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More on Audio Impedance Matching

I felt, this deserved it's own thread-

At the risk of more flames and becoming an outcast, I'd just like to flesh out a couple more points on this very contentious subject.

I read John Fields comments in the "common base thread" about the audio amplifier matching (thanks for that). But as I pondered it later on, I decided I needed to look at this further. It was keeping me up at night.

And audioph00ls just need to know. ;-)

Two nagging questions remained:

  1. What then is the role of the output transformer in a tube amplifier, if not for impedance matching?
  2. What is the design procedure for arriving at the turns ratio of the transformer? So I dug out a few old books on tube design. For the sake of discussion, consider a simple 1-tube class A stage:
+-----------------+-. ,-+---------+ __ /| | )|( -| | | | )|( -|__| | | +' '-+---------+ \| ------- | | | | +---+---- - - - - | | | --- | | |-----| - | | | --- | .-. +--+--+ - | | | .-. | --- | | | | | --- - | '-' | | --- | | | '-' | | | +---+--------+-----+------------+----+ | === GND The first question was quickly confirmed as an "impedance matching device". But on the 2nd question, the books stopped short of saying "match the impedance".

The statement that I _did_ find was:

"The step-down ratio n for an output transformer should approximate a value such that n squared times the load resistance equals the resistance to be _presented_ to the tube." [1]

So then the important question became:

What is the value to be "presented to the tube"?

I eventually found an answer in a problem's solution, stating:

"For maximum undistorted power output, the load impedance for a 3-electrode tube should be equal to about twice the plate resistance." [2]

Ah ha! So the key word is "undistorted"!

So under ideal conditions, you might match to the plate resistance Rp but in order to avoid distortion, you match to

2 * Rp. I am of course, ignoring real transformer properties here but this only slightly influences.

From this I have to conclude there is indeed impedance matching involved but it is purposely _not matched exactly_. For vacuum tube amplifiers the 2Rp mismatch is done for the sake of improved distortion characteristics. From a power perspective, this is suboptimal but the distortion factor overrules.

Comments?

As a secondary point, it would appear that by using a 4 ohm speaker on a

8 ohm speaker [tube] amplifier, you may in fact get more power delivered. This is because the presented load at the primary exactly matches Rp. BUT with higher distortion. Consider:

If Rp = 4k ohms, and RL=8 ohms, then n~=32

SQRT(2Rp/RL) ~= 32

Attaching RL=4 ohms, results in 4*32^2 ~= 4k ohms being presented to the tube. This matches Rp=4k for max power (but with more distortion).

Warren

[1] Chapter 5 Vacuum-tube Amplifiers, Section 9. Output Transformers and Other Coupling Arrangements for Class A Power Amplifiers, Radio Engineer's Handbook, by Fredrick Emmons Terman, McGraw-Hill Book Company, Inc, 1943 [2] Article 461 solution, in Radio Physics Course: An Elementary Text Which Explains the Principles of Electricity and Radio, 2nd Edition, by Alfred A. Ghirardi (1932).
Reply to
Warren
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For a tetrode or a pentode or a class-A operated transistor or mosfet, the "plate" impedance is much higher than for a triode. If you try to match that high impedance, it will clip and not deliver much power.

For a given power supply voltage, to get maximum power before clipping, the plate/drain voltage should be getting pretty close to ground at peak negative swing, and at that point the device should be conducting close to its maximum possible current (see the datasheet curves for details) so the transformer should present an impedance of VB+/Ipeak, something like that. It's easier to calculate with transistors or mosfets, since their collector/drain curves are so flat (ie, that have very high output impedances.)

Also, even for the triode, if there's negative voltage feedback you can worry less about soft distortion and can match closer to the maximum power case. Negative feedback pushes the output impedance of an amp down towards zero, at which point you stop thinking about impedance matching and start thinking about peak current capability.

Those old tube references (I have both of them!) usually addressed class A transformer coupled amps, like in an old radio, without feedback. Tubes, and gain in general, were expensive in those days, so it was unusual to use negative feedback in audio amps. So they just matched for max output at some tolerable distortion.

