Triode Amplifier Circuit Questions from a Mega NOOB

[Cochise Rembrandt]

Greetings and salutations.

I am working on getting a grasp on tube amplifier design and construction. At the basic level I am trying to figure out a few things about tube amplification and I'm hoping for some advice.

I am studying the input triode amplifier circuit in Robert Megantz's book and I'm trying to get a handle on a few issues.

Concerning the Load resistor and the supply voltage, if the plate current is held at zero by a negative grid voltage then the plate voltage = the supply voltage and nothing crosses the resistor.

Question #1 Does voltage act similar to say, osmosis where things move from higher concentration to lower concentration? In other words will the voltage only move across the load resistor if there is less voltage on the other side? When grid voltage increases then electrons move from the cathode to plate, decreasing the plate voltage. It also increases the current to the plate which moves voltage as well. Am I over thinking this? Is it more to do with the creation of current that is moving the voltage across the resistor?

Question #2 In terms of the input signal from the guitar. I understand that it is an AC signal of a certain frequency. It hooks up with the grid and lowers voltage to the grid allowing electrons to move to the plate, but how does the frequency then get passed up to the plate? Is it that the grid is modulating in a phasic manner to the AC signal? i.e. are the electrons moving in a wave like manner to recreate the frequency?

Question #3 Amplification. This is more to see if I understand this correctly. By pulling voltage across the load resistor is the AC signal coming up from the plate then hooking up with DC voltage and riding it? Or is the voltage getting pulled across in a phasic manner as the grid allows electrons across in a phasic manner? Its this passing on of the signal to higher voltage I find amazing.

And as far as I can tell the capacitor set up from there downstream would block out the DC voltage and allow an amplified AC signal to move on to volume and tone controls. Is that about right?

Thanks for your help and time. I attached a photo of the schematic from the book which is what I've been pondering.

[Gingertube]

On-Line resource

Look here for the explaination of how the triode amplifier works.
The Valve Wizard

The "fundamental" thing to know about a tube is that it is a "depletion" device - That means that unlike a transistor where you need to apply a "bias" voltage to turn it on (enhancement device), a tube with no bias runs flat out (maximum current) and you need to apply a "bias" voltage to (partially) turn it off.

Cheers,
Ian

[Cochise Rembrandt]

Thanks for the info. I will check out this other resource.

[exclamationmark]

Most tubes in general can be viewed as voltage controlled current devices. With 'controlling' voltage in this case being the difference between grid voltage and cathode voltage. The current drawn by the tube (ie, the current from plate to cathode) is converted into voltage gain by virtue Ohm's law when it flows across the load/plate resistor.

So a small voltage (0 volts in most cases!) difference between grid and cathode causes the tube to drawn it's maximum current, causing a huge voltage drop across the plate resistor. Meaning the voltage is at this point is very low (because you have dropped most of it across the plate resistor). Conversely when the tube has a negative voltage difference between grid and cathode (or even a positive voltage in some rare cases...) this causes the tube to draw very little current. And by following our trusty Ohm's Law we find that since the plate resistor has very little current flowing through it, very little voltage is dropped across it, and a high voltage is present.

Without a plate resistor or load of any sort we generally can't get any voltage amplification out of a tube. The plate resistor really is the essence of voltage amplification.

[Cochise Rembrandt]

Trusty Ohm's

I think I get it. I need to chomp on this a bit.

Still not sure how the DC voltage brought across the load resistor from the supply obtains AC signal character. I am going to read the pdf Ian offered up and hopefully things will be more clear.

[Enzo]

What is important is the concept of "circuit." No osmosis. There is a power supply and return. Return or common is often called ground. If a path for current exists between the two, then current will flow. In a tube, the path is through the plate resistor, tube, cathode resistor. Think of the tube as a switch or a resistor or something similar for a moment. If the tube stops conducting, the circuit is open - there is no path for current. The supply voltage is in the resistor, but no current can flow. Ohm's Law tells us current through a resistance causes a voltage drop. The voltage from one end to the other of a resistor varies with the amount of current and the resistance. SO zero current (tube in cutoff) means zero volts dropped across the resistor. Note that doesn't say NO voltage ON the resistor, just no voltage difference between ends.

As you turn on the tube allowing current to flow, the current through the resistor causes a voltage drop. SInce the upper end of the plate resistor is connected to the supply, that voltage will be steady. Let's use 300v for the supply as an example. As current increases through the tube/resistor the voltage ACROSS the resistor will increase. Don't worry about the math, but if we have a 100k resistor (100,000 ohms) and 1 milliamp of current flowing (0.001 amp), then 100v will be dropped across the resistor - there will be 100v end to end on the resistor. SInce the one end is steady at 300v, then the other end has to be 200v. 300v less the 100v. SO the plate of the tube sits at 200v in my exAmple.

If I reduce the current through the tube by making the grid more negative, then the voltage acros the resistor reduces as well. Say 1/2 a milliamp, 0.5ma, through that 100k resistor, that means only 50v will be dropped. SInce the resistor is stuck at 300v on the top end, then the other end must be at 250v. And going the other way, if I increase current through the tube and resistor to 2ma, then 200v is dropped, and that means the tube end of the resistor becomes 100v. (300v less the 200v.)

Current through a resistance creates a voltage differential across that resistance. That is Ohm's Law, and there is nothing more fundamental in electronics than Ohm's Law.

The signal is an AC voltage, but it does n't just lower or raise grid voltage. From its rest voltage, that signal alternates between going higher and lower. An audio signal waveform is both positive and negative changes from instant to instant.

Think of the above example with the grid changing tube current from 0.5ma to 1ma to 2ma, and how it changed the plate voltage. Now imagine it does this very fast, as the signal voltage does. When the signal moves more positive, the tube current increases, and that means the voltage across the plate resistor increases. But as that resistor voltage increases, that means the bottom end of it will be lower in voltage. And vice versa. SO when the grid voltage goes up, the plate voltage goes down, and vice versa there. When it does this at audio frequency, that grid signal is going up and down a mile a minute, and that means the plate voltage is varying similarly, though at higher voltages.

That plate voltage is always DC, it just varies at the same rate as the signal. In other words, the plate voltage varies from say 100v to 200v, rather than from -50 to +50. SO that is your signal impressed upon the DC.

In one sense, AC signal doesn't flow through the tube, it controls the tube current. Then the varying tube current leaves a varying voltage at the plate, by way of varying the voltage across the plate resistor. SO signal doesn't pass through the tube so much as it tells the tube grid to make a new, larger copy of itself at the plate. Kinda like the sound from your mouth doesn't go through an amplifier and come out the speaker, your voice controls the electronics that recreate a copy of your voice and send that out the speaker.

[Cochise Rembrandt]

Thanks Enzo. That makes a lot of sense to me. I was thinking of "voltage drop" as the difference between the plate and the supply rather than it being created by the load resistor pulling off the voltage and creating the drop. Cool.

[Enzo]

Ohm's Law. If you are going to learn anything in electronics, start there. I have been soldering almost 60 years, and chasing circuits around most of that time, and yet I still use Ohm's Law every day. Got a little pocket calculator right next to me to do the arithmetic.