Push-Pull transformer question

103

Hello,

I have a simple question about a push pull output transformer.

For example, when I go to Hammond, say one of their PP OPTs has "4000 Ohm ct" impedance on the primary side and 8 Ohm on the secondary side. My question is exactly where you get this 4 kOhm impedance on the primary side (assuming the secondary load is an 8 Ohm speaker). Is it plate to plate impedance (so that plate to center tap becomes 1 kOhm) or is it plate to center tap (so that plate to plate becomes 16 kOhm). Since plate-ct is the effective load to each tube, I am guessing that the transformer is 4 kOhm plate-ct (so you can directly use this value on a tube's load line) such that plate-plate is actually 16 kOhm, but I could not find a definite description...

Thanks!

mdpaudio

4K center tapped mean it is 2k for each end to the center tap. Each output tube sees a 2k load when an 8 ohm load is on the secondary.

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PRR

Paid Member
It is 4k plate to plate. (they used to say it this way.)

What it is to the CT depends on if the tubes work in A or B. Confusing. So the "plate to plate" interpretation is universally used in specs.

103

Thank you all!

Well, more confusion though.

4K center tapped mean it is 2k for each end to the center tap. Each output tube sees a 2k load when an 8 ohm load is on the secondary.

Does that mean with a transformer claiming "4 kOhm ct" there is actually no 4 kOhm impedance in the primary (if each end to the center tap is 2 kOhm, the plate to plate or end-to-end impedance is 8 kOhm, if I understand correctly)???

What PRR stated makes sense to me, however.

Thanks again, but I guess I need more inputs...

PRR

Paid Member
There is no impedance "in" a transformer. Only impedance ratios and optimum impedance.

Transformer impedance goes by square of turns ratio. A CT winding is 1:2 turns, so 1:4 impedance.

And before we fry brains..... why care? Look in the tube datasheet and get a transformer of the same number.

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103

PRR,

Sorry that I wasn't too clear, yes, I understood that an OPT is just(?) a piece of metal with bunch of windings and an impedance in each coil may not be defined unless a certain secondary load is connected to other winding(s). But in the case above, mdpaudio was stating that

4K center tapped mean it is 2k for each end to the center tap. Each output tube sees a 2k load when an 8 ohm load is on the secondary.

hence my confusion. I would take what mdpaudio says if that is how things are defined, but I just don't know.

All your statements make sense to me.

And if you want to design something new, I guess we need to fry our brains, right

Thanks.

nigelwright7557

Hello,

I have a simple question about a push pull output transformer.

Thanks!
The impedances come from reflected impedances.
If secondary is loaded with 8 ohms then turns ratio squared gives 4K at primary side.
In this case square root(4000/8) = 22 turns ratio

103

nigelwright7557,

Thanks! Yes, understood. My question is, say the secondary had 100 turns, then does a PP OPT claiming "4 kOhm ct" has 2236 turns plate to plate (thus 1118 end to center, giving 1 kOhm impedance end to center) or 2236 turns end to center (thus 4472 turns end to end)? Hammond transformers use this notation (X Ohm ct) for the winding with a center tap and I just don't know exactly what part is "X Ohm" (when 8 Ohm or whatever is connected to the single ended side).

Miles Prower

nigelwright7557,

Thanks! Yes, understood. My question is, say the secondary had 100 turns, then does a PP OPT claiming "4 kOhm ct" has 2236 turns plate to plate (thus 1118 end to center, giving 1 kOhm impedance end to center) or 2236 turns end to center (thus 4472 turns end to end)? Hammond transformers use this notation (X Ohm ct) for the winding with a center tap and I just don't know exactly what part is "X Ohm" (when 8 Ohm or whatever is connected to the single ended side).
When you see something like: "4KCT" that means 4K P-2-P. If you're running the finals in Class A, that means each final sees a plate load of 2K since it's a balanced, two-phase system. If operating in Class AB, where one final cuts off, then the plate load is 1K. This was the case with the Le Renard design: I had a junk box Stancor OPT that was rated 4K4 P-2-P with secondaries of 4R, 8R, 16R, 125R and 500R. It also included tertiaries for UL (since the screen voltage needs to be less than VPP) and was designed for 6L6 finals running Class A. Since I used it with Class AB 6BQ6s, the per phase load was 1K1.

