Questions about center-tapped transformer for tube amplifier (MC 275)

Question

I have a question about the center-tapped AC transformer used in the attached figure and its bias voltage generation. Mainly, my question is how to generate DC supply voltage for the 12AZ7 (V4) plate (ORN - RED/WHT - ORN/WHT inductor).

From the figure, the plate DC voltage of 12AZ7 should be defined by the voltage at the 'RED/WHT' tap of the 'ORN - RED/WHT - ORN/WHT' inductor. But this center tap is connected to another center tap of an AC power transformer shown at the bottom (RED/YEL tap of M-215).

Initially, I thought the current determined the plate voltage of 12AZ7 through a 36 kΩ resistor (18 kΩ + 18 kΩ).

This is because 12AZ7 is used as a cathode follower, its grid voltage is biased at -57 V, and the cathode voltage shall be close to -57 V. Since the DC voltage at the other side of the 36 kΩ resistor is biased at -150 V, the voltage difference between the 36 kΩ resistor determines the current, which determines the plate voltage.

But I thought this argument did not make sense because the plate-cathode current should be the function of anode voltage, not the other way around.

In short, I cannot see how the M-215 transformer (shown in the bottom) center tap (RED/YEL) is creating 205 V.

I would greatly appreciate it if anyone could help me understand this circuit better.

Answer 1

I'm answering my own question so that people who ask a similar question in the future can use my answer as a reference.

Most of full-wave rectifier example uses either (1) a transformer w/o center-tap + 4-diode full-wave rectifier or (2) a transformer with a center-tap + 2-diode full-wave rectifier. Since I couldn't find an example of using a transformer with a center-tap + 4-diode full-wave rectifier combination, I just created a schematic and ran the simulation.

The theory of operation is following. In the given schematic, the DC bias is provided by the reverse-biased diodes, not by the center tap. This way, the center tap generates a nearly DC voltage that is half the amplitude of the two ends of the transformer output (algebraic mean of voltages at two end tips of the transformer).

I think the Mcintosh engineer did this because they wanted to minimize the coupling between power supply voltage at V4 from all the other gain stages. V4 stage gain is nearly one due to its cathode follower configuration. Due to the signature 'unity coupled feedback,' the last stage also has a relaxed gain. However, the last stage must have a high power supply voltage due to its large swing and driving capability. V4 also sees a large swing, but its plate voltage requirement can be relaxed by a bootstrapping technique used in the V4 plate, as shown in the figure.