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In Reply to: Re: Resistor as parafeed plate load? posted by Kurt Strain on October 29, 2001 at 19:56:09:
i don't want to open a whole can of worms here on loadlines and power out, but isn't Rl=2Rp considered by many as the case where you get the maximum undistorted power output from triodes???those aren't my words... but found in reuben lee's transformer book....
dave
Follow Ups:
Yes, it is true, but only under the assumption that the plate voltage is fixed and low enough that there is no danger of exceeding the plate dissipation limit.Historically, the first need was for power gain, and the optimum load was RL=rp. This gives the maximum power for a given input (grid) voltage, assuming the grid voltage is small. This is still handy if you want to make a one-tube amplifier...
As tubes got more affordable, the main limitation became plate voltage - batteries were still common, and capacitors were expensive and of poor quality. In that case the maximum power occurs when RL=2rp. Example: a 45 at 180v. Plate resistance is 1650 ohms, so the load would be 3300 ohms and the current about 30mA (RCA says 2700 ohms and 31mA). Reich gives a good derivation of this. Notice that the plate dissipation is 5.6 watts in this case, well below what the tube can stand.
Eventually, power tubes with low plate resistance were developed, and the plate dissipation became the limiting factor. (Very old tube manuals often do not list a maximum plate power dissipation!) I've done the analysis myself because I've not found it except in fragmentary form in the usual books. Maximum power occurs when the plate voltage is maximum and the current is adjusted to result in maximum dissipation. In that case (and with a few other assumptions thrown in!) the load resistance is approximately RL=(Eb/Ib)-2.4rp. Example: a 2A3 running 300v (max rated) at 50mA (to make 15 watts plate dissipation), the load would be (300/.050)-2.4*800 or 4080 ohms.
Can you tell this is my favorite topic? :^)
RL=(Eb/Ib)-2.4rp. Example: a 2A3 running 300v (max rated) at 50mA (to make 15
watts plate dissipation), the load would be (300/.050)-2.4*800 or 4080 ohms.
could you give a bit more info on what that 4080 is????i guess my question is, if we look at the 4080 load as the reflected load and assume a 1K signal (to keep the inductance out of the picture) i see the tube putting out the max undistoretd signal as the curves go, but when we consider the turns ratio, we will have more of a stepdown so the voltage output at the load will be less than a turns ratio that presents a 2000 ohm load... i understand the 2K load will cause the loadline to rotate... possibly making the AC swing available less so we may actually end up with less overall power even given the smaller turns ratio, but i was wondering if anyone knows if this is the case???
so by reducing the turns ratio by 30% we will get 30% more voltage, but will the reflected load presented to the tube change the tubes power out by more or less than 30%???
again its just another one of these things that needs lots of asteriks to qualify the speciffic situation.
dave
Knowing Dave for a compleat eXperimenter, I'll just mention that everything from half this value to twice this value has been used successfully in practice by someone. Lower load impedances give more power and more distortion, gradually shading into clipping due to cutoff. Higher load impedance gives less power due to grid-current clipping, but lower distortion below clipping.I made a more complete presentation at VSAC last month. Sometime in the next few weeks I'll get the presentation cleaned up and Brad says he'll post it and/or put it on the DVD of the seminars. That will probably be the best way to explain the assumptions behind the 4080 value. Otherwise it would take a couple napkins to write on, and a few beers!
Hey Dave, I believe you are correct, but wouldn't half the power in this instabce be fed right back through the power supply? I think max power would be with an infinite plate load and a primary load equal to 2xRp....whadaya think?
Peace,
Me
In Reply to: Re: Resistor as parafeed plate load? posted by JeffreyJ on October 30, 2001 at 13:55:17:ooopps you are correct sir!!! (sound of hand smacking forehead) i wan't looking at the whole picture...
forgot about that whole voltage divider thing going on with parafeed... so while the tube would be putting out full power,
the output trannie would only see 1/2 of that power and the rest would go through the power supply...who cares about the tube putting out max undistorted power when 1/2 of that is tossed out the window???
in this case i suppose max "useable audio" power would be from an infinate plate load and an output trannie that presents 2X the Rp...
thanks for the correction.
dave
I thought it was where Rl = infinity. But I am unsure.Kurt
yup you are right .. see above... my bad... gave the correct info but the wrong application... i hate when i do that :-(dave
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