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Topic: Resistance of digital pin configured as INPUT on DUE (ARM based) board (Read 4703 times) previous topic - next topic

Jiggy-Ninja

What ever the value ( a practical value ) of R2 you use, you will always measure no voltage across it, because R1 is so high then virtually no current flows through it.
Getting back on track here from that diversion, I think the reason a spec isn't given for input resistance is because a CMOS input is not a simple resistance, especially with ESD protection diodes. Maximum leakage current is basically the only thing they can give there.

Wawa

Should you model the "input resistance" of an analogue pin to ground?
The "resistance" of an analogue pin seems connected to a virtual point of about  0.3*VCC.
A/D reading of about 350 from a "floating" pin, even on a pin with a cap to ground.
Leo..

Southpark

Yeah..... maybe input impedance. Could do some things to measure it. As everybody says here.... the resistive portion is going to be so large..... to not worry about.

Wawa

Last time I tried that it was >1000Megohm.
Try it yourself by measuring voltage between pin and ground and pin and VCC. DMM on 200mV.
The DMM is really a 10Megohm resistor with a voltmeter across.
Ohms law will give you leakage current and input impedance.
Leo..

Liam918

If you're considering something like that, go back to the drawing board.
As others have pointed out your concepts are just wrong.
A little bit harsh there, don't you think? ::)



The input resistance, is, for all practical purposes, infinite unless you use the INPUT_PULLUP instruction.
What ever the value ( a practical value ) of R2 you use, you will always measure no voltage across it, because R1 is so high then virtually no current flows through it.
My example might not seem practical to you. However, if we're talking about an electrical circuit (the one from my schematic above) which I could easily build on my desk right now - I consider this to be very practical. And if you will use a ≫1 MΩ resistor it will still behave in a certain way. It might not be what the overwhelming majority is concerned about, but I still like to think about what if. I really dislike the generalizing statement "never need to know the exact value for the resistance" - even if there isn't an exact value for this to begin with.

However, in between all this, you explained some things about the "composition" of digital input pins, which I found very interesting. I need to read up on that. So thank you for that. I appreciate it. :smiley-wink:

Jiggy-Ninja

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A little bit harsh there, don't you think? ::)
Yes I do. I chose that tone on purpose.
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My example might not seem practical to you. However, if we're talking about an electrical circuit (the one from my schematic above) which I could easily build on my desk right now - I consider this to be very practical. And if you will use a ≫1 MΩ resistor it will still behave in a certain way. It might not be what the overwhelming majority is concerned about, but I still like to think about what if. I really dislike the generalizing statement "never need to know the exact value for the resistance" - even if there isn't an exact value for this to begin with.
What we're trying to get through to you are two things:

1) The input to a CMOS circuit is not a simple resistance. The gate resistance is extremely high, there is gate capacitance, and if there are ESD protection diodes they will contribute their own reverse leakage currents.

2) The leakages are not characterized to have a specific value. The datasheet only gives a maximum value for the leakage current, there is no entry for minimum or typical. The manufacturer doesn't care what the exact value of the leakage current is, just that it's low enough to not cause trouble.

So if you think you can use the pin input in a resistive voltage divider, or that you can rely on the input maintaining a stable leakage characteristic while the chip operates, you are wrong on every front.

Plus there's the matter of the low currents being used. The maximum leakage current of an ATmega328P is 1 uA. Nanoamp currents are hard to work with because leakages around the circuit become so much more significant. The circuit has to be kept clean, otherwise skin oils, dust, or leftover flux from soldering can create parasitic resistances in parallel with your circuit elements and change it.

It is not practical. Due to the unpredictability, complexity, and sensitivity it will be a pain in the ass.

rmetzner49

I misunderstood the OP's question.  I though he WAS inquiring about the pullup resistors.  Most people already know CMOS inputs are such a high impedance as to be insignificant.

Grumpy_Mike

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And if you will use a ≫1 MΩ resistor it will still behave in a certain way.
In a practical circuit it is very difficult to get a resistor that high. Anything over 10M is quite hard to actually achieve. The resistance through a finger print is about this high.
And even if you could get such a high resistance then their is so much going on that you are not taking into account, your simple minded analysts falls flat on its face.

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