Thread: Math behind OT impedance rules-of-thumb?

1. Math behind OT impedance rules-of-thumb?

Hi. I've seen all over the forum(s) (and tube literature as well) some ROT , or typical design values for OT impedance for each tube plate resistance. And there are rules for each configuration as well (pentode/triode, SE/PP and combinations). Anyone knows the math behind those rule of thumb? For example, just to name, why a single ended tube "like to see" about 1/10th of ra? Why operating as a triode, Zout must be higher?

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2. Originally Posted by elushi
Hi. I've seen all over the forum(s) (and tube literature as well) some ROT , or typical design values for OT impedance for each tube plate resistance. And there are rules for each configuration as well (pentode/triode, SE/PP and combinations). Anyone knows the math behind those rule of thumb? For example, just to name, why a single ended tube "like to see" about 1/10th of ra? Why operating as a triode, Zout must be higher?

It all has to do with load lines and the plate characteristic curves. The curves in triode vs pentode configuration are different and thus will cross the load line at a different point.

If you don't understand load lines yet, you'll need to gain that understanding in order to illustrate it.

The class of operation will determine how much power will be dissipated by the valve itself during the transition period of the AC output signal (i.e. the period in which the signal is transitioning from positive to negative and vice versa). The load impedance and B+ voltage will determine how much power will be dissipated by the load (i.e. the OT) at the peaks of the swing.

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3. I'd recommend you give RG's site a read: http://www.geofex.com/Article_Folder...es/xformer.htm

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4. In addition to loadline analysis, there's another way to approach how to choose the proper OT primary impedance.

You start with the choice of tube and power supply voltage. You determine how much power the tube can dissipate (from the specs), and use Z=V/I to determine the primary impedance. Note you must keep peak and average power straight.

So, for example, let's say your power supply voltage is 260V and you're using a 6BQ5 in SE operation, whose max. plate dissipation is rated at 12W. In SE operation the quiescent plate current is usually set for maximum plate dissipation (to get maximum power output), so the quiescent plate current will be 260V/12W=.046A. So, V=260V, I=.0476A, and Z=5.7k ohms.

For PP outputs it's a little more complicated. Each tube uses half of the primary, so the impedance seen by each tube is 1/4 the total plate-to-plate impedance. Also, in class AB operation you don't bias quiescently for maximum plate dissipation. The best approach is to glue two sets of characteristic curves together, with one upside down, joined on the x-axis at the supply voltage. Then, plot a load line whose slope equals the OT primary impedance. It should sit within the maximum dissipation curves. Or, plot a load line that just touches the maximum dissipation curves (again, for maximum output) and the calculate its slope. The slope equals the desired primary impedance.

There are more complete explanations of this approach in the literature, including my book "Design and Construction of Tube Guitar Amplifiers."

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5. Originally Posted by elushi
For example, just to name, why a single ended tube "like to see" about 1/10th of ra? Why operating as a triode, Zout must be higher?
Triodes deliver maximum power when Zpri is equal to 2*ra; this is proved in any old textbook. However, they will still work perfectly well into a wide range of loads (but with less output power). Increasing Zpri will tend to reduce the harmonic distortion, so hifi designs will tend that way..

I have only seen the "Zpri = 1/10 ra" recommendation for pentodes on one website. As far as I know, it is nonsense. The ideal load for a pentode is one that passes through the knee of the curves, and that depends on the screen voltage, not ra.

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6. Originally Posted by TransLucid
There are more complete explanations of this approach in the literature, including my book "Design and Construction of Tube Guitar Amplifiers."
Page 56

I read your book in 2 days, very nice, congrats. I wish I had an introductory book like yours way back. I think you forgot to mention at what frequency that 5.7K impedance is measured at (1 KHz)?

Edit: In fact, both author's on this thread arrived together in the mail:

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7. Originally Posted by Merlinb
The ideal load for a pentode is one that passes through the knee of the curves
I thought the ideal load was one that passes just below the start of the knee for maximum voltage drop across the load no?

