OK, I soldered the wires in my mouse for the LeDS. I wanna put one Led on each side of the mouse. 5v+---Resistor (75ohm)---LEd (3.7 20ma)----Led (3.7 20ma)---ground 5v Does this looks right? thanks (I will have pics when I am all done )
My EE courses were A LONG TIME AGO, but it would seem adding one resistor in series would decrease the amperage not the voltage. Perhaps the lower amperage would prevent the 5 volt source from burning out the LEDs. To convert 5V into 3.7V with 20ma you'd need a circuit something like this: Caution: ASCII art Code: -------------------------- | | | | | 65ohm R | | 5V |--------------| | | | | | | | 185ohm R 3.7V | | | | |--------------| |------------------------| The resistor values would vary with the VDC source amperage. There's a online tool at http://home.att.net/~BillBowden/Applet.htm Not sure if this helps, but it was intersting to research. Cheers, Mike
Or better yet check out linear's LED calculator at http://www.bit-tech.net/article/68/. Amazing what you can find by reading around here...
If I used linear's tool correctly, it would take a 65ohm resistor for a 3.7V 20ma LED ona 5V source. 100ohm is the "next standard 5% resistor" so perhaps you're in the ballpark. I'm unsure how having the LEDs in series will affect the power to the second LED, it may be too dim. Of course, you could always hook it up and see if it blows up. That method works for me as long you have spares. <joking>
No it looks like you have them in series not parallel as I tried to depict. But it looks like form linear's tool, that a parallel resistor circuit is not needed. Just add a resistor in series (like you originally showed) and the lower amperage would take care of it. It looks to me like 75ohm is in the ball park. You may have to drop to a lower resistor because you have two LEDs in series. The 75 ohm resitor will drop the amerage to the first LED (V=IR) and there should be an additional amperage drop across the first LED further reducing the current to the second LED. As in all components there's an allowed tolerance so yo may find a happy medium that will power both LEDs. I imagine worst case you'll just blow one or both LEDs if you get it wrong. Cheers, Mike
You have missed the obvious, 2 x 3.7V leds will require AT LEAST 7.4V to work. You will need to wire the leds between 5V and 0V seperately, each with their own 65 ohm resistor. (or there abouts)
From the Basic rules of electronics sticky thread: Wouldn't that imply that voltage for series LEDs is not straight addition? Of course it may not be possible to stay with tolerance for the first LED and still power the second one in series --> back to the parallel circuit. Espescially if the drop is more toward the 5V range. Would a 3.7V LED produce a 3.7V drop?
The absolute voltage across an LED will vary with its current. Usually manufacturers quote a "typical" Vf along side a "typical" If. An LED with a "typical" Vf = 3.7V will drop "about" 3.7V across it when the current through it is "about"= If (usually 20mA). If you use these figures you won't go far wrong. Like most electronic component parameters they are only a guideline.
So i should solder 2 wires each to my 5v and ground wire and run 2 different circuits? so each led has its own 2 wires? L=LEd R=resistor (75) G= Ground \= spliced wire 5v | \ |\ | \ | \ | | R R L L | / | / | / | / | / | | G
The voltages do add up directly. The drop is not due to a resistance, but due to the way that an LED works. <semiconductor physics> There are several ways of generating light from an electric current. The first way, which is horrendously inefficient, is to basically use a piece of resistor wire. Pass lots of current through it, and it will get hot, and start to glow as it gets red-hot. Then it burns through. So, surround it in an inert gas (eg Argon) so that it can not burn through, and you can make it brighter by passing yet more current through it. Hold the argon in some glass around the wire, and lo, for we have a light-bulb. Very inefficient, as the wire still breaks eventually, they produce more heat than light, etc. The second way is to apply a large voltage to a tube full of gas, which causes the gas to emit electrons. Coat the inside of the tube with phosphors, and lo, for we have a flourescent tube. Still not brilliantly efficient, but better than a bulb. The third way is to take a completely different approach. By giving electrons in a semiconductor some energy (typically several eV, depending on the semiconductor and the colour of light you want to get), they are promoted to higher energy levels. However, they don't want to be at higher energy levels, so they fall down again. In doing so, they loose energy, which is emitted as a photon, which has its wavelength (and thus colour) determined by the energy drop of the electron. This is an LED. Since the energy drops of the electrons are always very similar, this is why the colour of light from an LED is always concentrated on one particular colour (wavelength). You can feel that they are much more efficient as they generate little to no heat in use, and they don't die like lightbulbs, but continue to work for many years, regardless of how often they are switched on and off. The specific voltage is required to make enough electrons be promoted to higher energy levels to make a reasonable light output. (There are also some other semiconductor physics reasons why you need a specific voltage, since you are dealing, after all, with a diode, which just happens to emit light also). </semiconductor physics> It is typically advisable to not go too near the maximum ratings. The voltage is required, but you don't need all of the current. If the specs are 3.7V and 20mA, calculate for 15mA, and go to the nearest E12 resistor value (E12 is the standard series of resistor values).
By the way, my tool will work this out for you. You just need to know to add the current for parallel combinations of LEDs, and add the voltage for series. If you enter Vf > Vs, I'll let you know about it...
