How to calculate push-pull impedence?

[TheKT88KilledJFK]

I've searched for this all over the internet, and there's many different answers in many different forms, so I shall attempt to gather one here in a format I'm I can comprehend. 

When looking at plate resistance for output tubes, I look at the voltage on the plate and the current running through the tube for a target power.  What do I use to get an apparent impedance at this operating point?  I know the screen bias has to affect the resistance in some way and I know that the higher the plate voltage, the rate of decrease in impedance increases, probably in a logarithmic manner so it doesn't seem I can just predict plate resistance as if the tube plate resistance isn't dependent on the screen voltage?

Basically, I'm trying to figure out the interface between the power tube and the OT primary and I'm trying to work out how to find the plate impedance of specific applications.  I know the data sheets have info, but they're normally for a specific application that isn't stated or is understood in a convention that I'm unfamiliar with.  I'm specifically working on sorting out a push pull, two tube 150w output transformer spec for a bass amp.  I want to be able to provide specs for custom winding or start winding myself.

For a push pull setup, it's my understanding that you want each side of the OT primary to match the tube attached to it, so this would mean the plate resistance for a specific operating point should match that side of the OT primary's impedance.  This has always been the point of maximum power transfer in other devices I've worked on, and deriving the maxima for power transfer always ends up with the variables being equal in 2 factor equations like V*I.  What I don't understand is I see 5k primaries being listed for tubes with 15k impedance listed on the tech sheets. Why? 

If I'm taking something like a KT150 with 10k listed plate impedance, shouldn't the OT have 20k impedence a-a?  Obviously this is WAY to much impedance for any reasonable B+ voltage, we would need to raise the current.  So where does the 10k impedance come from?  Is this the actual resistance and we're just using lower impedance OT primaries as a compromise?

[shooter]

 the "actual" value is dynamic, probably math beyond my 4function calculator.
data sheet are good for "close enough" operation, look at the "ratio" between plate volts and OT impedance
on the attached data sheet, also look at the last page, impedance is plotted on the X-axis 

[TheKT88KilledJFK]

the "actual" value is dynamic, probably math beyond my 4function calculator.
data sheet are good for "close enough" operation, look at the "ratio" between plate volts and OT impedance
on the attached data sheet, also look at the last page, impedance is plotted on the X-axis

Luckily that sheet lists the a-a load resistance.  I keep running into tubes that don't list this so I need to be able to determine it.  I do know fixed conditions I'm aiming for, like expected power output (50w each tube), B+ ( probably 520VDC since that's what I have on hand), and cathode current which is what I've calculated to be 94mA per tube.  If I just take 520VDC and divide it by .094A I come out with 5.5kohms which seems reasonable.  Is the a-a of the OT actually relative to each tube or just each side of the primary?  So if we have a 5.5k plate resistance then it would double to an 11k impedance for the a-a on the primary? 

Am I just adjusting the bias on the grid to match the apparent plate resistance to the OT primary anyways?
Are the OT primary ratings measurements per side?  If so, then things make a lot more sense. 
 

[shooter]

Are the OT primary ratings measurements per side?  If so, then things make a lot more sense. 
 

i'll give that a qualified yes  B+ is one "end"[middle] of a coil, plate the other, each tube gets one coil.
That's why I basically build SE, easier on the brain

[DummyLoad]

i have never seen a datasheet with published conditions for push-pull indicate 1/2 primary. all are anode to anode datum.


--pete

[tubeswell]

For guitar amps where 'close enough is good enough', you can use a simple '2-step method' for calculating optimal Push-Pull reflected load, thus:


1st step is to calculate what a single ended reflected load for centre-biased Class A operation would be for the type of output tube you are using, hence:


Zout= Va/(Pa/Va), where Va is the plate voltage (at idle) and Pa is the rated maximum plate dissipation of the tube.


