... To be honest and with no formal electronics training all l can assume is what l read in Merlin’s article on DC coupled CF. Without the diode, arcing inside the tube could destroy the PI tube. And the resistor is to block any switching issues. As far as Ohm’s law, l have a basic grasp but lack in-depth understanding. ...
No worries! I had the following concept explained to me early, and it should be a useful tool for you.
Ohm's Law:
- There's a "push" (Volts), a resulting flow (Current), and an impediment to current-flow (Resistance).
- Some of the "push" is lost when current flows through a resistance, which amounts to a voltage-drop across that resistance.
- We can understand Current in terms of Voltage & Resistance: Current = Volts / Resistance, where more-volts or less-resistance yields more-current.
- But there's a different way to look at Ohm's Law, as Volts = Current x Resistance. So no-current ---> no-voltage-drop across the circuit resistance.
The bit I was shown in the Navy was the "Ballon-Rock Theory":
- There is a resistor from tube plate to the B+ supply, which is like a balloon.
- There is a resistor from cathode to ground, which is like a rock.
- For normal idle conditions, the tube's plate & cathode are at an equilibrium that is neither B+ nor Ground.
- The normal plate current causes a voltage-drop across the plate load resistor, so the plate is at some voltage less than B+.
- The normal plate current causes a voltage-drop across the cathode resistor, so the cathode is at some voltage greater than the ground voltage (usually 0v).
When there is a fault that kills tube plate current, the "plate gets pulled up by the balloon" and "then cathode falls like a rock."
- Plate voltage rises to B+, cathode voltage falls to 0v.
Now look again at Merlin's circuit around this
direct-coupled cathode follower:
- When power is first switched on, the heaters have not warmed the tube-cathodes and there is no plate current.
- The stage before the cathode follower has its plate at B+.
- The cathode follower's grid is now also at B+, because there is no current to drop voltage anywhere.
- The cathode follower's cathode is at 0v, because there's no plate current to create a voltage drop to elevate its voltage above ground.
A 12AX7 may be rated for 330v plate-to-cathode, but there is now some large positive voltage between
grid and cathode (where the grid is normally negative of the cathode). If we look at the mica spacers holding the tube's elements in place, we also see a very close spacing of the grid & cathode (certainly closer than plate-to-cathode), which should make us worry that the tube cannot withstand this large voltage differential.
Merlin inserts a diode between grid & cathode (plus a resistor). The banded end of the diode is towards the cathode, so it conducts when the grid is more than 0.7v positive of the cathode (plus whatever voltage drop is created across the 47kΩ resistor).
- The prior stage's plate load resistor (value not provided, but often 100kΩ) and the series resistor between plate & grid (value not provided) form a string with the 47kΩ resistor, cathode follower load (also often 100kΩ) & diode from B+ to ground. So the diode causes some amount of voltage-drop across each of these resistors, bringing the follower's cathode upward and its grid downward to reduce voltage-stress across the tube. The grid-to-cathode voltage will like be below 70v and unlikely to cause arcing.
When the amp reaches a steady-state, the cathode will be positive of the grid, and the diode will no longer conduct. So the whole circuit arrangement is only active at startup, and to protect the tube from arcing.
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