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-   -   Understanding load on triode and impedance matching (https://www.diyaudio.com/forums/tubes-valves/338970-understanding-load-triode-impedance-matching.html)

itsikhefez 9th June 2019 08:32 PM

Understanding load on triode and impedance matching
 
Hi,

I've looked around and asked this question in various threads but still haven't fully understood this. The context of the question is that I'm trying to build a 45 based SE headphone amp. Let's assume a standard 5K:8R transfomer is used, and that it would have 2W available to 8ohm speakers. Now, what happens when 32ohm and 300ohm headphones are connected to the 8ohm tap?

This is my understanding:
If headphones are connected directly, 32ohm will have 500mW and 300ohm will have 50mW. The reflected load on the primaries should ideally be around 5K. When connected directly, the load will be 20K and 187K respectively.

OTOH, connecting a 12ohm/3W resistor across the output would "fake" the reflected load to 5.4K-7.2K, which seems better. In this case though, how much power is being "wasted" on the load resistor and how much remains for the HP's?

I've basically gotten conflicting answers on whether the load resistor is required or not, hence my confusion.

Lastly, I would like to understand what are the differences in Voltage swing for the headphones, if they were connected to an 8ohm tap vs. a 32ohm tap.

Thanks!

rayma 9th June 2019 09:00 PM

Quote:

Originally Posted by itsikhefez (https://www.diyaudio.com/forums/tubes-valves/338970-understanding-load-triode-impedance-matching-post5817883.html#post5817883)
how much power is being "wasted" on the load resistor and how much remains for the HP's?

Can you post the circuit? A resistor in parallel with the load will split the power according to 1/R for each, since power = V^2/R, and V is the same for both.

itsikhefez 9th June 2019 10:35 PM

The circuit is a Tubelab SE. I don't think there is any NFB.

OK so assuming the same 8ohm output, 32ohm HP - 12ohm load resistor gets ~72% of power and 32ohm gets 27%, HP will have 545mW. 300ohm HP gets ~3.8% which is 76mW.

If these calculations are correct, then overall power to the HP's seems the same with or without the load resistor, but the reflected load on the triode is different in both cases. Without load resistor, the tube is being "under" worked, while with the load resistor, it's being loaded to full power.
Is this correct? If so, which is the preferred scenario? There has to be a "correct" way to do it.

rayma 9th June 2019 10:39 PM

Quote:

Originally Posted by itsikhefez (https://www.diyaudio.com/forums/tubes-valves/338970-understanding-load-triode-impedance-matching-post5817983.html#post5817983)
Without load resistor, the tube is being "under" worked,
while with the load resistor, it's being loaded to full power.

Power varies with total loading, but George should take this further.

itsikhefez 9th June 2019 10:41 PM

Quote:

Originally Posted by rayma (https://www.diyaudio.com/forums/tubes-valves/338970-understanding-load-triode-impedance-matching-post5817987.html#post5817987)
Power varies with total loading, but George should take this further.

George and another source recommended the resistor approach and 2 other sources mentioned that loading with the resistor is not necessarily required. I'm trying to understand the theory behind both approaches.

It seems that in terms of overall power it doesnt make much difference, but there is a big difference in terms of the load on the triode. I simply don't know what are the tradeoffs of each approach

rayma 9th June 2019 10:45 PM

George designed the circuit, so you should best go with his approach. Changing the load moves the load line around, and may limit the available voltage swing, increase distortion, and/or be asymmetrical. Most output transformers are optimized for a particular impedance, and should be used with this in mind.

The power is not an issue, since headphones need little power. Also, a parallel resistor will tend to flatten the impedance curve seen by the amplifier and make it more resistive, which is a good thing, especially for a circuit with highish output impedance and no loop nfb. The voltage output with frequency will then be more flat with a flatter impedance curve.

itsikhefez 9th June 2019 11:15 PM

Thank you, this makes sense.
So, assuming a parallel resistor is used, what would be the difference between using an OPT that is tapped at 8ohm and one with 32ohm? With the 8ohm tap, a 10-12 resistor would be used to match impedance. ~47 ohm would be used with a 32ohm tap.

Power output would be the same in both cases, right? But, with the 32ohm tap there would be a higher voltage swing. Is this correct?

rayma 9th June 2019 11:24 PM

Quote:

Originally Posted by itsikhefez (https://www.diyaudio.com/forums/tubes-valves/338970-understanding-load-triode-impedance-matching-post5818019.html#post5818019)
Power output would be the same in both cases, right? But, with the 32ohm tap there would be a higher voltage swing. Is this correct?

Right on both. I suspect that you will prefer the sound one way or the other though, so it try both ways. Using the lower impedance tap (and lower resistor) will result in a flatter impedance curve for a given headphone. The higher impedance tap may have different advantages, like better hf.

itsikhefez 9th June 2019 11:27 PM

Thanks again!
I still haven't ordered the transformer.
The amp will primarily used with high impedance HP's which benefit from a high voltage swing, so I think I'll go ahead with a single 32ohm secondary.

Lastly, is there a simplified way to calculate the available voltage? Is it as simple as dividing the output voltage from the triode by the transformer ratio?

rayma 9th June 2019 11:31 PM

Quote:

Originally Posted by itsikhefez (https://www.diyaudio.com/forums/tubes-valves/338970-understanding-load-triode-impedance-matching-post5818036.html#post5818036)
is there a simplified way to calculate the available voltage? Is it as simple as dividing the output voltage from the triode by the transformer ratio?

If you know the rated output power into a particular load R, the power P = (Vrms^2) / R , and then Vrms = sqrt (P x R).

If output power is 8W into a load R of 32 ohms, then the rms output voltage is sqrt (8 x 32) = 16 Vrms.


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