2nd step is that the number you obtain from the 1st step becomes the plate-to-plate load for a push-pull amp using two tubes of the same type each with the same plate voltage as in the 1st step, but running in Class AB1. In the class A part of the signal cycle, (where both tubes are running), each tube 'sees' 1/2 of this (across each 1/2 of the OT primary between the plate and the CT). In the Class B part of the signal cycle, the load seen by the 'on-tube' will be 1/4 of this overall plate-to-plate load resistance (because only one half of the OT primary will be conducting into the total load reflected from the secondary winding). The point in the signal cycle at which Class B operation kicks in, depends on how hot or how cold you bias the output tubes, and how much output signal swing you get from the PI.


If you want to run a push-pull amp totally in Class A, then you should double the plate-to-plate reflected load obtained from the 1st step, because both tubes will be conducting (in opposing phase) all the time, and so you want each tube to see the load that it needs to function in Class A without exceeding its maximum Pa.




Another way is to use a 'rule of thumb' that relates the optimal reflected load to the plate resistance of the tubes you are using:


For triodes, this optimal band is within 2-to-4 times the plate resistance of the tube.
For tetrodes and pentodes, the optimal band is within 1/6th-to-1/10th of the plate resistance of the type of tube.
Use lower values for lower distortion, %THD.
Use higher values for higher output power.
(See the attached article by OTM)

[TheKT88KilledJFK]

For guitar amps where 'close enough is good enough', you can use a simple '2-step method' for calculating optimal Push-Pull reflected load, thus:


1st step is to calculate what a single ended reflected load for centre-biased Class A operation would be for the type of output tube you are using, hence:


Zout= Va/(Pa/Va), where Va is the plate voltage (at idle) and Pa is the rated maximum plate dissipation of the tube.


2nd step is that the number you obtain from the 1st step becomes the plate-to-plate load for a push-pull amp using two tubes of the same type each with the same plate voltage as in the 1st step, but running in Class AB1. In the class A part of the signal cycle, (where both tubes are running), each tube 'sees' 1/2 of this (across each 1/2 of the OT primary between the plate and the CT). In the Class B part of the signal cycle, the load seen by the 'on-tube' will be 1/4 of this overall plate-to-plate load resistance (because only one half of the OT primary will be conducting into the total load reflected from the secondary winding). The point in the signal cycle at which Class B operation kicks in, depends on how hot or how cold you bias the output tubes, and how much output signal swing you get from the PI.


If you want to run a push-pull amp totally in Class A, then you should double the plate-to-plate reflected load obtained from the 1st step, because both tubes will be conducting (in opposing phase) all the time, and so you want each tube to see the load that it needs to function in Class A without exceeding its maximum Pa.




Another way is to use a 'rule of thumb' that relates the optimal reflected load to the plate resistance of the tubes you are using:


For triodes, this optimal band is within 2-to-4 times the plate resistance of the tube.
For tetrodes and pentodes, the optimal band is within 1/6th-to-1/10th of the plate resistance of the type of tube.
Use lower values for lower distortion, %THD.
Use higher values for higher output power.
(See the attached article by OTM)

That clears up everything man!  Thank you so much!

[PRR]

> each side of the OT primary to match the tube attached to it

Never.

BY FAR the best way is to steal values from the tube datasheet, or a similar tube.

[TheKT88KilledJFK]

> each side of the OT primary to match the tube attached to it

Never.

BY FAR the best way is to steal values from the tube datasheet, or a similar tube.

Another practice by the community I am unfamiliar with!   I can see that there are a plethora of available OTs for any application or combination I can think of now. 

Just checking my simple math on the issue, specifically for a KT150 (which I know isn't that much better than a KT120), the 3k plate resistance sums to 6k on the OT a-a for two tubes in push-pull setup, but we can use a 2k a-a OT because previous scientists/engineers have proven that the quality of sound is better at a sub .4 ratio even if the power transfer isn't?  Interesting.

I assume that if I were to add tubes then the impedance of the plate resistance is calculated in parallel, so for a four tube pp setup I'd just say the apparent Rp is 1.5k then get a 1k a-a OT?

[PRR]

> KT150, the 3k plate resistance sums to 6k on the OT a-a for two tubes in push-pull

KT150, none of the pentodes, is 3K Plate Resistance.

Yes, I see that on the New Sensor "data" sheet; idiots, that is Load Impedance not plate resistance.