It's the same difference with center tapped PTXs, where one manufacturer might write "325-0-325" where as another might write "650VCT".

wg_ski

To overcome the confusion on what the plate impedance is it’s helpful to remember some hard and fast rules, then visualize what is really going on. A 4K CT end to end is always 1K end to CT. That’s physics, the impedance ratio is the square of the turns ratio. Run in class B, only one side is driven at a time so you see a 1K load. In class A, both sides are driven so simultaneously. The load on each rises to 2K BECAUSE of the simultaneous excitation by two parallel current sources. They are out of phase with each other, but driving opposite phases of the transformer. That puts them in PARALLEL. Two 2K impedances in parallel make 1K.

A 4K trafo in a class AB amp playing music at reasonable levels actually loads the output tubes with 2K most of the time. It drops to 1k for large signal peaks. The output stage voltage gain, which is what is really important, tends to remain rather constant. That gain is gm*RL. RL is high and gm is low down at quiescent bias. Gm rises as tube current increases (to a point) - so as one cuts off and the load goes to 1K the gm*RL product tends to stay reasonably flat. Some tubes it’s flatter than others, and there tends to be a sweet spot for bias current - where the gm is about half what it is at high power/current. All these possible “composite gm” curves are different - resulting in different harmonic (and IM) distortion signatures. One of the reasons why different push pull designs sound different from one another.

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jcalvarez

Paid Member
There is no impedance "in" a transformer. Only impedance ratios and optimum impedance.

Transformer impedance goes by square of turns ratio. A CT winding is 1:2 turns, so 1:4 impedance.

And before we fry brains..... why care? Look in the tube datasheet and get a transformer of the same number.
Datasheets not always show the optimal primary impedance for a push pull (or any other) setup. Some do, but most don't, especially for tubes not intended for audio.
Having said that, you are right, datasheets are the right place to look, the have the tube curves, and from there you can find the right transformer/tube combination.
And, of course, this site. There is a huge amount of information here!

rongon

To the best of my admittedly limited knowledge, the 'best' push-pull OPT primary impedance to choose for a particular output tube is a moving target. Or it's more of a range of acceptable targets. For example, there are push-pull EL34 amps that use OPTs with 3.5k pri Z, others with 5k pri Z, and still others with 6.6k pri Z.

Change the B+ voltage and the optimal ratio of primary to secondary impedance changes with it.
Change the class of operation (Class A, AB1, AB2) and again the optimal primary to secondary impedance changes.
Then again, what exactly does 'optimal' mean? Lowest THD at a given output level? Most power delivered into the load?

A lot of that is just this dummy's way of looking at what wg_ski said a couple posts ago.

MerlinB explains the relationships on his Valve Wizard site:
https://www.valvewizard.co.uk/pp.html

I suppose if you have a pair of pentodes that you want to use for a push-pull output stage, you could use the published plate curves, pick a primary impedance, and draw loadlines until you find something that looks like it will work well. Which is what jcalvarez just wrote!

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Alex T

Plus doing everything by book does not guarantee you sound you'd like to get

Miles Prower

To the best of my admittedly limited knowledge, the 'best' push-pull OPT primary impedance to choose for a particular output tube is a moving target. Or it's more of a range of acceptable targets. For example, there are push-pull EL34 amps that use OPTs with 3.5k pri Z, others with 5k pri Z, and still others with 6.6k pri Z.
It depends on a few things: what VPP are you using? Are you designing for lotsawatts or low distortion? The higher your RL the less distortion.

I suppose if you have a pair of pentodes that you want to use for a push-pull output stage, you could use the published plate curves, pick a primary impedance, and draw loadlines until you find something that looks like it will work well. Which is what jcalvarez just wrote!
Sometimes you have a suggested design, and sometimes you don't. The STC 807 Report lised a great many design suggestions for audio and RF applications. The data sheets for the 6BQ6GA/GTB have nothing for audio final use. That means doing loadlines. For designing for audio final use, hit the most linear part of the characteristic. In this case, VPP= 300 -- 400 looks like it will do that while remaining not too close to Class B conditions. Here, select VPP= 350V (since I had a junk box Stancor PTX with a 650VCT secondary that met the current demand). Going too high with VPP means operation well into Class AB. That would be OK if designing an RF power amp that can operate in Class B or Class C, but 700V would mean ESSSSS-Loads of X-over distortion. You also don't want VPP too low either as that leads to more vertical loadlines and more distortion.

For the first go at a loadline, I set IPq= 25mA that stuck with the rated PD. The THD estimate is a barely acceptable 5.0%. The second go increased IPq to 50mA. You can do this since the ratings of these TV HD finals is very conservative. That one change, everything else the same, greatly cuts the estimated THD. It definitely made an audible difference, with the hotter bias sounding much better. The difference being closer to Class A and reduced X-over.

