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I am still new to electronics so please note that I am still a bit confused about current/voltage etc.

I have a 74199n Shift Register by Segnetics and on its datasheet it says that the maximum input current is 5 mA, which appears to be really low. (Is it really the maximum or did I confuse something?)

Is low value because the IC is just quite old (The datasheet says ~1985)? Can I safely exceed this value and if not, how can I lower the current flow to not exceed the limit?

Here are the absolute maximung ratings I got the information from: (I cant find the datasheet again, you can try searching it up)

enter image description here

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    \$\begingroup\$ 5 mA of input current is really high, not really low. \$\endgroup\$
    – Hearth
    Nov 7 at 18:16
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    \$\begingroup\$ You have it backwards. Older ICs tend to have higher I/O current capability than newer ICs because they use larger transistors. \$\endgroup\$
    – DKNguyen
    Nov 7 at 18:18
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    \$\begingroup\$ Of course you can exceed a current rating. Expect the IC not to work afterwards of course. \$\endgroup\$
    – Neil_UK
    Nov 7 at 18:18
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    \$\begingroup\$ If you have a datasheet that is both online and not on some dodgy information-harvesting site, please edit your question with a link. Or cut and paste a picture of the specification you're looking at. \$\endgroup\$
    – TimWescott
    Nov 7 at 18:21
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    \$\begingroup\$ "Can I safely exceed the value" -- If it's not for safety, it would not have been called "maximum rated xxx". \$\endgroup\$
    – Mitu Raj
    Nov 7 at 19:06
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Study the datasheet - if 5mA is listed under ‘absolute maximum’ then that’s a hard limit, don’t ever exceed it. If the figure appears under ‘electrical characteristics’ then it’s the highest current that the input will draw when the pin and the Vcc pin are at the maximum normal operating voltage, which may be 5.5V or thereabouts. In the latter case, if you pull the pin a little higher than the supply voltage you might see a higher current, but this is generally not advisable and there are few situations where it could occur.

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IC's are designed to be powered from a voltage supply. Normally, you just hook them up to the correct voltage and do not worry about the maximum current limit. The IC will not try to use excess current.

The maximum input current rating would only apply to unusual situations where you somehow cause the IC input current to be unnaturally high.

Note: the datasheet is screwed up. Ioh is listed as -800 V which makes no sense. This causes me to distrust the entire datasheet.

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What the input current specification tells you is what current the inputs will draw when given a normal input. Think of current as being "pulled", not "pushed"--as long as you give the IC a normal range of input voltages and don't go outside its ratings, it won't pull more than 5 mA from whatever's connected to the input.

If you do exceed its ratings, all bets are off and the part will probably not survive, but you do have a better chance of the chip surviving if you limit its current somehow. Better to just not go outside the ratings, though.

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All outputs are current limited.

That means that nobody prevents you from sinking more current than recommended. You can even short an output.

You will get no more than 20/50 mA.

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I wouldn't worry too much about this spec. Inputs of TTL devices are usually floating inputs. The device will draw the necessary current required to meet the logic level as defined further in the DC Electrical Characteristics Il and Ih. Under normal circumstances you will never input more than 5mA into the input pins. However, if you were driving the pin with a constant current source which exceeds 5mA for some silly reason, you could damage the input.

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