Yes, if it loads OK in 3K SE, 6K may be a starting point for P-P. But probably not the best use of the tube.

_I_ would go to the 6550/KT88 grandfather. The GE sheet shows internal plate resistance is 15K (@ 250V 140mA), not 3K. It shows on page 3 known-good conditions up to 100 Watts, 600V, 5K loading. Because KT150 is rated near double the dissipation of 6550/KT88, we might hope for 200 Watts. To get there the supply voltage must go up by 1.414 times... 840V, and KT150 is rated 850V, OK.

Note however that one KT150 is $95, and 6550/KT88 can be had near $35. Paying 3X the price for 2X the power is poor economics. And 600V bites, I never want to work with 850V. (Also very awkward for 400V electrolytic caps.)

Having hefted 350W tube amps in my youth, I see NO point in tubes over 100W today. At 100W you can use common 6L6/EL34 tubes in quads with common Fender or Marshall replacement transformers. If you need more power, build a 20W amp and mike it into a 3,000W PA system.

[TheKT88KilledJFK]

> KT150, the 3k plate resistance sums to 6k on the OT a-a for two tubes in push-pull

KT150, none of the pentodes, is 3K Plate Resistance.

Yes, I see that on the New Sensor "data" sheet; idiots, that is Load Impedance not plate resistance.

Yes, if it loads OK in 3K SE, 6K may be a starting point for P-P. But probably not the best use of the tube.

_I_ would go to the 6550/KT88 grandfather. The GE sheet shows internal plate resistance is 15K (@ 250V 140mA), not 3K. It shows on page 3 known-good conditions up to 100 Watts, 600V, 5K loading. Because KT150 is rated near double the dissipation of 6550/KT88, we might hope for 200 Watts. To get there the supply voltage must go up by 1.414 times... 840V, and KT150 is rated 850V, OK.

Note however that one KT150 is $95, and 6550/KT88 can be had near $35. Paying 3X the price for 2X the power is poor economics. And 600V bites, I never want to work with 850V. (Also very awkward for 400V electrolytic caps.)

Having hefted 350W tube amps in my youth, I see NO point in tubes over 100W today. At 100W you can use common 6L6/EL34 tubes in quads with common Fender or Marshall replacement transformers. If you need more power, build a 20W amp and mike it into a 3,000W PA system.

KT88 sunn amps were the first I experimented with!  That's why I went with my username, late 60s sunn amps hahaha.  Luckey probably remembers me asking tons of questions about that a year or so ago.  Point being is that I'm just trying out a platform that I know nothing about to see if I can derive the proper general design parameters, but if KT150 has the same Rp in class ab1 as KT88 then I guess it doesn't serve that purpose.  I'm also building a 6 el34 amp, but only trying to get ~108 watts out of it. 

Tung-sol listing the load impedance as plate impedance is unfortunate indeed.  I don't think they're the only ones doing so.  I didn't want to assume any of the parameters of the KT150 based of the rest of the kinkless tetrodes but that slight mislabeling forces me to do so. 

[tubeswell]

Also see JM Fahey's extensive (and very well put) reply on this topic in this recent thread. http://music-electronics-forum.com/t45254/

[davidwpack]

I have some confusion on this topic. Anyway this thread can be saved in the archives?

[shooter]

this thread can be saved

here's some things I do to "achieve" for myself;
open the thread, then copy the link, I paste in word.  that way when it's on page 137, I click the link in word, walla found.
If it's really good, I highlight the text from each helpful reply and copy/paste in word.

pretty soon you have your own "tailored"  help file/folder now organizing...........

[davidwpack]

Yeah, I'll definitely do that. I figured it'd be beneficial to other people too. I'd say this question is bound to be asked again.

