How to determine desired OT primary impedance

[ktritty]

Hello. I've been tinkering with tube circuits in scant spare time for about 5 years now, and have somehow never come across how one determines what OT primary impedance to choose when building a power amp. So far, I seem to have lucked out, but I'd like to know if there is some sort of model/equation that can be easily applied. I've looked around a bit, but have not found an answer to this.

For example, let's say I have a power stage that has 2 6V6's (one push, one pull, pretty straightforward). Voltage drop from anode to cathode is 355 volts, biased to 36mA. What OT primary would I choose? So far, the method I have used is as follows:

Single Tube Z = U/i
Push-Pull Z = U/sqrt(2)*i
(Divide result by number of tubes per side)

In this case, 355/(1.414*0.036) = 6900 ohms. It occurs to me that this method is probably not correct at all, but the numbers seem close (serendipity?). Thanks for your help!

[Diablo]

Tube data sheets usually give some examples of typical operating conditions which includes the OT primary impedance.

http://www.tubebooks.org/tubedata/HB..._2/6V6-GTA.PDF

[ktritty]

Thanks. BUT... how did the author of the tech sheet determine the proper impedance? I'm looking to learn how to determine it using my own observations.

[J M Fahey]

Basically: Z:dV/dI in this case d meaning the Greek letter "delta" which I can't write with this keyboard (yes, I know, there's an alt+something code) .
Delta means "change"; in this case maximum minus minimum value.
First of all open http://www.r-type.org/pdfs/6v6gt.pdf on page 3 and have a good look to the "Average plate characteristics" graph.
Better yet, print it filling a sheet , have a sharp colored pencil on hand and a ruler.
I will give you the "coordinates" of two points, we'll solve it graphically, and check mathematically.
Mark the points "A":[360V/20mA] and "B":[50V/98mA]
Join them with a straight line.
1) The "A" point is where the plate rests at idle: 360V DC/20mA (I chose a quite cold bias current, because I'm maximizing output power).
2) The "B" point is the maximum saturation one, where the tube is passing the maximum current possible (98mA) , and "losing" the least voltage possible (50V).
3) That straight line is called the "load line", shows all the intermediate values of current and voltage between those two extremes we have just drawn.
4) Delta V is = 360V-50V=310V.
5) Delta I is = 98mA-20mA=78mA.
6) Optimum Z, which maximizes power, is: 310V/0.078A=3974 ohms. <- this is the impedance for a single tube, a single side of the push pull, it would be called "Za".
The usual spec for a transformer is Zaa , 4x that value= 16000 ohms.
Now, why the usual spec is 8000 ohms?
Because the "official" datasheet is conservative, and specs screen voltage as 250V.
Guitar amp designers (me included) *love* to abuse tubes (and everything else), and slam screens with almost plate voltage, an unheard of 300 to 350V in this case, which increases available current a lot, at the expense of distortion and durability.
*If* that 6V6 can supply, say, around 130 mA in such conditions, the acceptable load impedance lowers substantially.
We also play with another factor: a speaker's "Z" is close to its nominal value only between, say, 250 to 400Hz; most of the time, specially on low and high frequencies is *well* above it (look at any impedance curve) and we can safely bet that the reflected impedance will be much higher than nominal, most of the time.
That 8000 ohm nominal transformer will look quite close to 16K to the tubes, most of the time.
Well, that's basically it.
As you see, datasheet curves are excellent, but we seldom respect them.
Rockn'nroooollllll !!!!!!!!!!!
In fact, we *should* measure our own tubes, under real conditions (I do )

[Steve Conner]

I think the manufacturer's recommended load impedance is based on actual experiments done in his own labs. (Well, was, when Philips, RCA etc. were making the original tubes back in the 50s and 60s.)

There's no easy way of calculating the optimum load for pentodes and beam tetrodes. For a start, maximum power and minimum distortion don't occur at the same load, so how do you decide which is "optimum"? JM's method is the best I know, and the same one I use, but it can only give a ballpark figure.

For instance, I have an amp with 6550s that gives the same power into either 8 or 16 ohms, in spite of having no taps on the OPT, which I think was originally a 6.6k to 16 ohm unit. The tube drop, power supply sag and so on, all just happened to cancel each other out.

The fact that those two impedances give the same power implies a maximum in between. If I could find a 12 ohm speaker, I'd probably get more power still, but as JM points out, the impedance of an 8 ohm one might well be nearer 12.

[ktritty]

Thanks everybody. Sorry my first reply was groggy and unclear. I had never even thought to look on spec sheets for impedance data! JM's approach is a heuristic one, which is exactly what I was hoping for. There is no 'real equation' for anything tangible. Van Der Waals gas equation, Newton's laws of motion, all of these are simply approximations that are generally good enough for our needs. I was hoping that somebody had developed an approach like JM. I believe that analytically, Z=∂V/∂i (on a mac, you just hit alt-d). It seems to me that there ought to be some way to develop a 'Van der Waals' type of equation for tube output impedance (one equation, but with a table of parameters that are plugged in depending on tube type, configuration, power vs distortion, etc.) Which is basically what JM did graphically. I think that every builder has done this in their head in some fashion anyway, it's just up to newbies like me to figure it out, especially isolated (no schooling or amp-building buddies) newbies like me.

My background is chemical engineering. Your 'load line' is pretty much just like an operating line for a distillation column. Just with different units, and resulting in better music.

[ThermionicScott]

Load lines may not be the most "rock and roll" way to do things, but if you have characteristic curves for the desired screen voltage, they're the best way to get the most power. I think they're fun, but then, I enjoyed long division from an early age.

