Understanding load on triode and impedance matching
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rayma
diyAudio Member

Join Date: Apr 2011
Quote:
 Originally Posted by itsikhefez 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.

Last edited by rayma; 9th June 2019 at 08:03 PM.

 9th June 2019, 09:35 PM #3 itsikhefez   diyAudio Member   Join Date: Aug 2015 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
diyAudio Member

Join Date: Apr 2011
Quote:
 Originally Posted by itsikhefez Without load resistor, the tube is being "under" worked, while with the load resistor, it's being loaded to full power.

itsikhefez
diyAudio Member

Join Date: Aug 2015
Quote:
 Originally Posted by rayma 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

 9th June 2019, 09:45 PM #6 rayma   diyAudio Member   Join Date: Apr 2011 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. Last edited by rayma; 9th June 2019 at 10:01 PM.
 9th June 2019, 10:15 PM #7 itsikhefez   diyAudio Member   Join Date: Aug 2015 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
diyAudio Member

Join Date: Apr 2011
Quote:
 Originally Posted by itsikhefez 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.

Last edited by rayma; 9th June 2019 at 10:29 PM.

 9th June 2019, 10:27 PM #9 itsikhefez   diyAudio Member   Join Date: Aug 2015 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
diyAudio Member

Join Date: Apr 2011
Quote:
 Originally Posted by itsikhefez 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.

Last edited by rayma; 9th June 2019 at 10:41 PM.

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