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I can't understand the functioning of the circuit of a NOT gate using a triode (vacuum tube) and a resistance (big rectangle connected between the triode's anode and the positive power supply). For information, here A equals non-B.

enter image description here

ocrdu
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Aleksei Koshelev
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    B should be a "negative" voltage ... – Antonio51 Feb 08 '24 at 15:34
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    I don't know who drew that schematic, but for sure they don't know what technical communication is meant to achieve. If the rest of the book/article is as bad as this circuit drawing, you'll do best to not waste your time with it. It'll only confuse you. And it's not a working circuit anyway, just a rather poorly illustrated working principle. – Kuba hasn't forgotten Monica Feb 09 '24 at 06:47

2 Answers2

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Short answer: when used for logic, vacuum tubes behave like voltage controlled switches. The grid vs. cathode voltage turns the tube on or off. The challenge in making logic out of tubes is knowing how tubes behave and working with their characteristics.

Let's begin by revisiting your diagram and adding some details:

  • ground the cathode
  • set the supply to +150V
  • 'B' is the grid voltage (input)
  • 'A' is the anode voltage (output)

Note: the tube anode is also called the plate.

Now, we'll create a sim showing the tube basic behavior (simulate it here):

enter image description here

Here's how the tube behaves when we manipulate the grid (B) voltage:

  • 0V at grid: the tube will be ‘on’, cathode-to-plate current will flow. There will be a voltage drop across the load resistor, so the plate voltage (A) will be lower than 150V. How much lower? Depends on the tube characteristics and the size of the load resistor (in this sim, 75V.)

  • -20V at grid: the tube will be ‘pinched off’, little or no current will flow, plate voltage (A) will be close to 150V.

  • +1V at grid: tube will be 'on', and some additional current will flow from grid to cathode (grid and cathode behave like a diode.) You want to avoid a lot of grid current, but a small amount won't hurt. In this sim, the plate is at 51V.

Result? By swinging the grid between +1 and -20V we have a circuit that 'inverts', to a swing of 51 to 150V. So we have this weird negative input and high-voltage positive output. How to get a usable output from it? And further, how do we use it as logic?

IBM did it this way in 1958, for the 604 keypunch machine:

enter image description here

From here: https://hackaday.io/project/20733-hot-logic/log/57045-tube-logic/

Since we can, let's simulate what IBM did (simulate it here):

enter image description here

The IBM designers chose to ground the cathode, define 'high' as 140~150V and 'low' as 50V. Then, they used a long-tail voltage divider at the grid input to level shift the 150V/50V output from the input make the 0 / -20V negative grid voltage needed to achieve 'pinch-off' at the output.

With the signals conditioned like this they have a basic logic block with a defined input-output scheme for connecting logic elements.

Using more circuit variations, we can create a whole machine like the ENIAC: http://archive.computerhistory.org/resources/text/Knuth_Don_X4100/PDF_index/k-8-pdf/k-8-r5367-1-ENIAC-circuits.pdf

Here too: https://www.cs.drexel.edu/~bls96/eniac/ch1.pdf

And finally, some more vacuum tube logic ideas: https://hackaday.io/project/169147-low-voltage-vacuum-tube-logic-gates/

hacktastical
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  • "Vacuum tubes used as gates have rather unfriendly biasing". A high-mu triode, like a 6N7 (or an 811 if you want to be emphatic) should work without needing level shifting. – TimWescott Feb 08 '24 at 20:48
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    IBM's tube type was similar to a 6J6, according to the manual I linked. All tubes need a negative grid bias to pinch off, regardless of mu. So some level shifting is always needed to achieve that. It's also possible to lift the cathode as is done with amplifiers, but IBM chose to do it by using a long-tail pulldown. – hacktastical Feb 08 '24 at 21:29
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A tube is a voltage driven current source. If the voltage A raises compared to the cathode, the current through the tube raises. As this current passes through the resistor it lowers the anode voltage of the tube.

greg
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