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8. It's all about voltage and current.

An output tube has a maximum anode voltage it can withstand to have a reasonable chance of not arcing over. In a Class AB amp, each tube pulls its primary connection down and this forces the "off" tube anode above the power supply by transformer action. The sum of power supply and transformer-supplied peak voltage has to be less than the tube arc-over limit. So the tube arc-over and the operational class limit how big the maximum power supply voltage can be.

The output tube has a (sloppily defined) maximum current it can pass for a given set of conditions. It has a certain residual voltage across it when this is happening. The tube won't allow more current through (in general) and won't pull that much current below what would be called a saturation voltage in a transistor.

The references to the "knee" are describing ways to take the sloppy definition of the conduction knee into account. You can pick a place above, in, or below the knee, and this allows more or less power at more or less distortion as a tradeoff.

But the impedance which gives the most power out for a given power supply is the impedance which is the difference between the max power supply voltage and the knee voltage, divided by the current at the knee. "Knee" here really means "wherever you choose the knee to be" because it's a sloppy, ill defined region. That calculation makes the tube work between its voltage maximum and its current maximum, and that gives the max power out.

If you pick a different point on the knee, or a different power supply, things change a bit. But the best power impedance doesn't change all that much with small changes in "knee point" or with small changes in power supply.

The distortion output changes as well. The load impedance for most power is generally lower than the load impedance for lowest distortion.

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10. Originally Posted by Merlinb
Triodes deliver maximum power when Zpri is equal to 2*ra
So it means that for a 6BQ5 Zpri should be 76K

Originally Posted by Merlinb
I have only seen the "Zpri = 1/10 ra" recommendation for pentodes on one website. As far as I know, it is nonsense
I read that in the AX84 theory pdf. But it was just to name one.

Originally Posted by R.G.
It's all about voltage and current.
An output tube has a maximum anode voltage it can withstand to have a reasonable chance of not arcing over. In a Class AB amp, each tube pulls its primary connection down and this forces the "off" tube anode above the power supply by transformer action. The sum of power supply and transformer-supplied peak voltage has to be less than the tube arc-over limit. So the tube arc-over and the operational class limit how big the maximum power supply voltage can be.
The output tube has a (sloppily defined) maximum current it can pass for a given set of conditions. It has a certain residual voltage across it when this is happening. The tube won't allow more current through (in general) and won't pull that much current below what would be called a saturation voltage in a transistor.
The references to the "knee" are describing ways to take the sloppy definition of the conduction knee into account. You can pick a place above, in, or below the knee, and this allows more or less power at more or less distortion as a tradeoff.
But the impedance which gives the most power out for a given power supply is the impedance which is the difference between the max power supply voltage and the knee voltage, divided by the current at the knee. "Knee" here really means "wherever you choose the knee to be" because it's a sloppy, ill defined region. That calculation makes the tube work between its voltage maximum and its current maximum, and that gives the max power out.
If you pick a different point on the knee, or a different power supply, things change a bit. But the best power impedance doesn't change all that much with small changes in "knee point" or with small changes in power supply.
The distortion output changes as well. The load impedance for most power is generally lower than the load impedance for lowest distortion.
Jokes apart, thank you RG, think I kinda got the idea. Man!...what a great site you have!

Originally Posted by kg
Thank you Ken! By the way, a few days ago I read Building Your Own Tube Amp | ken-gilbert.com. Great article!

PD: Now I'm confused with something else. impedance changes with frequency right? so, the values for Zout in datasheets and manuals are with respect with what frequency?

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11. Originally Posted by elushi
Now I'm confused with something else. impedance changes with frequency right? so, the values for Zout in datasheets and manuals are with respect with what frequency?
Another good question for someone who's learning.