John

Reply to
John Larkin

You are oversimplifying a subject that has a lot of "gotchas" (as J.L. pointed out in part with his pentode counter-example).

In general, when you build a power amplifier you're interested in getting the most power for the overall cost, with decent sound. Or you're interested in getting the best damn sound for the overall cost. In neither of those cases are you interested in chasing some will-o-the-wisp goal of matching the impedance of the load to the output impedance of the amplifier.

Impedance matching is what you do when distortion and efficiency aren't a problem, but getting the best power _gain_ is. _Then_ you want to impedance match. A true impedance match doesn't guarantee good efficiency -- in fact, for a true Thevinin source, a perfect impedance match guarantees that half of the power ends up being dissipated in the source.

So the real purpose of the transformer in a tube (or old transistor) amp is to match the speaker impedance and desired power level with the voltage supply and current-carrying capabilities of the output tubes. The actual impedance match may be _horrid_. In fact, there's a great deal to be said for an amplifier with as close to zero output impedance as possible because the fidelity tends to be better when speakers are driven by perfect voltage sources.

--
www.wescottdesign.com
Reply to
Tim Wescott

"Warren"

** Its not contentious - just widely misunderstood.
** But it is "matched" for best power transfer - not simple equal value matching.
** The output tube characteristics and DC supply voltage are where you start.
** Begs the question.

** Means gross distortion - ie clipping.

A well designed tube output stages operates with somewhere between 1% and

10% harmonic distortion below the point of clipping and BEFORE any loop feedback is applied.
** NO.

The best load impedance is that which produces the maximum power output. Output tubes have cathode current limits which can be approached but not exceeded - so if at the practical limit of the plate voltage DOWN swing the current limit is approached, then there is no more undistorted power to be had.

** Nonsense.

Lose the mad idea that all impedance matching is EQUAL VALUE matching

The famous max power transfer theorem ONLY applies to non current limited sources - so it has little application in electronics.

** Hogwash.

** You are way off off the mark.

.... Phil

Reply to
Phil Allison

alue

g the

e

also IRL efficiency is also important in many cases, and is definitely not served by equalising impedances.

For once I agree with Phil. Making it up on the fly has a habit of not reaching a clear understanding.

3 of the costliest components in an audio power amp are the output transistors and supply transformer, and the size of those is primarily down to total power output in the case of transformer, and power dissipation in case of trannies. It follows from that that amps aren't designed to grossly waste power by 1:1 impedance matching.

That's just a simple snapshot of one aspect of it all.

NT

Reply to
NT

"Warren"

** A class A, single tube output stage is a special case.

  1. The output transformer is actually an inductor with a deliberate air gap and a max primary DC current rating that depends on core saturation.

  2. Whatever the bias current chosen for the tube, the current when operating cannot exceed double this value or there will be gross distortion.

  1. The heat dissipation in the tube is greatest when there is no drive signal.

The output transformer is normally designed to transform the impedance of the speaker load up to such a value that the current in the tube will be close to double the idle value when the plate voltage is at its practical lower limit. This voltage is rather higher for triodes than for beam tubes or power pentodes.

Eg. Using a 6V6 tube with plate supply of 270 volts and a DC idle current of 40mA ( 10.8 watts ) the plate voltage will drop to about 70 volts minimum. The voltage across the transformer primary will then be 220 volts and the additional current flow 40 mA. Makes the ideal impedance about

200/0.04 = 5000 ohms.

For an 8 ohm load, the turns ratio needs to be the sq rt of 5000/8 = 25:1.

The max ( core saturation free) current for the transformer needs to be at least 80mA.

The turns ratio umber needs to be tweaked a bit to allow for any losses in the transformer.

.... Phil

Reply to
Phil Allison

Phil Allison expounded in news: snipped-for-privacy@mid.individual.net:

Yes.

Yes.

I don't get this point. Idle and average current- are they not the same?

Ok, I see what you mean.

Yes, interesting.

Agreed.

Of course.

For class B do you calculate in the same way based on one half's peak current into half of the split primary?

Warren

Reply to
Warren

John Larkin expounded in news: snipped-for-privacy@4ax.com:

..

