But getting DC on the heaters requires a DC power supply with its rectifiers and smoothing capacitors. And as anyone who has ever tried to hotrod an existing 6.3 heater winding into a DC power supply will testify that it seldom works as well as one had hoped. (DC power supplies have losses associated with the conversion of AC into DC.) An extra transformer could be added, but at some expense and effort. No, free is still best.
Free DC voltage can be had, if DC voltage was being wasted in the amplifier, that is. Where to look? The push-pull output stages often use a common cathode-bias resistor to set the idle current of the output tubes. This resistor is bypassed by a large valued capacitor to shunt away the AC signal to ground. Say a pair of KT88 type output tubes are used with 75 mA of idle current per tube, then 150 mA of free DC is available for the heaters. Two 12AX7s or 12AU7s or 12AT7s in series need 150 mA of heater current and 25.6 volts of heater voltage, which the KT88s will easily provide. The excess voltage needed to bias the tubes can be made up with a small valued cathode resistor in series with the heaters.
A further refinement would be the use of a three terminal voltage regulator instead of the cathode resistor. The regulator can be transformed into a current source with the addition of one small valued resistor. This current source would then work to set the both the correct heater current and the correct bias voltage for the output stage.

How far can this free voltage trick be taken? Here is an example of the extreme limit. Imagine an SE OTL amplifier that held thirteen 6AS7s all wired in parallel, 26 triodes in all. The idle current for each triode is set to 96 mA and the total amplifier current draw through the output stage is 2.5 amps, which is also the amount of heater current required per 6AS7. Thirteen 6.3 heaters in series require 82 volts and equal a resistance of 32.8 ohms.

By stringing thirteen heaters in series, we have defined a resistive load for the thirteen cathodes in parallel. In other words, we have created one large Cathode Follower that uses its own heater string as a cathode load. The output impedance of such an amplifier is around 5 ohms, which applying 20 dB of feedback will drop down to 0.5 ohms. The power output would be about 25 watts. If an isolation transformer were used as the power transformer, then the 120 VAC would equal 170 DC, which would give the 6AS7s a cathode-to-plate voltage of 88 volts and the total dissipation for the amplifier would be 425 watts. The resulting efficiency is only about 5 percent; no one ever said Class A was efficient, except advertising departments. (Oops, I almost reopened that can of worms.)
If 13 tubes are just too much, then the 6082 might be a better choice. This is the same tube as the 6080, except that it has a heater voltage of 25.6 volts and a heater draw of 600 mA. Three 6082s wired in parallel with an idle current of 100 mA per triode will give us the required 600 mA for the heater string, which will require 76.8

 pg. 4