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Circuit description
This month's circuit uses a Split Load phase splitter, a Cathode Follower, and a split rail power supply. The dual polarity power supply is crucial to this circuit as it allows us to take advantage of null in noise at the midpoint of the power supply.
Only one half of the Split-Load Phase Splitter is used: the inverting half. It gives us the option of phase reversal. Usually the plate and cathode resistors are matched in this circuit, but here the cathode resistor is made purposely smaller to ensure the gain is unity at the plate. The formula for its gain is the same as that for a Grounded Cathode amplifier: GAIN = ΅Ra / [Ra + rp + (΅ + 1)Rk]. The 2 meg and 1 meg resistors serve to define a voltage dividor that yeilds a one third division of the the peak to peak voltage potential of the power supply, which in this case yields -34 volts. This voltage is used to bias the Split Load Phase Splitter's triode, which will see one third of the total power supply voltage, i.e. 66 volts.
The Cathode Follower is DC coupled at its grid to the ground voltage, i.e. 0 volts. Its cathode is loaded by two resistors in series with each other, a 200 ohm and a 10k resistor. The last resistor is attached to the negative supply rail. The connection of these two resistors is at roughly 0 volts and serves as the output of this circuit.
A switch, either toggle or rotary, is used to select the input for the Cathode Follower. When no phase reversal is desired the switch path would be directly to the volume control potentiometer; and when it is desired, to the output of the Split Load phase splitter.
The Trick
Imagine two resistors wired in series and spanning the two power supply rails. If both resistors are equal in value, the voltage division is one half. Thus if the two rail voltages differ only in polarity, the voltage at the resistors' connection will be 0 volts. One half of DC +100 and -100 is 0. The math holds up to the presence of noise on the power supply rails, which means that as long as the anti-phase relationship and amplitude of positive and negative power supply rail noise match, the output noise from these sources will equal zero at that same connection. One half of AC +1 and -1 is 0.
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