Andrew,
Of course, it IS immediately clear that I AVOID impedance matching from a linearity point of view, as impedance matching results in decreased linearity.
I sincerely apologize to Jan for this line: "On what bases is that immediately clear???"
Of course, it IS immediately clear that I AVOID impedance matching from a linearity point of view, as impedance matching results in decreased linearity.
effebi,
imagine water flowing in a pipeline, a large and constant diameter providing little resistance is advantageous.
imagine water flowing in a pipeline, a large and constant diameter providing little resistance is advantageous.
Well, I do not do plumbing usally, so it is hard to me to speak in non- electronics words. I should say , however, that any REAL pipeline (i.e. current source) has some vortex that return the flow (parallel resistor) So is advantageus if you have a big smooth pipe (low load resistance) and few vortex.
But I would appreciate if you can confute my previous posts with electronics argumentation, if any.
effebi
Well, that's only true if the output impedance of the source really is a physical resistor right there in series with its output. In the general case, the output impedance tells us nothing about the losses inside the source.
For example, a current source amplifier with an output impedance of say 100 kiloohms may (hopefully!) have an efficiency of more than 0.01%. 😉
Jan,
Yes, it is the logical consequence of what I said previously. I apologize again.
Yes, in the case of voltage transfer.
effebi,
the analogy is perfectly valid, there´s no need for intricate mathematical models. Likewise, it should be easy to realize that widening the pipeline anywhere (decreasing the resistance) will cause a pressure (voltage) drop, giving lower distortion but also a smaller dynamic range. Aiming maximum energy transfer in audio is not an exciting idea, again, it can result in seriously increased distortion.
Note: for maximum power dissipation in the load, the load resistance needs to be equal to the amplifier intrinsic resistance according to the Thevenin/Norton theorem.
Nice to discuss with you.
the analogy is perfectly valid, there´s no need for intricate mathematical models. Likewise, it should be easy to realize that widening the pipeline anywhere (decreasing the resistance) will cause a pressure (voltage) drop, giving lower distortion but also a smaller dynamic range. Aiming maximum energy transfer in audio is not an exciting idea, again, it can result in seriously increased distortion.
Note: for maximum power dissipation in the load, the load resistance needs to be equal to the amplifier intrinsic resistance according to the Thevenin/Norton theorem.
Nice to discuss with you.
Wrong, 100%. For one moment I thought you got it. Too bad.
Again: max power transfer, minimum power loss, is when the source has a minimum output impedance (minimum loss) so all of the power can be absorbed by the load.
If the source and load have matched impedances. they each dissipate half of the power, you you lose 50% of power from source to load.
Power = I^2 R. The I is the same in source and load, so if R is also same (matched), each has the same dissipation, half of the total.
If you have source with 0.1R and load with 10R, again, with same current (necessarily), source loss dissipation is I^2 * 0,1R; dissipation in load is I^2 * 10R, or 100 times as much. Loss is 1%, versus 50% in the case of matched impedances.
You get it now? I get tired to repeat it. Stay focussed!
jan didden
[BTW You remind me of a former member from scandinavia. Are you sure you are a new memeber here?]
jan didden
you have changed the conditions to suit your definition.
That is not what he meant initially. At least I think it's not what he meant.
Hmmm, now that you mention it...who was that guy...it's on the tip of my tongue...
why did we stop displaying addresses in the corner of our posts?
Was it something to do with security or lack of it?
GK springs to mind, but I do like his contributions. And all the other experts that have been chased away by our bickering.
Was it something to do with security or lack of it?
GK springs to mind, but I do like his contributions. And all the other experts that have been chased away by our bickering.
I miss Glen...🙁
But that is NOT who I'm thinking of. The x-member I'm thinking of, his name rhymed with Mumba...
😀
True
This is also true but it is not the maximum output power the amplifier can produce. Assuming a linear amplifier, the maximum power the amplifier can produce is when the source output Z is equal to the load Z; 50% of power is lost in load and 50% in the source. Of course half the output voltage is also lost in the source so that we generally try to have a much lower source Z.
If efficiency is the goal, then Jan is on the right track. It looks like you guys are trying to define two different situations.
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originally he posed the idea that impedances should be equal to minimise the loss.
I understood that to mean that we have a certain amount of power available and the best way to extract that power is to feed it into a load that has the same impedance as the source.
Various references were made to RF practice where this is accepted as being the case.
If that holds for the RF situation then the science does not change when considering lower frequencies. The science is the same and the rule is the same.
Now you are trying to change his statement into something he did not say.
You have changed the conditions that make his statement appear untrue and deliberately biased towards making your definition appear true.
You are considering a different situation, just like considering voltage transfer is a different case.
I think he is considering THE situation - where both the output impedance and the load impedance determines the amount of power delivered.
Hey, wait a second now! Sorry for repeating myself, but here it goes again:
The output impedance of a circuit actually says nothing about the power dissipated within! If this is not immediately obvious then these examples may help:
For example, an 100W class AB audio amplifier with a 0.01 ohm output impedance driving a 8 ohm load at full power dissipates a lot more than just a hundred milliwatts, obviously. Conversely, an amplifier that has a current-source characteristic will not have a power loss that is enormous just because the output impedance is high.
It follows that the power loss can not be predicted from the output impedance.
The output impedance of a circuit actually says nothing about the power dissipated within! If this is not immediately obvious then these examples may help:
For example, an 100W class AB audio amplifier with a 0.01 ohm output impedance driving a 8 ohm load at full power dissipates a lot more than just a hundred milliwatts, obviously. Conversely, an amplifier that has a current-source characteristic will not have a power loss that is enormous just because the output impedance is high.
It follows that the power loss can not be predicted from the output impedance.
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Speaking for me, I assumed voltage (not power), since the original subject was the VAS driving the output stage. It seems to me, power is not the issue, voltage is.
Where it went off track was when you said 'for power' and yes, for max power transfer, impedance should be matched.