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.
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.
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.
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.
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.
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.
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.
Yes, but forget it.
.
Ok, pick one, WHICH is better?
5k or 7 k ?
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".
A lot of people feel a *need* to answer "something" , even if no clue, so that is not surprising.
But ... do you *have* a 32 ohm tap?
In any case, available voltage will be even higher, so you´ll need to attenuate even more.
1) It would require an 8ohm load if a 5K:8Ohm OPT is used. I dont have one yet so that part is theoretical
2) Some 300ohm are fine with 100mW. The Sennheiser HD650 spec max current at 500mA. The LCD-4 is 200ohm and has "recommended" power of >500mA
3) Why forget it?
4) Since a range of headphones can be used, 5K indicated the reflected load if a 32ohm HP is used, and 7K if a 300ohm HP is used. In both cases, the 5-7K range is much closer to the ideal load versus no load resistor
I don't have an OPT yet. I am ordering a custom one and can spec any secondary tap I want.
Hence the thread... I want to make the right choice
2) Some 300ohm are fine with 100mW. The Sennheiser HD650 spec max current at 500mA. The LCD-4 is 200ohm and has "recommended" power of >500mA
3) Why forget it?
4) Since a range of headphones can be used, 5K indicated the reflected load if a 32ohm HP is used, and 7K if a 300ohm HP is used. In both cases, the 5-7K range is much closer to the ideal load versus no load resistor
I don't have an OPT yet. I am ordering a custom one and can spec any secondary tap I want.
Hence the thread... I want to make the right choice
I have run my 45 based TSE-II directly into a pair of headphones from an 8 ohm OPT with excellent results. I stated that this was a possible avenue to try. That was done in response to some OTL experiments done in this thread:
The all DHT SET Headphone Amp
This topic has been brought up is several threads, so I will state what I know NOW. I need to do some testing on a live amp before stating anything concrete, and my amp is in a state of disassembly st the moment.
I ran my amp with a small Edcor 5K to 8 ohm OPT. and wired my 32 ohm cans directly across the output. The amp put enough power into those cans to make sound heard across the room. It was quiet enough to hear nothing when there was ni music playing.
itsikhefez has 300 ohm phones. Simply wiring 300 ohm phones across an 8 ohm OPT isn't going to load the output tube properly, and I'm not sure that enough power will be available for the 300 ohm phones.
We have an ideal 2 watt amp playing into an 8 ohm load. It is producing 4 volts RMS of audio signal to do that. Feed the same 4 volts into a pair of 32 ohm phones, and you get 500 milliwatts, which is quite loud when it's strapped over your ears. Apply the same 4 volts to a 300 ohm load, and you only get 53 milliwatts. Is 53 milliwatts enough?
Generally speaking increasing the load on a triode without feedback will lower the distortion up to a point, where it may begin to increase. And of course increasing the load will reduce the power output. Sticking 32 ohm phones across an 8 ohm OPT does work nice on my particular setup. It may not work so well on other OPT's. This can be fixed by putting a resistor in PARALLEL with the phones to keep the load on the amp where it wants to be, which is probably higher than 8 ohms.
A 300 ohm load on an 8 ohm OPT is a gross mismatch. The mismatch could be fixed with a parallel resistor, but the power would drop slightly.
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Speaking as one with considerable experence most transformer manufactures will bend over backwards to accomodate your needs.(There is nothing in this world apart from money that a good database of applications is valued.) I live in uk and have found transformer manufacturers most willing to help.
Starting this thread I thought there was a simple, clear cut answer to this question. I have searched hours through threads to find applications of similar people. Multiple taps is an option, but based on my findings there are quite a few people from this forum, and other commercial products that use a single 32ohm tap to power 32-300ohm headphones... and I think that is what I will do as well.
I've created a spreadsheet and plugged in numbers for various load resistors, series resistors and HP impedances. I'm still unsure what to actually use.
Without any load at all, 300ohm cans on the 32 tap will get roughly 200mV which is more than plenty. Reflected impedance in this case is 54K or 30x Rp. Is that still a considered a gross mismatch?
Ofcourse, putting a load resistor helps here but reduces power for 32ohm cans significantly (less so for the 300ohm cans) Using 20, 47, 80, and 220 ohm values I've been able to come up with some combinations that may work. The issue is that I would have no way to measure the end result, so was hoping the answer here was straightforward.
I've created a spreadsheet and plugged in numbers for various load resistors, series resistors and HP impedances. I'm still unsure what to actually use.
Without any load at all, 300ohm cans on the 32 tap will get roughly 200mV which is more than plenty. Reflected impedance in this case is 54K or 30x Rp. Is that still a considered a gross mismatch?
Ofcourse, putting a load resistor helps here but reduces power for 32ohm cans significantly (less so for the 300ohm cans) Using 20, 47, 80, and 220 ohm values I've been able to come up with some combinations that may work. The issue is that I would have no way to measure the end result, so was hoping the answer here was straightforward.
If the turns ratio is 30x then the reflected impedance will be load resistance * 30 * 30.
So for 8 ohms it would be 7200 and for 300 ohms would be 270,000.
I think it would be best that you crack a book open and get to understand the fundamentals of how tubes operate against various loads... Tubes are NON-LINEAR and using simple first order equations and assumptions of linear scaling of power is not close to reality... This thread will grow to about 400 pages of 8 million opinions.... Also when dealing with transformers, moving the reflected loads all about will greatly alter the flux density as well as the frequency response.... SO you can see it will turn into a Train Wreck in no time...So learning and forming your own ideas based on analysis will be more beneficial for you in the short and long run..
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