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The Real Thing:
Ultra-Linear OTL Power Amplifiers
Only the brave need apply. OTL amplifiers have a well-deserved reputation for breaking down and often damaging loudspeakers in the process. This makes sense, as the protection from the lethal voltage within a tube amplifier is usually the isolation provided by the output transformer. For example, if a cathode-to-plate short occurs in a transformer coupled push-pull amplifier, nothing much happens to the speaker even though the tube is blown (and probably, a handful of resistors, diodes, and wire). An OTL amplifier, on the other hand, is like performing a trapeze act without a net. To get an idea how much potential energy is inside an OTL amplifier imagine how many watts your loudspeakers would have to endure if one of the output shorted plate to cathode. The math is simple voltage squared divided by resistance:
Power = VēR.
Assuming 170 volt rails in the amplifier, this means 3612 watts into an 8 ohm speaker! Protection circuitry helps, but there is always the chance that it will not be fast enough. Adding an output coupling capacitor looks like a better idea than first imagined.
Having scared away the weak-hearted, we can move on to describe what an ultra-linear OTL power amplifier would look like. Actually, a power amplifier would not look too different from the headphone amplifier examples already given. More output tubes, beefier tubes throughout, much higher power supply voltages, certainly, but not fundamentally different in design. The first step is to forego any temptation to build a Class A ultra-linear OTL amplifier. Here is why: 100 watts would require 850 watts of plate dissipation alone. We must add to this figure the heater and support circuit dissipation. A stereo amplifier would trip your house's circuit breaker at turn-on. Five channel surround sound would require the power company to set up one step-down transformer for just your house.
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Once we accept a lean Class AB as the only practical choice (along with using feedback to keep down the distortions that result from Class AB crossover problems [or Gm doubling effects] and to lower the output impedance), we must determine how many output tubes we will need. Even the most muscular sweep tube can only conduct a peak current of about 2.2 amps (the 8908) and this peak current is only possible with a grid 2 voltage of 250 volts. The 6LF6 limit is about 1.2 amps. The problem with these tubes is that they are almost impossible to find at a reasonable cost. This leaves the EL509/6KG6 as the only choice.
The EL509 is the only sweep tube in current production as far as I know and its cost is under $30. Although its maximum recommended continuous cathode current draw is only 500 mA, it can put out peaks of over 1 amp. Now the question is how many watts we want from our amplifier. The formula for peak current demand based on RMS watts into a load is:
Ipeak = Ö(2 x Watts x R)
R.
Assuming 8 ohm speakers, if we want
100 watts, we will need a peak of 5 amps;
80 watts, 4.5 amps;
60 watts, 4.5 amps;
50 watts, 3.5 amps;
40 watts, 3.2 amps;
30 watts, 2.7 amps;
20 watts, 2.2 amps;
16 watts, 2 amps.
Thus, a 100 watt amplifier would need 10 EL509s to play it safe, although 8 would probably prove adequate. This number of output tubes results from dividing the peak amperage by 1 amp and doubling that number, as the amplifier must be able to swing the peak amperage negatively as well as positively. Of course, we should always round up to build in some safety margin. Besides, the more output tubes, the lower the output impedance.
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