance where impedance is highest. Ohms law.
 

pogo

Major Contributor
Joined
Sep 4, 2020
Messages
1,242
Likes
382
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?
 

Geert

Major Contributor
Joined
Mar 20, 2020
Messages
1,938
Likes
3,526
But how does this fit with the measurements in post #237, which actually point to a lower SPL?

The measurements show (or mention) 0,4 dB higher output with an 8 ohm load (red line) opposed to a 2 ohm load (blue line). So at speaker resonance > higher impedance > higher output (SPL).

(Taking into account the graphs resolution is to low and is missing a decent legend and description to make an accurate analysis of these measurements).
 

pogo

Major Contributor
Joined
Sep 4, 2020
Messages
1,242
Likes
382
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!
 

Geert

Major Contributor
Joined
Mar 20, 2020
Messages
1,938
Likes
3,526
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!

As I said, due to the graph missing a legend and insufficient info in the accompanying text we're not sure what we're looking at. Can you tell for certain what the 4 lines in the graph refer to? We're not even sure what they're measuring exactly. We see a 0,4 dB and 6 dB of difference, ... of what parameter and between which scenario's?

Next to that, you don't exactly know what effects our at play in these measurements. They measure an amp with a switch to change DF, but you don't know how this feature is implemented and which side effects it might have. It's probably implemented by changing the negative feedback gain, as more negative feedback typically leads to a higher DF. This would mean you're not measuring the effect of different DF in isolation, instead your measuring the whole amplifier working in another tuning.
 

pogo

Major Contributor
Joined
Sep 4, 2020
Messages
1,242
Likes
382
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.
 
Last edited:

SIY

Grand Contributor
Technical Expert
Joined
Apr 6, 2018
Messages
10,386
Likes
24,752
Location
Alfred, NY
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.
Yes, we're aware that you don't understand the basic voltage divider. And apparently, Ohm's Law also eludes you.
 

pogo

Major Contributor
Joined
Sep 4, 2020
Messages
1,242
Likes
382
I don't see my error in thinking and trust the measurements made, i.e. the SPL is 0.4dB lower at 2ohms/df=60 as opposed to the higher df.
If I understood KSTR correctly, the SPL should actually be higher at low df values!?
 

xnor

Active Member
Joined
Jan 12, 2022
Messages
193
Likes
207
But how does this fit with the measurements in post #237, which actually point to a lower SPL?
Let's say the ideal amp (with Zout=0) outputs 1V, regardless of load.

Add a resistor with R=0.0114 ohm as output resistance and connect an 8 ohm load, such that DF=700.
The 1V will drop in parts over the output resistance, the rest will drop over the load.
In this case you get a 0.012 dB drop.
Decrease the load resistance to 2 ohms and now the drop increases to 0.049 dB.


Increase the output R to 0.1333 ohm, using an 8 ohm load, such that DF=60.
Now you get a 0.143 dB drop.
Decrease the load resistance to 2 ohms and now the drop increases to 0.56 dB.

The difference between the two is the observed ~0.4 dB.


Obviously, as you lower the load resistance, a higher proportion of the amp's (unloaded) output voltage will drop across the output resistance.


This can be directly translated into a complex load, such as a woofer:
Imagine that at the resonance frequency of 100 Hz Z=8 ohms, but at 1 kHz and up it's just 2 ohms.
At the resonance frequency, the drop will be 0.143 dB. At 1 kHz the drop is 0.56 dB.

=> That's a +0.4 dB higher voltage at the resonance frequency relative to 1 kHz. Effectively, it's a +0.4 dB bass boost.
 
Last edited:

pogo

Major Contributor
Joined
Sep 4, 2020
Messages
1,242
Likes
382
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.
 

xnor

Active Member
Joined
Jan 12, 2022
Messages
193
Likes
207
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.

DF, afaik, is usually calculated with a nominal 8 ohm load impedance.
Obviously, if you pick impedances at various frequencies from a speaker's impedance plot and calculate a DF from that, then you will get wildly varying numbers.

In my example above the DF drops from 700 to 175 and from 60 to 15 for each case respectively when load impedance is lowered from 8 to 2 ohms.

Again, obviously, as you lower the load resistance (or increase the source resistance and thus lower the DF), a higher proportion of the amp's (unloaded) output voltage will drop across the output resistance. Less voltage across the load.
 
Last edited:

pogo

Major Contributor
Joined
Sep 4, 2020
Messages
1,242
Likes
382
@xnor, Tnx for the confirmation.

Again from post #239:
Actually, an amp with low DF should sound like less low frequency power regarding the frequency response measurement. But exactly the opposite happens!

What explanation do you have for this observation?
 
Top Bottom