Amplifier Output Transformers and Output Impedance

Since 4 ohm and 8 ohm speakers are common today, you see amps with 4 ohm and 8 ohm output taps. I've assumed that the 4 ohm tap would present a 4 ohm impedance to the load; likewise with the 8 ohm tap. However, recently I've noticed a few amps measured by Stereophile that have a much lower (like around 1 ohm) output impedance than their output tap indicates.

While this will lessen the interaction with the speaker's impedance, doesn't it lessen the amount of power delivered to the speaker?

Any thoughts on the designer's rationale?
Hi Bob,

No, the 4 ohm and 8 ohm tap designations do not signify that the output impedance of the amplifier is 4 ohms or 8 ohms, and in most cases the output impedance will be significantly less than those numbers. What they signify is just that the tap is intended for use with speakers that have those impedances, at least nominally.

Keep in mind that damping factor, which for an 8 ohm tap would be defined as output impedance divided into 8 ohms, would be extremely poor (exactly 1) if the amp's output impedance on that tap were 8 ohms.

Also, at audio frequencies the power delivered to the load is not maximized when output impedance = load impedance. That is a concept that is applicable to rf frequencies, where impedance mismatches result in some fraction of the signal delivered to the load being reflected back to the source. At audio frequencies, if an 8 ohm source impedance is connected to an 8 ohm load, one-half of the power generated by the output stage of the amp would be dissipated across the output impedance, and therefore wasted.

Think of it in terms of the amplifier being an ideal voltage source (zero output impedance) in series with a resistor equal to the output impedance, with the speaker impedance being connected between the output side of that resistor and ground. So what you have is essentially a voltage divider, resulting in progressively less voltage being placed across the load as the output impedance increases.

On the other hand, though, keep in mind that highish output impedances can be advantageous with some speakers, as you've no doubt seen in Ralph's "Competing Paradigms" paper.

Best regards,
-- Al

The function of the output transformer is to make the small speaker impedance appear to be much larger to the output tube(s). How much larger it appears is determined by the ratio of the primary to secondary windings.

A 4 or 8 ohm tap does not mean that is the output impedance of the amp. It means that hooking up a speaker with that impedance will present the proper load for the output tubes as determined by the amp's designer.

Hi Al.
At audio frequencies, if an 8 ohm source impedance is connected to an 8 ohm load, one-half of the power generated by the output stage of the amp would be dissipated across the output impedance, and therefore wasted.
Yes, that's correct. I agree that the source impedance and the load form a voltage divider. Based on that I worked a few examples and the power dissipated by the load is a maximum when the source impedance equals the load impedance. As the load impedance decreases relative to the source impedance the power dissipated by the load decreases; likewise when the load impedance increases relative to the source impedance. I assumed a constant voltage from the source in all cases.
Herman, thanks for the insight. I don't know why I never considered that before. Using an output transformer to make the load impedance appear larger will cause the amp to behave more like a constant voltage source than it would without the transformer. Isn't that correct?
Al, from here:
The justification for designing an amplifier to have a high output impedance is that the maximum transfer of power occurs when the output impedance is equal to the load impedance. An 8 ohm impedance feeding an 8 ohm load transfers more power to the load than would a greater or smaller impedance.
Does the ratio of primary/secondary windings change for the different taps?

I assume there are more turns in the secondary than in the primary. Is that correct?
Thank you to you both for pointing out that the output tap of the transformer does not equal the output impedance.
Hi Bob,

Yes I see what you are saying about the examples you calculated, which is correct as far as it goes. But try some examples in which, rather than raising and lowering the load impedance relative to the source impedance, you keep a fixed load impedance and lower the source impedance.

That will result in a greater fraction of the source voltage appearing across the load, while simultaneously resulting in increased current flow (since the total of the two resistances will be reduced). Ergo, more power to the load.

In the examples you tried, you probably kept the source impedance constant, resulting in reduced current + higher voltage across the load when you increased the load impedance, or increased current + reduced voltage across the load when you reduced the load impedance. Reducing the power delivered to the load in both cases, relative to the equal impedance situation, as you found. But those examples reflect a different variable than you were asking about.

BTW, I concur with Herman's well stated comment.

