Understanding Electron Flow in Vacuum Diodes

[Abimbola1987]

Dear Sirs

I have been going through the "radiotron designers handbook", but I can't seem to find the answer to the doubt I have about the electron flow of the source Vp. Kindly consider the below two circuits, the difference is the polarity of Vf.

Question: In both circuits, the electrons that exits the negative (-)terminal of Vp, where do they flow?

Kind regards
Abim

 

[scottdave]

The Vf in your circuit provides current through the cathode filament to heat it up and start them up if emission. The electrons are free to leave the metal but they without an electric field, they will not go in any particular direction. rity n. VP provides the driving electric field and the negative terminal is the source for electrons. The electrons leave the cathode in a "beam" through the vacuum toward the anode. You may like reading this, as well.

Note: the below was original link that I posted, but I recently tried and doesn't work. I found the Internet Archive link to the same NEETS document.
http://www.fcctests.com/neets/NEETS...nic-Emission-Tubes-and-Power-Supplies.pdf.pdf

 

[BvU]

In both circuits: from the cathode to the anode
https://www.physics-and-radio-elect...es-and-circuits/vacuum-tubes/vacuumdiode.html

 

[Abimbola1987]

The Vf in your circuit provides current through the cathode filament to heat it up and start them up if emission. The electrons are free to leave the metal but they without an electric field, they will not go in any particular direction. rity n. VP provides the driving electric field and the negative terminal is the source for electrons. The electrons leave the cathode in a "beam" through the vacuum toward the anode. You may like reading this, as well.
http://www.fcctests.com/neets/NEETS...nic-Emission-Tubes-and-Power-Supplies.pdf.pdf

Dear scottdave,

Thank you for your reply, I love those military educational papers, they are written so well that every draftee can understand them.

VP provides the driving electric field and the negative terminal is the source for electrons.

Yes, that is for an indirectly heated tube, if you look at my drawing it is directly heated, in the paper you linked to, it reads:

"Because the filament is the only heated element in the diode, it is the ONLY source of electrons within the space between filament and plate."

And the heating element (filament) is powered by Vf, so by which rule/law/phenomena do electrons flow from the negative terminal of Vp to the cathode?

 

[Abimbola1987]

In both circuits: from the cathode to the anode
https://www.physics-and-radio-elect...es-and-circuits/vacuum-tubes/vacuumdiode.html

Yes, that is for an indirectly heated tube, if you look at my drawing it is directly heated, and that makes me still in doubt as to where electrons flow from the negative terminal of Vp.

 

[BvU]

Makes no difference, it's just that the potentials of filament and cathode are forced equal -- the potential difference between cathode and anode is not affected

 

[Abimbola1987]

Makes no difference, it's just that the potentials of filament and cathode are forced equal -- the potential difference between cathode and anode is not affected

OK, so the electrons go from Vp(-) to the cathode, but by which rule/law/phenomena? Kirchhoff's current law?

The conceptual problem I have then is that of CIRCUIT B; Electrons are flowing from Vf(-) to Vf(+) and at the same time electrons are flowing in the opposite direction through the same wire from Vp(-) to the cathode?

 

[BvU]

You mean there are electrons flowing in opposite directions within the same wire ? No. Kirchoff current law also rules at the node in the line that connects V_f and V_p to the filament, both for left and for right picture.

The potential difference between cathode and anode is important for I_p and it is correct that the two pictures difffer in that respect: the one on the left has an average cathode potential of V_f/2 and the one on the right -V_f/2

In general practice, V_p >> V_f , though, which is probably why we didn't see your problem.

 

[QuantumQuest]

And the heating element (filament) is powered by Vf, so by which rule/law/phenomena do electrons flow from the negative terminal of Vp to the cathode?

Yes, that is for an indirectly heated tube, if you look at my drawing it is directly heated, and that makes me still in doubt as to where electrons flow from the negative terminal of Vp.

Let's say that the filament in the directly heated tube is powered by an AC voltage ($V_f$), so it continuously changes polarity going from diagram $A$ to $B$. What does this change mean? AC applied voltage will cause small increases and decreases of the temperature which result in a small increase and decrease of emitted electrons, which may or may not be of significant importance, depending on the application at hand - in many circuits it is not. But the voltage $V_p$ does not change polarity, so there will be no change in the flow of current as long as you don't give plate a negative potential to drive the tube to a cut-off state.

 

[Abimbola1987]

Dear BvU,

You mean there are electrons flowing in opposite directions within the same wire ?

I don't see how they could, but that was my understanding of your previous reply, I'm sorry if I misunderstood you.

Again, my conceptual problem of CIRCUIT B remains, you have a connection point, cathode(right side) Vf(+) and Vf(-), three wires with same potential connect. Vf(+) > Vf(-) so electrons flow from Vf(-) through the cathode to Vf(+). Now where do the electrons that exits the negative (-)terminal of Vp flow? If they flow to the cathode(right side) they would flow opposite the Vf(-) -> Vf(+) electron flow, so the only way they could flow is to the Vf(+), but do they?

 

[Abimbola1987]

Let's say that the filament in the directly heated tube is powered by an AC voltage ($V_f$), so it continuously changes polarity going from diagram $A$ to $B$. What does this change mean? AC applied voltage will cause small increases and decreases of the temperature which result in a small increase and decrease of emitted electrons, which may or may not be of significant importance, depending on the application at hand - in many circuits it is not. But the voltage $V_p$ does not change polarity, so there will be no change in the flow of current as long as you don't give plate a negative potential to drive the tube to a cut-off state.

