6L6 PP with ~420V B+ where to find the right OT ??

[kagliostro]

A friend has build a Fender style amp with a pair of 6L6 as power tubes and B+ at 420V

as OT he is using a 2K OT with an 8ohm speaker connected to the 4ohm out so the tubes see a load of 4K

with this arrangement he has a low power (near 20W)

from the datasheet a pair of 6L6 in PP with ~420V B+ requires a load of 5.6K

http://www.mif.pg.gda.pl/homepages/frank/sheets/093/6/6L6GC.pdf

but we are not able to find an OT with such primary impedance

is it to be considered a custom OT to be wind on purpose ?

Councils, indications ??

Thanks

Franco

[PRR]

4K output should not be 20 Watts.

How is the bias? How are you testing?

[HotBluePlates]

And what is the screen voltage in the amp?

[kagliostro]

Bias 37mA

About test method and G2 voltage I'll ask


Actual PT Hammond EDB 290 EEX

http://www.diyitalia.eu/download/file.php?id=12105

Actual OT Hammond 1750W

https://www.hammfg.com/files/parts/pdf/1750W.pdf

Thanks

Franco

[kagliostro]

The requested data:

bias @ 37mA (each tube)

B+ @ 421V

G2 @ 418V


the power was measured ......... ahem ..... by ear 

---

My fried borrowed a 4.2K / 8ohm OT and now all is fine with the amp

I think he was doing something wrong with the connection of the other OT, otherwise I can not explain the problem that he had

Many thanks to all

Franco

[drgonzonm]

Using an 8 ohm speaker in on a 4 ohm tap will result in a lower volume.   Using the power formula, P=V²/R, the apparent power will be half.  Look at the data sheets from hammond, you will see that the 4 ohm tap has the sqrt of 2 lower voltage than an 8 ohm tap, and half the voltage of a 16 ohm tap. 

To a realistic reading even by ear, hook the 8 ohm speaker on a 8 ohm tap. 

[HotBluePlates]

Using an 8 ohm speaker in on a 4 ohm tap will result in a lower volume.   Using the power formula, P=V²/R, the apparent power will be half.  ...

This is logical, but it assumes "all else unchanged." Unfortunately, all else does not remain unchanged.

If the supply voltage and operating point allow, the plate voltage swing ("V" in the formula) gets bigger with a higher load impedance. If the output tube could manage the same plate current swings, then with a larger load impedance Ohm's Law shows the plate voltage swings will get bigger (same-I * bigger-R = bigger-V).

So in your power formula, R got larger (power goes down) but V also got larger (power goes up). But the same-current assumption above may not hold...

We can't know the net effect without more information. What we can know is that if the original load impedance was the one impedance for the tube, supply voltage and class of operation which delivers maximum output power, then any deviation above/below will result in less clean output power.

...  Using the power formula, P=V²/R, the apparent power will be half.  Look at the data sheets from hammond, you will see that the 4 ohm tap has the sqrt of 2 lower voltage than an 8 ohm tap, and half the voltage of a 16 ohm tap. 

This also seems to be logical proof an 8Ω speaker on a 4 Ω tap will give half-power (if you do the math for the indicated secondary voltages, 8Ω on the 4Ω tap does give half-power compared to 8Ω on the 8Ω tap).

The flaw here is that the secondary voltages assume the same voltage swings occur on the primary regardless (the secondary voltages are equal to the primary voltages transformed by the step-down ratio from primary to secondary).

However, the changed secondary impedance (8Ω on the 4Ω tap) result in a change of reflected primary impedance, which then changes how the output section performs. Again, we can't conclude any rigid relationships without more info on the output section supply voltage, current capability and class of operation.


The approach used for the wrong assumptions above treated the output section somewhat like a transistor output stage... That is, it assumed the output section could throw any current required to maintain the output voltages indicated at the secondary of Hammond's diagrams.

By using that key assumption, you would also conclude that attaching a 4Ω load to the 8Ω tap gives double-power. And those kind of games get played with transistor power amps (attaching lower loads to get louder/more-power output), right up to the point the output transistors melt down.

Tubes don't work that way, so going lower or higher from optimum will result in less output power. This is graphed on the more complete data sheets, even though the specific numbers (loads, power output, distortion %) really only apply for the single operating conditions listed in the graph. However, the general principle always applies.

[drgonzonm]

http://www.radioremembered.org/outimp.htm

See above link, that explains current changes as secondary load changes. 

[HotBluePlates]

http://www.radioremembered.org/outimp.htm

See above link, that explains current changes as secondary load changes.

Yes, the link includes a large number of simplifying assumptions to:
A) Make the math obvious to the reader, and
B) Demonstrate (only) the concept of reflected impedance.

The assumption that halving the load resistance will always double the primary/secondary current (or vice versa, doubling load & halving current) is not true in practice. Assuming an initial ideal load impedance for maximum output power, a tube will run into some other constraint before it can behave the way the simplified math predicts. For example, the output tube's screen voltage dictates what current the tube can pass; if there are 0v from screen-to-cathode, the tube will pass 0mA no matter how low the load impedance.

It is also the very reason we have output transformers (because tubes function well into loads very much higher than a speaker impedance) and transistor amps don't (because even a 2Ω load is a manageably-high impedance to a transistor output stage).

I point this out because on another forum, the idea is taken to its logical extreme where folks say you can mismatch the OT tap & speaker load to have the same result as a power attenuator. And in reality, the volume from the speakers is hardly changed. Try it for yourself & see. And be careful about reading too much into simple examples. As in this case, accurately applying an oversimplified example often means failing to have an accurate model of the behavior you want to understand.

[drgonzonm]

I purposely avoided the complex number solutions to this problem.  As there was several available, all starting with the assumption that the secondary impedance was known.   Those solutions also suggested that speaker data would be needed, something not available, as calculation at a single parameter, say 1 kHz, keeps the quantitative results very qualitative

First assumption made, is we are using the nominal impedance of the speakers. 
Second assumption made, is the test frequency used on the o/t's typically 1 kHz, when, for guitar amps 400 Hz, might be a better test frequency. 

And then if current was not a big concern, then why is it that many O/T manufacturers put larger wire on the 2 and 4 ohm secondary windings, (They by design carry a heavier current). 

HPB, I do not disagree with your approach, but sometimes, simple answers are better.    The question was, why does it appear the power seemed to about half the power expected.  not a lot of information was provided, so a semi-qualitative answer is appropriate.

[HotBluePlates]

... HPB, I do not disagree with your approach, but sometimes, simple answers are better.    The question was, why does it appear the power seemed to about half the power expected.  not a lot of information was provided, so a semi-qualitative answer is appropriate.

Since Kagliostro's friend correctly attached an 8Ω load to a 4Ω tap to achieve the correct reflected impedance, you overlooked the simplest answer: A probable poor connection.

A friend has build a Fender style amp with a pair of 6L6 as power tubes and B+ at 420V
as OT he is using a 2K OT with an 8ohm speaker connected to the 4ohm out so the tubes see a load of 4K
with this arrangement he has a low power (near 20W)

My fried borrowed a 4.2K / 8ohm OT and now all is fine with the amp

I think he was doing something wrong with the connection of the other OT, otherwise I can not explain the problem that he had