How many mA are required for a LOW and HIGH signal on a digital CMOS gate.

Page 3 of the pdf data says:

Input leakage current: +/- 1uA max. at 5.5 volts and 25C.

....... Phil

Hi, Mr.JD. The data sheet tells you.

On the bottom of p.3, it says that input leakage current is +/-1uA max at 25C, and +/-10uA over the temp range. It also says on the top of p.4 that input capacitance is 4pF typical.

You should note that a typical input, like all CMOS logic inputs, has an input current of nearly zero uA.

CMOS inputs in general don't have any DC input current -- you look at them as small (pF level) capacitive loads. That can cause problems with transition times, which can be an issue under some circumstances.

Good luck Chris

Almost none. All you really have to consider is the desired rise time vs input capacitance (4p) and the leakage current (10uA in that datasheet). The input rise/fall time would dominate if you wanted to run the device really fast, but even then something

Thanks for the insults, LOL. I knew that CMOS, because they are mosfet driven, and not transistor driven, that they did not need a current. However, you misread what I asked. What meant was does the CMOS see a

not very much just leakage current, 0 to 0.01 ma

Colin =^.^=

You have to provide up to 10 uA of input leakage current plus whatever is needed to charge the input capacitance of 4 pF as fast as the want the input to change states. I=C*(dv/dt), where dv is the logic swing (actually the swing needed to get from the minimum low state input voltage to the maximum high state input voltage, or vice versa) and dt is the rise or fall time you require.

" How many mA must be applied to signal pins to make a LOW or HIGH signal? "

The data sheet has the answer to both.

........ Phil

If it's CMOS, essentially none (something like 1/1,000,000,000,000 amp). You have to charge or discharge the gate capacitance, which is a few tens of pF (I think). Then no more electrons flow.

Not reliably because there could be some leakage. There should always be a path to ground, even if it is through many megohms, if you want the input to see "low" reliably.

You do *not* "apply" current; inputs are from *voltage* sources.

...and only because it is easy and *fast* to test for 1uA; typical is in the low nanoamp region.

You did *not* ask or say that originally.

It cares only about voltage.

Graham

Thanks Mr. Captain Obvious.

If it was so obvious, why did you post the original question.

You come across as either a not-too-bright troll, or someone short of understanding, but hates to be told anything.