[dennyg]

I had a similar thread on this topic just a couple weeks back.  After much input from forum members and research, I extracted (and likely oversimplified) calculation for PP load impedance:
* determine what plate voltage you're going to run - set approximately by the PT you plan to use
* determine the plate current that gets you to 70% of the dissipation for the tube
* divide your plate voltage by double the bias current (since your OT is looking at plate-to-plate) and that's the target RL you want for OT
I just used this on my latest build: Using a hammond 270FX that yields loaded B+ around 385V (using 115V primaries in the US with 120V wall voltages in my house) and I want option to use both 6V6 and 6L6GC pairs if possible.  Here's the calcs: 
---70% bias current for 6V6 at 385V is 26ma x 2 = 52ma.  385/.052 = 7.4Kohms so an 8K OT will do. 
---70% bias for 6L6GC at 385V = 54ma x 2 = 108ma.  385/.108 = 3.6K.  So I'll need 4K RL. 
* So i ended up buying a 4K OT with 4/8/16 secondaries.  I use 1-to-1 speaker connections for the 6L6's (i.e. 4ohm speaker into 4ohm jack)
And when I run 6V6's i just run speaker load double the Z of the jack - e.g. 8ohm speaker into 4ohm jack yields 8K load. 

If I'm off base I'm sure someone will correct me!

[TheKT88KilledJFK]

I had a similar thread on this topic just a couple weeks back.  After much input from forum members and research, I extracted (and likely oversimplified) calculation for PP load impedance:
* determine what plate voltage you're going to run - set approximately by the PT you plan to use
* determine the plate current that gets you to 70% of the dissipation for the tube
* divide your plate voltage by double the bias current (since your OT is looking at plate-to-plate) and that's the target RL you want for OT
I just used this on my latest build: Using a hammond 270FX that yields loaded B+ around 385V (using 115V primaries in the US with 120V wall voltages in my house) and I want option to use both 6V6 and 6L6GC pairs if possible.  Here's the calcs: 
---70% bias current for 6V6 at 385V is 26mv x 2 = 52mv.  385/.052 = 7.4Kohms so an 8K OT will do. 
---70% bias for 6L6GC at 385V = 54mv x 2 = 108mv.  385/.108 = 3.6K.  So I'll need 4K RL. 
* So i ended up buying a 4K OT with 4/8/16 secondaries.  I use 1-to-1 speaker connections for the 6L6's (i.e. 4ohm speaker into 4ohm jack)
And when I run 6V6's i just run speaker load double the Z of the jack - e.g. 8ohm speaker into 4ohm jack yields 8K load. 

If I'm off base I'm sure someone will correct me!

Mathematically this works out to the same understanding I have, but OTs spec'd for these tubes rarely match these numbers.  For example, the JCM800 2204 with two el34s running at 70%dissipation at 480VDC plate voltage, and 36mAx2 works out to be about 6.7k a-a, but we don't see marshall putting 6.7k primaries in, they're 3.7k.  According to the paper posted up above, this SEEMS to be because of the desired signal transfer characteristics of running the tubes against half or less of their apparent Rp.  I'm still trying to make sense of it as well. 


Looking into it more, if they really do this for the sound, then it seems like there should be consistency in their approach.  Looking at the 100 watt model at 70% dissipation, the current is ~144mA, Pvdc=480VDC, so the Rp works out to ~3.3k.  They use a 1.7k OT. So, 1.7/3.3 = .51 and  3.7/6.7=.55 ratios of Apparent Rp to OT impedance, which are not too far off from each other in proportion so the frame of thought seems plausible. 

Edit:  I read that Fahey author'd link above.  During class A operation, the tubes are acting like they're in parallel instead of in series, so the apparent resistance halves.  This leads me to believe that manufacturers are matching their OT's to the class A part of operation even for class AB1 amps? 

[shooter]

but we don't see marshall putting 6.7k primaries in, they're 3.7k. 

I'm way better with meters n scopes than slide-rules and calculators, I like understanding the theory, but at the end of day, the datasheet, scoping the speaker side signal, and listening, gets you north of 90% good enough for both musician and audience 

[TheKT88KilledJFK]

but we don't see marshall putting 6.7k primaries in, they're 3.7k. 

I'm way better with meters n scopes than slide-rules and calculators, I like understanding the theory, but at the end of day, the datasheet, scoping the speaker side signal, and listening, gets you north of 90% good enough for both musician and audience 

Hahah I would gladly surround myself with numerous test Tx's just to see the results but my wallet limits me to simple mathematical speculation. 