Yes, I can understand that, thank you. But I don't feel that it answers my question though; the electrons that exits the negative (-)terminal of Vp, where do they flow?

 

[QuantumQuest]

Yes, I can understand that, thank you. But I don't feel that it answers my question though; the electrons that exits the negative (-)terminal of Vp, where do they flow?

If you understand what I wrote and the other posts by @scottdave and @BvU then let me ask you: have you studied the diode circuit yourself? Do you know how to analyze a circuit regarding voltages and currents? I don't ask in an offensive way; I'm trying to help your understanding. If you have done all of the above, you should be able now to answer the question yourself. In any case, let me ask you: what are the individual circuits formed in a diode tube circuit and what is the flow of current in each of them?

 

[tech99]

This is a directly heated (filament cathode) vacuum diode. One side of the filament has slightly more anode voltage than the other, because the filamant supply adds to the B supply on one side. But the difference is very small.
In the case of a triode, there is an issue because the filament voltage is enough to provide significant grid bias. In fact, taking the grid leak to the negative end of the filament may provide sufficient bias for a voltage amplifier.

 

[Abimbola1987]

If you understand what I wrote and the other posts by @scottdave and @BvU then let me ask you: have you studied the diode circuit yourself? Do you know how to analyze a circuit regarding voltages and currents? I don't ask in an offensive way; I'm trying to help your understanding. If you have done all of the above, you should be able now to answer the question yourself. In any case, let me ask you: what are the individual circuits formed in a diode tube circuit and what is the flow of current in each of them?

Dear QuantumQuest

I don't ask in an offensive way; I'm trying to help your understanding.

Don't worry, I'm not offended in any way, I find this a civilized forum. i.e. stick to the rules and everything will be just fine. And I appreciate every input I get from PF members, I actually learn or re-learn new things here.

have you studied the diode circuit yourself?

Yes, hence my question, I reached a point where I doubt in my knowledge.

Do you know how to analyze a circuit regarding voltages and currents?

No, I haven't done textbook circuit analysis since college, and I haven't used it since so it is well forgotten.

If you have done all of the above, you should be able now to answer the question yourself.

I obviously have an idea about how I see the circuit, but is it correct? Well, I'm in doubt, so I thought I would turn to PF to obtain a straight and correct answer.

In any case, let me ask you: what are the individual circuits formed in a diode tube circuit and what is the flow of current in each of them?

First part of your question: I see two individual circuits the first being Vf and the cathode, the second being Vp cathode and anode. Since the cathode is common for both circuits I'm in doubt about the second part of your question.

 

[Abimbola1987]

Dear tech99

Thank you for your reply.

This is a directly heated (filament cathode) vacuum diode. One side of the filament has slightly more anode voltage than the other, because the filamant supply adds to the B supply on one side.

If "anode" in "slightly more anode voltage than the other" is not a typo, then I'm not sure what you mean by that.

"because the filamant supply adds to the B supply on one side." I assume you mean "Vf" (in my drawing) when you write "B supply", so my question is, why and how does it do that?

 

[QuantumQuest]

No, I haven't done textbook circuit analysis since college, and I haven't used it since so it is well forgotten.

Yes, that's understandable but the question you ask can be properly answered only through this. But for a simple, qualitative way see below.

First part of your question: I see two individual circuits the first being Vf and the cathode, the second being Vp cathode and anode. Since the cathode is common for both circuits I'm in doubt about the second part of your question.

There are two series circuits. The filament power supply and the filament itself form a series circuit (filament circuit). The path of the second series circuit is from one side of the filament, across the space to the plate, through the resistor and source $V_p$, then back to the filament (plate circuit). Also, a part of the filament circuit is also common to the plate circuit. This part enables the electrons coming from the filament to return to the filament. There are also some basic facts in the operation of a such circuit. First, plate current flows only when the plate is positive. If plate is negative there is no current (cut-off state) and the plate current flows only in one direction: from the filament to the plate.

 

[tech99]

Dear tech99

Thank you for your reply.
If "anode" in "slightly more anode voltage than the other" is not a typo, then I'm not sure what you mean by that.

"because the filamant supply adds to the B supply on one side." I assume you mean "Vf" (in my drawing) when you write "B supply", so my question is, why and how does it do that?

There are several explanations. For instance, the two batteries can be looked upon as being in series. If the B battery is 100V and the filament battery is 3V, then one end of the filament will have 100V to the anode and the other end will have 103V to the anode.

 

[Abimbola1987]

Yes, that's understandable but the question you ask can be properly answered only through this.

Yes, you are absolutely right.

But for a simple, qualitative way see below.

There are two series circuits. The filament power supply and the filament itself form a series circuit (filament circuit). The path of the second series circuit is from one side of the filament, across the space to the plate, through the resistor and source $V_p$, then back to the filament (plate circuit). Also, a part of the filament circuit is also common to the plate circuit.

Yes, that is what I meant.

This part enables the electrons coming from the filament to return to the filament.

That is the part I was missing, the electrons into Vp(+) equals electrons out of Vp(-) and they obviously have to return, though I would think the electrons return to Vf(+), rather than going opposite Vf(-) -> Vf(+) flow to reach the physical filament strand inside the tube.

Thank you.
Abim

 

[Abimbola1987]

There are several explanations. For instance, the two batteries can be looked upon as being in series. If the B battery is 100V and the filament battery is 3V, then one end of the filament will have 100V to the anode and the other end will have 103V to the anode.

OK, that makes it clear to me, thank you very much.

Abim