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Understanding load on triode and impedance matching
Understanding load on triode and impedance matching
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Old 10th June 2019, 05:56 AM   #21
Hearinspace is offline Hearinspace  Canada
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Understanding load on triode and impedance matching
Quote:
Originally Posted by itsikhefez View Post
Sorry but I'm missing something in your explanation here.
You're saying that it doesn't matter if the reflected load on the tube is 5K or 100K? I forgot where I saw it but one resource mentioned that ideal load was 3*Rp (which is around 5K for the 45), and in another, there was a graph that showed the distortion/power at a given impedance (I couldn't find such a graph for the 45). Based on the graph, there was an "ideal" load to put on the tube for lowest distortion.
Or is that what you are saying?
First of all, that 3x rule is only ball-park. Sometimes given as 3-5 times or 2.5-5 times. . . . more power out with less and often less distortion with more.

Also, rp changes with operating point.

You mentioned reading and you mentioned graphs, maybe this will interest you. "Load impedance and operating points for single ended amplifier stages" by Paul Joppa. It's in three parts, all of which are contained in the compilation linked here.
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Old 10th June 2019, 06:46 AM   #22
itsikhefez is offline itsikhefez  United States
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Understanding load on triode and impedance matching
Quote:
Originally Posted by PRR View Post
Tube amps can usually *idle* without a load. OVER-drive, clipping, without a load causes large kick-backs which damage transformer insulation.

Best power in a transformer coupled triode is often near 2*rp (depending on ratings. Best distortion is at *infinite* load, but that is zero power. Do not worry about loading in 10X or 20X rp if you get enough power output.
Quote:
Originally Posted by Hearinspace View Post
First of all, that 3x rule is only ball-park. Sometimes given as 3-5 times or 2.5-5 times. . . . more power out with less and often less distortion with more.

Also, rp changes with operating point.

You mentioned reading and you mentioned graphs, maybe this will interest you. "Load impedance and operating points for single ended amplifier stages" by Paul Joppa. It's in three parts, all of which are contained in the compilation linked here.
Sorry for being dense here but I want to make sure I understand this correctly.

For sake of example, this is a 16ohm tap with 32ohm load resistor and Rp of 1700ohms (which is the Rp of the 45 at 275V) With the load resistor, reflected load is 5K-9K (3x-5.5x) with 32-300ohm loads connected. Without the load resistor, reflected load is 10K-93K (6x-54x) with same load. In both cases, there is more than enough power.

So I guess that in this case, it makes more sense to put the load resistor, correct?

Thanks for the link-- will read and attempt to understand.
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Old 10th June 2019, 04:19 PM   #23
DF96 is offline DF96  England
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As a general rule, the higher the load resistance seen by a triode the lower the distortion. However, higher load resistance also means less power so you need to compromise.

In addition, higher load resistance with a given OPT etc. usually means poorer LF response. Again you need to compromise.

Adding a parallel resistor will probably increase distortion, reduce maximum output, but also means better LF response. Best option is to use a transformer which roughly matches the load.
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Old 10th June 2019, 07:27 PM   #24
itsikhefez is offline itsikhefez  United States
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Understanding load on triode and impedance matching
Thanks for the explanation. Combined with the rest of the responses here, I think I have a good understanding now on the trade-offs. When an 8ohm speaker is used, the "ideal" load of 5K represents a point of high output power with acceptable distortion.

For my specific application then, seems like a 32ohm tap without a load should be good. This gives more than enough power to 32-300ohm headphones, with lower distortion as the headphone impedance increases. Series resistors can be added to further pad the output and reduce noise if necessary.
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Old 10th June 2019, 10:52 PM   #25
itsikhefez is offline itsikhefez  United States
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Understanding load on triode and impedance matching
How is Zout determined ?
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Old 10th June 2019, 10:53 PM   #26
nigelwright7557 is offline nigelwright7557  United Kingdom
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Zout is determined backwards from speaker.

So if you use 8 ohms then multiply that by output transformer turns ratio squared and you get Zout at output valve.
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Old 10th June 2019, 10:56 PM   #27
itsikhefez is offline itsikhefez  United States
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Understanding load on triode and impedance matching
I meant the Zout that is typically spec'ed for headphone amps.
Some have lo/high Z switch.

Last edited by itsikhefez; 10th June 2019 at 11:01 PM.
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Old 10th June 2019, 10:58 PM   #28
nigelwright7557 is offline nigelwright7557  United Kingdom
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Low will be 8 ohms and high 300 ohms.
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Old 11th June 2019, 12:37 AM   #29
johnsurnamerobinson is offline johnsurnamerobinson
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Just to muddy the waters (no pun intended) it is also important especially for single ended output that the transformer has enough iron to avoid saturation otherwise you will see some very nasty distortion by this I mean your headphones will be a very small power load the main thing you will need to account for when sizing the o/p transformer is the output valve bias current.
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Old 11th June 2019, 01:42 AM   #30
JMFahey is offline JMFahey  Argentina
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Quote:
Originally Posted by itsikhefez View Post
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!
1) if amplifier requires an 8 ohm load, by all means provide it.
Specially on a fussy circuit such as a Tube amp, even worse single ended.

2) 2W into 8 ohm means 4V RMs

Which mean 500 mW RMS, too much for them and which need attenuation; and 50mW into 300 ohm, which is probably fine.

The amplifier will be properly loaded and work as expected.

Quote:
When connected directly, the load will be 20K and 187K respectively.
Yes, but forget it.
Quote:
OTOH, connecting a 12ohm/3W resistor across the output would "fake" the reflected load to 5.4K-7.2K, which seems better
.
Ok, pick one, WHICH is better?
5k or 7 k ?
Quote:
In this case though, how much power is being "wasted" on the load resistor and how much remains for the HP's?
Not exactly wasted, you are loading the amplifier with what it likes, it can provide 2W which is way too much for headphones, one way or the other you NEED to "waste" excess,a.k.a. "Attenuation".
Quote:
I've basically gotten conflicting answers on whether the load resistor is required or not, hence my confusion.
A lot of people feel a *need* to answer "something" , even if no clue, so that is not surprising.
Quote:
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.
But ... do you *have* a 32 ohm tap?
In any case, available voltage will be even higher, so youŽll need to attenuate even more.
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