Best regards,

-- Al

P.S.: I wrote the above before seeing your post with the link to JA's statement. All I can say is that he must have been having a bad day. Consider the example of an ideal voltage source (zero output impedance) driving an 8 ohm load, and compare the power delivered to the load with the situation where the same ideal voltage source drives the same 8 ohm load, but through an 8 ohm resistor (representing an output impedance). Only 1/4 as much power will be delivered to the load when the 8 ohm resistor is present compared to when it is not, because both the current flowing through the load and the voltage appearing across the load will be cut in half.
Does the ratio of primary/secondary windings change for the different taps?
Yes, so that the output tubes see the same load impedance for an 8 ohm speaker load as for a 4 ohm speaker load.
I assume there are more turns in the secondary than in the primary. Is that correct?
No, that would step up the voltage to the speaker, relative to the voltage swing at the plates of the output tubes (the opposite of what is wanted), and would make the output tubes see a load impedance that is smaller than the speaker impedance rather than larger. The impedance seen by the output tubes (i.e., the impedance looking into the transformer primary) is equal to the speaker impedance times the square of the ratio of primary turns to secondary turns.

Al, JA's comment and my examples are not referring to a design that maximizes the power to the load. Obviously, to do that you want zero loss across the source impedance.

For a fixed source impedance the power delivered to the load is a maxima when the source impedance equals the load impedance. Assuming the output impedance of an amp is relatively constant (at least relative to the impedance of a typical speaker), I think the model is valid.
Hi Bob,

I believe that we are in agreement on two things:

1)For a given source impedance (i.e., a given amplifier output impedance), the power delivered to the speaker load will be maximized if the load impedance equals that given source impedance (assuming everything else is equal, and also assuming that the source impedance is high enough to be equalled by a reasonably designed speaker).

2)For a given load impedance, the power delivered to the load will be maximized if the source impedance is as low as possible (assuming everything else is equal).


1)The wording of JA's statement that you quoted, as I read it, clearly indicates that the power delivered into a given speaker impedance, such as 8 ohms, can be maximized by raising the source impedance to equal that 8 ohm or other given speaker impedance. Which of course is completely wrong, as I think you agree.

2)Your original question, which my answers were intended to address, was "I've noticed a few amps measured by Stereophile have ... around 1 ohm output impedance; doesn't this lessen the amount of power delivered to the speaker?"

I think that we are now in agreement that the answer to this question is "no," per point 2 above.

Best regards,
-- Al
Hi Al,

I'm finally on the same page with you. Sorry for being slower than usual and thanks for your patience.

The wording of my original question is just embarassing. I had done those fixed source impedance examples many months ago and unfortunately what seems to have stuck was "equal impedance => max power."

I also agree with you about JA's comment. I guess one drawback of being editor is that there's no one to edit your writing.

Thanks for the info on the relationship between primary/secondary turns and impedance. I started working on a derivation of that this morning over breakfast. Stand by for questions. :-)
Another question...

Assume the output impedance of an amp design turned out to be 2 ohms. I see two plausible (to me) uses of an output transformer.

1) Make a 4 ohm speaker look like a 2 ohm load on one tap and do the same for an 8 ohm speaker on another tap.

2) Make a 4 ohm speaker look like a 10 ohm or 20 ohm (or whatever X >> 2 ohm) load on one tap and do the same for an 8 ohm speaker on another tap.

Are there any other cases to consider?

Is case 1) an instance of impedance matching? Does it have any merit?

What's the merit of case 2)?
Thanks for your patience.
No problem. I've always found your posts to be informative, and I've learned some good things from them, in other subject areas. And I appreciate your interest in developing as good an understanding as possible of the technical aspects of audio.

Not sure I completely understand your last set of questions. Power tubes, like most vacuum tubes, are designed to operate at high voltages and relatively low currents, and therefore into high impedance loads (e.g., thousands of ohms). Speakers are designed to operate at lower voltages and at much higher current levels, and to present very low impedances to whatever is driving them. As Herman indicated, the transformer's main purpose is to match all of that.

So the transformer has to be designed such that the square of the ratio of primary turns to secondary turns, for the 8 ohm tap, results in the power tubes seeing an input impedance looking into the transformer primary of that several thousand ohm figure, as chosen by the designer based on the selected output tube, the overall topology, and the tube operating points he has selected. Likewise for the 4 ohm tap, for which the turns ratio (primary/secondary) will equal the ratio for the 8 ohm tap times the square root of 2 (so that a 4 ohm speaker connected to the 4 ohm tap will result in the same impedance in the plate circuit (on the primary side of the transformer) as an 8 ohm speaker connected to the 8 ohm tap.

