Karl,
You are right: there is an easy formula for determining the optimal load impedance for the maximum power output:
  Rl = (Vp - 2rpIp) / Ip
where the assumption made is that the tube's rp is constant. It isn't. Still, this is a good starting point.

One point that you might be missing is that in an SE amplifier the idle current can equal its maximum plate current. Since what our SE amplifier is amplifying is music, we are safe in assuming that the tube can relied upon to occasionally deliver current at twice its maximum continuous limit, as it will get a rest when the plate swings positively (the plate current drops on the positive swings).

Plugging your tube parameters into the formula yields:
  (600 - 26000.2) / 0.2 = 1800 ohms,
which is what KR recommends for that tube, if remember correctly. These values yield an output wattage of 36 watts. This will not work, of course, as the plate dissipation will be exceeded. If we back off on the current to 140 mA, then the dissipation will drop to 85 watts and the math looks like this:
  (600 - 26000.14) / 0.14 = 3085 ohms,
which would still yield a good deal of output power. But I would not run the tube that hard; backing off just a little can dramatically extend the life of the tube. My choices would be the following: Vp = 550, as that is what you get when a 400 VAC transformer winding is rectified; Ip = 100 mA; Wp = 55 watts. The math now looks like this:
  (550 - 25500.1) / 0.1 = 4400 ohms
  0.1(Rl / 2) = 22 Wrms
As you already know, the transformer will not transfer the full potential watts to the load, because of the DC resistances of the windings, the eddy currents within the stack, the electromagnetic constriction losses. But to this list of loses must be added the non-symmetrical voltage swing the triode undergoes. In other words, do not be too surprised to find that the finished amplifier only puts out 18 watts.

Still, the difference between 22 w and 18 w is barely hearable; whereas, a difference in output impedance of 1 ohm (depending on your speaker) can be decisive. The 1800 ohm impedance transformer will yield an output impedance of 2.66 ohms, while the 4400 ohm, 1.1 ohms. Good luck!

Editor

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Subject: SE Amp Output Transformer
Question:  Is there a rule of thumb, or better yet an actual engineering formula, for determining what the primary impedance (range?) of an output transformer should be when you want to get maximum RMS power out of the tube? 

Tube parameters: Ep-max = 600V Ip-max = 200 mA        rp = 600 Ohm      gm = 6000    mu = 4

Reasons for asking:

1.  First you should know that I'm a first time DIYer.  I listened to 300B amp on my system 6 months ago and want that SE sound.  I purchased a copy of the RCA Radiotron 4th Edition and have been reading and looking at schematics because I want to know how to design these things, not just put kits together.

2.  I'm concerned that there is a "real" limit to the impedance after which the distortion (or something else) goes so out of whack it becomes useless as an AF amplifier.

3.  My listening room is large.  The 300B amp couldn't fill it.  So I bought a pair of KR VV52 (from Welbourne on closeout) because they were listed at 24W with a place max plate dissipation of 85W.

4.  I now read in the RCA book that 95% of the RMS power developed across the transformer primary will be transferred to the speakers.  RMS power can be found by taking peak plate current (Ipp) and using the formula RMS pwr = RL(Ipp / (2 *
Ö 2)

5.  The max current on the VV52 is 200 mA.  Plugging in 100 mA gets 0.0050* RL
     To get 17W RMS (~24W peak) needs an RL of 3400 ohm, if the full 100 mA can be swung without crossing 0 Vg.  I have plotted some load lines and know that this is not possible which means I need a higher impedance
to achieve the higher output power.

6.  The rp of the tube is listed at 600 ohm.  The RCA book says primary impedance should 2 - 4 time the rp = 1200RL - 2400RL.  Plug in 2400 and I get 12 watts at 100 mA.

Karl

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