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Conversely, the tube whose cathode was forced 0.5 volts positive sees a -0.5 volt decrease in its grid-to-cathode voltage and conducts less current. You can readily see that the mid-point ground halves the voltage presented to each triode, effective halving the Gm of the triodes. But as the triodes are effectively in parallel with each other, the total Gm sums to unity.
"But wait, there's more!" Remember what you just read in the section on the Broskie Cathode Follower, "When a signal is applied to the cathode rather than the grid, we have more than just a change in the grid-to-cathode voltage, we have a change in cathode-to-plate voltage as well. Do not forget that the triode, unlike the transistor or MOSFET, has rp, a change in cathode-to-plate voltage will mean a change in the flow of current through the triode."
So it would seem that the Gm of the triodes in this circuit is effectively increased by (mu + 1)/mu because of the signal is being applied to the cathode rather than the grid. But this is not quite right. For right now, let us forget the addition of the plate transconductance and move our attention to the grid-to-cathode voltage relationship.
We do not receive the full transconductance from both triodes, as the grid's voltage is fixed and the applied voltage is split between cathodes. In other words, only half the Gm per tube is available to buck the applied voltage, so we would expect the Gm to be halved to
Gm = mu/2rp.
"But wait, there's more!" The power supplies are floating in this circuit and represent (ideally) zero ohms impedance to AC signals and they are attached to the cathodes and move with the cathodes. Thus, while one triode's cathode was being forced 0.5 volts positive by the addition of the battery, its plate was being forced 0.5 volts negative by the relayed signal through the power supply connected the other tube's cathode.
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Thus, the total plate transconductance (Gp) of the triodes must be added to the half the effective Gm of the tube:
Gm“ = mu/2rp + 1/rp.
While this circuit is still being run in strict Class A, the triodes are effectively in parallel, as the power supplies represent zero impedance to AC signals. Consequently, the Gm is doubled to:
Gm“ = 2( mu/2rp + 1/rp)
which reduces to
Gm“ = (mu+2)/rp.
Warning, do not be fooled into believing that the Circular/Bridge amplifier has in some respect departed from the more conventional totem pole configuration. It hasn't. In a properly designed Totem Pole amplifier, the same effective transconductance obtains:
Gm“ = (mu+2)/rp.
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In the case of the 6DJ8, the increase in Gm“ will be slight as the mu (33) will predominate, but in the case of a 6AS7 or a 12B4 or 6C33, the increase will be dramatic as the these tubes have very low mu's.
In spite of the complexity of the circuit, the math for the output impedance is simple:
Zo = 1/Gm“
or what is equivalent,
Zo = rp/(mu +2).
If multiple pairs of output tubes are used, then
Zo = rp/[n(mu +2)],
where n = the number of pairs used.
For example, if one 6AS7 is used to supply both triodes for this circuit, the output impedance would be 280/(2 + 2) or 70 ohms.
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