next topic »

[on:]

If I am deploying a stand alone projevt and only use 1 of the analog inputs, does it provide any benefit to enable the pull up resistors on the unused pins?
I am thinking of stability more than  anything else.

[Reply #1 on:]

It probably won't hurt. The analog reading is actually done "on command" (ie. when you do an analogRead), it isn't just reading away for its amusement. But setting unused pins to a known value is probably going to help keep noise out of the processor.

[Reply #2 on:]

Thank you Nick, So following what you said, I shouldnt add anything to the current draw either? The device is battery operated.

[Reply #3 on:]

Well you could put in pull-down resistors to tie them low. That shouldn't consume much, if anything. Pull-ups will add to the current drawn (as you would expect) according to my measurements here:

http://www.gammon.com.au/forum/?id=11497

[Reply #4 on:]

You can save substantial amounts of power by just running from 3.3V rather than 5V, BTW.

[Reply #5 on:]

Wow, that is a great article, thank you.
I am using my own board in my device si it has no USB,

[Reply #6 on:]

Well you could put in pull-down resistors to tie them low. That shouldn't consume much, if anything. Pull-ups will add to the current drawn (as you would expect) according to my measurements here:

http://www.gammon.com.au/forum/?id=11497

Actually, I'm not sure I would expect pull-ups to add to the current drawn, if nothing else is connected to the pins. Where is the current flowing to?

In fact, in my experimenting with sleep modes, I make all the pins inputs and enable pull-ups before going into power-down mode, and supply current waffles around between 0.1µA and 0.2µA.

[Reply #7 on:]

According to my measurements on the page quoted above, the difference is about 0.9 uA.

I presume where the current is "going" is making the input MOSFETs "work harder" by amplifying noise.

See page 43 of the datasheet:

If the input buffer is enabled and the input signal is left floating or have an analog signal level close to VCC/2, the input buffer will use excessive power.

and:

An analog signal level close to VCC/2 on an input pin can cause significant current even in active mode.

[Reply #8 on:]

According to my measurements on the page quoted above, the difference is about 0.9 uA.

I presume where the current is "going" is making the input MOSFETs "work harder" by amplifying noise.

See page 43 of the datasheet:

Right, I saw the measurements, and hence the question as my results seem different. Shouldn't having the pullup turned on reduce noise? As opposed to a floating pin?

Datasheet section 14.2.6, p81, recommends enabling pull-ups to provide a well-defined level for unconnected pins.

Recently I've been thinking about starting each sketch with the following in setup():

Code:

PORTB = PORTC = PORTD = 0xFF;    //enable all pullups

[Reply #9 on:]

an analog signal level close to VCC / 2

A pullup would put the signal level close to VCC not close to VCC divided by 2.

My results have been similar to Jack's.  For unconnected pins, I have not measured a difference between input with pullup, output low, or output high.  But my Extech multimeter may not be up to the task.

[Reply #10 on:]

Er, yes you both make good points. The pull-up on the face of it would result in not-floating pins.

However to check, I re-measured using the board shown on the page I linked above.

First with this sketch:

Code:

#include <avr/sleep.h>
void setup ()
{
  for (byte i = 0; i <= A5; i++)
    {
    pinMode (i, INPUT);    // changed as per below
    digitalWrite (i, LOW);  //     ditto
    }  
  // disable ADC
  ADCSRA = 0;  
  set_sleep_mode (SLEEP_MODE_PWR_DOWN);  
  sleep_enable();
  // turn off brown-out enable in software
  MCUCR = _BV (BODS) | _BV (BODSE);
  MCUCR = _BV (BODS);
  sleep_cpu ();  
}  // end of setup

void loop () { }

Results:

Code:

Meter: 0.11 uA
Using Dave Jones' uCurrent device: 120 nA

So, around 110 to 120 nA with pins as input and no pull-ups.

Now with the revised sketch:

Code:

   digitalWrite (i, HIGH);  //    enable pull-ups

Results:

Code:

Meter: 75 uA
Using Dave Jones' uCurrent device: 75 uA

I admit that neither figure agrees with my previously-posted results (maybe the fuses were different).

I can't explain why my figures are so different to yours, except that we must be using different hardware, and maybe different fuse settings.

[Reply #11 on:]

However let's assume you aren't planning to sleep. Then the results are a bit different. This sketch:

Code:

void setup ()
{
  for (byte i = 0; i <= A5; i++)
    {
    pinMode (i, INPUT);   
    digitalWrite (i, LOW); 
    }
}  // end of setup

void loop () { }

Current measured: 15.3 mA.

Changing to set all pins to pull-ups gives 11.8 mA.

So in that case, we saved 3.5 mA by turning the pull-ups on.

[Reply #12 on:]

However let's assume you aren't planning to sleep. Then the results are a bit different. This sketch:

Current measured: 15.3 mA.

Changing to set all pins to pull-ups gives 11.8 mA.

So in that case, we saved 3.5 mA by turning the pull-ups on.

Pretty sure I observed something similar recently, unfortunately did not record the results. But hence the idea of starting a sketch with all pullups on. That plus a couple of my projects which have been occasionally acting a bit funny, like they're responding to capacitive effects or maybe static electricity.

[Reply #13 on:]

However let's assume you aren't planning to sleep. Then the results are a bit different. This sketch:

Current measured: 15.3 mA.

Changing to set all pins to pull-ups gives 11.8 mA.

So in that case, we saved 3.5 mA by turning the pull-ups on.

Pretty sure I observed something similar recently, unfortunately did not record the results. But hence the idea of starting a sketch with all pullups on. That plus a couple of my projects which have been occasionally acting a bit funny, like they're responding to capacitive effects or maybe static electricity.

On an unaltered Arduino (I have a dozen or so Pro Mini's lying around, some adulterated, so virginal ...) I see pins floating to the voltage of other pins in the same port.  For digital pins, I see them signal close to whatever is nearby -- most often around line frequency.  So, I'm guessing they are sensitive to capacitive effects (parasitic capacitance on the die, maybe?) as well as RF.

[Reply #14 on:]

Good point. Probably to get defined behaviour you should have defined inputs into the pins. I would have thought that tying a pin low through a 10K resistor would do that at minimal cost, but maybe I'm wrong.