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Why can't we light an LED with a single positive wire on the anode, leaving the cathode unconnected?

Let's suppose I connect a wire with a positive voltage to the anode of the LED. Now, I think, current will flow past the light emitting diode. This should, by my understanding, make it glow, even though the cathode isn't connected to anything. However, experimental evidence and common knowledge show that this isn't the case.

Is it possible in any situation(IDEAL) that we calculate current and voltage value so that this circuit will work, and automatically flow to ground after passing through the circuit? If so, do we need a connection to ground in the real world?

Please point out my mistake if you notice a simple error I've made, and let me know what is right.

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    \$\begingroup\$ I'm.... not entirely sure this is even a valid question. Please, work on your grammar. \$\endgroup\$ Mar 12, 2011 at 13:31
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    \$\begingroup\$ @Fake Name - Not everyone is a native English speaker! Zain's profile says they're from Pakistan. Fixing little things like this is what the "edit" button is for. \$\endgroup\$ Mar 19, 2011 at 1:54
  • \$\begingroup\$ @reemrevnivek - I would have tried to fix it if I could figure out what the question was about. \$\endgroup\$ Mar 19, 2011 at 7:05
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    \$\begingroup\$ This question was about basic circuit theory. \$\endgroup\$
    – Zain
    Mar 19, 2011 at 7:12

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The idea of a closed circuit works for low frequencies - where the corresponding wavelength is much larger than the components and wires. Kirchoff's laws hold.

Things get tricky when the frequency is higher. A sudden change in voltage propagates at the speed of light (or some good fraction of it in cables, transmission lines) and there is more current at one point than at another.

In theory, you could put a sharp-edged voltage pulse on one lead of an LED, have nothing connected to the other, and for a tiny instant in time, as the pulse passes through the LED, have enough current for it to glow. But it would be extremely brief.

So what if you send a series of pulses?

A good rule of thumb to remember is at light speed, one nanosecond is about one foot (30+ cm). LEDs and the pulse-pushing circuitry I imagine would be a few inches (or cm) and so things happen on a scale of maybe tenths of nanoseconds. You'd have to work with frequencies at several GHz.

Another problem - every positive pulse you put on the anode lead will go through the LED and add positive charge to the non-connected cathode lead. Each positive pulse will add more. That charge has nowhere to escape to - just a tiny bit can flow back as leakage current, no diode being perfect. From a physics point of view, so what? Just let the whole contraption develop a positive charge. Figure a few milliamps lasting for say 50 ns, times 5 billion times per second (just making up numbers), you quickly get to coulombs of charge, and many volts in just seconds. At a practical level, it's not very practical at all.

I wonder if it would work better to have two LEDs wired anti-parallel, and feed GHz pulses to one end of the pair and leave the other end disconnected? (I leave that thought for others to discuss.)

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  • \$\begingroup\$ @Kortuk heh. I was just saying that a question that may seem obvious to most electronics experts, can have answers that are very enlightening. I meant no offense \$\endgroup\$
    – Earlz
    Mar 16, 2011 at 1:12
  • \$\begingroup\$ @Earlz, it came off differently, no harm done. \$\endgroup\$
    – Kortuk
    Mar 20, 2011 at 3:53
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Yes, you could do that. If you use a high-voltage generator to suck charge out of one metal sphere and deposit it onto another, you would then have a potential without a complete circuit, and you could connect the two spheres together to get a current.

charge flow (current) from one charged sphere to another

This would only last for a short time, however, because the charge imbalance would decrease as you discharge them. If the spheres are oppositely charged at first, they will become neutral after the current flows. if one sphere was neutral and the other charged, then the equilibrium they reach will be half as much charge on each, instead of becoming neutral.

But what if you want charge to flow continuously from one sphere to the other? Then you need a conveyor of some type to force the charge back to the other side. In this case, you've completed the circuit with a voltage or current source.

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  • \$\begingroup\$ Wouldn't that still a complete circuit, just with some large capacitors? \$\endgroup\$
    – Nick T
    Mar 12, 2011 at 17:05
  • \$\begingroup\$ I don't think most people would consider it a circuit, since the spheres are not near each other or connected in any obvious way. The spheres could be charged in isolated rooms and then brought together, for instance. \$\endgroup\$
    – endolith
    Mar 12, 2011 at 20:59
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    \$\begingroup\$ @endolith, you are effectively charging the spheres as a capacitor in a complete circuit(taking charge from one and depositing on the other). You are then shorting it. Noone may consider it a circuit because it has no value. It is still a case of applying potential to the two elements as a capacitor, even if they are both capacitors in reference to the "ground potential. \$\endgroup\$
    – Kortuk
    Mar 15, 2011 at 6:04
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    \$\begingroup\$ @endolith, the photoelectric effect was using high enough frequency electric energy to be able to use the free space to couple energy. I am not saying you are wrong, but I think this explanation may give the wrong idea to someone new in the field trying to understand it. You are not a slouch, you know you theory, and I think you are able to use this to explain an open and not have any loss of the fact that for standard operation you are going to have a solid conductor connecting the circuit in a complete loop. \$\endgroup\$
    – Kortuk
    Mar 16, 2011 at 14:25
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    \$\begingroup\$ @Kortuk: Actually, I answered like this because I think this explanation increases understanding for someone new to the field. Current is just flows of charge, and it's important to have a gut feeling for how charge works in the absence of complete circuits in order to understand better how it works in circuits. webcache.googleusercontent.com/… \$\endgroup\$
    – endolith
    Mar 16, 2011 at 14:41
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In order for current to flow between two points (in general) the fundamental requirement is that ther must be some potential (voltage) difference between those points.

Current is physically the effect of electrons moving through your circuit (as prescribed by electromotive forces given rise to by the aforementioned existent potential difference).

If those electrons have no place to go (i.e. back to ground), and current were somehow flowing into a floating node like the LED situation you described, then you would get a build-up of electrical charge on the on the floating node that would violate energy conservation principles.

Edit

In response to the comment, think of it like this. Current will only flow from a higher potential to a lower potential (again electromotive force governs this). If a current were able to flow into a floating node, charge would build up there, and the result of this would be that the potential (voltage) at that node would increase. Eventually it would increase to a point where there was no potential difference between that node and the source node, and current would no longer flow.

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  • \$\begingroup\$ Please explain the line then you would get a build-up of electrical charge on the on the floating node that would violate energy conservation principles.How the violate energy conservation principle will voilate.thanks! \$\endgroup\$
    – Zain
    Mar 12, 2011 at 6:49
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There would be some capacitance between the unconnected terminals of the LED and battery, so a little current would have to flow before the unconnected terminals before their potentials was equal to those of the connected terminals. In practice, however, that level of capacitance would likely be so small as to be impossible to measure much less use.

Note that while a one farad capacitor can accept a whole coulomb of electrons (one amp for one second) before building up a volt of potential difference, that doesn't mean a coulomb of electrons flowing in with none flowing out. In order for one side of a capacitor to accept a negative charge, the other side must have a balancing positive charge. Essentially what happens is that as electrons flow into the negative side, their nearby presence makes electrons want to leave the positive side. Likewise as capacitors leave the positive side, their relative absence makes electrons want to enter the negative side.

It's possible for substantial AC electric currents to flow without a complete circuit, if there's enough capacitance. Two leads some distance apart, however, aren't going to have enough electrostatic effect to yield any useful capacitance, though.

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