Is it possible to build logic gates (at least only OR, AND gates) by just using electric wires and resistors with no semi conductors (no diodes, no transistor or any other semi conductor material or device)?

The reason I'm asking this is paper circuits, by using only graphite pencil (graphite pencil lead contains some amount of graphene which is semiconductor but not a detail to focus since its in minimal amounts and don't affect it) and paper (recreating electronic components to some extent, and to try to challenge myself) making capacitors, resistors are simple but what about logic gates.

So I thought something like this, a power source that only supplies a consistent current (even if its shorted) e.g.: bench supply.

  • drawing 2 lines connecting them together somewhere before the output.
  • making output on end and to have higher resistance by drawing paper faintly, less current can pass due less graphite = less conductive.
  • electricity takes shortest path that does have less resistance.

if a: false b:true then current going to rush through a and return to supply no electricity passing also due everything connected it should affect the output, output should be negative after this point.

Will something like this work? Here is thing susposted to drawn to paper.

I know this is short circuit technically but as stated earlier using a source that supplies fixed amount of current. By doing that at least its not going to burn or destroy the supply. I revised it a bit.


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    \$\begingroup\$ Logic gates are nonlinear devices and so cannot be built of purely linear devices such as wires and resistors. You will need to add a nonlinear element if you want a logic gate. \$\endgroup\$ Commented Feb 4, 2022 at 2:32
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    \$\begingroup\$ Would relay-based logic meet your definition of using only conductive material? \$\endgroup\$
    – Dave Tweed
    Commented Feb 4, 2022 at 4:09
  • \$\begingroup\$ it does but goal is make components out of graphite and paper so far I made very primitive capacitors, resistors tried to make inductor semi-worked though @ Dave Tweed \$\endgroup\$
    – ragedcoder
    Commented Feb 4, 2022 at 4:11
  • \$\begingroup\$ Logic is based on "analog rules" defined in specs. Schematics are also called Logic Diagrams which also follow rules which yours do not comply. Yet I still understand. Try to learn and follow the rules of "logic diagrams by looking at some good ones. Ground is on the "lower level" not vertical , reading is from left to right, even if you write Chinese or Farsi. \$\endgroup\$ Commented Feb 4, 2022 at 13:41

2 Answers 2


No, this is not possible with only resistive (or capacitive/inductive) elements like your graphite tracks. The reason for this is that resistive elements are passive and can only attenuate signals; they can't amplify them. Amplification requires an extra power source that can "inject" additional energy in a way that is controlled by the input signal. You need amplification to build logic circuits - otherwise the signals would only get weaker and weaker with every stage until they're undetectable.

Another way to look at the same problem is this: Given that you only have resistors and voltage/current sources to work with, you can apply the superposition principle to any circuit that you're able to build. This, in turn, means that the output voltage of any circuit that you could possibly build with graphite tracks is just a linear combination of the input voltages. In other words, the only kind of circuit you can build is one that takes its input voltages, makes them weaker by a certain factor, and then adds them up. The output voltage will always be less than the largest input voltage, too. You can't physically achieve any switching behavior.

This is also what the example circuit you drew does: It's a simple voltage divider that outputs the average of the two input voltages.

Current in a resistive circuit doesn't "switch" between paths once one of the paths becomes "full". Instead, the current divides between all of the available paths in proportion to their conductivity. In your case, that means there's less current flowing through the faint track and more flowing through the thick one. The ratio between these two currents will always be the same.

Here's yet another way to think about it: Logic gates are non-linear because they switch abruptly (their output voltage doesn't increase or decrease proportionally to the input voltage). Resistors, capacitors and inductors, however, are linear elements. Any circuit that only contains linear elements will itself be linear, which means it can't exhibit any switching behavior. Therefore it can't contain any logic gates either.

If you want to build logic circuits without semiconductors, you could use electromechanical relays. (Or vacuum tubes - but these are expensive and harder to use.) In general, you need some kind of non-linear amplifying device.

