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will be much less than unity.
An Alternative Configuration
By switching the ground to T1's cathode, we regain a Cathode Follower low output impedance of the output stage but at the cost of less total gain and an inverted output signal. But then the use of a feedback loop becomes an option and not a necessity in this configuration. (If feedback is desired, then it should be applied at the grid of the first stage, and not the cathode as was done in the previous examples.)
The same phase splitter modification is used to balanced the output tubes, but in this case it works by subtracting the difference in signal from T1's input voltage.
The output impedance of this variation is much better than the previous example:
Zo = rp / 2(mu + 1).
The easy way to determine the Zo of a circuit is to force the output one volt more positive and then figure out (or actually measure) the net increase or decrease in current flow into that point. (This same technique can be used to figure out the impedance of any point within the amplifier.) For example, if a one volt battery was placed across the outputs of the amplifier shown at the left, the resulting current that would flow through the battery would equal 2Gm, where Gm is the transconductance of each output tube.
To visualize this result, imagine that T2's cathode is quickly forced to +1v. Since its grid is fixed at the phase splitter's cathode voltage, which has not moved with the forced 1 volt offset at the amplifier's output, its grid-to-cathode voltage has decreased by 1 volt, as the cathode is now 1 volt more positive than the grid than it was at idle. This will cause its current flow to decrease by 1 times its Gm. T1's cathode does not see the introduction of the 1 volt battery, as its cathode is fixed at 0 volts. T1's grid, on the other hand, does see a 1 volt increase in its grid-to-cathode voltage, as the 1 volt boost at the output is relayed through the capacitor C to the phase splitter's cathode resistor, which now has 1 volt less voltage across its leads, which is matched by the plate resistor. And as the B+ voltage has not changed, the plate voltage must climb 1 volt more positive. T1 now has 1 volt more on its grid relative to its cathode than it did at idle, which causes its current to increase by 1 times its Gm. Thus when we add up the total
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