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flow. Thus net change in current is the absolute difference in each tube's change in current, as the top tube is now only conducting 5 mA and the bottom tube is conducting 10 mA more current than the top tube this extra current must flow into the amplifier causing the pulse. Consequently, the solid-state power amplifier must be able to source 10 mA of current or the pulse will not be sustainable. Now we can figure out the output impedance:
V / I = R
1 volt / 0.01 amps = 100 ohms.
Had the circuit consisted of one triode configured as a Cathode Follower, the result would have been the same. A 1 volt pulse at its output brings the cathode to a voltage 1 volt greater than the grid, which effectively is the same as driving the grid 1 volt negative relative to the cathode. The change in current would be equal to the full 1 volt, as there is no resistor network diving the voltage, times the transconductance of the tube:
V x Gm = I.
In the case of a 6922 with a Gm of 10 mA per volt, the current will decrease by 10 mA. If the idle current was set to 10 mA, then the tube would stop conducting altogether and the solid-state amplifier would be shouldered with the task of maintaining the idle voltage plus the 1 volt pulse across the cathode resistor. In other words, the solid-state amplifier would see the same 10 mA current draw at its output as in the previous example.
In a nutshell, our balanced converter has two tubes, but as they only see half of the voltage disturbances at the output, only half off each tube's Gm can be called into play to fight the disturbance. Since we have two tubes, we have twice times one half, which yields one.
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To bypass or not to bypass
In our analyses of the circuit's output impedance, the cathode resistors were assumed to be bypassed, but in the schematic of our balanced converter, the cathode resistors are left unbypassed. What happens to the output impedance if the resistors are unbypassed?
Normally, an unbypassed cathode resistor will greatly increase the output impedance. In a Grounded Cathode amplifier the effective increase in the value of rp is equal to the (mu + 1) times the value of the cathode resistor. Here the value one cathode resistor is simply added to what the output impedance would be with bypassed cathode resistors. The formula for the output impedance with bypassed cathode resistors is given by:
Zo = [ 2rp / (µ+1) ] || (2rp / µ).
And with unbypassed cathode resistors:
Zo = [ 2(rp / (µ+1) + Rk) ] || [ 2(rp + (µ+1)Rk) / µ ],
or roughly
Zo = 1 / Gm + Rk.
In the case of a 6922 with a Gm of 10 mA per volt and a cathode resistor of 200, the Zo will equal 100 ohms with bypassed cathode resistors and 300 ohms with unbypassed cathode resistors.
The result is basically the same as a single Cathode Follower with the same tube. In many ways what we have created is one super tube out of two.
Re-Balancing the input impedance
Early, the mismatch in input impedance was mentioned. Because the top resistor network attaches to the output, the effective impedance of the resistor string is magnified by the
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