It's important to remember:
transformers do not have impedances - they have ratios.
A transformer is a lever, or gearbox to allow the high voltage but low current tubes to move high current but low voltage speakers. So they have an impedance ratio that matches (for instance) an 8 ohm speaker to a 4400 ohm plate to plate impedance for 6L6s. That same transformer would happily convert a 16 ohm speaker to 8800 ohms at the 6L6 plates, or a 10 ohm resistor to 10*(4400/8) = 5500 ohms at the tubes. The transformer doesn't much know or care what you hang on the output, it just does the fixed-ratio conversion.

[For the people who carefully check technical matter in forum postings: yes, I'm aware that the inner workings of the transformer work better with some range of external impedances than others, so there is a usable impedance range, but lighten up and don't confuse the people who are trying to learn. 8-) ]

When a maker quotes a transformer "impedance" they will usually also note the speaker load they mean it to be driving when it delivers that impedance to the plates.

And you are correct, impedances change with frequency. The transformer people quote their specs at mid-band, an indeterminate frequency in the middle of the transformer "good band" where you can ignore the transformer's imperfections for the most part. The transformation ratio quoted for the transformer holds inside the quoted frequency range for the transformer.

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12. Originally Posted by R.G.
transformers do not have impedances - they have ratios
When a maker quotes a transformer "impedance" they will usually also note the speaker load they mean it to be driving when it delivers that impedance to the plates.
I do know that, but I (maybe incorrectly) assumed that the primary impedance was always quoted with an 8-ohm load at the sec.
But now comes to my mind another question (sorry!): if primary inductance change (signal swings from the lowest freq to the highest freq. reproduced by a guitar).
Why doesn't it mess the bias? I mean, with a high frequency Zpri will be high and Ra will be low because it's capacitive...but things work.

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13. oops I forgot why Ck was there for...

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14. Originally Posted by elushi
I do know that, but I (maybe incorrectly) assumed that the primary impedance was always quoted with an 8-ohm load at the sec.
A lot of time it is, but it pays to check.
But now comes to my mind another question (sorry!): if primary inductance change (signal swings from the lowest freq to the highest freq. reproduced by a guitar).
Why doesn't it mess the bias? I mean, with a high frequency Zpri will be high and Ra will be low because it's capacitive...but things work.
The primary inductance doesn't change (... much - it's very much a second order effect) and the output transformer is used under conditions where it thinks it's working only on an AC signal, the two half-primaries each doing a half-cycle and the DC through the transformer cancelling out in them. This is kind of impedance independent. Since bias is a DC condition that is mostly cancelled out for the transformer, the AC conditions can't affect the bias.

oops I forgot why Ck was there for...
Ck helps with crossover issues in cathode biased output stages, but the AC versus DC conditions reasoning holds in either fixed bias or cathode biased stages as far as the output transformer goes.

There is a bias-change effect in resistor/cathode biased output stages. Biasing in Class AB is intended to make the DC bias current small compared with the peak AC signal current. When the signal is zero, all you have is DC bias. But when you run a signal through it, both output tubes start conducting more current peaks than their DC bias condition, and the sum of the currents for the two tubes does go up. Each tube pulls opposite directions on the output transformer, but each delivers its signal current the same direction to Rk, so the average current there goes up with bigger signals. The resistor/cathode biased stage does get biased more off by bigger signal peaks that last longer than the Rk/Ck time constant.

Anyway, Ck is there for different things.The bias shifts, but it's because of overall signal level, not because of signal frequency.

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15. Originally Posted by elushi
But now comes to my mind another question (sorry!): if primary inductance change (signal swings from the lowest freq to the highest freq. reproduced by a guitar).
Why doesn't it mess the bias?
"Bias" only applies to the static DC current which flows through the transformer under zero signal conditions. Since impedance doesn't apply to DC and transformers cannot pass DC, the speaker on the secondary draws nothing from the primary when there is just static DC on the primary. This means that when the valves are in a quiescent state (i.e. idle, or "zero signal conditions"), the valves do not see the reflected impedance from the secondary load and only see the "DC resistance" of the primary coil itself, which has nothing to do with the transformer's impedance ratio.

