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I'm hoping someone can help a noob out here. I've been reading a lot of data sheets of late looking for chips that will work for a small project. In short I need to drive a number of outputs (mainly LED) at 5 V 20 mA. The load is between the IC and ground (source current?) and it can't be changed. There is a ready made solution and the supplier insists it will be OK but having read the data sheet I don't think it will.

My understanding is that is has a max of 35 mA per pin, but a total of 70 mA for the IC. Am I wrong? Would really appreciate some input.

Data sheet is here:https://datasheet.octopart.com/74HCT573N,652-NXP-Semiconductors-datasheet-9888300.pdf

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ You are correct. Get on to your supplier and ask them to justify their thoughts. \$\endgroup\$
    – Andy aka
    Commented Jun 9, 2018 at 12:03
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    \$\begingroup\$ They said: "70mA is the supply current (chip) not the output current. As is stated in the below table the output max current is 35mA for each output, being the nominal current 20mA, enough to control a LED in every output." \$\endgroup\$
    – Nik Middleton
    Commented Jun 9, 2018 at 12:42
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    \$\begingroup\$ And that supply current is also the main part of the current that feeds outputs. \$\endgroup\$
    – Andy aka
    Commented Jun 9, 2018 at 15:00
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    \$\begingroup\$ Could you post a schematic showing the IC and LED's? This website has a built-in schematic editor. You can edit your question and then click on the icon that looks like a schematic symbol, and it will load the schematic editor. I would like to know the total number of LED's and about how much current is ACTUALLY going through each one. Maybe you will be OK afterall, or maybe a simple part change can fix everything up. \$\endgroup\$
    – user57037
    Commented Jun 9, 2018 at 17:04
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    \$\begingroup\$ Hi, basic schematic added. Q0-7 are 8 outputs and attach to equipment that requires 20 ma. It's not always LED's but they are non inductive. I have no control over end devices. My belief is that turning on all 8 outputs at 20 ma exceeds the IC max current.(70 ma). My solution is to use P channel mosfets to switch the power. I'm loosing some voltage (4.85v - 4.9v) is what I'm getting as a output but that's ok \$\endgroup\$
    – Nik Middleton
    Commented Jun 11, 2018 at 13:07

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You are not wrong. The IC has a global limit of 70 mA.

However, driving a LED at 20 mA is often unnecessary today. You should make sure that the design actually tries to do that. The current supplied to the LED depends (most likely in this case) on the current limiting resistor and the LED IV-curve. A simple indicator LED will be fine at 5 mA.

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  • \$\begingroup\$ Thank you. Glad to know i'm not going nuts. I should have added that there are 2 LEDS being driven from each output. It would have been so simple to use a current drain driver but sadly it's not an option. \$\endgroup\$
    – Nik Middleton
    Commented Jun 9, 2018 at 12:49

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