Impedance matching / bridging between tube gear and modern audio interfaces

[abbey road d enfer]

Maximum power transfer sounds like it is the most efficient case.

It only "sounds" like it. Actually it's only 50% efficiency. "Transfer" is a deceptive term. The accurate description is "maximum power theorem", which has been insisdiously transformed in "maximum transfer theorem". Anyway "maximum" should not be read as "optimum".

I’m thinking about for a given amount of power.

There is no "given amount of power". Power delivered by the source depends on the load. Most of the sources in audio are voltage sources with an internal impedance. When the load is high compared to the source Z, current is small, as is power. When the load is small, most of the energy is lost in the source impedance. There is a point where power delivered to the load is maximum, when Zs=Zl.
This is assuming that the source can deliver enough current in all conditions. In practice the active stages are often limited by their own drive capability, or the power supply, or a dedicated protection system.

As the source impedance goes down and the load impedance goes up there is less current transfer and more voltage transfer.

That is true.

For a given amount of power.

See above.

As source impedance goes up and load impedance goes down there is more current transfer and less voltage transfer.

Partially true. Less voltage transfer, yes. Current transfer is debatable, since current in the load is equal to current delivered by the source, current transfer is always 100%.

 

[abbey road d enfer]

Why do equal source and load impedance’s minimize losses?

This is simply not true. When Zl=Zs, losses are 50%. Maximum efficiency (i.e. power to the load as close as possible to power delivered by source) is when Zs is very low and Zl is very high. Since current is very low, the system power is minimal.

 

[ruffrecords]

This is simply not true. When Zl=Zs, losses are 50%. Maximum efficiency (i.e. power to the load as close as possible to power delivered by source) is when Zs is very low and Zl is very high. Since current is very low, the system power is minimal.

And the reason for this is that the losses ae I squared R where R is the source impedance. As R is fixed, maximum efficiency requires minimising I. This is one reason why electricity pylons run at such high voltages.

Cheers

Ian

 

[Cqwet Dbdfte]

This is simply not true. When Zl=Zs, losses are 50%. Maximum efficiency (i.e. power to the load as close as possible to power delivered by source) is when Zs is very low and Zl is very high. Since current is very low, the system power is minimal.

"... losses are 50%.." Do you mean voltage or power?

Like CMMR pointed out in his example, POWER transfer efficiency is maximum when source and load are matched, i.e. equal.
And CMMR also pointed out, for audio signalling matching impedance is not important, with high impedance loads only voltage is important, and a higher load impedance will not drag down the voltage from the source, as to not induce non-linearities.
Use CMMR's example of 5V and 50Ohm and work out the math.

Impedance matching in audio is not an issue. Filters are calculated for specific load and source impedances, and would be buffered at input and output to avoid dependency on connected equipment.

Yes, an unloaded 50 Ohm source will have a higher voltage, but who cares. Unused 50 Ohm sources are terminated with 50 Ohm termination resistors, to preserve system performance. Line reflections were a problem on telefaxes, way back when in style, over long distances, and special termination circuits were needed to restore signals to get good pictures.

 

[abbey road d enfer]

"... losses are 50%.." Do you mean voltage or power?

Since current is the same in the source and in the load, both voltage and power are equally shared.

Like CMMR pointed out in his example, POWER transfer efficiency is maximum when source and load are matched, i.e. equal.

I don't support the name "transfer", which I consider an deceptive term, almost an abuse.

Impedance matching in audio is not an issue.

There are several examples where matching is required in audio. So-called "passive" EQ's and passive x-over filters require some kind of matching. Many pieces of equipment that use output xfmrs are sensitive to loading.

 

[Cqwet Dbdfte]

When Zl=Zs, losses are 50%. No, voltage from the source is half, power in the source and the load are the same.
Overloading has nothing to do with matching. Under-loading a signal source will not distort the signal. Over-loading any signal will most likely ruin the signal.
Did you do the math?
Run a simulation in LTspice.

 

[5v333]

if U2/U1 = 1/2
then U2*I /U1*I = 1/2 aswell
which is
P2 / P1
so
P2 / P1 = 1/2

 

[Cqwet Dbdfte]

You should add resistance into the math to clarify impedance match.
Again, matching impedance for audio signalling is not important. Over-loading has nothing to do "matching". I may be a great "match" for my wife, but we have essential differences.

