A single LED on an dual DC supply?

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Thanks guys for your invaluable inputs! I've been doing some thinking & my thoughts go like this.

Electrons in a conductive material travel freely,yet the movement of these molecules depend or restricted by the inherent friction of this conductive material. This friction,then can be called the "resistance" -R- of that particular conductive material. Now, if we then apply a force to this material, then it would move even faster & more force we inrodduce, faster will these electrons move, yet the resistance of the material remains constant right?

Lets say that this force is V,voltage. So, when we apply more voltage, then the electrons too should move faster. However, the resistance will not allow all the voltage to pass through & consequently a certain amount of voltage or is dropped across this resistor. So my question is, where does that dropped voltage go & what happens to it? As HEAT perhaps?

Another simple analogy will be the lowly light bulb! What makes a 60w bulb brighter than a 40 watt one? Doesn't a 60w consume more power (W)= current x voltage than a 40w? Why? Then wouldn't this have the same effect as using a more current demanding brighter LED than a less brighter low current one?
 
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...yet the resistance of the material remains constant right?
At a constant temperature, yes, the resistance may remain constant. The resistance is caused by charge carriers colliding with other fixed particles in the conductor. I'm not 100% certain so don't quote me.
...where does that dropped voltage go...?
I mentioned it before in post #39... current isn't across a component and voltage isn't through a component. You have those concepts backwards. Voltage across points A and B allows current through from A to B.
The electrons don't move faster; the voltage causes more electrons to move. Once you understand the above, that voltage is across two points, you may better understand that it doesn't really go anywhere.
Perhaps the simplest example uses a capacitor. Placing a voltage across the two plates causes an excess of electrons on one plate (the negative side) and a deficit of electrons on the other plate (the positive side). If a short is then placed across the capacitor terminals, the excess electrons on the negative side will move to fill in where electrons are lacking on the positive side. The voltage hasn't gone anywhere, it has only more or less balanced itself among all the available atoms.
An electron must acquire energy to jump from one atom, and releases energy when it falls into a hole in another atom. At least some of that energy is released as heat.
What makes a 60w bulb brighter than a 40 watt one?
With both operating at the same voltage, it is an increase in current that results in the increase in watts. With more current the lamp filament can glow hotter and/or brighter. LEDs are different than incandescent because in this case, when an electron falls into a hole with an LED, some energy is released not as heat but as a photon of light. More current means more electrons falling into holes.
This is all in the physics category moreso than benchtop electronics. Look for basic semiconductor tutorials online or in a textbook. It will explain better than I and will show why certain elements like silicon and germanium are best suited for semiconductor electronics, and why other elements such as boron or arsenic are best suited for achieving desirable semiconductor characteristics.
 
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The OT asked for a single multicolour LED to be placed in a dual DC supply. Now my question is a bit simpler I hope. I'd like to add one single "power on" led to a dual supply like this one: P05-Mini

Where in this schematic can I put one single power on LED? Between V+ and GND I assume (with a resistor). Before or after the regulators?
 
The OT asked for a single multicolour LED to be placed in a dual DC supply. Now my question is a bit simpler I hope. I'd like to add one single "power on" led to a dual supply like this one: P05-Mini

Where in this schematic can I put one single power on LED? Between V+ and GND I assume (with a resistor). Before or after the regulators?

You can put the LED (with a resistor) from"+" to GND, from GND to "-" or from "+" to "-". Before or after the regulator(s).
A LED you don't want to shine too bright takes 2-3mA, no more. The LED current is unimportant for the power consumption.
The current controlling resistor may take a bit more than expected power loss because almost all the voltage will be across it. About 3mW per volt.
 
thanks fauxfrench! so it seems that it really doesn't matter where i put it? but will it cause an "imbalance" between the rails if i "take" voltage / current only from one rail?

In theory yes, but why do you think that the rails are loaded exactly equal (within 3mA) with the normal load? Only with two same value precision resistors on the two outputs will the loading be the same.
Any amplifier you use on the rails will not draw (exactly) the same current from "+" and "-".
Such symmetrical regulated power supplies perform as two substantially independent power supplies. A positive and a negative. Say we use 1.5A LM317/337 regulators - you can draw 1A from one rail and no current from the other for long without a problem. The voltages remain constant and numerically the same if designed for equal values.
Emotionally, I understand your feeling of asymmetry/imbalance and if you make a virtual unregulated midpoint of a higher voltage, asymmetrical loading will influence the midpoint voltage.
 
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The electrons don't move faster; the voltage causes more electrons to move.


Whoops, mistake there, all charge carriers react to the applied electric field and move faster for a stronger field on average. Actually the random thermal motion is orders of magnitude larger, but only the concerted net drift in the direction of the field is visible externally as a current(*), and this "drift velocity" is an average.


(*) actually the thermal motion is visible as Johnson noise, but this is a tiny.
 
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