You probably have the third one I looked at as well:

Radiotron Designer's Handbook My copy is 1953.

That's a good point about NFB.

Warren

Reply to
Warren

"Warren"

** Only makes the load on the power supply constant.

Think about it.

** Yep.

But the idle current may be very low and then only max tube dissipation finally limits power output.

Given that class B has a max efficiency over 70%, one can get more output power than the sum of the tube's dissipation ratings. Up to double.

Eg.

One pair of EL34s ( 25watt diss each ) can output 100 watts with a plate supply of 700 volts and screen held at half that.

Max plate current is 200mA at 600 volts drop = 3000 ohms per side or 12,000 ohms plate to plate.

So for an 8 ohm load, the overall turns ratio is sq rt 12,000/8 = 38:1.

The effective output impedance of such a stage is at least 10 times higher than 8 ohms until feedback is applied.

.... Phil

Reply to
Phil Allison

Tim Wescott expounded in news:RfmdnVQfLYdJlvnTnZ2dnUVZ snipped-for-privacy@web-ster.com:

..

I think you'll agree that it's a common accepted practice to start with a simple process and then refine it.

Sure, I understand about trade-offs.

The idea of impedance matching was simply a means to achieve that goal. No will-o-the-wisp intended. But I do see the problem with that approach now.

Phil A explained that downstream rather well, I thought. It's starting to sink in. The concrete numbers really helped.

I'll have to chew on that one for another day. ;-)

Warren

Reply to
Warren

Phil Allison expounded in news: snipped-for-privacy@mid.individual.net:

I think I get it now- if the tube is sitting idle, there is no power going through the transformer into the load. So the tube dissipates 100% of its "working power".

But at full signal the load is ideally consuming 50% of that while the tube dissipates the remaining 50%.

Good point. This is because the tube's power rating is for continuous duty, while class B operates at 50% duty cycle.

So if I just take one side, I would calculate 19 turns (from 3000 ohms). Times 2 for the other half of the split winding agrees with your 38:1.

Without NFB, the secondary expects about 80 ohms? Wow.

Warren

Reply to
Warren

"Warren"

** No - there is no difference in the load.

But without loop feedback, the amp exhibits high output impedance.

The signal voltage at the output would increase by a factor or 10 if the load were removed.

This is not uncommon with tube guitar amps, some have no feedback and others some very little.

.... Phil

Reply to
Phil Allison

I have the old black one and the later, huge red one. I forget the dates just now. They're not very useful these days, but the old stuff is fun to read now and then.

John

Reply to
John Larkin

John Larkin expounded in news: snipped-for-privacy@4ax.com:

Mine is red. I had a difficult time acquiring this copy for the price I wanted. I especially liked chapter 5 on transformers.

Warren

Reply to
Warren

Phil Allison expounded in news: snipped-for-privacy@mid.individual.net:

That's what I was thinking (guitar amps).

Thanks, Warren

Reply to
Warren

Phil Allison expounded in news: snipped-for-privacy@mid.individual.net:

...

The 70 volts minimum begs a question now that I think about it.

How did you arrive at that minimum voltage? Experience or rule of thumb?

This value obviously has a large bearing on the computed ideal impedance.

Warren

Reply to
Warren

"Warren"

** Look at the published plate voltage / plate current curves.

A graph is worth 500 words

Plus I see the reality of this stuff in my everyday work.

Plus I have a clue about how tubes work and you do not.

You are getting very tedious.

... Phil

Reply to
Phil Allison

Phil Allison expounded in news: snipped-for-privacy@mid.individual.net:

Ok, good enough. The particular resource I was just looking at was missing this but I have seen them.

Tis the nature of anything technical.

Thanks for your help, Warren

Reply to
Warren

It is sort of an ancient-lore kind of thing.

Actually, tubes are pretty tedious. I'm glad I don't have to deal with them any more.

A mosfet will swing down to within a few volts of ground. And they don't need filament supplies! Or periodic replacement.

John

Reply to
John Larkin

[snip]

AHHH! NO! It doesn't "beg a question". Please look up what "begging the question" means. Thank you.

Bob

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Reply to
BobW

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