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The McIntosh Unity coupled circuit.

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Posted on August 31, 2013 at 01:02:59
Michael Samra
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 First of all this is my last weekend to horse around before I go back to my real job on Tuesday.Oh how I love getting summers off but teaching is very rewarding as long as you have cooperative students.I have been fortunate for the most part but I won't jinx myself until I see what I'm handed upon my return.
Anyway,I promised myself and others that I would take measurements on the Mac Iron..On the Mc60s,the OPT transformers measure 595 ohms which we might as well call a 600 ohm primary impedance.The Mc275 is also reading that same primary impedance on the 16 ohm tap.It takes exactly 6.1vac in to get 1vac out on the 16 ohm which 6.1 divided by 1 equals 6.1 and then squaring the 6.1vac gives you 37.21vac times the 16ohm tap and that gives 595.32.
The Mac circuit is very different in the sense that no voltage amplification is done in the output stage.All the voltage amplification is done in the prior stages to where there is 325vac on the grids of the KT88s.Now,some think there is a two to one voltage gain going on in the output stage and I assure you there is not.This is a unity coupled transformer and unity gain circuit and what it does is swing the voltage in the output stage from the driver,but it doesn't amplify it in any way,shape or form.I had a long talk with Doc Hoyer and he verified what I already found out from measurements..Now,the output stage does increase current with respect to idle current in the PS,and with the voltage it swings thru the low impedance output stage,that is how it develops its power,IxV=P.You can actually remove the output tubes in a Mac amp and you will hear your speakers but they won't have much bass.I just verified tonight about low impedance primary.This is one of the most remarkable circuits of all time as was the Deuce.I guess we could say that Frank McIntosh and Stu Hegeman were no idiots.
We always hear of amp builders striving for lower impedance in their driver and output stages but,to have a 600 ohm impedance loading two high power tetrodes,that is remarkable.
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken

 

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RE: The McIntosh Unity coupled circuit., posted on August 31, 2013 at 06:10:07
Audioquest4life
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 Wow, that is incredible. I am not an in depth tinkerer in electronic circuitry like you; however, I appreciate excellent design topologies.

When I first read about the Autoformer technology from Mcintosh, I was really interested in their use of the concept. Ultimately, after about 30 years in the audio hobby, I have been able to afford a pair of MC2301 tube amps, and can't be happier. Thanks for the detailed analysis.
Ciao,
Audioquest4life

 

The load presented to the finals isn't as low as your meas seem to indicate...., posted on August 31, 2013 at 07:39:20
Steve O
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 ...In this case there is ~ 600 ohms loading both the anode and the cathode but if you refer the cathode load to the anode as if this were a more conventional output stage, the equivalent impedance is ~2,400 ohms P-P. 2,400 ohms while seemingly low is probably on the mark for class B 6550s.

 

RE: The McIntosh Unity coupled circuit., posted on August 31, 2013 at 09:26:37
Stuben
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 Mike,

Took a look at the schematic...looks curiously similar a cathode follower OP along with similar characteristics like unity gain and low OP impedance.

Yet the plates are loaded.

Are the inductors on the cathodes part of the OPT primary?

Looks like these guys figured out how get best of both worlds..

Very nice

Stuben

 

RE: The McIntosh Unity coupled circuit., posted on August 31, 2013 at 09:45:33
Tre'
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 "Are the inductors on the cathodes part of the OPT primary?"

Yes and no. They are a second, separate primary winding.

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: The McIntosh Unity coupled circuit., posted on August 31, 2013 at 11:05:40
Eli Duttman
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 IIRC, the windings in "Mac" O/P "iron" use a bifilar technique.


Eli D.

 

RE: The McIntosh Unity coupled circuit., posted on August 31, 2013 at 11:23:20
Steve O
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 Most Mac unity coupled OPTs used bifilar windings in the primary (anode of final "A" against the cathode of final "B"). They were double insulated due to high DC potential difference betw the two "filaments". The MC 75/275 OPT used a trifilar primary to accommodate the voltage limitations of the bootstrapped 12AZ7 drivers. IIRC, the MC3500/MI350 used a quadfilar primary winding.

 

Tre..You are right, posted on August 31, 2013 at 11:56:15
Michael Samra
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 But it also reads the exact same 600 ohms when I apply the AC to the cathodes where its 6.1vac to get 1Vac out which I thought was unusual being the cathode winding looks like it wouldn't inductively couple to the secondary with the plate primary in front of it.
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken

 

RE: The load presented to the finals isn't as low as your meas seem to indicate...., posted on August 31, 2013 at 12:02:21
Paul Joppa
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 Actually it's 2400 plate to cathode, for each tube. That would be 9600 ohms plate to plate if it were in class A, but it's a class B amp basically so the per-tube impedance is where the tube is operating.

