RE: Can -- Sampling time point of -- 87.5%

Hi John,
 
You need to get your head around the method that a CAN bit is organized.
It's simple math really.  If a bit is divided into 16 segments use your
calculator to get the result of  at sample point at segment 14;  14/16.
 
John Dammeyer
 
 

"ELS! The Solution"
Automation Artisans Inc.
http://www.autoartisans.com/ELS/
Ph. 1 250 544 4950


-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of John
cliffer
Sent: Tuesday, December 04, 2012 9:48 PM
To: CANLIST
Subject: [CANLIST] Can -- Sampling time point of -- 87.5%


But still it tell's about how to calculate all the segments of an bit
......  but not what will be sample rate for this calculation as shown in
examples.


Can open standard say try to achieve about  87.5% sample rate for
acceptable performance on CAN bus.

Any suggestion for this ?



//John 



On Wed, Dec 5, 2012 at 3:40 AM, YAP <[email protected]> wrote:


On Tue, Dec 4, 2012 at 8:29 PM, John cliffer <[email protected]> wrote:
> Hi,
>
> I am going through this good application note from philips AN1798.pdf.
>>> 4.1 Step-by-Step Calculation of Bit Timing Parameters
>
> Here no where it is mentioned about achieving correct sampling time
point.
> Canopen suggest to use 87.5% sampling point
>
> 1> How shall we start the calculation of time quanta for segments to
achieve
> -- 87.5% sampling time point?
> 2> What is the formula to calculate sample rate of bit ?
>
> Please suggest.
>
> //John



IMO good paper here
http://ww1.microchip.com/downloads/en/AppNotes/00754.pdf


/Ake

--
 ---
Ake Hedman
eurosource, http://www.eurosource.se
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