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Understanding signals question ! — Parallax Forums

Understanding signals question !

tobon48tobon48 Posts: 22
edited 2006-03-13 19:48 in Learn with BlocklyProp
Hi!

In Chapter 4 R/C Circuits and Variable Resistors page 51 you can read:

... when the I/O pin was set to an input, thereby making it high impedance ...

should't it be low impedans ? how can the R/C net discharge trough an high impedans ?

regards
tobon48

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Quidquid latine dictum sit, altum sonatur.

Whatever is said in Latin sounds profound.

Post Edited (tobon48) : 3/13/2006 6:06:15 PM GMT

Comments

  • Martin HebelMartin Hebel Posts: 1,239
    edited 2006-03-13 16:08
    The capacitor discharges through the resistor, therefore the discharge time is function of R and C and not current flow due to the pin.

    -Martin

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    Martin Hebel
    Perform an Employer's Survey of Electronic Technologies Graduates· - Click here!
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  • tobon48tobon48 Posts: 22
    edited 2006-03-13 16:12
    Hi !

    But you got to have a closed loop for the current to flow !

    tobon48

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔

    Quidquid latine dictum sit, altum sonatur.

    Whatever is said in Latin sounds profound.

  • Martin HebelMartin Hebel Posts: 1,239
    edited 2006-03-13 16:18
    I don't have the text in front of me, but typically it looks like:

    Pin -------------
         |      |
         C      R 
         |      | 
        Vss    Vss



    Pin goes high as output, charging the capacitor, so the cap has + on the top.

    Pin goes to input to measure RCTIME. This allows the capacitor to discharge from + (top) through R to Vss.

    The time it takes the capacitor to go from 5V (1) to 1.4V (0) is what is measured for RCTime.

    That help?
    -Martin

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Martin Hebel
    Perform an Employer's Survey of Electronic Technologies Graduates· - Click here!
    Personal Links with plenty of BASIC Stamp info
    and SelmaWare Solutions - StampPlot - Graphical Data Acquisition and Control
  • tobon48tobon48 Posts: 22
    edited 2006-03-13 16:22
    Hi again !

    That's right ,but it means that "Pin" must be close to Vss for the current to flow.

    tobon48

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔

    Quidquid latine dictum sit, altum sonatur.

    Whatever is said in Latin sounds profound.

  • Martin HebelMartin Hebel Posts: 1,239
    edited 2006-03-13 16:27
    No.. the discharge path is from the top of the capacitor through R to Vss back to the (-) side of the capacitor.· think of it like a battery.· It will work even if Pin is disconnected.· The large the value of R and/or C, the longer time it will take.

             --------
         |+     |
        ___     R
         -      |   
         |------| 
        Vss    Vss

    -Martin
  • tobon48tobon48 Posts: 22
    edited 2006-03-13 16:46
    Hi !

    But look at your scematic , you have a closed loop.
    If it function like you said R wouldn't have any effect .

    tobon48

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔

    Quidquid latine dictum sit, altum sonatur.

    Whatever is said in Latin sounds profound.

  • Martin HebelMartin Hebel Posts: 1,239
    edited 2006-03-13 16:54
    The discharge path is through the resistor.

    The larger the resistor is, the lower the current will be: I = V/R

    Low lower the current is, the longer it takes the capacitor to discharge.

    Or, the large the capacitor is, the more charge there will be so, again, longer time.

    t = 1.1RC

    As R or C increases, t (time) increases.

    -Martin
    ·
  • tobon48tobon48 Posts: 22
    edited 2006-03-13 17:01
    Hi !

    ··· Pin
    ···· !
    ···· !
    ···· C
    ·····!
    ···· R
    ···· !
    ··· Vss
    If Pin not connected or high impedans no current can flow trough R , you must have a closed loop for current to flow.

    tobon48


    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔

    Quidquid latine dictum sit, altum sonatur.

    Whatever is said in Latin sounds profound.

  • Martin HebelMartin Hebel Posts: 1,239
    edited 2006-03-13 17:09
    The source of the current is not the Pin.· The pin only charges up the capacitor.· The capacitor will then act as a voltage source.· The 'closed-loop' for current flow is:

    ·(+)side of capacitor --> Resistor --> to (-)side of capacitor

    Someone will have to take it from here, I apparently can't state it in a way to be understoop.

    -MH
  • tobon48tobon48 Posts: 22
    edited 2006-03-13 18:31
    Hi !

    I saw now on your schematic that the resistor and capacitor was coupled in parallell, but in the text in Understanding Signals the resistor and capacitor are coupled in series.

    tobon48

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔

    Quidquid latine dictum sit, altum sonatur.

    Whatever is said in Latin sounds profound.

    Post Edited (tobon48) : 3/13/2006 6:39:44 PM GMT
  • Beau SchwabeBeau Schwabe Posts: 6,568
    edited 2006-03-13 19:18
    tobon48,

    On page 51 you said .... "... when the I/O pin was set to an input, thereby making it high impedance ...". This is a correct
    statement, if you read further on the same page you will also see...."Different resistor sizes affect the charge or discharge time",
    however there has not been a reference to a schematic yet.


    On page 52 (ACTIVITY #1) ....Here you are not using the RCTIME function yet, instead the goal is to measure the resistor/capacitor
    time constant with the circuit you indicate. Notice, the code available for this activity (page 54) does not set the pin as an INPUT,
    but instead keeps it an OUTPUT or "low impedance" so that you can see the charge/discharge curves of the capacitor through the
    resistor on a scope.

    On page 57 (ACTIVITY #2) ...Introduces you placing a resistor in "parallel" to the capacitor to determine the discharge time of an
    RC network. Here, the code is available on page 59, which does make use of the RCTIME command.


    The description that you refer to on page 51 tries to explain the inner-workings of what the RCTIME command does when
    you use the command in your program.


    Reference:
    www.parallax.com/dl/docs/prod/sic/Signals.pdf

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Beau Schwabe

    IC Layout Engineer
    Parallax, Inc.

    Post Edited (Beau Schwabe (Parallax)) : 3/13/2006 7:21:12 PM GMT
  • Martin HebelMartin Hebel Posts: 1,239
    edited 2006-03-13 19:37
    Hi again,

    As I initially stated, I didn't look in the text for the exact figure, and Parallax shows it 2 different ways depending on the text, but the principles are the same.· The Power supply itself acts as the path for the closed loop.· Sorry for the confusion.

    So, from +Cap --> resistor --> Vdd --> P/S -->Vss --> -Cap

    -Martin
    tobon48 said...
    Hi !

    I saw now on your schematic that the resistor and capacitor was coupled in parallell, but in the text in Understanding Signals the resistor and capacitor are coupled in series.

    tobon48

  • Xanadu VectorXanadu Vector Posts: 4
    edited 2006-03-13 19:40
    tobon48 said...
    Hi!
    In Chapter 4 R/C Circuits and Variable Resistors page 51 you can read:
    ... when the I/O pin was set to an input, thereby making it high impedance ...
    should't it be low impedans ? how can the R/C net discharge trough an high impedans ?
    regards
    tobon48

    I think when the I/O pin is an input it's high impendance because the R/C will discharge through VSS not the I/O pin thereby making it "high impedance".
  • tobon48tobon48 Posts: 22
    edited 2006-03-13 19:48
    Hi !

    I was reading on page 51 and looked on page 52 [noparse];)[/noparse]

    tobon48

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔

    Quidquid latine dictum sit, altum sonatur.

    Whatever is said in Latin sounds profound.

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