Impedance of digital inputs

Would it be safe to say that ALL digital inputs (pins on microprocessors, microcontrollers, gates, etc.) are always high input impedance? i.e. When doing circuit analysis is it okay to model a digital input as a very large (100K or higher) resistance to ground? Thanks in advance...

Reply to
Patrick
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Sounds good. What about digital inputs of other logic families, e.g. TTL? If I remember correctly TTL inputs draw a few micro-amps of current with a logic HIGH (5V) input.

Reply to
Patrick

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No.  Take a look at old 7400 TTL and you\'ll see that it takes 1.6mA
max to pull the input of a gate low.
Reply to
John Fields

Essentially yes, if they are *CMOS*. They look like, say, 1p-10p capacitance, with a very high dc impedance. For the most part they can be considered open circuits for DC. A more accurate model would be to include diodes from the inputs to the supply rails to model leakage.

Kevin Aylward snipped-for-privacy@anasoft.co.uk

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Reply to
Kevin Aylward

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Do it matter?

His question was:
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Reply to
John Fields

err....do people still use TTL?

Kevin Aylward snipped-for-privacy@anasoft.co.uk

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Reply to
Kevin Aylward

They draw a hell of a lot more when LOW.

Graham

Reply to
Pooh Bear

Actually, they're _sourcing_ current when low. ;-) ;-)

Cheers! Rich

Reply to
Rich Grise

I use the occasional 74F38, which is almost TTL.

John

Reply to
John Larkin

They're sourcing _conventional_ current, Kevin.

--
John Fields
Professional Circuit Designer
Reply to
John Fields

Yeah, yeah, I know ! Didn't want to spoil the flow of it though.

Graham

Reply to
Pooh Bear

The way kewl thing about this is, it doesn't matter! To switch from conventional to electron flow, just change all the pluses to minuses, and vice versa. ;-)

The thing that really weirds me out [but The Philosophizer loves it] is that, in all of the equations that relate electricity and magnetism, you can swap those two, and the answers still come out right!

I'm working on a 3-D diagram of the two 7-dimensional hypertoruses that the Universe is derived from...

Cheers! Rich

Reply to
Rich, Under the Affluence

Well, actually, no they are not. There are sinking electrons *into* the input!

Kevin Aylward snipped-for-privacy@anasoft.co.uk

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Reply to
Kevin Aylward

I thought 74F was analogue ! ;-)

Graham

Reply to
Pooh Bear

On the OP question, I just had an interesting experience with a Xilinx part. In *unprogrammed* mode, the bus hold circuitry is active in series with an equivalent resistor of about 25k to the I/O pin.

I caught that just as I was connecting things up to it (because I wanted to leave it unprogrammed for initial testing of other parts of the board) and had to whip up a simple piece of verilog to force inputs to *be* inputs and outputs to be at the suitable level. It didn't do anything else, so I simply had a bunch of assignments.

The old rule of thumb that FPGA IO pins are inputs when unprogrammed doesn't always hold ;)

Cheers

PeteS

Reply to
PeteS

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So\'s the square root of minus one, but that doesn\'t stop us from
using it.
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Reply to
John Fields

They can't. Conventional current is imaginary...

Kevin Aylward snipped-for-privacy@anasoft.co.uk

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Reply to
Kevin Aylward

Isn't everything analog(ue)?

John

Reply to
John Larkin

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Doesn\'t seem to be once you get down to the nitty-gritty.
Reply to
John Fields

One exception is the Hall effect.

that one depends on the sign of the charge on the charge carriers and has been measured to have the opposite sign in P-type semiconductors than it does in regular metals and N-type semiconductors.

huh?

Bye. Jasen

Reply to
Jasen Betts

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