The Tube CAD Journal: The White Cathode Follower

Push-pull output stages are basically difference amplifiers, which is why a phase splitter is needed. Presented with the exact same signal, the amplifier does not amplify. This works wonderfully to reduce power supply noise in a conventional transformer coupled push-pull amplifier, as the noise is common to both phases (or at least we hope it is). The problem with inline push-pull amplifiers is that the noise is not always equally imposed on the two input signals. This discrepancy gives rise to noise at the output. So in order to reduce the noise at the output of an optimally designed White cathode follower, we must ensure equal noise, equal in both amplitude and phase, superimposed on the signal at each grid. Yet we only one input, the top grid, available to us.

Just what is the existing noise distribution in an optimally designed White cathode follower? For the top triode, the answer is easy enough: its grid sees whatever is the noise present at the previous stage's output. But the bottom triode sees half the B+ noise at its grid by necessity. This results from specifying that the plate resistor equal the reciprocal of the triode's Gm. A +1 volt pulse applied to the B= voltage will be relayed to the bottom triode's grid. The bottom triode will then fight this positive pulse by increase its connected by this +1 volt against its Gm, which in this case is 10 mA per volt. Thus the increase in conduction for both triodes is 10 mA. Now 20 mA (the idle plus the pulse current) through a 100 ohm resistor equals 2 volts. And as the resistor has just been pulled up one volt, the extra 1 volt developed is displaced by the that pulse. In other words, they cancel, leaving the top triode's plate voltage unchanged. Wait a minute if the pulse is needed to cancel itself, how can it, once cancelled, work to cancel itself? It can't. The best it can do is tie and leave half of the pulse present on the top plate and in turn on the bottom grid.

Another way of seeing the mechanism at work is to find the effective (not the nominal rp) impedance at the top triode's plate and then figure out the voltage division between the plate resistor and the rest of the circuit. Because the coupling capacitor relays the any voltage variation on the top plate to the bottom grid, the effective Gm for the entire circuit is equal to the bottom triode's Gm, 10 mA per volt. The inverse of the Gm equals the impedance. Thus the effective impedance of the top plate is 100 ohms. Since this 100 ohms impedance finds itself in series with a 100 ohm resistor, the voltage division is 50%.

Making the first stage perform this voltage division requires that the plate resistor equal the sum of the rp and the cathode resistor magnified by the mu + 1. Putting this into a formula:

Rk = (Ra - rp) / (mu + 1).

Obviously, the plate resistor must exceed the rp and the cathode resistor must not be capacitor bypassed. Using the 6DJ8 as an example, a 20k plate resistor should find a 500 ohm cathode resistor to match.