- Scott

[opacheco]

Basically: Z:dV/dI in this case d meaning the Greek letter "delta" which I can't write with this keyboard (yes, I know, there's an alt+something code) .
Delta means "change"; in this case maximum minus minimum value.
First of all open http://www.r-type.org/pdfs/6v6gt.pdf on page 3 and have a good look to the "Average plate characteristics" graph.
Better yet, print it filling a sheet , have a sharp colored pencil on hand and a ruler.
I will give you the "coordinates" of two points, we'll solve it graphically, and check mathematically.
Mark the points "A":[360V/20mA] and "B":[50V/98mA]
Join them with a straight line.
1) The "A" point is where the plate rests at idle: 360V DC/20mA (I chose a quite cold bias current, because I'm maximizing output power).
2) The "B" point is the maximum saturation one, where the tube is passing the maximum current possible (98mA) , and "losing" the least voltage possible (50V).
3) That straight line is called the "load line", shows all the intermediate values of current and voltage between those two extremes we have just drawn.
4) Delta V is = 360V-50V=310V.
5) Delta I is = 98mA-20mA=78mA.
6) Optimum Z, which maximizes power, is: 310V/0.078A=3974 ohms. <- this is the impedance for a single tube, a single side of the push pull, it would be called "Za".
The usual spec for a transformer is Zaa , 4x that value= 16000 ohms.
Now, why the usual spec is 8000 ohms?
Because the "official" datasheet is conservative, and specs screen voltage as 250V.
Guitar amp designers (me included) *love* to abuse tubes (and everything else), and slam screens with almost plate voltage, an unheard of 300 to 350V in this case, which increases available current a lot, at the expense of distortion and durability.
*If* that 6V6 can supply, say, around 130 mA in such conditions, the acceptable load impedance lowers substantially.
We also play with another factor: a speaker's "Z" is close to its nominal value only between, say, 250 to 400Hz; most of the time, specially on low and high frequencies is *well* above it (look at any impedance curve) and we can safely bet that the reflected impedance will be much higher than nominal, most of the time.
That 8000 ohm nominal transformer will look quite close to 16K to the tubes, most of the time.
Well, that's basically it.
As you see, datasheet curves are excellent, but we seldom respect them.
Rockn'nroooollllll !!!!!!!!!!!
In fact, we *should* measure our own tubes, under real conditions (I do )

JM,
Is this method valid for SE or PP only???......

Thanks
Opacheco.

[J M Fahey]

Hi opacheco.
The basic method I sketched implied an AB class push-pull.
There I consider the plate swinging from +B down to saturation voltage (usually 50/60V) and current swinging up from idle to max. available (under puny plate saturation voltage, which is a hard feat to accomplish).
There's something I'd like to point before somebody finds it (in good faith) an error.
In fact the plate voltage reaches practically twice (2X) the steady +B voltage, during part of the cycle, while the tube is off and "the other" is saturating.
Why do I ignore it?
Because we are in the "dead" half of the cycle, where the non conducting tube has full screen voltage, double plate voltage, but it's still receiving an out of phase driving signal at its grid, which adds to the negative bias .....
If you have, say, -40V bias, in that moment you may have another -40V added.
With -80V on its grid, no tube will conduct, at all.
Not many people consider or care about this, while it's significant.
A very common misconception is that Class AB biased tubes go from max to idle current, but not lower ... it's not the case.
Even more important, *real* Class A biased power amps, also reach cutoff at some part of the cycle, opposite to the common widlely spread misconception that they somehow "always" pass current, and never go into cutoff.
Well, they might barely get there, but they sure do.
And in an overdriven guitar amp, where driving signal may be (usually is) more than just needed to saturate it, we may get high negative voltages as I described above.

If you need the SE method, just ask.
Starting a typical weekend full of anxious Musicians, but Monday or Tuesday will do.

[opacheco]

J M Fahey,

Of course!!....please let me to know the SE method!!!!!....

Thanks for your help.
Opacheco.

[Gingertube]

Patrick Turner has a lot of info on load matching on his website.
loadmatch4-pp-beamtetrodes for push pull, there is a section on SE as well.
Cheers,
Ian

[J M Fahey]

Excellent info, very complete.
Anyway when I have 1/2 hour free I'll write my simplified version; both ways match of course.
A very interesting fact he points out (on which I did not want to insist just to avoid useless flame wars), is the real world power out calculation.
2x6550 for *32W* with very reasonable 400V, needing hairy 560V to deliver 56W .... a bath of reality.
There are so many "60W"amps out there with a pair or 6L6 or EL34; so many "120W" with 4.
Oh well.
Thanks again, excellent data.

[opacheco]

Gingertube,
Thanks for this Link...that is fantastic theory!

J M Fahey,
I am waiting for your method!, Thanks for your comments and help.

Opacheco.

[Gingertube]

For "quick" method you can also look at Kevin O'Connor's "Principles of Power (POP)".
This has the calc method to work out power supply voltages and currents and output tranny primary impedance given a tube type and the number of tubes (2 , 4, 6 etc).
It also has the "reverse" method of calculating power supply voltages and currents and tube power dissipation starting with a given output transformer primary impedance.

That is a tube driven calc and a transformer driven calc are both there. THis was alsom in one of the "TUT"s - TUT2 IIRC.

Cheers,
Ian

[opacheco]

Thanks Girgetube!!......I will try to found this book locally!
Opacheco