  • \$\begingroup\$ goal being to decrease current to a point that it looks like switching but it doesnt it just seem to do it thats it (like not enough current to light up led most of it goes to other track, it looks like closed but still there is some current going in) also for drops what if I added a capacitor to circuit and charged enough and a resistive path to slow down it then sent the output from it to next logic gate wouldnt it work? capacitor discharges if its 0, if its high enough charges. \$\endgroup\$
    – ragedcoder
    Commented Feb 4, 2022 at 1:55
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    \$\begingroup\$ There is no way at all for you to create logic gates using only resistors (graphite tracks), capacitors and inductors. All circuits you can build from these elements are passive and linear, which means that the superposition principle applies. This, in turn, tells us that the output of every possible passive circuit is a (possibly frequency-dependent) weighted average of the input values. In short: All circuits that you can build without active elements are just averaging circuits. The set of possible circuits does not include logic gates. This can be proven mathematically. \$\endgroup\$ Commented Feb 4, 2022 at 2:06
  • \$\begingroup\$ you are right but still even though this wont be logic gate will there going to be some difference when both of inputs are true (im beginner due that I could been misinterpreted some stuff) if i hooked up any indicator on the output like a led it would light up dim if first one negtive second one positive, if both of positive its going to be brighter, right? also you said on alternatives you can use relays, arent relays passive too? \$\endgroup\$
    – ragedcoder
    Commented Feb 4, 2022 at 2:26
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    \$\begingroup\$ @TonyStewartEE75 Hmm, I'd like to disagree back! ;) - You said that logic signals don't have to be amplified - but then you gave the example of switches and relays. Relays do amplify signals: you can switch a large current with a small one and inject additional energy from a supply into the logic circuit. Synchronous logic circuits, for example, absolutely do require amplification because the signals "go around in circles" with every clock cycle. Without amplification, the signals would just dissipate after a few cycles and the circuit would stop operating. \$\endgroup\$ Commented Feb 4, 2022 at 17:53
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    \$\begingroup\$ @TonyStewartEE75 A logic gate's output signal levels must be able to drive another gate; otherwise you wouldn't be able to build logic circuits with it. This is why there's an amplifying element at the output each (possibly compound) gate in diode logic. All diode logic circuits necessarily contain these amplifying elements. It's a philosophical question whether this amplifying element is "part of the gate" or "between gates", the fact is that it's there (and has to be there for the circuit to work). The diode logic article you linked even discusses this in the "Restoration" section. \$\endgroup\$ Commented Feb 4, 2022 at 18:21

Question: Will it work?

Answer: No

Historically using voltage and switches form primitive logic with ON/OFF to make tandem wired-OR logic with AND in series. Exclusive-OR (XOR) is done in-home stairway switches with 3 wires between two SPDT switches to have each switching toggle the function of the other's position (Up/Down) for On/Off.

In some cases, negative logic was used so that a Normally Open (NO) switch could have been a logic "0" for "Off" with a pullup resistor to V+ and a closed switch state to ground flowing current as "1", although this must be declared by design as we assume normally logic "0" = near 0V.

The 1st (electric) computers used this logic with magnetic Relays, then vacuum tubes, diode logic, then resistor-transistor logic or RTL then TTL then NMOS briefly then CMOS.

But this assumption is false for DC.

"electricity takes shortest path that does have less resistance"

The correct assumption is DC conducts the path of least resistance, regardless of distance.

While high voltage potential (HVDC such as a cloud) may arc across the shortest gap or insulation path with lowest breakdown voltage, which is often the shortest path (or tallest tree or antenna) which then turns the insulation into a conductor of the least resistance.

Thus two traces of graphite are independent of distance but resistance is determined by the amount of conductive material.

Then if the path to the short is assumed to be lossless relative to the short, that short will have the full voltage across it's path with some current and the output will be at the potential in between half-way which is neither 0 or V+=1.

Ground is always defined locally wherever you decide is 0V. But if the traces leading to the short are also graphite, "the short" no longer has 0V on side side with a network of resistance paths which is again ambiguous.

Reason why it won't work

So in logic we define thresholds with margin or V_low(max) or V_hi(min) for both outputs and inputs for each state of lo or hi (0 or 1) and the grey zone in between is undefined.

Your schematic was in between "0" and "1" at best with missing specifications. (Again if you don't understand, be specific and try to find out why with a question then research it, or ask and be specific)

  • \$\begingroup\$ This should make you ask more questions to yourself to then motivate you to find more answers. \$\endgroup\$ Commented Feb 4, 2022 at 5:44
  • \$\begingroup\$ I know you stated this is additional commentary, but as that is the case, it doesn't answer the question that was asked. -1 \$\endgroup\$
    – MCG
    Commented Feb 4, 2022 at 8:07
  • \$\begingroup\$ @MCG if you understand the question was "Will this work" and how I explained that it won't work so your conclusion is fallacious -1, Logic is governed by rules with analog thresholds and margins as I explained Also his assumption was wrong about the rules of DC and the shortest path. That's conditional for insulators and BDV. Bye-the-bye , you never point out suggestions for the answer or improvements. -2 or even asked a question -3 and are just being myopic \$\endgroup\$ Commented Feb 4, 2022 at 13:15
  • \$\begingroup\$ In short ( pun intended) he was demonstrating a very thin analogy to "wired or" logic but without specs, so it will never work \$\endgroup\$ Commented Feb 4, 2022 at 13:23
  • \$\begingroup\$ @MCG Here's a question for you. What do all these different binary logic examples I gave have in common to make them "work"? ( at least most of the time) \$\endgroup\$ Commented Feb 4, 2022 at 13:27

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