It's 2010 and what's your fav VAS transistor?

Jump to Latest
#241
note, that post206 requotes both views that I drew attention to.
It depends on your viewpoint.
To preserve as much of the voltage or to preserve as much of the power.
To preserve as much of the power then equal source and load impedances do just that.
That was my original interpretation. But I could see an alternative where we might want work with the available voltage and minimise the loss of that instead. That was not what I interpreted from his first statement.
 
#242
Only the resistive component of an impedance is lossy. It is common to have the load impedance equal the source impedance without large losses, transmitters are the prime example. While transistor audio amplifiers give the appearance of low source impedance this value is useless for calculating losses. They are actually variable impedance sources with the impedance becoming the lowest at clipping.

That is where class D shines, always clipping with a mostly reactive source impedance
 
#243
AndrewT said:
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.
Click to expand...

The issue with the RF is that to avoid reflections and interference, you need to have a source, cable and load impedance that is the same. If they are not the same, the reflections etc cause power losses, obviously you don't want that. In THAT situation, the minimum loss and max power transfer is when all impedances are matched. In THAT situation, you dissipate 1/2 power in the source Zout and 1/2 in the load.

Audio doesn't concern itself with reflections and cable mismatch so there is no power loss due to reflections etc, therefore no need to match source, cable and load impedances. In THAT situation, max power transfer (min loss) is when source is voltage source with ideally zero Zout.

Edit: In ANY situation, max power transfer is with zero source impedance. It's just that in RF, to avoid reflections, the source impedance is matched to the load and THEN you lose 50% of your power in the source impedance.

jan didden
 
Last edited:
#247
cliffforrest said:
I think possible confusion comes from when one cannot have zero source impedance - a classic being a tube amp with an OPT. In this case maximum power IS when Zout = Zin. Hence TX taps to match the speaker.

This is "how it used to be" before SS amps became the norm.
Click to expand...

That is not max power in terms of losses/transfer. It is max power for the given supply voltage and bias current. If you mismatch you get asymmetric clipping which limits the max undistorted power with a given bias. Entirely different case.

jan didden
 
#248
janneman said:
Well, do the sums. I did it in an earlier post.

jan didden
Click to expand...

I saw the sums and I2R is not the loss in an impedance it is the loss in the resistive component of an impedance. Many impedances have a small resistive component and a corresponding low matching loss.

For a given load impedance power transferred is at a maximum with zero source impedance but for a given source impedance maximum power transferred is when load impedance equals source impedance for zero source impedance power transferred approaches infinity as load impedance approaches zero. Loss doesn't come into it unless there is a resistive component in the impedance.
 
#251
AndrewT said:
I did the sums and posted a selection in 219.
Did I not present them in a fashion that all could see that Rs=Rload for maximum power delivery from a source that already exists?
Click to expand...

Andrew, I didn't see that post. I agree with the case you presented.
My point was that in audio you don't need to match source- and load impedance, like in RF where you are concerned with reflections and absorption. Like in a preamp, where you want low source impedance to be able to drive a (capacitive) cable, it would be nonsense to try to match the load impedance to the source. Your power amp example is correct, but you wouldn't try to get a 0.1ohms speaker.

jan didden
 
#252
Jan,
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?]
Click to expand...
Could you rather present detailed, relevant and valid calculations please? Do you really think that you can upset firmly established physical principles?
 
#256
As previously stated,
For maximum transfer of power, the output impedance must be the same as the input impedance.
For maximum power dissipation in the load, the load resistance must be equal to the amplifier intrinsic resistance according to the Thevenin/Norton theorem.
Maximum power transfer has nothing to do with efficiency.

Useful equations:
Joule's law P = V I
Ohm's law
I = V / R
V = I R
R = V / I
Also
P = l² R
P = V² / R

Equal source and load resistances

V (source) 5.5V
R (source) 1 Ohms
I (source) 5.5 A
P (source) 30.25 W

V (load) 5.5V
R (load) 1 Ohms
I (load) 5.5 A
P (load) 30.25 W (reference value)

V (all in all) 11V
R (all in all) 2 Ohms
I (all in all) 5.5 A
P (all in all) 60.5 W

10 times higher load resistance

V (source) 1 V
R (source) 1 Ohms
I (source) 1 A
P (source) 1 W

V (load) 10V
R (load) 10 Ohms
I (load) 1 A
P (load) 10 W ¤¤¤

V (all in all) 11V
R (all in all) 11 Ohms
I (all in all) 1 A
P (all in all) 11 W

10 times lower load resistance

V (source) 10 V
R (source) 1 Ohms
I (source) 10 A
P (source) 100 W

V (load) 1 V
R (load) 0.1 Ohms
I (load) 10 A
P (load) 10 W ¤¤¤

V (all in all) 11 V
R (all in all) 1.1 Ohms
I (all in all) 10 A
P (all in all) 110 W
 
#259
janneman said:
But that's wrong. If I have a system with 50 ohms source and 50 ohms load you lose half the source power.
If you have a 1 ohm source with 100 ohm load you lose 1% power.
Click to expand...

You use 10/505 of available power, and lose the rest.


Source voltage = V;

Current = V/(100+1 Ohm).

Load voltage = 1 Ohm * V/(100 +1 Ohm)

Load power = V^2/101

In case of equal resistances:

Current = V/(1+1 Ohm)
Load voltage = 1 Ohm * V/(1+1 Ohm)

Load power = V^2 /2

I.e. instead of 50% power transfer you get 1/101, i.e. less than 1 %.

dadod said:
I agree with Workhorse, for impendance matching read some text books and let us have VAS discussion here.
Click to expand...

I don't understand what kind of discussion are you going to have about VAS if don't consider power transfer.
 
Last edited:
#260
I agree. At some level, power is transfered to from the VAS to the next stage, it has to. Even Fets require some amount of power to be transfered, at higher frequencies. Best that it is as little as possible, but it is still a measurable value.

Perhaps Kanwar's avitar smoke is going to his head.😀

Just picking at you Kanwar🙂
 
You must log in or register to reply here.
Top Bottom