It also applies to the static negative control grid voltage as well. The input signals at the control grids swing above and below that static negative DC voltage value. Think of it as an AC signal being superimposed onto the static DC voltage at the control grid.

As the signal is swinging positive/negative, plate current is swinging above and below that static DC bias current. You now have an alternating DC current (I say "alternating DC" because while it is an "alternating current" in the sense that it is constantly alternating in value, it's still positive going current in terms of flow direction so the direction of flow is not "alternating" like it would be in pure AC) through the primary, which the transformer does pass, thus the speaker draws from that and the valves now see the reflected impedance of the transformer that is provided by its impedance ratio.

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16. Originally Posted by Wilder Amplification
Since impedance doesn't apply to DC
I know what you're saying, but this is sure to confuse him if he goes on and learns more. Impedance in general as a concept applies to all frequencies, including DC; it's just that at DC inductive impedance goes to zero and capacitive impedance goes to infinity, leaving on the resistive parts to consider at DC. Then as you note:

and transformers cannot pass DC, the speaker on the secondary draws nothing from the primary when there is just static DC on the primary.
which is exactly correct.

As the signal is swinging positive/negative, plate current is swinging above and below that static DC bias current.
... in a class A amplifier where the bias is such that the plate current in either output tube never goes to zero. In a Class AB amp, the plate current swings lower until it hits zero, while the opposite tube is conducting more current.

You now have an alternating DC current (I say "alternating DC" because while it is an "alternating current" in the sense that it is constantly alternating in value, it's still positive going current in terms of flow direction so the direction of flow is not "alternating" like it would be in pure AC) through the primary, which the transformer does pass, thus the speaker draws from that and the valves now see the reflected impedance of the transformer that is provided by its impedance ratio.
Yes. The transformer combines alternating DC pulses on opposite sides of its primary into an alternating AC current, which can then be transformed. Any DC offset is lost as far as transforming through to the secondary.

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17. What are you guys talking about! Now my head is hurting.

The important thing is that Ra of a pentode or beam tetrode is very high and more or less meaningless. You can't use it to calculate the optimum load impedance. You have to take a large-signal approach that looks at the maximum current you can shove through the tube, the maximum voltage, power dissipation and so on. There is a lot of latitude: depending on the supply voltage and load impedance, a pair of EL34s can produce 20w of luscious Class-A cathode biased power, or 100W of harsh, horrid Class-B mush.

The physics behind Ra of a triode: The plate voltage influences plate current in just the same way as the control grid voltage. You can prove that for maximum power output the load impedance should be equal to Ra.

But in a tetrode or pentode, this effect is stopped by the screen grid, which launches electrons towards the plate almost irrespective of its voltage. Ra is almost infinite compared to a triode.

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18. Originally Posted by Wilder Amplification
This means that when the valves are in a quiescent state (i.e. idle, or "zero signal conditions"), the valves do not see the reflected impedance from the secondary load and only see the "DC resistance" of the primary coil itself, which has nothing to do with the transformer's impedance ratio.
well, they reflect impedance as long as it stay close to an ideal transformer...It's less confusing to me to think that it transform voltage, and with the speaker connected, it induces a current that can drive it (which is large because the speaker has a very low impedance)...Am I right? well...in fact V, I and R will all three make a "balance" according to Ohm's law.

Originally Posted by Wilder Amplification
"Bias" only applies to the static DC current which flows through the transformer under zero signal conditions
It just came to my mind that impedance question because I thought that, with signal of frequency f Hz applied to grid, inductive impedance of OT would be 2.pi.f.L and capacitive impedance of the tube would be 1/2.pi.f.C (with C the tube internal capacitance) and so the impedance load would also change and this in turn would made the tube instable. but I was just raving.

