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Why do amplifiers with a high output impedance have trouble delivering power/current to low impedance speakers/headphones?

virtua

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Hi everyone,

I've read about damping factor, and how it relates to the control of drivers, and the corresponding changes that it can make to a frequency response based on the impedance curves of the amp and drivers.

However, I've always noticed that amplifiers which have a high output impedance and corresponding low damping factor were never able to deliver much power/current to low impedance speakers/headphones. Is this at all related to damping factor, or is there other variables involved here? What causes this to happen? I feel like this is a really simple question but I haven't really found a simple answer to it yet.

Also, beyond this, I've found examples such as the SPL Phonitor X, which despite having an output impedance less than 0.4ohm (measured by Amir) was still unable to provide enough power/current to low impedances. This especially made me question things, this to me looked similar to the characteristics I've commonly found high output impedance amplifiers to have. Is this to do with a lack of current able to be supplied by the power supply?

Thanks,
Virtua
 

fpitas

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I'll kick this off with an article about DF:

 

unpluggged

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It's because their output impedance limits the current they can deliver to the load. As basic as Ohm's law.

On the second question, some amplifiers with very low output impedance can use current limiters as means of protection. I'm not familiar with the SPL, but RME ADI-2 Pro/DAC's "Extreme Power" HP amp is an example.
 
OP
virtua

virtua

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I'll kick this off with an article about DF:

Thanks, I've read through the article and feel like I have a better understanding of the losses that occur with a low damping factor now. So from what I understand, you could build a very powerful amplifier, however if the internal impedance of the amplifier is too high compared to the headphone/speakers, making for a low damping factor, it will cause great amounts of signal loss (signal being voltage and related current) vs an ideal lower internal impedance design. So even with an amplifier and headphone/speakers both with flat impedance curves, it will cause great signal loss uniformly (which would be shown for example in Amir's low output impedance distortion vs power graphs being limited in maximum level before clipping), but in most cases will cause varying signal losses at different frequencies which inherently change the FR. Would this be about right?

It's because their output impedance limits the current they can deliver to the load. As basic as Ohm's law.

On the second question, some amplifiers with very low output impedance can use current limiters as means of protection. I'm not familiar with the SPL, but RME ADI-2 Pro/DAC's "Extreme Power" HP amp is an example.
That's really interesting about the current limiting, I think it could definitely be the case here. It seems there has been some idiosyncratic design choices on the Phonitor X, so I just find it a bit interesting in quite a few of the measurements since it's able to deliver a huge amount of power into high impedance loads but very little into low impedance loads.

Thanks for the help everyone!
 

Sokel

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It's because their output impedance limits the current they can deliver to the load. As basic as Ohm's law.

On the second question, some amplifiers with very low output impedance can use current limiters as means of protection. I'm not familiar with the SPL, but RME ADI-2 Pro/DAC's "Extreme Power" HP amp is an example.
Yep,mine also limits current to 38A,it's common for high power class D ones.
 

Sokel

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38A. That's funny.
That's what they say.
If they are wrong so am I :)

38A.PNG


(it's the one inside Peachtree that Amir tested )
 

jsilvela

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I think the very simple answer is to see your amp as a battery in series with a resistor, as per Thévenin's theorem.

So, you have an ideal battery (pure voltage source) connected with two resistors in series. The amp's impedance, and the load's impedance.
If the load's impedance is equal to the amp's, it gets half the voltage, and 1/4 the power.
If the amp's impedance is much lower than the load's, then most of the voltage goes to the load.
 

restorer-john

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I think the very simple answer is to see your amp as a battery in series with a resistor, as per Thévenin's theorem.

So, you have an ideal battery (pure voltage source) connected with two resistors in series. The amp's impedance, and the load's impedance.
If the load's impedance is equal to the amp's, it gets half the voltage, and 1/4 the power.
If the amp's impedance is much lower than the load's, then most of the voltage goes to the load.

