This formula gives us the idle current per output tube as well as the peak output current per tube. Now we take the desired peak output current into the load and divide it by the idle current per output tube. For example, given a B+ of 80 volts, a rp of 3300 ohms, a 9 volt span across a current source, the idle current per triode should be 10.75 mA, as
   10.75 = (80 - 9) / (2 x 3300),
which we round down to 10 mA.
   Thus a peak of 30 mA into the load requires three triodes. This will bring the tube envelope count up to 4, thus halving the battery playback time. But then no one ever said that Class A, single-ended, OTL was cheap. (Well, no one other than advertising departments).
   How was the value of 9 volts picked for the current source voltage displacement? It comes from multiplying the peak output current against the load impedance, 30 ohms. This value is, in practical terms, insufficient. The FET within the current source stops acting as a current source when the voltage drops too far, as it enters its triode region. Even if a more elaborate current source is used, say 3-pin terminal voltage regulator configured as a current source, at some point a dropout voltage will be encountered.

    In the second example, the DCR of the choke serves to bias the output tube. Both circuits will just barely drive a 300 ohm Sennheiser headphone certainly not a 32 ohm Grado. More triodes are needed. How many? How much current do wish to deliver into load? Each triode section can yield up to 10 mA with a 6DJ8. How was this figure determined? The best approach is to examine the plate curves of the tube. But for rough estimating, the following formula is useable:
    Current = Vp/2rp