Why does the NFB drive the Cathode instead of the Grid of that triode.?
Another Reason:
The feedback loop is not just the wire from the speaker and the 22kΩ resistor. If you only had a wire from the speaker to the feedback injection point, you get 100% feedback and turn the amp into an oscillator.
The amount of voltage fed back has to be reduced some amount. So the 22kΩ series feedback resistor
and the 1.5kΩ cathode resistor form a voltage divider. If the amp outputs 4w into the 4Ω speaker then there will be √(4w*4Ω) = 4v RMS across the speaker terminals. The 22kΩ and 1.5kΩ divide that down to [1.5kΩ/(22kΩ+1.5kΩ)] * 4v = 0.255v RMS applied to the cathode.
If you apply the feedback to the grid, you now need another resistor to ground to make your feedback voltage divider. There already is one in the form of the resistance between the Volume control wiper and ground, but that changes with the volume setting (and not in a good way). So you'd need to add a cap between the volume control and the new resistor to ground at the tube grid.
So injecting at the cathode means you save a resistor and a cap. Simple economics.
When you input the Cathode...the plate will still be out of phase with the cathode...right.?
No. A single input to the grid results in a plate output of opposite polarity as the input signal. A single input to the cathode results in a plate output of the same polarity as the input signal.
Anyone who has converted an AB165 Bassman to AA864 specs knows this; the feedback injection point moves from the phase inverter grid to under the phase inverter tail resistor (in the AA864 Bassman circuit).
If you have a 5F1 Champ, add a temporary cap between the volume control and 2nd gain stage, add a couple-kΩ resistor to ground and hook up the feedback to the grid. What happens?
(You can fix it by swapping OT leads, but your experimental result should prove the plate signal polarity changes if you swap input points.)
Topic: Schem question (Read 3791 times)