[shooter]

my wallet limits me

  I got sent back to '91 '84  wages not to long ago so I just draw n color alot

[PRR]

> If I'm off base

I'm not sure you are in the ballpark.

Even allowing for the mV versus mA mixup.

The Idle current has about nothing to do with the Max Power conditions.

The "70% rule" is a dubious thing.

Moshing them together does get you "near" a near-A set of number, more by chance than analysis.

Tube-books, and known-good commercial designs, are our gold standard.

Plotting loadlines is only for the truly desperate, or terminally bored.

[dennyg]

> If I'm off base

I'm not sure you are in the ballpark.

Even allowing for the mV versus mA mixup.

The Idle current has about nothing to do with the Max Power conditions.

The "70% rule" is a dubious thing.

Moshing them together does get you "near" a near-A set of number, more by chance than analysis.

Oh man,  the student just got a well-deserved drubbing from the prof!! The mv typo was so embarrassing that i corrected it. I should've referred to Tubeswell's link before drawing such an elemental conclusion.  The first thing Fahey says is enlightening:

Class AB is a confusing label, so let´s split it into a "pure Class B" section and a "pure Class A" one.
1) pure Class B: it´s the main component and for maximum power you should order your transformer based on this value.

then he goes on to calculate it...but surprised at the result for a 6L6 (assuming 19W version and not a 30W 6L6GC):

so optimum plate load impedance:
* Zp: 380V/0.14A=2700 ohms ... that´s for a single plate. Za
Since it´s a push pull, plate to plate impedance Zaa is 4X that (because plate to plate transformer is 2X voltage=4X impedance) so Zaa=10800 ohms.

Is the conclusion to above that a pair of 6L6's with a plate voltage swing of 380 has optimal OT primary of 10K?

[tubeswell]

Is the conclusion to above that a pair of 6L6's with a plate voltage swing of 380 has optimal OT primary of 10K?

You need to read to the end of the post

[dennyg]

I did read the end where he states that for Class A section, Zaa is half or 5.5K.  But he starts by saying the OT should be chosen based on Class B Zaa which is 11K.    What am I missing?

BTW I found this thread  - populated with some excellent content by Tubeswell and PRR on the OT Z topic:

http://el34world.com/Forum/index.php?topic=18963.msg195595#msg195595

[TheKT88KilledJFK]

After playing around with voltage controlled current sources driving transformer models all day,   It looks like plate impedance has a small effect on frequency response and power output.  Further more, the ranging the OT primary impedance from 3k to 8k doesn't really make too much of a difference either.  It looks like the largest effect of the frequency response of the system is the inductive load on the secondary which is the speaker. 

A higher turn, therefore higher henry, OT can improve low end and general power output slightly for large jumps in inductance value.   

Someone please let me know if I am in error.  Also, all the Tx's on Hammond's site are rated center tap to anode, not anode to anode.  I'm not saying that's right or wrong, just pointing out a source of confusion for myself in understanding.

[DummyLoad]

Also, all the Tx's on Hammond's site are rated center tap to anode, not anode to anode.

what rating? please be more specific:  current? impedance? DCR? inductance?

--pete

[TheKT88KilledJFK]

Also, all the Tx's on Hammond's site are rated center tap to anode, not anode to anode.

what rating? please be more specific:  current? impedance? DCR? inductance?

--pete

The tables with the Tx listings have a column named Primary Impedance that gives a number like 7k CT, but the CT isn't and indication of 7k impedence at the center tap.  Their datasheets list the impedance as anode to anode, regardless.

We buy P-P OTs by the plate-to-plate impedance. For a CT winding this is *4 times* the one side impedance. So far we pencil 4*1.87K= 7.5K.

This made no sense to me until I realized the # of windings is reduced by half, and the impedance is the square of that ratio, so any reduction would be squared.  Felt like a clot after this!

[PRR]

> 7k impedence at the center tap.

In most amps, the "impedance at the CT" is dead zero: the power supply capacitor.

The number is end to end.

Also SPICE has current sources, which may save time over voltage sources and VCCS parts.

I'm not sure that playing with linear sources tells you anything about maximum power output with limited supply voltage, tubes of limited current, and limited dissipation.