 

RE: The load presented to the finals isn't as low as your meas seem to indicate...., posted on August 31, 2013 at 12:47:57
Tre'
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 That's what I thought. But I looked and the windings are not in series, per se.

It's 2 separate windings. One center tapped for the plates with the center tap feed by the B+ and a second center tapped for the cathodes with the center tap grounded.

Does that put them in parallel? Or are they in series with the tube as part of the chain? What does that do to the load seen by the tube?

No answers, just questions.

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: The McIntosh Unity coupled circuit., posted on August 31, 2013 at 13:30:54
Michael Samra
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 Stu
I was able to inject the AC voltage at the cathodes and get identical readings which I thought was odd.
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken

 

Hmmmmm..., posted on August 31, 2013 at 14:15:35
Steve O
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 ... My understanding of the Macintosh unity coupled output circuit is derived from the Radiotron Designers Handbook volume four. The Macintosh circuit is discussed (and somewhat dismissed) in some detail on pages 594 through 596. On page 595 it seems to be fairly clearly stated that the overall equivalent plate to plate impedance seen by the output tubes is four times that of either the individual plate to plate or cathode to cathode windings. I can verify that Mike's raw measurements are essentially correct for the 75/275 @ 600 ohms/overall cathode or plate winding, thus my 2,400 ohm equivalent overall plate to plate figure. Have I misinterpreted the RDH4 discussion or did they get it wrong?

 

RE: The McIntosh Unity coupled circuit., posted on August 31, 2013 at 14:30:17
Michael Samra
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 Steve
you are exactly right..All three windings have the same number of turns and that's why the 600 ohm impedance no matter where it's measured.Imagine winding the quadfilar..According to Doc,the Mc60 and Mc275 are essentially the same transformer only the Mc60s have C cores and they are bifilar as you stated.
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken

 

Doc implies the 75/275s aren't double C-cores? (nt), posted on August 31, 2013 at 14:39:58
Steve O
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RE: Doc implies the 75/275s aren't double C-cores? (nt), posted on August 31, 2013 at 14:44:50
Michael Samra
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 They aren't according to him..The 30s and the 60s are but didn't make sense to me either.The 275 is trifilar the 60s are bifilar..How did you guys come with 2400 ohm P to P..I know the tube conducts from cathode to plate so in reality,I would have measure the plate to cathode windings some how.If I measure 600 ohms plate to plate and 600 ohms cathode to cathode,wouldn't we just sum the two and get 1200 ohms?
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken

 

RE: The load presented to the finals isn't as low as your meas seem to indicate...., posted on August 31, 2013 at 15:03:08
Michael Samra
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 Paul
How did you measure from plate to cathode to come up with 2400 ohms? The cathode winding is inductively coupled to the secondary as is the plate but are both primary windings..How could we effectively measure,a summation of cathode and plate windings at the same time.When I measured plate to plate and cathode to cathode,it was 600 ohms total for the plate and 600 ohms total the cathode.Are you working in the FB winding also?
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken

 

RE: The load presented to the finals isn't as low as your meas seem to indicate...., posted on August 31, 2013 at 15:32:36
Michael Samra
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 I understand that part but,wouldn't we add the cathode impedance to plate impedance to get the total load impedance the output tubes would see?
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken

 

RE: The load presented to the finals isn't as low as your meas seem to indicate...., posted on August 31, 2013 at 17:08:13
Tre'
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If we look at the 600 ohm winding in the plate circuit as being in series with the 600 ohm winding in the cathode circuit, windings in series don't add that way.

1/2 of a winding is 1/4 the total impedance so 600 + 600 = 2400

Now I'm not saying that those 2 two 600 ohm windings are in series, because there is a tube in the meddle, but if they are then it's 2400 ohms.

Let me put it this way.

If the center tapped cathode winding was 2 separate windings and we took one of them and wired it in series with one half of the plate winding (and did the same for the other side) and then built a normal PP amplifier with the cathodes grounded, the load for each Class B tube would be 2400 ohms and the P to P impedance would be 9600 ohms. (the Class A load for each tube would be 4800 ohms)

I have a pair of Freed output transformer from Krone-Hite LDS-102R amplifiers and the cathode windings are separate and are the same impedance as the separate plate windings and can be used the way I described and the impedance will be as I have described.

The same goes for the 2 identical secondary windings.

