While the circuit above contains all the essential components, the version below is more practical in that it can provide an output voltage much more negative than the Op-Amp power supply voltage. The 2.5 voltage reference is an IC that can be readily had and is very accurate. The Op-Amp can be any FET input Op-Amp, such as the LF411 or AD711. The power supply that feeds the Op-Amp must be unregulated, as regulation by presenting a constant voltage, would prevent the monitoring of the B+ fluctuations. The circuit works because of the Op-Amp's feedback constantly striving to maintain an equal voltage on its positive and negative input pins. Thus, since the positive input is shorted to ground in the above circuit, the negative input will also be at ground potential. And since the 12.5k resistor connects to the -2.5 volt voltage reference, exactly 2.5 volts is what is developed across this resistor, which means exactly 0.2 mA of current must be flowing through this resistor. In other words, we have created a constant current source out of a resistor, a voltage reference, and an Op-Amp.  And this current can flow into only two paths: the 60k and the 8k resistors.

   Given a B+ voltage of 400 volts and an Op-Amp power supply voltage of 6 volts, the 60k resistor can absorb 0.1 mA of current and remaining 0.1 mA must go into the 8k resistor. In order for the 8k resistor to draw only 0.1 mA of current the voltage at the cathode of the output must be 0.8 volts, which implies that its current draw is equal to 80 mA. 400 volts times 80 mA equals 32 watts.
   If the wall voltage drops and the Op-Amp's power supply voltage drops to 4.8 volts, then the 60k resistor path will only provide 0.08 mA of current and the remaining 0.12 mA must come from the 8k path, which means that the output tube's cathode must be at 0.96 volts, which implies that its current draw has increased to 96 mA. The new idle current times the new lower B+ voltage (320 vdc: 400 x 4.8/6) equals 30.72 watts. If the wall voltage climbs and the Op-Amp's power supply voltage climbs to 7.2 volts, then the 60k resistor path will now provide 0.12 mA of current and the remaining 0.08 mA must come from the 8k path, which means that the output tube's cathode must be at 0.64 volts, which implies that its current draw has decreased to 64 mA. The new idle current times the new greater B+ voltage (480 vdc: 400 x 7.2/6) also equals 30.72 watts.
    Notice that a total of a 180 volt change in the B+ has not shifted the output tube's dissipation far from its original 32 watts. Notice also that the dissipation decreases slightly with either an increase or a decrease in B+ voltage. In other words, this circuit is not a perfect constant power servo. But it is very close to being so at a up to 20% drift about the nominal B+ voltage.
    This circuit can be readily used by push-pull amplifiers and would only require doubling up of the Op-Amp circuitry while sharing the voltage reference.
    Remember, if you give this circuit a try, send us some feedback.

                              //JRB