2
\$\begingroup\$

I'm planning to use a ULN2003A 7-channel Darlington driver (datasheet) to control 4 to 7 LEDs in common-anode configuration from a Teensy MCU:

enter image description here

(The "trident" is a common-anode RGB LED).

The ULN2003A has a set of "clamp diodes" or "free wheeling diodes" which I gather are present to protect against voltage spikes when switching inductive loads. Those diodes are common-cathode out to pin 9 on the IC:

enter image description here

My questions are:

  • Do I need to connect pin 9 to anything if I'm just driving LEDs? (My guess is no.)
  • If I were driving motors or something and needed diode clamping here, what would I connect pin 9 to? The positive voltage of the motor side?
\$\endgroup\$
0

1 Answer 1

4
\$\begingroup\$

It's in the datasheet - https://www.ti.com/lit/ds/symlink/uln2003a.pdf?ts=1596498695539&ref_url=https%253A%252F%252Fwww.google.com%252F

8.4.1 Inductive Load Drive When the COM pin is tied to the coil supply voltage, ULN2003A device is able to drive inductive loads and suppress the kick-back voltage through the internal free-wheeling diodes.

8.4.2 Resistive Load Drive When driving a resistive load, a pullup resistor is needed in order for ULN2003A device to sink current and for there to be a logic high level. The COM pin can be left floating for these applications.

Your LEDs are not 'inductive', so you don't need to use that pin.

Motors are one possibility. Driving relay coils is probably the more likely scenario they were considering.

\$\endgroup\$
4
  • 2
    \$\begingroup\$ The phrase "the datasheet" is doing interesting work here, but thanks! \$\endgroup\$ Commented Aug 4, 2020 at 0:01
  • 1
    \$\begingroup\$ @RussellBorogove Ah! You're right, it's not shown in the ST datasheet! I had linked the Texas Instruments version. LOL You can be sure they're gonna effectively be the same. ST dropped the ball there, didn't they? \$\endgroup\$
    – Kyle B
    Commented Aug 4, 2020 at 0:04
  • 1
    \$\begingroup\$ Yeah, but doubtless this is something that's obvious to an experienced EE -- I'm a novice hobbyist, so just glad that my guesses were on the right track. \$\endgroup\$ Commented Aug 4, 2020 at 0:05
  • \$\begingroup\$ @RussellBorogove Your "guesses" were right on the money ;) Well done! \$\endgroup\$
    – Kyle B
    Commented Aug 4, 2020 at 2:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.