6 replies to this topic

[John West]

So I have an external pushbutton wired to my design. I have now learned that I need to have a pullup resistor to prevent spurious readings (a good article about this is http://www.ladyada.n...no/lesson5.html). I changed my code to the following:

button2 = new InterruptPort(Pins.GPIO_PIN_D8, true, Port.ResistorMode.PullUp, Port.InterruptMode.InterruptEdgeHigh);

However, this didn't seem to work. However, when I added this line:

button2.Resistor = Port.ResistorMode.PullUp;

It worked. Is this expected? If so, then I'm assuming that the constructor is used to tell it what mode is being used, but has no bearing on whether the resistor is external or internal. In order to actually turn the internal resistor on, you have to use the second line.

Is this correct? I'm posting because I think it is, and if so, it might be handy for others. I haven't seen an example showing the second line of code, so others might have missed it.

Can anyone confirm these findings with certainty? Has it been posted elsewhere and I've just missed it?

[Stefan]

Hi John West, If you haven't got a pullup/pulldown circuit next to your button, then yes, that's as expected. For more details, you should read http://en.wikipedia....ull-up_resistor

[Philip]

Hi John West,

If you haven't got a pullup/pulldown circuit next to your button, then yes, that's as expected.
For more details, you should read http://en.wikipedia....ull-up_resistor

Hi Stefan,

I'm puzzled by your statements above regarding the software statements. What does:

button2.Resistor = Port.ResistorMode.PullUp;

achieve that

button2 = new InterruptPort(Pins.GPIO_PIN_D8, true, Port.ResistorMode.PullUp, Port.InterruptMode.InterruptEdgeHigh);

does not? (apart from the reference to the interrupt). I'm not convinced that setting something twice in software can make up for a (real) missing (hardware) resistor. I'd be interested in why you expect it to be that way. Is it something to do with the configuration of an interrupt, rather than a basic I/O pin?

Phil

[Stefan]

Actually, that is strange I only set such values in the constructor and never had problems with it. How does your schematic looks like?

[Philip]

Actually, that is strange
I only set such values in the constructor and never had problems with it. How does your schematic looks like?

Hi Stefan

No pull-up resistors on my board. I had assumed that setting Port.ResistorMode.PullUp did this for me. Then I had a problem with digital inputs on my project - started reading, and found your post.

Turns out that my problem lay elsewhere in my code.

Once I have configured my inputs as Port.ResistorMode.PullUp, I can leave them open-circuit for a logic 1 and ground them for a logic zero (which confirms my original understanding).

I would be *really* interested to know what the internal pull-up value is - does anyone know? I suppose I should read the Atmel data sheet . . . .

Let me know what you think

Phil

[Dan T]

Quick check of the datasheet I have ("Prelim" from Feb 09) says pullup current is 60uA max with 3.6V... so 60kOhm?

[Stefan]

It's about 10-11kΩ See also: http://www.atmel.com...nts/doc6120.pdf page 610.