I'm a newbie, and am trying to drive several independent LEDs using a transistor array in a 5-volt CMOS circuit. Here's what I've come up with, showing only the first LED:


In the schematic, LED_ENABLE is the output of some logic IC in the circuit. I'm assuming that the LED has a voltage drop of 1.8 V and requires 20 mA of current.

Does this seem sane, or have I missed something? Specific questions:

  1. I've tied LED_ENABLE directly to the base pin, because my understanding is that the ULN2003A's internal resistor is designed to let me do so. Is that safe to do without thinking further? It seems like the value of the resistor should need to depend on \$I_{CE}\$ and the transistor's gain, so I'm not sure how this can be one size fits all.

  2. I chose the resistor voltage by calculating \$\frac{5 V - (0.9 V + 1.8 V)}{20 mA}=115 Ω\$ and rounding up, since the data sheet for the ULN2003A says that the collector-emitter voltage at saturation is 0.9 V for \$I_C=100 mA\$ (the lowest listed current). Is this the right calculation?

  3. If I understand correctly, I don't need the COM pin. Is it sane to leave it floating?

  4. And how about leaving the other base and collector pins floating? Maybe they should be tied to ground?

  5. Anything else obvious I've missed?

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    \$\begingroup\$ Everything you say you said and did sounds fine. Either a component is dead OR you atre noy conncting as you think you are. \$\endgroup\$ – Russell McMahon Apr 22 '15 at 11:56
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    \$\begingroup\$ I'd just like to say, thank you for a well formulated, coherent, and to-the-point question. \$\endgroup\$ – Nick Johnson Apr 22 '15 at 14:00

To answer your questions:

  1. Correct: it is ok to tie directly to tbe base pin, given the internal resistor. The reason one size fits all, is the combined gain of the darlington pair is so high, and that the output pins' current ratings mean any current at the output within specs will be sustained by the internal resistor at the base.
  2. Yes your calculation looks good.
  3. Look at the circuit diagram below which shows how the darlington pairs are implemented. You can leave COM floating.
  4. You can leave the other pins floating. They should have no bearing on your circuit, since as per the diagram below, they are independent (as long as you do not pull COM low).
  5. Perhaps you have not wired it up correctly, or the chip is dead. For the sake of troubleshooting, try connecting COM directly to ground, which would be equivalent to all transistor pairs saturated on. The LED should light. Perhaps the LED's polarity is reversed. Sanity check by connecting pin 1C directly to ground.

enter image description here

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    \$\begingroup\$ "Try connecting COM directly to ground". COM pin should not be connected to GROUND! It's supposed to be connected to the 'Load Supply Voltage'. \$\endgroup\$ – m.Alin Apr 22 '15 at 11:08
  • \$\begingroup\$ Thanks for the quick point by point answer! To be clear: I'm designing, not trouble-shooting; I haven't built this yet. I just wanted to make sure there wasn't some problem I missed in my design. \$\endgroup\$ – jacobsa Apr 22 '15 at 11:09
  • \$\begingroup\$ @m.Alin That would have no effect. No current would flow anywhere, given the diode polarity \$\endgroup\$ – CL22 Apr 22 '15 at 11:09
  • \$\begingroup\$ @Jacobsa, in that case, your circuit looks perfect. \$\endgroup\$ – CL22 Apr 22 '15 at 11:10
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    \$\begingroup\$ @Jodes The diode is there to protect the transistor in case you are switching an inductive load, such as a motor. It's called a flyback diode. In case of driving a LED you could leave the COM pin unconnected, but you should not connected it to GROUND! \$\endgroup\$ – m.Alin Apr 22 '15 at 11:20

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