Question about mismatched loads on a output transformer.

[Stevem]

Let’s say I have a OT that’s rated for 60 watts rms and has a 4 and 8 ohm secondary tap.


I run my calculations and find that the 8 ohm tap has a turns ratio of 11.6 and the 4 ohm taps ratio is 16.4.

For either of these turn ratios to generate 60 watts the pair of output tubes I choose must together must produce 236ma, and of course the OTs primary must be rated for at least that level of drive current.

Now here’s my question.

If I hook up 8 an ohm load on the 4 ohm tap is my finding correct that only 167ma will be called for from the output tubes ?

Conversely if I place a 4 ohm load on the 8 ohm tap then the tubes will be called upon to produce 331ma?

Actually here’s a part B of my question.

Is in the case of having a 4 ohm load on the 8 ohm tap the output stage will only be able to produce 36 watts of rms power, and then with the 8 ohm load on the 4 ohm tap only some 42 watts can be produced?

Thanks in advance for help on these questions.

[pdf64]

Expecting the amp to output 60W because the OT is rated at 60W seems wrongheaded?
Whatever, drawing the loadlines would help answer your queries. These days that's much easier https://www.vtadiy.com/loadline-calcula ... alculator/

[Stevem]

Thanks for that link but I don’t have the patience for all the pop up adds that site has.

[R.G.]

+1 on the transformer. A power rating on a transformer is only as good as the string of footnotes describing the operating conditions and frequency range. The 60W number is a capability, not a forced condition. And the "impedance" is instead a ratio assuming a slew of conditions on loading. It's easy to under-use an output transformer.

Tubes in general are high impedance devices, so they won't give more current than their Vgk drives them to. All those horizontal lines in the plate characteristic show that for, say, Vgk = -5V, the current will be that level, period. Changing the load resistance doesn't shift the actual plate current at a given Vgk by much, if any. the load can't suck more current out of the tube than the tube wants to give. Changing the slope of the load line does change the plate-to-cathode voltage for a given current, so it changes the internal power dissipation, because the current at a given Vgk is fairly fixed, but lower load impedances make the Vp-k higher. To me this is what a plate-characteristic load line chart says.

I'll go look for it in my archive of "really old electronics books and magazines", but there is a good chart of generic output power versus load impedance I remember. The chart has output power on the vertical axis and plate-to-plate load on the horizontal axis. The curve of output power versus p-p load has a broad round hump in the middle.

Apparently tube >designers<, the guys that actually specified the innards of the tubes, would run such charts to figure out what p-p load they would specify for advertising the best and most power loads for their new tubes. They would design the tube, get the guys on the line to make a one-off run for them, then go test them.

For example, a new 6l6 prototype would ideally have the peak of the power-vs-load hump at 4K to 4.4K or so.

On the same chart, they would plot 2nd +3rd harmonic distortion. This tended to have a broad valley as 2nd harmonic went down and 3rd went up. For a 6L6-ish device, the valley for distortion would be about 6K to 6.6k. This is how the guitar world came to use 4.4K for guitar amps and 6.6K for hifi amps on 6L6-ish tubes.

[R.G.]

Thanks for that link but I don’t have the patience for all the pop up adds that site has.

Use Firefox, install AdBlocker Ultimate. It's really quite effective.

I just clicked on the site. The only popup I got was a polite request to please allow ads. Apparently they try to detect ad blockers and ask for mercy.

[Astronomicum]

Actually here’s a part B of my question.

Is in the case of having a 4 ohm load on the 8 ohm tap the output stage will only be able to produce 36 watts of rms power, and then with the 8 ohm load on the 4 ohm tap only some 42 watts can be produced?

Are you using 236mA for these two calculations?

[Stevem]

Yes I am using 236ma for both.

My reason for my post was to confirm the added tube life of having placed a 8 ohm load on a 4 ohm tap and also having a 4 ohm load on a 8 ohm tap will try to suck the life out of the output tubes .

I was not asking to get the normal 60 watts out of the output stage when mis loaded.

And yes I know that with a pair of output tubes that can provide a combined 236 ma that they will have to also be able to deal with 480 to 500 volts on there plate to produce 60 watts.

[Helmholtz]

Doubling load impedance lowers plate current but significantly increases screen dissipation.
So likely lowers tube life compared to optimum load.

[Stevem]

Yes I see that now from your kind response to me in a post I responded to some two weeks ago.

Oh and I forgot to note that this OTs primary impedance is 4.3K center taped.

[Astronomicum]

Oh and I forgot to note that this OTs primary impedance is 4.3K center taped.

I thought I was with you up to part B, but now, not so sure. I calculate the primary impedance based on your turns ratio in your initial post as 1076 ohms, which then yielded your 236mA, 167mA, and 331mA calculations. I am now lost. Sorry.

[Helmholtz]

An OT having an Raa of 4.3k means a tube load of Ra = Raa/4 = 1075 Ohm across each half-primary in class B.

[Astronomicum]

Knowing that the turns ratio provided is just half the OT helps. Thanks.

[Helmholtz]

Knowing that the turns ratio provided is just half the OT helps. Thanks.

Just guessing myself.

[Stevem]

Divide the published 4.3k by 4=1075.

Then for the 8 ohm tap take the square root of (1075/8) =11.6, the same done for the 4 ohm =16.4.