 

[5v333]

Voltage divider:
U2 = U1* R2/(R1+R2)

if R1 = R2 = R
then U2 = U1* 1/2
and U2/U1 = 1/2

The current I is the same through both R1 and R2. So I is the only invariant.

 

[Cqwet Dbdfte]

But as R changes so does the power.
Not sure what you are trying to say...

 

[5v333]

I was just trying to show that the power loss is 50% when Z1 = Z2.

 

[5v333]

maybe i missunderstood something...

 

[Cqwet Dbdfte]

The power in the source is the same as in the load when their resistance numbers are the same. There is no power loss then.
But voltage is half with load connected. A characteristic impedance has a constant relationship between voltage and current, in RF selected for a compromise between conduction losses and voltage tolerance. 50 Ohms is a happy medium, for low and high power.

 

[abbey road d enfer]

The power in the source is the same as in the load when their resistance numbers are the same. There is no power loss then.

What a strange notion! The source delivers V.I, and the load receives V/2.I. Isn't it loss?

 

[Gold]

Thanks for the answers. I had never heard of the Maximum Power Theorem. This wiki quote is still a bit of a mind bender for me.

“The theorem results in maximum power transfer from the power source to the load, and not maximum efficiency of useful power out of total power consumed. If the load resistance is made larger than the source resistance, then efficiency increases (since a higher percentage of the source power is transferred to the load), but the magnitude of the load power decreases (since the total circuit resistance increases).[2] If the load resistance is made smaller than the source resistance, then efficiency decreases (since most of the power ends up being dissipated in the source). Although the total power dissipated increases (due to a lower total resistance), the amount dissipated in the load decreases.”

 

[abbey road d enfer]

Run a simulation in LTspice.

I don't really need to run a sim to calculate current, voltages and power in such a simple circuit.
If you did run a sim, you could see that the generator delivers 250mW and both resistors receive 125mW each.
The load receives exactly half of the power the source delivers.

R1+R2=100r hence I=5/100=50mA.
Source delivers 5v.50mA=250mW
Voltage acros R1=voltage acros R2=50r.50mA=2.5V (funny how it is equal to V/2, hah)
R1 and R2 receive 2.5V.50mA=125mW.

 

[5v333]

Even though the current is constant through the whole circuit, impedances consumes energy.
And since energy is proportional to power every resistor is developing power.
It is clear when you think about the fact that a resistor can become hot from current going through it.
That heat is energy which leaves the circuit.
This is kind of why there is a voltage drop in the first place.

 

[5v333]

Here is a mathematical demonstration.
lets say that Rload = Rsource times a variable k and voltages across source and load is U1 and U2.

voltage division gives
U2 = U1* Rload/(Rload+Rsource) = U1*k*Rsource/(k*Rsource+Rsource) = U1*k / (k+1).
Current I is the same for both resistors. Try multiply U2 = U1*k / (k+1) with I and you get the relationship of power.
U2*I = U1*I*K / (k+1) which is
P2 = P1*k /(k+1)
which becomes the ratio
P2/P1 = k / (k+1)

Now are there any highest or lowest points of effincy wih respect to the variable k?
lets diffrentiate P2/P1 with respect to k and try find a point of zero
d(P2/P1) / dk = 0 is the same as
1 / (k+1)^2 = 0

This has no local point of zero. the left side only becomes zero if k goes to plus or minus infinity. But k surely must be a positive number. And when k goes to infinity Rload goes to infinty and P2 / P1 goes to 1 (=100%).
We have now showed that maximum efficiency is achieved when k is as large as possible.
or in other words Rload is much much greater that Rsource.

 

[Cqwet Dbdfte]

The power delivered from the source has an impedance which also describes the optimum load impedance. The source impedance is not a loss term, it is a source characteristic, important for optimum power transfer, but largely irrelevant for audio signals where only the voltage matters, and input devices have high or very high impedance, thus ignoring power transfer.
Again, matching impedances for audio signalling is moot.

 

[Gold]

Even though the current is constant through the whole circuit, impedances consumes energy.
And since energy is proportional to power every resistor is developing power.

I thought this discussion might bump up against this. I have a feeling this gets very complicated very fast with reflected impedance losses, voltage and current being 90 degrees apart and a lot of other power quality stuff I know nothing about.