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Tubes / Valves All about our sweet vacuum tubes :) Threads about Musical Instrument Amps of all kinds should be in the Instruments & Amps forum 

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3rd January 2018, 04:45 PM  #21 
diyAudio Member
Join Date: Nov 2008
Location: tehran

one of important point in building good tube amp is how our circuit answer to 1khz or 10khz sin wave .need scope and audio generator.so use 8 ohm resistor for calculate output power .

3rd January 2018, 05:36 PM  #22 
diyAudio Member
Join Date: Nov 2002
Location: Hungary

No, It is not so simple!
Do not forget it has to drive a loudspeaker with 8 or 16 Ohm resistance. Output power = Irms2xR!
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"Replacing any coupling capacitors with a well designed and correctly matched driver or interstage transformer yields an almost shocking improvement in sound..." Peter Qvortrup 
3rd January 2018, 06:40 PM  #23 
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Join Date: Feb 2011

Correct! It it not at all simple. In case of triodes, you have to equal its plate resistance to the load in order to obtain a optimum power delivery. Normally this is not the case with OTL circuits. That is why a simple low impedance headphone is not suitable as load.

4th January 2018, 08:30 AM  #24 
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Join Date: Jun 2016

mario13,
Your Post, # 4. Usually a schematic is helpful, but always appreciated. I believe your description of the amp gives us enough idea to answer the question of power. It seems that you are using Ultra Linear mode, correct? Ultra linear is good. I do not use it, but many do. I never did like the term “distributed load” when it was used to describe Ultra Linear. It is like putting a ton on the car’s top (plate), and loading the trailer with a pound (screen). Several of the posts in this thread have given lots of good information. I often start with the tube curves and a load line. But I wanted to look at it using another method: Impedance goes to the square of the turns ratio. 2:1 turns = 4:1 impedance. 10 mA screen current at 50% winding (1/4 impedance, i.e. 2000 Ohms) 40 mA plate current at 100% winding (whole impedance i.e. 8000 Ohms) Peak Power = (I)squared R. i.e. the current can only decrease from 40mA to 0mA. Actually, it is better to have some current, not 0, 0 is clipping, and is beyond class A for Single Ended. Although it can go from 40mA to More than 80mA in the other direction, that additional power is mostly 2nd harmonic distortion. .01 squared x 2000 = 0.2 Watts Peak from the screen .04 squared x 8000 = 12.8 Watts Peak from the plate You can see, the purpose of Ultra Linear is Not to get more power than Pentode mode. 13W versus 12.8W. But for a sine wave, the RMS power is 1/2 of the peak power (6.5W). One purpose of Ultra Linear is to get a more linear output signal than Pentode mode (less distortion). The other purpose of Ultra Linear is to get a much better damping factor (lower impedance output) than pentode mode. Negative Feedback is almost always required for Pentode mode amplifiers. Negative Feedback is often needed for Ultra Linear amplifiers, especially if the primary load is equal or less than the tube’s effective plate resistance, rp, (the plate impedance is a function of the particular tube, plate voltage, %UL tap, and plate current in the Ultra Linear mode). A plate rp of 2k Ohms, and a primary of 4k Ohm might find negative feedback useful, but with an 8k Ohm primary might not need it. Now lets estimate the loss of the transformer: With the plate voltage at 375V, and the B+ at 395V, that is a 20V drop. 20V/0.04A = 500 Ohms. The DCR of the 8000 Ohm primary is 500 Ohms. (No, I did not include the much smaller voltage drop contribution of the much smaller screen current across 1/2 of the primary; just wanted to introduce the concept of power loss due to primary DCR). If they Calculated the 8000 Ohms primary from the square of the turns ratio primary:secondary, Then the real impedance is 8000 Ohms + 500 Ohms DCR. If they Measured the actual primary impedance properly, then the winding impedance is 7500 Ohms + 500 Ohms. In any case that causes a transformer loss due to DCR. There is also a secondary DCR loss too. If the turns ratio gives a calculation of 8000 Ohms, and the DCR is 500 Ohms, we have the voltage divider of 8000/(8000 + 500) ~ 0.94. With power = (V) squared/R, we have a power loss of 0.94 squared = 0.89. That is 1 dB. 6.5W x 0.89 = 5.79Watts. That agrees closely with one of the other members who posted in this thread. 
4th January 2018, 10:14 AM  #25 
diyAudio Moderator

You are responding to an 8year old post. Great information nevertheless...

4th January 2018, 10:26 AM  #26 
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4th January 2018, 11:23 AM  #27  
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Join Date: May 2007

Quote:


4th January 2018, 12:22 PM  #28  
diyAudio Member
Join Date: Nov 2002
Location: Hungary

Quote:
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"Replacing any coupling capacitors with a well designed and correctly matched driver or interstage transformer yields an almost shocking improvement in sound..." Peter Qvortrup 

4th January 2018, 12:47 PM  #29 
diyAudio Moderator

The question posted was how to calculate the output power, nothing about limiting... If you actually have a specific circuit in mind, perhaps it would be better if you start your own thread.

4th January 2018, 01:59 PM  #30  
diyAudio Member
Join Date: Nov 2002
Location: Hungary

Quote:
Forexample a ca. 10.5Wrms amp on 8 ohms speaker need to give 9.19Vrms and 1.14Irms By transformer coupled amps the transformer do the "trick",(lowering high voltage and converting low current to high) but what what about OTL amps? How will this be enough current from high voltage?
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"Replacing any coupling capacitors with a well designed and correctly matched driver or interstage transformer yields an almost shocking improvement in sound..." Peter Qvortrup 

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