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Using the example half way down a web page titled "Using the BBB to turn on a transistor" I wanted to know if a similar layout could use a Darlington to drive multiple LEDs instead of individual 2N3904 transistors? I'm thinking based on other forum posts I can and this is what I would need. Can someone please confirm?

I believe to mirror the examples layout of the LED after the transistor this Darlington would work.

OR

I could put the LED before the transistor and use something like this?

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  • \$\begingroup\$ Normally, you would put the LED and resistor in series, between the collector of the transistor and the positive supply. The ULN 2803 is often used for this task, but you don't really need the current capability of a darlington just to drive an LED. The Toshiba part you linked is a high side driver, and typically requires a 5 volt input as a High - so it isn't suitable for use with most microcontrollers. \$\endgroup\$ – Peter Bennett Feb 1 '14 at 16:59
  • \$\begingroup\$ So without getting into use I2C is there another solution that is simplistic as a transistor but comes as an IC? I was looking at the LM3046 but it is the wrong form factor for my proto board. ti.com/lit/ds/symlink/lm3046.pdf I would imagine you could do this with a MOSFET, if I understand correctly how they work but don't know if those come in the form of an IC? \$\endgroup\$ – ScottEH Feb 1 '14 at 17:26
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Whether using a darlington or not I wouldn't use the circuit you show - see below - your circuit alongside a better way to drive the LED: -

enter image description here

Looking at the right hand picture, the transistor is easily turned on to saturation by the GPIO output because the emitter is directly grounded with the LED and series current limiting resistor in the collector.

On your design, to switch on the transistor sufficiently you have to overcome the LED forward voltage (about 2V) before the base-emitter junction can start to conduct then, you need another 0.6 to 0.7V above that. If your GPIO is 3.3V then you'd get away with it but anything lower than 2.7V and you aint gonna get the brightness from the LED you think you should.

Another option is to use your circuit and reverse the position of current limiting resistor and LED - you won't need the base resistor for this either so it's a bit simpler. Now, the GPIO sets a voltage on the base (say 2.7V) leaving 2v across the newly placed emitter resistor - if that resistance is 100 ohm, the 2V will ensure that 20mA flows thru the LED in the collector and providing the collector voltage is about 4v or greater, 20mA will always flow - don't make collector voltage too high though as this will warm up the 2N3904 because it is now current regulating and (say) on a 12V supply it will be dropping 8v across it which means a power dissipation of 8 x 20mA = 0.16watts - not too bad but don't go much higher if no heatsinking is used.

The same applies when using a darlington except the transistor "imposes" about another 0.7V between collector and emitter when conducting meaning it will still work on a 5V supply but not much lower.

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The ULN2803A (or ULN2003A) is a simple and cheap way to drive up to 7 or 8 relatively high current loads.

Note that a darlington drops about 0.8V at 100mA. enter image description here

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In searching for the right Darlington for my project I found a TI model ULN2003LV designed for low voltage/lower power applications. The data sheet specifically mentions that it is compatible with 3.3v and 5v microcontrollers.

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