Max Power Delivery / Load Impedance

[pdf64]

Given that max power delivery from source to load occurs when the source impedance = load impedance, why does the output voltage of a tube amp, when operating open loop, way more than double when the load is removed? The source impedance of the tube amp maybe 10 x higher than the output impedance rating but still it delivers max power into the rated load; higher or lower (eg 16 or 4 ohm load on 8 ohm tap) and the max power drops.
I'm thinking it must be something to do with getting the optimum load reflected back to the tubes but it still doesn't seem to square up with the basic theory.
Thanks - Pete.

[loudthud]

Output tubes are not voltage amplifiers, they are tranconductors. They put out a current controlled by input voltage. When you remove the load, they still try to put out the same current. Since there is no place for the current to go, they try to increase the voltage in an attempt to maintain the current. Eventually they reach a voltage limit. The rules of "matching" impedance don't apply to audio amplifiers. Think of the load as the "intended" impedance. In audio amplifiers we like the the actual output impedance to be quite a bit lower than the load impedance so that the amplifier looks like a voltage source to the load. Even in guitar amplifiers where the small signal output impedance can be high, when the output is clipping, it's more of a weak voltage source.

[g1]

The rules of "matching" impedance don't apply to audio amplifiers. Think of the load as the "intended" impedance. In audio amplifiers we like the the actual output impedance to be quite a bit lower than the load impedance so that the amplifier looks like a voltage source to the load.

OK, I'm not sure if I'm getting lost here or misreading what you are saying. We use output transformers to match impedances so we get maximum power transfer. If we mismatch, we lose power. So I don't see what doesn't apply to audio amplifiers.
My understanding is along the lines of the bottom of this page: Maximum Power Transfer Theorem

[kg]

we're mixing terms that shouldn't be mixed.

we have a theory that states that maximum power transfer occurs when source and load impedances match.

then we have an OP which asks "why does the output voltage of a tube amp, when operating open loop, way more than double when the load is removed?"

any discussion of power transfer completely falls apart when talking about an open circuit!

[Steve Conner]

There are two effects at work.

First of all, if you tried the experiment with triode tubes, the output voltage probably would just about double. The output impedance of a triode does actually look like a resistance equal to the plate resistance, and the maximum power theorem applies.

Pentodes and beam tetrodes have a very high and non-linear output impedance that can't be used as the basis for an impedance matching calculation. To a first approximation, the output impedance is infinite, and that's why the voltage goes through the roof into an open circuit. (Stored inductive energy in the output transformer helps the voltage on its way to infinity.)

Non-linearity is the key to the whole thing. A non-linear device can have different characteristics depending on the amplitude of the signal you examine it with. The high impedance of pentodes is a small-signal property: a small change in plate voltage leads to a much smaller change in current, so we say the impedance is very high.

But on the other hand, you can define the large-signal impedance in terms of the plate supply voltage used, and the maximum current that the tube can safely handle.

This is one way of choosing load impedances for pentode amps. The other way is to search empirically for the impedance that gives lowest distortion, if low distortion is what you want.

[J M Fahey]

b) driving a load (putting power into it) in a safe, efficient, long lasting way.
We are not talking tone, triodes vs pentodes, distortion, etc. ; and rules about "matching impedance" *do* apply, usually trying to satisfy condition (b) rather than condition (a).

Let's see tha basic book example:
Imagine that I have a 1.5V AA battery. So I already know the maximum generator voltage.
Meaning the "unloaded" voltage.
Now I short it into an ammeter. It provides 1,5A.
Now I know the maximum generator current.
And, I know the generator internal impedance.
1.5V/1.5A=1 ohm.
By the way, these are quite realistic values. A fresh AA battery supplies somewhat over 2A shorted, uomto 2.8A a very good one; a quite worn one less than 0.5A. I use that quick test to check whether to replace them or not, much trustier than a simple Voltage check.

