Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: dude on May 04, 2023, 04:42:11 pm
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What does this diode do in this circuit’s PI.
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What does this diode do in this circuit’s PI.
Protection. Read and understand, especially the section called "A Useful Mod' for DC-coupled cathode followers:"
http://www.valvewizard.co.uk/dccf.html
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Thanks for the link, very technical reading which leads to studying other info from the Wizard to help understand DC-coupled CF. I think the the best learning curve is understanding OHM's Law, first and foremost, I got that down. I think PRR posted that somewhere, very good advice, thanks PRR.
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Thanks for the link, very technical reading ...
Let's have a "Check on Learning"...
Did you understand the bit Merlin said about, "... on start up, the grid will immediately rise to HT potential while the cathode is still cold, at ground potential"?
What did that mean to you? How is it directly related to Ohm's Law?
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Give me a day or so to answer, appreciate your question.
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Thanks for the link, very technical reading ...
Let's have a "Check on Learning"...
Did you understand the bit Merlin said about, "... on start up, the grid will immediately rise to HT potential while the cathode is still cold, at ground potential"?
What did that mean to you? How is it directly related to Ohm's Law?
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.
Perhaps, l asked a question that l wouldn’t understand the answer? Most Merlin’s info on DC coupled CF, is beyond my current ability. Perhaps if explained in layman’s terms, l might grasp the concept..? But l thank you for trying to educate me.
I can build an amp or follow a circuit from a schematic but understanding “why” it works is another thing, therefore, l could never design an amp…
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“why” it works is another thing,
:laugh:
I spent many years sitting in classes of "why", I passed all the tests so I could get on with earning money. within a couple years I learn't more than all the class hours combined, made a good living, "fixing that which was designed by those that understood "why".
grasping the basics of ohms law is a good thing to understand, it can save you many $$'s from burnt parts, can save your life from burning down the house.
ohms law is basic Algebra, not some magical sci-fi stuff. DO the math, understand the math, save a house :icon_biggrin:
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… Merlin’s article on DC coupled CF. Without the diode, arcing inside the tube could destroy the PI tube…
That’s the second reference you’ve made to a PI, presumably phase inverter.
The circuit in question is a direct coupled cathode follower.
The diode prevents the circuit pulling the cathode follower’s grid more than slightly positive of its cathode. eg at power up.
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… Merlin’s article on DC coupled CF. Without the diode, arcing inside the tube could destroy the PI tube…
That’s the second reference you’ve made to a PI, presumably phase inverter.
The circuit in question is a direct coupled cathode follower.
The diode prevents the circuit pulling the cathode follower’s grid more than slightly positive of its cathode. eg at power up.
Yes, l did reference a CF to the PI, two totally different things, usually a CF is before a tone stack, basically it attenuates the signal to better service the tone stack.., correct..?
I did get what the diode does but not the “why” as Shooter mentioned. Forgive my ignorance, l do read a lot of amp theory but understanding everything I read is another thing. I do need to research DC coupled CF and normal Fender CF and probably others.
Regarding the PI, phase inverter, my understanding is there are two signals, 180* apart and PI balances those signals as one..?
You have help me in the past many times and l thank you for that. You guys know your stuff, most of you have years working in electronics. Some here came to this forum naive like me but soaked up more than l, l admire that.
I’m just a guitar player who is a tone freak, l now can at least satisfy myself obtaining the tone l hear in my head, thanks to this forum.
I also appreciate HBP, asking me a question that l really can’t answer logically, in trying to educate me.
Thank you all, especially Sluckey who has taken the time to answer my sometimes obvious and ridiculous questions, you my friend are a kind individual as most all members are, thanks again to all here.
I forgot the most important member, Mr Doug Hoffman, without him none of this info would be here, thank you Doug
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I did get what the diode does but not the “why” as Shooter mentioned.
Look at the link I posted in reply #1 again. Read only the paragraph titled "A Useful Mod' for DC-coupled cthode followers:". Merlin explains the "why" in simple layman terms.
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... 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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HBP, I appreciate the simplified answer. I liked the analogy, thanks for taking the time to explain. Now I need to read up on how vacuum tubes actually work. I guess you can teach an old dog new tricks, I'm am old dog, lol