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19. Originally Posted by Steve Conner
Hi Steve! my bad! I'm afraid I change myself the subject of the thread...

Originally Posted by Steve Conner
You can't use it to calculate the optimum load impedance.
In fact, I started the thread because I thought it was some kind of formula with Ra and Rl in parallel...

Originally Posted by Steve Conner
The physics behind Ra of a triode: The plate voltage influences plate current in just the same way as the control grid voltage. You can prove that for maximum power output the load impedance should be equal to Ra.
So it means that with the same impedance load that you use with a pentode (i.e. the same OT) when you switch the tube to a triode (for example, with a switch that connects screen grid to OT) you are not driving full power any more (because the tube sees 1/4 of the primary and this is usually far less than Ra), right?

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20. Originally Posted by elushi
well, they reflect impedance as long as it stay close to an ideal transformer...It's less confusing to me to think that it transform voltage, and with the speaker connected, it induces a current that can drive it (which is large because the speaker has a very low impedance)...Am I right? well...in fact V, I and R will all three make a "balance" according to Ohm's law.
That is precisely how one should think about it! One thing that most people have the misconception of is that OTs somehow are different from a power transformer. But in reality, an OT is just a power transformer that operates at audio frequencies. It transforms high voltage/low current to low voltage/high current just like a step down power transformer does. Impedance is just a figure that is calculated off of these two things.

Amplifiers are just modulated DC power supplies...an AC power inverter if you will. As Steve Connor once mentioned, a push pull output section is basically just a full wave center tap rectifier circuit flipped upside down and backwards.

Originally Posted by elushi
It just came to my mind that impedance question because I thought that, with signal of frequency f Hz applied to grid, inductive impedance of OT would be 2.pi.f.L and capacitive impedance of the tube would be 1/2.pi.f.C (with C the tube internal capacitance) and so the impedance load would also change and this in turn would made the tube instable. but I was just raving.
One thing one must remember is that idle bias current only exists when the amp is in a quiescent state. Once there's a signal present, plate current and plate voltage fluctuate with the input signal as well as with the impedance of the transformer so the "static bias current" no longer applies except to mark where the "zero crossing" point of the output signal will be on the primary side of the OT.

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21. Originally Posted by Wilder Amplification
As Steve Connor once mentioned, a push pull output section is basically just a full wave center tap rectifier circuit flipped upside down and backwards.
Yes! Totally! I didn't noticed that!

Originally Posted by Wilder Amplification
Once there's a signal present, plate current and plate voltage fluctuate with the input signal as well as with the impedance of the transformer so the "static bias current" no longer applies except to mark where the "zero crossing" point of the output signal will be on the primary side of the OT.

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22. Originally Posted by Steve Conner
a pair of EL34s can produce 20w of luscious Class-A cathode biased power, or 100W of harsh, horrid Class-B mush.
The 25W max plate dissipation that shouldn't be exceded is Vp * Ip, with Vp=Vpp-Vk the voltage drop of the tube, right?

Output power is Ip^2*Rl with Rl = Zout/4 (for push pull OT center-tapped at B+) and also can be calculated as Vpp-Vp * Ip, right?

I'm confused because if I plot a load line for a EL34 tube with 450V B+ supply and 5k p-p load and the line exceeds the 25W limit curve almost all the time.
Just to got an idea of the slope, these are two points:
First point of the line: 450V @ 0A (that's not the bias point), second point of the load line 0V @ 450V/1250 = 0.36A

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23. Well, the 25W figure is an average. Unlike puny transistors that can be damaged by high power peaks, the plate of a tube takes several seconds to heat up.

In particular, it's a push-pull amp so the tube is only conducting half the time, therefore the average dissipation is half what the diagram implies. Actually a little more than half, to the extent that the amp runs in Class-AB rather than pure B.

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