This transformers are great because you can wire the windings any way you want. Right now I have all the primary windings in parallel and both secondary windings in series and a 13 ohm load on the secondary reflects 325 ohms for my circlotron output stage. 4 6as7gs give about 20 watts Class A. Not BS Class A, but real Class A.

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

Can you elaborate on how you arrive at the 9,600 ohm figure?.., posted on August 31, 2013 at 19:12:00
Steve O
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 ...It seems clear to me that the equivalent P-P load for the stated conditions is ~ 2400 ohms. My logic: We know that the cathode and anode windings have a 1:1 turns ratio compared to each other and by measurement, a 6.1:1 turns ratio referenced to the 16 ohm secondary. If the two primary windings are placed in series, the turns ratio pri:sec is now 12.2:1 and the equivalent primary impedance reflected from the 16 ohm tap = (12.2)^2*16ohms = 148.8*16 = 2,380 ohms. This analysis appears to be in full agreement with that given in RDH4. Since both you and Paul come up with 9,600 ohms under the same conditions, I must be missing something really fundamental here.

 

RE: Can you elaborate on how you arrive at the 9,600 ohm figure?.., posted on August 31, 2013 at 19:25:17
Tre'
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 " If the two primary windings are placed in series, the turns ratio pri:sec is now 12.2:1 and the equivalent primary impedance reflected from the 16 ohm tap = (12.2)^2*16ohms = 148.8*16 = 2,380 ohms."

And if those two primary winding are placed in series with the other two primary windings we get 12.2 + 12.2 or 24.4:1 24.2 squared is 595.36 * 16 = 9525.76 ohms across the whole primary.

P.S. I should re-read Mike's post but I understood him to say that each half of each winding (plate and cathode) measured 600 ohms impedance.

If I'm wrong and each whole winding measures 600 ohms then that would be 2400 ohms for the whole thing (the whole plate winding and the whole cathode winding) in series.

Edit, I re-read Mike's posts and found this "When I measured plate to plate and cathode to cathode,it was 600 ohms total for the plate and 600 ohms total the cathode"

That means it's 2400 ohms for the whole thing and only 600 ohms for each tube.

I'm not sure how that works, a 6550 in Class B loaded at 600 ohms?

That doesn't sound right to me but I don't spend much time with anything but true triodes without feedback.

Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"

 

The McIntosh Unity coupled circuit. is the only creation Tim de Pavaricini says he wished he had invented~nT, posted on August 31, 2013 at 19:33:39
Cleantimestream
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 ~!
The Mind has No Firewall~ U.S. Army War College.

 

I see the point of confusion..., posted on August 31, 2013 at 20:21:43
Steve O
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 ...while the bifilar wound unity coupled scheme utilizes four separate physical primary windings, schematically Mac draws them as two center tapped primary windings: cathode to cathode and anode to anode. IOW there are two 600 ohm composite windings, each built up from two physical windings of 150 ohms each. What Mike measured @ 600 ohms was the value of two physical windings in series: From cathode to cathode Mike measured ~ 600 ohms and from anode to anode he also measured ~ 600 ohms. So ... There are either 4 x 150 ohm primary windings or 2 x 600 primary windings depending on how you choose to look at the situation. Mike measured from the 2 x 600 ohm perspective and I'm in agreement with his numbers. From this I arrive at the 2,400 ohm figure for the equivalent, single end to end primary impedance.

Note that the cathode to cathode winding "center tap" is grounded and anode to anode winding "center tap" is fed from B+.

Edit: see schematic of similar MC240. Mike measured from "yel to yel/wht" and from "blu to blu/wht".



 

It's ingenious at multiple levels..., posted on August 31, 2013 at 20:28:25
Steve O
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 ...but real world implementation is complex which is off putting to many.

 

Got it. Thanks for sticking with it and straightening me out. (nt), posted on August 31, 2013 at 21:14:56
Tre'
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 .
Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: It's ingenious at multiple levels..., posted on August 31, 2013 at 21:51:41
Michael Samra
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 but real world implementation is complex which is off putting to many.

It is a brilliant piece of engineering but very few know how that output circuit works in actuality.It is like nothing else out there and we have limited access to any technical data that could even vaguely explain it.We know the principle of what happens but as far as what goes on inside that transformer,it would take Frank Mcintosh,Sidney Corderman,or Stu Hegeman to explain it.Doc Hoyer explained quite a bit but I would love to see the inner workings of the unity coupled output transformer in action.
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken

 

RE: It's ingenious at multiple levels..., posted on September 1, 2013 at 03:56:55
jazbo8
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 From the horse´s mouth...