Back to our test: I'm too lazy to draw a Voltage Vs current (or Vs load impedance) graph, I'm on my way to a charcoal grilled beef, lamb and chicken BBQ, plus "empanadas " (savoury meat pies) and assorted delicacies (today's our National Holiday)

but I'll provide a couple examples:
1) Load too high: 10 ohms.
we'll get 1.36V across it (the remainder is lost inside the battery or across its internal impedance).
Power into the load: 1.36x1.36/10= 0.186W=186mW
Power dissipated/lost into the generator = (1.5-1.36)squared/1 ohm=20mW
Good, only around 10% of available power lost heating the battery.
2) Load too low: 0.1 ohms.
we'll get 0.136V across it (the remainder is lost inside the battery or across its internal impedance).
Power into the load: 0.136x0.136/0.1= 0.186W=186mW
Power dissipated/lost into the generator = (1.5-0.136)squared/1 ohm=1.86W=1860mW
TERRIBLE: around 90% of available power lost heating the battery, only 10% useful power.
3) "Optimum" load: 1 ohm.
Easy, voltage split evenly: .75V across the load; .75V lost inside the generator.
Power into the load: .75V squared/1 ohm= 0.56W=560mW
Power dissipated (heating) into the generator=same=560mW.
Yes, this is the maximum power transfer available from this generator.
The million dollar question is: is this *good*?
Usually not: we overheat our generator and waste a lot of energy.
That's why in general "it's better to go *up* in impedance than go *low* "

Now to a very different problem: what's all this talk about plate impedance, damping, voltage feedback, current feedback, triodes vs. pentodes, etc?,
Fact is, these are all *active* devices and because of that they "behave as if" they were of a certain internal impedance, but let us remind that they are not the generators themselves (which really is the Power Supply) but they *control* the output of said PSUs in a certain way.
And will continue doing so while said PSU can supply what they ask, but not beyond.
So in the original problem posted here, a triode will behave like it has a certain internal plate impedance, a pentode like it has another, much higher one, an SS amplifier, through clever use of feedback will *behave like* it has whatever we want, from almost perfect voltage generator (very low internal impedance) to almost perfect current generator (*very* high internal impedance) to but will continue to do so only within the voltage/current the PSU can supply.
That's why RG, Steve, and others , sometimes refer to amps as "that thingie hooked between the PSU and the speaker" ... and they are right of course.

Back to the OP question, it's common for many tube amps to "behave as if" they have damping=1, meaning its internal impedance is roughly equivalent to load impedance , and we find that to sound good, SS amps and their mixed feedback also mimic that, but really their internal "raw" impedance is less than that, under penalty of overheating and having terrible efficiency.

In 1972, when I switched from Tubes to SS (long story) one of the first things I noticed when measuring one of my last Twin clones was what you found : unloaded voltage was twice the loaded one.
I studied and experimented a lot until I could achieve the same.
I might say I independently invented what much later was called "Valvestate" more than 10 or 15 years before it was used in Guitar amps.
Go figure.

[diagrammatiks]

well you do lose power but not that much.

all that matters is the reflected impedance.

if you have an 8ohm to 4k transformer then putting 16ohm on that tap gives you 8k.

That difference will change your loadline a bunch and definitely effect how the amp sounds but the actual swing potential of the output tube is not changed that drastically.

[kg]

Back to our test: I'm too lazy to draw a Voltage Vs current (or Vs load impedance) graph, I'm on my way to a charcoal grilled beef, lamb and chicken BBQ, plus "empanadas " (savoury meat pies) and assorted delicacies (today's our National Holiday)

damnit. now i'm hungry.

[Chuck H]

No proper carnivore can read the words "(savoury meat pies)" and not begin to drool. Hope it was a fun day Juan. And thank you for the "every man" explaination.

[Gingertube]

Guys,
THis website has some useful tech info on matching loads to output tubes and some other interesting reading. The guy is a HiFI Amp builder but basic info remains the same.
education+diy

Cheers,
Ian