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I need to display the temperature acquired by my LM35 on a 2 digit 7 segment display. What I have is LTD383-R2 Common Anode Display. Datasheet here.

The display has 1 Anode pin + 14 Cathode pins for he 14 segments.

Any idea on the best way to drive this (with least components)? I am using Atmega328.

The options that came to mind were

  1. Use a decoder. Google search came up with DS8669 from TI, but it seems that IC is now obsolete.
  2. Go the usual route of using a ULN2803 darlington array IC + 74HC595 shift register. But this would mean using 4 IC's in total (2 for each digit of the display)

Thanks !

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  • \$\begingroup\$ If you don't have enough high-drive outputs to directly drive it from the Atmega, you might want to consider a different display. You could also use the '595s to directly drive the segments. Variants exist with DMOS output transistors (not all that cheap). \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Sep 27, 2014 at 7:18

3 Answers 3

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Another 16 bit shift register with constant current sinks is the TLC 5926, datasheet available here

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  • \$\begingroup\$ Single ic solution, though I'm partial to the i2c version myself. \$\endgroup\$
    – Passerby
    Commented Sep 27, 2014 at 20:12
  • \$\begingroup\$ Thanks ! Looks like just what I need :). Whats the part number for the i2c version of this chip? \$\endgroup\$
    – Ankit
    Commented Sep 28, 2014 at 12:12
  • \$\begingroup\$ I don't know whether it's second-sourced or not, but from TI, it's the TLC59116. \$\endgroup\$
    – EM Fields
    Commented Sep 28, 2014 at 17:09
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Options:

  • drive the segments directly from your uc. using the 200mA figure from the absolute maxima (which is strictly speaking wrong to do) you have ~ 14 mA per segment. You won't see the difference between 14 and 20 mA.

  • if you have the uc pins but want to use the full 20mA you can use two ULN2003 chips.

  • if you don't have enough uc pins you can go the 595 route. It has a total supply current limit of 70 mA (again: only mentioned in the wretched maxima section), so you have 10 mA per segment. Would be OK for me.

  • if you really want this display AND you have only 3 uc pins AND you want the full 20 mA per segment, sigh, you must go the 4 chips solution (or something equivalent, using a chip that is probably more expensive and more difficult to get).

  • ditch the display and get a more standard one with 2 anodes and shared cathodes (or the other way round), so you can multiplex. If you get a low-current version (use high-brightness displays that work perfect with 1 or 2 mA per segment) you can drive it directly from 10 uc pins.

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  • \$\begingroup\$ Thanks for laying out the options. Ideally would love to g with the last one, but unfortunately here in India they are not easy to acquire unless in big quantities. \$\endgroup\$
    – Ankit
    Commented Sep 27, 2014 at 10:58
  • \$\begingroup\$ Why not go with the first option? The difference in brightness will be barely noticeable. \$\endgroup\$
    – Wouter van Ooijen
    Commented Sep 27, 2014 at 12:45
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I personally use the STP16CP05 for plain shift-register operation, or the TLC59116 if I want PWM control of each segment.

The STP16CP05 is the same kind of shift register as a 75HC595 but each output is a constant current sink.

The TLC59116 is an I2C PWM driver which again is a constant current sink.

Both have 16 outputs for driving all 16 segments of your LED display with only 1 resistor and a couple of deccoupling capacitors. Can't get fewer components than that really.

Here's an example with the TLC59116 I sell on eBay.

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  • \$\begingroup\$ How about some full disclosure, that's your auction Majenko... \$\endgroup\$
    – Passerby
    Commented Sep 27, 2014 at 20:16
  • \$\begingroup\$ I said I use them didn't I? \$\endgroup\$
    – Majenko
    Commented Sep 27, 2014 at 20:19

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