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Topic: Connecting 5 V to pin? (Read 663 times) previous topic - next topic

yen

Hi *,

I'm currently thinking about connecting multiple analog and/or digital inputs to a <= 5 V source - without any resistors if anyhow possible.
I measured the input resistance of an analog pin with a normal multimeter, and according to that one, it's > 200 kOhm (which should be save); measuring the current from 5 V to A0 returned 0 mA, too. Still, I learned not to trust those measurement :P and since I didn't find any useful information, here's the question: What's the input-resistance of any analog/digital pin - so, any chance of blowing the 328p by connecting inputs directly to a voltage source?

Thanks in advance,
Sebastian

retrolefty

Quote
What's the input-resistance of any analog/digital pin - so, any chance of blowing the 328p by connecting inputs directly to a voltage source?


The 328p datasheet expresses it as input current leakage (1 ua max) and is equivelent to around 5 meg ohms minimum of input resistance.

Keep in mind that it only applies if the pin is setup (or as default from reset condition) to be an input pin. Once you change the pin to be an output pin, it represents a very low impedenace to either ground or Vcc and a short circuit on a output pin will damage the pin from over current flow. All pins must not be wired to voltages above Vcc +.5vdc or less then ground - .5vdc as that will forward bias the clampling diodes and cause damage.

Quote
IIL
Input Leakage
Current I/O Pin
VCC = 5.5V, pin low
(absolute value)
1 ?A
IIH
Input Leakage
Current I/O Pin
VCC = 5.5V, pin high
(absolute value)
1 ?A

yen

:-\ Sounds risky; but thanks for the reply :-)

retrolefty


:-\ Sounds risky; but thanks for the reply :-)


The risk in inversely proportional to your knowledge of electronics and the AVR microprocessor chip. So for a beginner there is lots of risks as he/she learns, for the experienced user, almost no risk. It's all about where you are on the old learning curve.  ;)

Lefty

James C4S

Quote
I measured the input resistance of an analog pin with a normal multimeter,


Using a multimeter to measure resistance of an active circuit generally won't yield the results you expect.  You should never use a multimeter to measure resistance if power is applied to a circuit.
Capacitor Expert By Day, Enginerd by night.  ||  Personal Blog: www.baldengineer.com  || Electronics Tutorials for Beginners:  www.addohms.com

yen

#5
Sep 01, 2011, 04:33 am Last Edit: Sep 01, 2011, 04:46 am by yen Reason: 1
The risk in inversely proportional to your knowledge of electronics and the AVR microprocessor chip.

True; but trusting the controller to work as expected is yet another topic imho XD. And I tend not to trust the software that much in this case since I neither read the bootloader-code nor did I read the whole datasheet...  ;)

You should never use a multimeter to measure resistance if power is applied to a circuit.

Sure, but if I remove power from the controller, I'll probably get even worse results... And the resulting - distorted - value should only be lower than the real one, shouldn't it? So if I get a very high resistance-value, shouldn't there be a pretty good chance that the input-resistance really is high? (Sure, there are cases where this won't apply, especially when it comes to AC/alternating signals in common...)
Edit: Oh, and of course, when the pin is set as output, the measure will also be "a bit crazy"  :D

Grumpy_Mike

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So if I get a very high resistance-value, shouldn't there be a pretty good chance that the input-resistance really is high?

No.

A meter is a source of current, it tries to force it round a circuit. Using this to measure a circuit with voltage in it already can damage the meter or even the circuit. At best this practice is totally meaningless, at worst it is damaging.

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