1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

-Suppose that the distance between "D1" and "D2" is too large.

-Does "D2" give a delayed response compared to "D1"?

\$\endgroup\$
  • 4
    \$\begingroup\$ I would say yes. For that purpose the conductors should be though of as wave guides, while the wave is propagated with a finite speed. \$\endgroup\$ – Eugene Sh. Oct 11 at 18:06
  • 1
    \$\begingroup\$ @jsotola "no" to what? And this is way too simplified. Electrons in the conductors are set in motion by electric field which is propagated as a wave. \$\endgroup\$ – Eugene Sh. Oct 11 at 18:31
  • 1
    \$\begingroup\$ @jsotola, if we take the schematic arrangement as indicative of the physical arrangement, the ground wire is longer than the positive voltage wire. \$\endgroup\$ – The Photon Oct 11 at 18:39
  • 1
    \$\begingroup\$ @Jundullah The slow LED response will conceal any lightspeed-delay, unless the wires are hundreds of feet long. We CANNOT DETECT the delayed response, since the speed of typical LEDs is ~100 nanoseconds, while the circuit-delay is from EM waves, roughly 1000x faster. (Try laser diodes instead. Then it becomes possible to measure the fast pulse-edge timing.) \$\endgroup\$ – wbeaty Oct 11 at 19:05
  • 1
    \$\begingroup\$ @jsotola actually electrons are irrelevant here, since propagation delays would be the same for thin tubes of salt water, or even for acidic proton-conductors. Any waves coming from the signal-source will fly outward along both the pos and the neg wires at the same time. That's how 2-wire transmission lines work, regardless of charge-carrier polarity or signal polarity. D1 would light first, if the connecting wire is shorter. (perhaps use comm laser-diodes and 40GHz Finisar modules to perform nanosecond measurements on centimeter wire lengths.) \$\endgroup\$ – wbeaty Oct 11 at 19:27
1
\$\begingroup\$

In the case shown, with both diodes near the power supply but with a very long wire between them, there will be no delay. However, in the case shown below, there will be a delay:

schematic

simulate this circuit – Schematic created using CircuitLab

The reason there's a delay in this case and not in the other is related to the fact that information must travel at the speed of light or slower, so the information that the power supply has been turned on has to travel through the Extremely Long Cable™ at no faster than the speed of light.

For further research, I recommend websearching the term "transmission line".

\$\endgroup\$
  • \$\begingroup\$ What is the result of comparing electric current with water? @Hearth \$\endgroup\$ – Jundullah Oct 12 at 10:11
  • \$\begingroup\$ @Jundullah You mean the hydraulic analogy? Well, in water, pressure changes propagate at the speed of sound, so consider that the current flow at the end of a long pipe would be delayed relative to the beginning of the pipe. \$\endgroup\$ – Hearth Oct 12 at 11:20
0
\$\begingroup\$

Yes, D1 will emitt light a very short time earlier than D2. The current flowing through D1 will have to load the the very long cable's capacity before the same current will flow through both LEDs.

\$\endgroup\$
0
\$\begingroup\$

No, for an ideal diode (or LED) there is no capacitance or inductance and the current through both is equal. (and in the ideal world wires are made out of superconductors)

For a real world diode there is a slight amount of parasitic capacitance and inductance that might cause a slight difference if you're looking at very small timescales, for most applications this would be negligible.

If you wanted to analyze this for a distance that is 'too far' then model it as a transmission line ( a long line of RC or RLC filters) or insert the calculated of resistance of the wire or trace between diodes between diodes

\$\endgroup\$
  • \$\begingroup\$ The question asks us to "Suppose that the distance between "D1" and "D2" is too large." This means (at least to me) we can no longer assume the lumped circuit approximation holds. \$\endgroup\$ – The Photon Oct 11 at 18:38
  • \$\begingroup\$ @neil_UK Sorry, I'm going to relearn the correct grammar one of these days! \$\endgroup\$ – Voltage Spike Oct 11 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.