 

Thanks Jaz, posted on September 1, 2013 at 16:24:08
Michael Samra
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 I have the hard copy of this magazine and I never knew the rest of the article was in it.
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken

 

My mistake!, posted on September 1, 2013 at 17:45:23
Paul Joppa
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 Sorry all; I thought the 600 ohms was from one end of a winding to the centertap of that winding. I understand now, it's from one end to the far end.

As for the windings in series, I simplified by assuming class B operation, so when one side is conducting the other isn't doing anything. The power supply has essentially zero impedance at audio frequencies, so B+ is the same as ground - at audio frequencies. Then the signal current loop ia B+ to half the plate winding to the plate through the tube to the cathode to half the cathode winding to ground which is B+. So the two half-windings are in series.

 

RE: My mistake!, posted on September 2, 2013 at 07:22:27
Tre'
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 "So the two half-windings are in series."

Yes and, the way I understand it now, those two half windings are 150 ohms each so each output tube is loaded with only 600 ohms assuming Class B operation.

Tre'


Have Fun and Enjoy the Music
"Still Working the Problem"

 

RE: The McIntosh Unity coupled circuit., posted on September 2, 2013 at 08:00:57
Eli Duttman
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 Vintage "Mac" amps should be refurbished. They exist to play music, not occupy space on a shelf. OTOH, parting a "Mac" out in order to obtain "iron" for experimentation is (IMO) an act of vandalism.

There is an option for the DIYer who wants to "play" with unity coupled O/P stages, Plitron's PAT-1070-UC toroidal transformer. Warning, this is expensive "iron".


Eli D.

 

errr....No, posted on September 4, 2013 at 05:07:49
PakProtector
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 roughly speaking the voltage gain is 2:1. input signal at the grid is followed by the final's cathode, and that same current also moves the plate in the opposite direction. Very much like a split-load PI circuit...:)

And of course the 600R of plate is combined with 600R int he cathode, so a 2k4 Ohm load is what the final's actually see. Can't help thinking that if Doc Hoyer actually thinks what you report him to that he should re-examine the circuit.
cheers,
Douglas

Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

THERE IS NO VOLTAGE GAIN WHAT SO EVER IN THE OUTPUT STAGE., posted on September 5, 2013 at 03:16:50
Michael Samra
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 DOUG
If you don't believe this you need to talk to Doc Hoyer who knows the circuit better than anyone in the country now,including those at Mcintosh.

"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken

 

Why not just make a few measurements and settle the matter?..., posted on September 5, 2013 at 05:00:43
Steve O
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 ...if you do, be sure to report your results here.

 

Someone has already done it..., posted on September 7, 2013 at 02:10:01
jazbo8
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 Doug's calculation is closer (in fact his 2:1 estimate was merely a shorthand I believe), the gain of each tube's gain is determined by its bias, but typically less than 1, in Jan's example shown on the link below, it turned out to be 0.85, so for the whole output stage 0.85x2=1.7. The McIntosh Unity Couple moniker is more for marketing rather than a description of its electrical property... From the original article by McIntosh & Gow (see link above) - [i]"...almost the entire gain in this final stage is lost by virtue of the feedback resulting from this method of loading..."[/i] Note it did not say that the final stage has unity gain. Also the unity output design is patented by McIntosh and NOT taken from RDH4.

 

The suggestion to have Mike make the measurements..., posted on September 7, 2013 at 07:57:44
Steve O
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 ...was primarily intended as a means of enhancing understanding of the output stage operation. i.e. "Trust but verify". My own personal discussions with "Doc" Hoyer indicate he does indeed understand the operation of the unity coupled output stage quite well and this. understanding is essentially consistent with what Doug stated (my understanding also). I think Mike may have misinterpreted what he believes the Doc to have stated.

For the record, RDH4 presents a relatively clear and concise if abbreviated analysis and critique of the unity coupled output circuit. The discussion fully attributes the circuit to Gow and McIntosh/McIntosh labs and in no way claims any creative involvement with its development and realization.

 

RE: The suggestion to have Mike make the measurements..., posted on September 7, 2013 at 08:14:48
jazbo8
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 Glad it is all sorted out ;-)

 

It is indeed simple...:), posted on September 9, 2013 at 04:32:56
PakProtector
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 The stage has ~2:1 votlage gain, and with more detailed numbers it is indeed likely to turn out somewhere between 1.5 and 1.8:1 depending on which tube you stick in there.

The lesson ought to be, examine it yourself and see what is going on instead of trusting a statement.
cheers,
Douglas

Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

 

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