My current project... need some advice. :)

[Marks256]

I need to make a LED display thingy. All i need to know, is will i beable to turn an LED on RELIABLY on the end of about 20' of cable running it on DC, or should i use AC?

If you need me to explain more, please say so. I am in a rush right at the moment, so i can't get into detail.

 

[audioguru]

An LED operates on DC. An ordinary size LED has an absolute max current of 30mA so 20mA to 25mA is usually used. 20' of cable has nearly no resistance so will be fine. Use a power supply voltage higher than the voltage of the LED and calculate a current-limiting resistor for it.

A red LED is usually about 2.0V, a green ordinary LED is about 2.5V, and a very bright green, blue or white LED is about 3.5V.
Therefore if a 6V battery or 6V regulated supply is used to power a single LED, the current-limiting resistor value should be 160 ohms for a red LED, 150 ohms for an ordinary green LED and 100 ohms for a very bright LED.

 

[the dude of dudes]

An LED operates on DC. An ordinary size LED has an absolute max current of 30mA so 20mA to 25mA is usually used. 20' of cable has nearly no resistance so will be fine. Use a power supply voltage higher than the voltage of the LED and calculate a current-limiting resistor for it.

A red LED is usually about 2.0V, a green ordinary LED is about 2.5V, and a very bright green, blue or white LED is about 3.5V.
Therefore if a 6V battery or 6V regulated supply is used to power a single LED, the current-limiting resistor value should be 160 ohms for a red LED, 150 ohms for an ordinary green LED and 100 ohms for a very bright LED.

I am still learning, and would really appreciate it if you could show your math. I am especially interested in the effect resistors have on voltage (aside from a resistor network). In my mind 6v / 100 ohms = 60ma. Thanks for your time!

 

[audioguru]

The LED is 3.5V and has a 100 ohm current-limiting resistor in series with it. 6V is across both of them in series so the resistor has 6.0V-3.5V= 2.5V across it. Therefore the current is 2.5V/100 ohms= 25mA. If there is no LED then the 100 ohm resistor would have the full 6V across it and the current would be 60mA.

An LED is a diode. Diodes have a fairly constant voltage across them. It is about 0.7V for an ordinary diode, about 2.0V for a red LED, about 2.5V for an ordinary green LED and about 3.5V for a very bright green, blue or white LED.

If you connect an LED to a strong voltage source then either the voltage is too low so the LED won't light, or the voltage is too high causing an extremely high current which will blowup the LED. A higher voltage with a current-limiting resistor must be used.

 

[Sceadwian]

Diode's aren't generally pictured as resistors, but do have a constant forward voltage drop which can be viewed as a resistance. If you take your original 6 volts, and subtract the 3.5 volt's that the diode drops you end up with only 2.5 volts on the other side of the diode, the 100 ohm resistor will give you 25ma's of current flow through that diode. What thickness wire were you planning on using? There are plenty of sites with links to the ampacity of various wire thickness's. You're going to get a certain amount of voltage drop over 20 feet of wire. For example, 24AWG wire (Standard UTP ethernet cable) Has a resistance of, .0302 ohms per foot. If you use the same gauge wire on both VCC and GND you're going to end up with .0604 ohms per foot, or 1.2ohms for a 20 foot VCC/GND loop. On a 100ma load the voltage drop will be about .05 volts, not a lot.

 

[Marks256]

Ok, that is all i needed. Thanks.

Oh, and i know that LEDs are DC. That doesn't mean you still can't pulse power to it... like, say, 60hz or even 100hz.

 

[Sceadwian]

No, but if you violate the diode junction temperature you get smoke. And at it's extreme's for a spike diodes will fail fast and not smoke. The semiconductor materials oxidize under the instantanious combustion to the point where they become non-conductive. Because the power cycle is so slow compared to the voltage failure mode of a diode you don't get flames. Screw up once and your diodes will fail short circuit and vaporize all it's material, light the ceiling on fire and generally just destroy people and materials as fast as it can. Hacks kill.

 

[hjames]

I don't know about the "oxidize and become non-conductive", but I've definitely had a few LEDs where the little bitty bond wire vaporizes leaving everything else intact. More commonly though, overheating, charred leads and cracked epoxy are the usual remnants. That and burnt fingers.

 

[Hero999]

I doubt they oxidise because there's no free oxygen. Normally the PN junction melts and fuses together or the connecting leads melt.

 

[Roff]

Ok, that is all i needed. Thanks.

Oh, and i know that LEDs are DC. That doesn't mean you still can't pulse power to it... like, say, 60hz or even 100hz.

If you use AC, you need to realize that the reverse breakdown voltage of LEDs is on the order of a few volts, and you need to prevent significant reverse current (a few microamps) from flowing. To do this, you can either use a series diode (with breakdown voltage higher than your AC peak), or you can add an antiparallel shunt diode (or another LED), which will roughly double your power dissipation vs the series diode (generally not a problem).

 

[Hayato]

Ok, that is all i needed. Thanks.

Oh, and i know that LEDs are DC. That doesn't mean you still can't pulse power to it... like, say, 60hz or even 100hz.

Yes, you can use pulsed DC to make some sort of energy saving.

But as Ron said, you need to add a series diode (1N4007) to prevent the reverse current causes a breakdown (specially if you are using the 60Hz from the energy lines).

 

[Sceadwian]

Even if the average power through the device is within it's operating range, voltage or current spikes that cause even momentary heating beyond the junction temperature allowance will destroy the device.