Adding to what Jim has already said:
In a typical vacuum tube audio chain (E.g. a Johnson Ranger), additional filtering is included on the B+ supplied to lower level stages. This additional filtering takes the form of a series resistor added to the B+ line feeding each tube, plus a bypass capacitor to ground on the tube side of that resistor. This additional filtering accomplishes two purposes: it
reduces residual hum on the B+ applied to the low level stages, and it isolates the stages (sharing the same B+ bus) from each other.
For example, if you look at the Johnson Ranger schematic ... and the first audio stage (microphone preamplifier - V7a), there is a 220kohm resistor added in series with the 300V B+ supply bus (R20), bypassed to ground by a 0.1uF capacitor (C51). Since the average plate current in V7a is low (about 0.33 mA), the voltage drop across the 220kohm resistor is less than 75 volts. The associated RC time constant is 0.1uF x 220kohms = .022 seconds. This means that this RC circuit is a low pass filter whose 3dB rolloff frequency is 7Hz. As a result, there is about 24dB of additional reduction of 120Hz ripple in the plate voltage supplied to V7a versus the 300V B+ supply bus. Some recommended mods to the Johnson Ranger (e.g. WA1HLR's mods
http://www.amwindow.org/tech/htm/ranger.htm) increase this extra ripple filtering on the B+ of V7a by increasing the value of C51 from 0.1uF to more than 10uF (along with other changes).
Separately, with respect to the ripple on the HV supply's output: (for those who are comfortable with a little math...)
For a fixed power supply output current, I (for example: I= 120mA), you can calculate the ripple at the output of a capacitor input, full wave rectifier-based power supply as follows:
The ripple waveform is approximately a sawtooth waveform that has a 120Hz repetition rate. In each cycle, the voltage across the output capacitor drops by: dV (volts) = (I/120Hz)/C. At the end of each cycle, the voltage rises back up to its full value.
For example, if C=50uF, and I=120mA... then the peak-to-peak amplitude of the sawtooth (hum) waveform is dV(volts) = (120ma/120Hz)/50uF = 20 volts. If the average output voltage of the supply is 600VDC, then the peak to peak ripple will be 20V/600V = 3.3%.
[Note: As Jim has already pointed out: for a choke input, full wave rectifier-based power supply, with the same value of C, and delivering the same
fixed 120mA current, the residual 120Hz ripple would be considerably smaller... but the unfavorable effects of low frequency audio components in the
modulated current being drawn from the supply could be greater]
Although not answering your question directly ("acceptable power supply ripple on the modulator")... when I have modulated the B+ on homebrew plate modulated transmitters (also Class E FET-based transmitters), I have found that keeping the peak to peak 120Hz ripple on the unmodulated B+ supply below 2% of the average B+ supply voltage makes the residual power supply hum inaudible to people listening to my signal.
Stu
