Why does triode voltage drop change as a function of grid voltage?

Imagine a thermionic diode connected in series with a resistor $$\R\$$. The circuit is hooked up to a voltage source $$\V\$$. The diode conducts a certain current $$\I\$$ and $$\\Delta V_\text{diode}\ + \Delta V_\text{resistor} = \Delta V_\text{diode} + IR = V\$$. That is: the voltage drops across the diode and the resistor sum to the total applied voltage.

Presumably the voltage drop $$\\Delta V_\text{diode}\$$ arises from the acceleration of electrons as they travel from cathode to anode. The kinetic energy of the electron velocity is lost as waste heat when the electron slaps against the anode.

Now, imagine replacing the diode with an equivalent triode. Imagine the voltage drop from cathode to the control grid is 0V. The grid is effectively superfluous and nothing about our circuit has really changed yet.

Finally, imagine lowering the grid voltage. This should reduce the current $$\I\$$ that conducts through the diode. Thus $$\\Delta V_\text{resistor} = IR\$$ also decreases. Thus the the voltage drop $$\\Delta V_\text{diode}\$$ should increase, since $$\\Delta V_\text{diode} + \Delta V_\text{resistor}\$$ must remain equal to the applied voltage $$\V\$$.

My question is: why/how is energy per coulomb lost as charge travels through the triode?

Because of the grid voltage, only electrons emitted at high-enough velocity can "punch through" the grid field and arrive at the anode. (This is why the current conducted by the tube drops.) My intuition was that these electrons that do succeed at passing from cathode to anode arrive at the anode with greater velocity, and thus more energy is lost as heat.

But then I thought: if one integrates the electric field from cathode to anode (the kinetic energy added due to the potential difference between cathode/anode), doesn't the work being done per electron remain constant (regardless the grid voltage)?

In summary: when the grid voltage is at a lower voltage than the cathode, the triode current drops. But what explains why $$\\Delta V_\text{triode}\$$ rises?

Examples

Example of voltage drop across resistor with no voltage applied to grid:

Example of voltage drop across resistor with negative voltage applied to grid:

Note that the voltage drop across the resistor has decreased (and the voltage drop across the triode has increased).

• You know? I spent my early hobby days studying little else than vacuum tubes. And I struggled at first to tear down wrong concepts and build up new, better ones. But even I'm confused by your writing. Your final "where does the voltage go?" just leaves me completely without a clue, now. I know you are confused. But I don't know where or how. I'm confused by your confusion, I guess. And colloquial language, perhaps. Could you take some narrow case and completely walk me through every single detail of your thoughts? Maybe that would help.
– jonk
Sep 19 '20 at 4:04
• Hi! I've edited my question to try to be more clear. Fundamentally, my question is: why/how does the voltage drop across a triode increase when a negative voltage is applied to the control grid? Sep 19 '20 at 4:27
• Okay. So, take "Now, imagine replacing the diode with an equivalent triode. Imagine the voltage drop from cathode to the control grid is 0V. The grid is effectively superfluous and nothing about our circuit has really changed yet." Why do you think "nothing ... has really changed yet"? Imagine the 3D vector field $\mathscr{E}$ in volts per meter in the diode case. Now, imagine inserting a grid half-way in between (let's say it is a fine grid with 5% of the area blocked with screen wire) at the exact same potential as the cathode. Does the 3D vector field change in no way at all? Nada?
– jonk
Sep 19 '20 at 4:35
• Hmm. I am uncertain. I'm not sure what would change, or what possible change might be relevant to my question. I've updated again with some images that hopefully help explain what I am asking. Which is: in the second image, the voltage drop from cathode-to-anode has increased. But how are electrons moving through the triode losing more energy than before the grid voltage was lower? Sep 19 '20 at 5:00
• I see you are certain and that you've added images from a program that I hate (a lot) to bolster your thoughts. In reply, I'm going to recommend that you take a look at a method discussed by J. Clerk Maxwell in "Electricity and Magnetism," 3rd ed., Vol. I, Sec. 203, Cambridge, London, 1904. Although vacuum tubes weren't even invented then, his analysis of the electrostatic field with respect to a shielding screen of parallel wires there can be readily applied to analyzing a triode's field. I think I'll let you be certain and let others address your question. There's a barrier between us.
– jonk
Sep 19 '20 at 5:02

2 Answers

@jonk may speak harshly to me for over simplifying the model [ :-) :-( ] but the following may help.

Consider the triode as a voltage variable resistor = Rt.
Variation of resistance with voltage is non linear but that does not affect the model.

Below

Vg = applied grid voltage (Vgrid)
Vt = triode anode-to-cathode voltage
Rt = triode effective resistance
It = Anode current
Rl = load resistor = anode resistor

Starting with Vgrid = 0 and then applying an increasingly NEGATIVE voltage to the grid then Rt increases. For an anode resistor Ra and a supply voltage Vs the triode anode Voltage Vt is
Vt = Vs x Rt / (Rt + Ra).
As Vg goes increasingly negative Rt increases and so Vt increases - because Rt becomes larger compared to Ra so drops a larger Voltage.

Similarly -
It = Vs / (Rt + Rl)
So, as Rt increases with negative-increasing Vg, It will fall.

Which is what you see.

• This is actually probably a very accurate way to describe vacuum tubes at a basic level. Sep 19 '20 at 15:24
• Great explanation. Just wonder how Vg affect art … just an idea or there is certain mathematical relationship (on a simple level). Also as It is changed, and when filter out the direct current so the “ripple” can be sent to the control grid of the second tube, is the variation of voltage significant higher than the first tube control grid. And if so, is it because b+ is much larger or what …. Sorry still in the mist. Aug 9 at 23:13

Thanks, all. I think I understand now. I can give an explanation which hopefully also explains my original confusion.

Imagine two parallel plates: one has net charge $$\-Q\$$ and the other has net charge $$\+Q\$$.

If each plate is assumed to have infinite area, then the electric field induced by each of the plates has uniform magnitude $$\E\$$ (always directed away from each plate).

Thus, given the two plates, the field strength is $$\2E\$$ between plates, and zero everywhere else.

The voltage difference between plates can be calculated by integrating the electric field over the distance between the plates. This is proportional to the distance between the plates. It is $$\2Ed\$$.

Next, imagine slipping a third, middle plate with charge $$\-Q\$$ halfway between the first two. The electric field is no longer uniform. It is halved to $$\E\$$ between the negative plates. It grows to $$\3E\$$ between the middle plate and positive plate.

Still, the potential difference between the original plates is $$\2Ed\$$. In fact, it doesn't matter what charge is on the middle plate. Its effect on the electric field is always anti-symmetric and thus cannot change the potential difference between the two original plates.

When I was considering the triode, I was thinking in these terms. I was stuck. I thought: interposing a negatively charged grid shouldn't change the voltage difference between cathode and anode.

What I forgot is that the charge on the cathode/anode plates is not fixed! If the grid voltage is lowered, less current will flow. At a lower current, less energy is lost by electrons passing through the subsequent resistor. To restore equilibrium, there is capacity to (temporarily) increase current by building up a greater charge difference across cathode/anode (similar to a capacitor charging).

As the charge difference builds, the voltage across the cathode/anode increases. This happens until $$\\Delta V_\text{triode} + IR = V\$$ again.

When equilibrium is regained, we see that electrons moving through the triode are subjected to a greater electric field, and thus more energy is being spent accelerating the electrons.