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Good Day,

I am trying to build an IR system that works via the Arduino. So far, it works great. However, I want to have one IR Led on the back of my device and one in the front. However, the Arduino pin only supplies 40mA max.

Hence, when I used on IR Led, the range was still pretty decent. However, upon connecting two of the IR Leds to the same pin, it was very weak (1m range only). I was thinking that to solve this problem, I can get a cheap 7400 series AND gate. Then, I can feed the IR Line from the Arduino and split them into the two AND gates, and the AND gate will drive the IR Leds, and both will get full power (I am not sure of the rating output of a 7400 AND gate)

Is this a good idea? Any idea what a 7400 series AND gate output max current is?

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While @JimDearden's answer provides a suitable approach for operating the two infrared LEDs in parallel as desired, my answer addresses the specific question asked, to resolve some possible misunderstandings that may have motivated the question:

  • An AND gate is used for taking two or more inputs, and providing a single output representing the logic AND operation on those inputs. The purpose underlying the question is the opposite: To take a single input (from the Arduino pin) and drive two outputs. Hence, no rationale for using an AND gate
  • The classic 7400 series TTL logic gates have recommended operating conditions of merely 16 to 22 mA per output (current sink) depending on manufacturer. See the 7408 AND gate for instance. In other words,

    1. They will not sink enough current for two IR LEDs if the output is used as a low-side switch, with the LED between output and the (presumably) 5 Volt supply. (actually they will sink up to 55 mA in short-circuit, but that is beyond recommended safe operating area)
    2. Many 74 series gates will not provide a high side output of any significance for turning on LEDs connected between output and ground - they are designed to be used with a pull-up resistor on the output pin.

    Note that some variants of the classic 74 series logic parts can both sink and source current, and there are even some that can deliver outputs in the ampere range. For instance, the 74595 shift register has an equivalent, the TPIC6595, which sinks 1.5 Amperes pulsed current per pin.

  • If the intention is to operate LEDs from the output of a 74 series gate, this won't work the way the output of the Arduino works: The LEDs would have to be connected on the Vcc side, with cathode at the logic gate output, and the output logic will need to be inverted in code.

So, in short, the 7408 AND gate is neither a viable nor a justifiable approach to solving the problem.

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  • \$\begingroup\$ Can I ask.. Would it be better to use an op-amp. Say an LM358. I would be setting it at unity gain, so what would the max current of an LM358 be? \$\endgroup\$ – Raaj Jul 2 '13 at 16:01
  • \$\begingroup\$ You can use a buffer like an SN7407, able to sink up to 30 mA per output. Or SN74ABT125, able to sink up to 64 mA per output. Notice that most digital buffers are able to sink more current than they can source, and the '07 is actually open collector so it cannot source current at all. \$\endgroup\$ – The Photon Jul 2 '13 at 17:21
  • \$\begingroup\$ @Raaj LM358 can source 30 mA and sink 20 mA (see the datasheet) - still not enough. Also an op-amp is just not the right tool for this job, a 50-Paise 2n2222 BJT is far more suitable. \$\endgroup\$ – Anindo Ghosh Jul 2 '13 at 17:21
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Why not just drive the LEDs with a transistor (NPN or MOSFET) - they have a much higher current drive capability.

enter image description here

Values shown give about 70mA total (35mA per LED)

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  • \$\begingroup\$ LEDs in parallel.... gross! \$\endgroup\$ – vicatcu Jul 2 '13 at 13:52
  • \$\begingroup\$ @vicatcu I don't see a problem here especially as they may not both be run in series from 5V that easily (depends on forward voltage of course). \$\endgroup\$ – Andy aka Jul 2 '13 at 14:03
  • \$\begingroup\$ and what happens if the supply is 3.3 volts? \$\endgroup\$ – JIm Dearden Jul 2 '13 at 14:14
  • \$\begingroup\$ I don't see a problem even with a 3.3 Volt supply. Typical IR LEds have a Vf of 1.2 to 1.4 Volts. The Vce(sat) of the 2n2222 is specified in the datasheet as a maximum of 0.3 Volts at Ib = 15 mA, Ic = 150 mA. Thus there will still be over 1.5 Volts headroom beyond Vce(sat) + Vf for a parallel connection. Even a series connection would work (1.4+1.4+0.3 = 3.1 V), if we are restricting this to infrared LEDs, not visible colors. Not much headroom, but it'll work. \$\endgroup\$ – Anindo Ghosh Jul 2 '13 at 14:28
  • \$\begingroup\$ @vicatcu Since each LED is separately current limited with individual resistors, where would you see a concern with running them in parallel as in this schematic? \$\endgroup\$ – Anindo Ghosh Jul 2 '13 at 14:30

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