Firstly, the 'pure voltage source' is not pure.
Secondly, the amplifier's impedance varies with frequency and output current.
Thirdly, the load impedance (resistance and phase) varies with frequency.
 

jsilvela

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Firstly, the 'pure voltage source' is not pure.
Secondly, the amplifier's impedance varies with frequency and output current.
Thirdly, the load impedance (resistance and phase) varies with frequency.
Right, that's true.
The OP was asking for a simple answer, and I was going for that, crude as it is.
You could apply my reasoning "per each single frequency".
 

theREALdotnet

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Firstly, the 'pure voltage source' is not pure.

When modelling a real voltage source you come up, as the first approximation, with a pure (ideal) voltage source and a series resistor. Thévenin's theorem states that all networks comprised of DC voltage sources, DC current sources and resistors can be reduced to this model. Not sure why it’s a theorem with its own name, it follows logically from Ohm’s law (or Maxwell’s equations, if you like).

So, the pure voltage source is a theoretical construct that does not exist in reality on its own.

Secondly, the amplifier's impedance varies with frequency and output current.
Thirdly, the load impedance (resistance and phase) varies with frequency.

That’s where things get interesting. Without frequency dependency of source and load impedance neither would be of much importance. The source impedance would simply limit the maximum (short circuit) current that can be delivered. For any desired output power and load impedance one would dimension the source voltage and resistance to suit.

But given the frequency dependency of both (second approximation), the source and load impedance act as a frequency-controlled potentiometer, causing linear distortions in the output power.

The only defence against that (other than minimising the frequency dependency) is making the source impedance as low as possible, orders of magnitude lower than the load, so that variances of either don’t matter much anymore.
 

tomtoo

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Its just the behavior of a ordinary voltage divider that the system internal amp resistance and load resistance is.
In easy words if the amp has high inner resistance it needs high outer resistance(load) to deliver voltage to the load. If you have very high inner resistance and very low outer, all the voltage is over the inner resistance.
 
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tomtoo

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Firstly, the 'pure voltage source' is not pure.
Secondly, the amplifier's impedance varies with frequency and output current.
Thirdly, the load impedance (resistance and phase) varies with frequency.

Sure. But if i would ask you does a 8% alc beer or a 2% beer makes more drunk. You would answer it depends on............ :)
 

theREALdotnet

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Sure. But if i would ask you does a 8% alc beer or a 2% beer makes more drunk. You would answer it depends on............ :)

The frequency, in glasses per hour?
 

SSS

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Hi everyone,

I've read about damping factor, and how it relates to the control of drivers, and the corresponding changes that it can make to a frequency response based on the impedance curves of the amp and drivers.

However, I've always noticed that amplifiers which have a high output impedance and corresponding low damping factor were never able to deliver much power/current to low impedance speakers/headphones. Is this at all related to damping factor, or is there other variables involved here? What causes this to happen? I feel like this is a really simple question but I haven't really found a simple answer to it yet.

Also, beyond this, I've found examples such as the SPL Phonitor X, which despite having an output impedance less than 0.4ohm (measured by Amir) was still unable to provide enough power/current to low impedances. This especially made me question things, this to me looked similar to the characteristics I've commonly found high output impedance amplifiers to have. Is this to do with a lack of current able to be supplied by the power supply?

Thanks,
Virtua
On the SPL website there are almost no technical specifications for the Phonitor X available. Therefore it is not possible to comment regarding the ability to drive headphones depending of the needed power or voltage or current. The damping factor does not tell you much on this. Whether a headphone will be loud enough depends on the headphone impedance (dynamic speaker, electorstatic ones are totally different) and the voltage/current capability of the driving amplifier. As some other participants here already stated, it just is a question of electricity formulas for a voltage divider between amplifier impedance and headphone speaker impedance. By the way, the damping factor is derived from the load impedance compared to the amplifiers output impedance. So if the headphone impedance is not known the damping factor cannot be calculated.
 
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