I’m an electronics newbie, so I’m trying to use a few IC's to check my understanding.

At the moment I’m trying to use a 74LS259 8 bit addressable latch. I’m trying to get it working without a micro-controller first to ensure that I understanding the wiring 100%.

But I am having a problem understanding the E and C pins … (see below)

I’ve wired it as follows:

  • A0-A2 - Address inputs - wired to a switch and the +5V Q0-Q2 - Outputs
  • Each wired to a LED and a 220ohm resistor
  • Q3-Q3 - Outputs - Not Wired (Is this a problem should I wire them to the -5V instead)

The issue I has is understanding how to wire the E (enable) and C (clear) pins.

The datasheet says that to select the “addressable latch” mode I should have:

  • E - Low
  • C - High

So I’ve wired E to 0V and C to +5V.

I then try to “write” a bit by taking A0 high and then D (data) high but this does not work.

I suspect that I have wired E and C wrong. I’ve searched around but I can’t find any circuit diagrams that could help me…

Can anyone assist? What am I doing wrong here?

Here is the circuit diagram at CircuitLabs:

enter image description here

  • \$\begingroup\$ Did you wire to -5V or to 0V as in the diagram? The circuit diagram is correctly wired. \$\endgroup\$
    – jippie
    Jan 2, 2013 at 8:58
  • \$\begingroup\$ A picture of your setup would be a good addition. \$\endgroup\$
    – jippie
    Jan 2, 2013 at 9:00
  • \$\begingroup\$ You really dont want to see a picture :). All the electronics stores have been closed over the holidays so I cobbled a bread board together out of two old IDE cables. Once I get a breadboard I'll post a pic of the working version. \$\endgroup\$
    – Andre
    Jan 2, 2013 at 9:19
  • \$\begingroup\$ As per Oli (below) I was incorrectly calling it -5V rather than ground. So then it seems like it is wired correctly apart from perhaps E & C \$\endgroup\$
    – Andre
    Jan 2, 2013 at 9:26

1 Answer 1


You don't show the connections on your schematic, but I'll assume you actually have them wired up to the switches you mention.
It looks as if you have C and E the wrong way round - they are active low inputs, so you need to tie C to +5V, and E to GND.
Do not wire the unused outputs directly to a power rail (e.g. +5V or Ground) as this could cause a direct short if the logic level is set to the opposite rail (e.g if you have the output wired to ground, and you set it high, then a lot of current will flow and potentially damage your chip)

So set your data pin D to the desired logic level for your output (e.g. high to light your LED), then A0 and C high, then pull E low. Q1 should light up (if it doesn't check the LED is the right way round, power is present, etc)

Also, if you only have a single supply like is shown, then the low side is generally referred to as ground, not -5V.

The bottom section is the relevant table in the datasheet for addressable latch mode:


You can see you must keep the clear pin high (tied to +5V) otherwise it will keep the outputs low (i.e. "cleared")

Also note this bit (the mention of Enable and Clear being active low - denoted by the bar above the letter):


  • \$\begingroup\$ Re ground: Thanks, I'll remember that. Re unused: Ok, I'll leave them unconnected for now Re C and E & D: I'll give that a go. I'm still a bit confused about the datasheet and the mode selection as it contradics this I think. I'll try this again tonight and see what happens \$\endgroup\$
    – Andre
    Jan 2, 2013 at 9:23
  • \$\begingroup\$ BTW I also need to take D high right? So I take A0 high to address Q0, then D high to set the addressed pin (Q0) high and then I take E low to activate the change? \$\endgroup\$
    – Andre
    Jan 2, 2013 at 9:29
  • \$\begingroup\$ Sorry, yes you set the D (data) pin to the desired logic level for your output, set the address pins, then pull E low to activate the change. I updated the answer to mention D. \$\endgroup\$
    – Oli Glaser
    Jan 2, 2013 at 9:32
  • \$\begingroup\$ Fantastic thanks. I think I'm begining to see what the datasheet means now. I'll give this a go as soon as I can this evening. \$\endgroup\$
    – Andre
    Jan 2, 2013 at 9:42

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