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103

Thanks everyone!

So I guess when I see the notation of "primary impedance is 4 kOhm ct" for a PP OPT, that means 4 kOhm impedance plate-to-plate (end to end), see attached 1st image. Hope I got it right. My original question has been solved then!

Well, now I understood what Miles Prower said, but the explanation wg_ski gave me is another confusion to me... Pretty much everything makes sense to me (especially about the impedance change when switching between A and B takes place in class AB operation), but he said in class A, because of the two "parallel" current sources (= tubes, I am guessing?) the load appears in 1/2. That is where I am not too sure about.

See attached 2nd image. I thought in a push-pull topology, in class A operation, because both tubes are simultaneously conducting (in red), the 2 tubes and the OPT are in "series" from audio signal view point (AC, not DC), see attached. As a result, the current is same as if each tube sees only 1/2 the OPT impedance (Zpp)...??? Whereas in class B operation, one tube is non-conducting or off (in blue), so the audio signal leaving the cathode of the conducting tube has to return through the capacitors and the center tap of the OPT to the plate (not through the other tube), so the impedance is 1/4 of the plate-to-plate impedance (I think this part is fine).

Again thanks!

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PRR

Paid Member
I just try not to think about it too hard.

As for non-audio tubes not having the number on the sheet: find an audio tube of similar voltage and current ability and use that as hint. Some early sweep tubes were literally 6L6GC with top-cap. Others were "6L6-like" with lower Mu(g2), lower useful V(g2), and higher zero-grid current.

wg_ski

Yes, in class A both tubes (current sources for AC analysis) are delivering power simultaneously to the same load. Not very efficiently because of all the standing current. But both are delivering power. At the same delivered voltage, the current goes to 1/2 for each.

Early sweep tubes were just adaptations of audio outputs. Take a 6L6-oid and use a top cap and it can stand just as much off state voltage as anything else so it can be used as a horizontal output. A 12L6-oid run in A2 as a triode is pretty linear, making a decent vertical amp. But then the race for bigger and bigger picture tubes required more and more power to run the horizontal sweep (and generate enough HV, as well as run the switch mode supply that powers everything else). That required a different tube design. Lots of heater power, lots of current capability, low required g2 voltage. Linearity as an audio amp was all over the map - and you can’t always find curves to do a loadline with. If you are looking for sweep tubes to run as audio amps, some of the modern (late 50’s to 60’s) vertical sweep tubes have nice curves which lend themselves to low distortion when run as pentodes - with power output north of what you get with EL84’s.

rongon

Looking at data sheets for horizontal sweep tubes, I almost always see a plate resistance specification. Doesn't that give you a good clue as to what load impedance might work well for it?

For instance, the 6V6 data sheet says its 'approximate' rp = 50k to 80k ohms.
For EL84 rp = 38k ohms.
For 6L6GC rp = 33k with Vp = 350V and Vg2 = 250V.

For 6AV5GA, it says rp = 14.5k ohms.
For 6GE5, rp = 18k ohms.

Does that mean horizontal sweep tubes work better than audio output pentode/beam tubes into lower load impedances?

kodabmx

It depends on a few things: what VPP are you using? Are you designing for lotsawatts or low distortion? The higher your RL the less distortion.

Sometimes you have a suggested design, and sometimes you don't. The STC 807 Report lised a great many design suggestions for audio and RF applications. The data sheets for the 6BQ6GA/GTB have nothing for audio final use. That means doing loadlines. For designing for audio final use, hit the most linear part of the characteristic. In this case, VPP= 300 -- 400 looks like it will do that while remaining not too close to Class B conditions. Here, select VPP= 350V (since I had a junk box Stancor PTX with a 650VCT secondary that met the current demand). Going too high with VPP means operation well into Class AB. That would be OK if designing an RF power amp that can operate in Class B or Class C, but 700V would mean ESSSSS-Loads of X-over distortion. You also don't want VPP too low either as that leads to more vertical loadlines and more distortion.

For the first go at a loadline, I set IPq= 25mA that stuck with the rated PD. The THD estimate is a barely acceptable 5.0%. The second go increased IPq to 50mA. You can do this since the ratings of these TV HD finals is very conservative. That one change, everything else the same, greatly cuts the estimated THD. It definitely made an audible difference, with the hotter bias sounding much better. The difference being closer to Class A and reduced X-over.
This is my loadline when I use 6BQ6... 320V B+, 60mA, 2k2 - 3k load (3k shown).

rongon

That's for triode-wired, no?