I don't think it is correct to say that the output transformer can make a 4 ohm speaker look like a 2 ohm speaker to match an amp's 2 ohm output impedance. All it can match is the 4 ohm speaker impedance that is connected on the secondary side to the impedance that is desired on the primary side, to properly match the characteristics of the output tubes and their associated circuits.

The relation of amplifier output impedance to the impedances of typical speakers (as well as their impedance variations with frequency), and the effects of those relations on both power transfer efficiency and frequency response, are certainly factors to be considered in the design. And they are factors that will be influenced by the turns ratio and other characteristics of the transformer. But the primary criterion that must be satisfied (pun intended) is that the plate circuits of the output tubes must be properly loaded.

Best regards,
-- Al
Good thread with great information.

Al or anyone else: Besides using the best tap to match up with a speaker, when your amp has the ability to adjust the bias and the manufacturer gives you a range of say 220 to 300 ma, would you want to bias more toward the upper end of that range if the speaker you are using likes current, say like a Magneplanar?
Couple of points that appeared earlier but it looks like they need to be restated. To get the maximum power out of a given amplifier you do need to make the load match the output impedance, but that is not goal with an audio circuit. With RF amplifiers like radio transmitters it is imperative to make that match. It not only maximizes the power to the antenna but also minimizes standing waves which will cause serious problems. In an audio circuit standing waves are not of concern. With audio we need to deliver adequate power to the speaker but also need a relatively low output impedance from the amp to control the motion of the cone i.e. we need a high damping factor so we absolutely do not want an impedance match. So Bob, for audio circuits just wipe the idea of an impedance match out of your mind.

For an audio circuit using a tube (not OTL) that tube needs to see a relatively high impedance typically in the thousands of ohms. There is going to be some ideal value that the designer shoots for to give a good operating point for the tube that is within the operating limits of the tube (voltage across and current through it) and gives a linear response. This is not going to be a value that maximizes power transfer, it is the point that sounds the best. After you figure out that ideal load for the tube then you pick a transformer that "transforms" the low speaker impedance into the higher impedance the tube needs.

Onemug, The bias current through a tube is not proportional to the current it can deliver to the load. I hesitate to say completely unrelated but close. It has more of an effect on distortion. If you bias too high or too low then you risk running into the limits of the tube at the positive or negative peaks of the signal in an unbalanced manner. You also need to bias it so you are operating in a linear portion of the tubes operating range. So if you can go from 220 to 300 mA you need to listen to it and pick a spot that sounds best to you.

Go here for a few more ideas on output impedance


Thanks Herman,

I'm using my ears but trying or correlate what I'm hearing with some knowledge.

My amp sounded great at around 240 mA with some low 90 db/8 ohm speakers. Hooking it up to the Maggie 3.6's (84 db/4 ohm) it sounded a little cold with some distortion in the upper freqs. I upped the bias to 300 and it got that magic back. I've been backing it down to see where I lose it again (experimenting as I type).

Do the Maggies, with their contrasting load, take the tubes out of their previous linear range and therefore have to find a "new" sweet spot?
The transformer does indeed transform impedance. So if the power tubes want to see 3000 ohms plate to plate, then a specific tap on the secondary side is going to be the one that works. So if you load the 4 ohm tap with an 8 ohm load, the load on the tubes will be significantly higher so they won't be able to make full power and the transformer will be able to ring (distort). Plus, the inter-winding capacitive effects will increase, degrading the frequency response.

If you put a 4 ohm load on the 8 ohm tap, the tubes are now going to see something significantly lower, and they will not be able to make full power, plus there will be more distortion.

For this reason the taps have to be used correctly to prevent distortion and maintain bandwidth.

Not all tube amps are designed to be voltage sources. In fact, there is a strong argument for current sources, not the least of which is that they are likely to sound better. If you measure such an amplifier using voltage source rules, the amp may well measure poorly, although it may sound fine.

for more info see:
Al,, sorry for disappearing, but work has taken over. I'll have time this weekend to continue this thread. I'll try to do a better job asking questions.

Thanks to all.