Bit Timing Theory: the length of unterminated stubs

[Dimitri Karch]

Hello!

In Philips application note for PCA82C250/251 CAN Transceiver (AN96116) the following formula is given to compute the "unterminated cable drop length" (see Chapter 6.5).

The length of some unterminated stub should be:

Lu < t_prop_seg / (50 * t_p)

t_p is the specific line delay per length unit (e.g. 5 ns/m)
t_prop_seg is the duration of the propagation segment

Does anybody know how was the formula deduced? Does the 50 mean the signal reflections disappear in 25 periods?
And why the length of propagation segment is used? I though the propagation segment should include only the round trip delay of signal edge. The reflections appear after signal edge and shell disappear during the phase buffer segment 1. Is it right?

Thank you in advance!

Dimitri Karch

Technical University Hamburg Harburg

P.S. Link to AN96116:
http://www.icbase.com/pdf/PHI/PHI08860106.pdf
--
Archives and useful links: http://groups.yahoo.com/group/CANbus
Subscribe and unsubscribe at www.vector-informatik.com/canlist/
Report any problems to <canlist-owner@...>

[Steve Corrigan-2]

Hello Dimitri,

Since stub-lines are un-terminated, signal reflections can develop in a stub
that drive signal levels back and forth across a receiver's input
thresholds, creating errors. Bit-sampling occurs near the end of a bit, so
it is mandatory that all signal reflections in a CAN stub-line be attenuated
before or during the propagation delay segment to provide an adequate margin
of safety.

To minimized reflections, stub-line length should not exceed one-third of
the line's critical length (about the same as Phillip's recommendation).
Beyond this stub-length, many variables come into play since the stub is no
longer considered to be a lumped parameter. This is the maximum length that
a stub remains invisible to a transmission line.

The critical length of a bus line occurs at the point where the
down-and-back propagation delay (tprop(total)) of a signal through a line
equals the transition time(tt) of a signal (the greater of the rise or fall
times).

Network Critical Length = tt = tprop(total)

Therefore, a typical CAN driver may have a 50 ns transition time, and when
considering a typical twisted-pair transmission line prop delay of 5 ns/m,
the down-and-back delay for one meter becomes 10ns/m. The critical length
becomes 5 m (50 ns / 10ns/m = 5 m), and the max un-terminated stub length
for the network is 1/3rd of the critical length, or 5/3 m (1.67 m).

When critical length is taken into consideration, driver slew-rate control
becomes a valuable design asset. The Standard recommends a maximum
un-terminated stub length of 0.3 m with a 1 Mbps signaling rate, but with
slew rate control, reduced signaling rate, and careful design, longer stub
lengths are easily obtained.

For example, if a 10 kW resistor is applied for slope-control at the Rs pin
(pin 8) of the  PCA82C251 CAN transceiver, an ~150 ns (as I recall) driver
transition time increases the maximum stub length to 15/3 m or 5 meters.

Cheers,
Steve C.


-----Original Message-----
From: canlist-owner@...
[mailto:canlist-owner@...] On Behalf Of Dimitri
Karch
Sent: Monday, September 14, 2009 9:49 AM
To: canlist
Subject: [CANLIST] Bit Timing Theory: the length of unterminated stubs


Hello!

In Philips application note for PCA82C250/251 CAN Transceiver (AN96116) the
following formula is given to compute the "unterminated cable drop length"
(see Chapter 6.5).

The length of some unterminated stub should be:

Lu < t_prop_seg / (50 * t_p)

t_p is the specific line delay per length unit (e.g. 5 ns/m)
t_prop_seg is the duration of the propagation segment

Does anybody know how was the formula deduced? Does the 50 mean the signal
reflections disappear in 25 periods?
And why the length of propagation segment is used? I though the propagation
segment should include only the round trip delay of signal edge. The
reflections appear after signal edge and shell disappear during the phase
buffer segment 1. Is it right?

Thank you in advance!

Dimitri Karch

Technical University Hamburg Harburg

P.S. Link to AN96116:
http://www.icbase.com/pdf/PHI/PHI08860106.pdf
--
Archives and useful links: http://groups.yahoo.com/group/CANbus
Subscribe and unsubscribe at www.vector-informatik.com/canlist/
Report any problems to <canlist-owner@...>


--
Archives and useful links: http://groups.yahoo.com/group/CANbus
Subscribe and unsubscribe at www.vector-informatik.com/canlist/
Report any problems to <canlist-owner@...>

[Dimitri Karch]

Hello Steve,

Thank you a lot for your detailed answer!

It took some time till I could get all you have wrote :) I should read about critical length of a transmission line and now I believe to understand what did you mean:
The reflections became not critical till the bus line (the way between two communicated transceivers) is shorter then the critical length:
L < l_crit / 3 = tt / (2 * r) / 3 = tt / (6 * t_p)
tt is the transition time (50 ns)
t_p is the propagation delay (5 ns/m)

But only thing I can get on this way is L < 50 / (6*5) = 1.67 m. The L is then the length of bus line between transceivers. It doesn't explain the relationship to the length of PROP_SEG.

I think the Phillip's formula is not SO complicated,
L_u < t_prop_seg / (50 * t_p)
2 * L_u * t_p < t_prop_seg / 25
So it just says, the round trip time in unterminated stub (2 * L_u * t_p) shell be 25 times less, then the duration of PROP_SEG. I though, the round trip time is then equal to period of reflection's oscillation, that shell be attenuated in 25 periods.
Is here something wrong?

Thank you in advance!

Dimitri Karch

Technical University Hamburg Harburg







-----Original Message-----
From: "Steve Corrigan" <steve.corrigan@...>
To: <canlist@...>
Date: Mon, 14 Sep 2009 11:08:45 -0500
Subject: RE: [CANLIST] Bit Timing Theory: the length of unterminated stubs

> Hello Dimitri,
>
> Since stub-lines are un-terminated, signal reflections can develop in a stub
> that drive signal levels back and forth across a receiver's input
> thresholds, creating errors. Bit-sampling occurs near the end of a bit, so
> it is mandatory that all signal reflections in a CAN stub-line be attenuated
> before or during the propagation delay segment to provide an adequate margin
> of safety.
>
> To minimized reflections, stub-line length should not exceed one-third of
> the line's critical length (about the same as Phillip's recommendation).
> Beyond this stub-length, many variables come into play since the stub is no
> longer considered to be a lumped parameter. This is the maximum length that
> a stub remains invisible to a transmission line.
>
> The critical length of a bus line occurs at the point where the
> down-and-back propagation delay (tprop(total)) of a signal through a line
> equals the transition time(tt) of a signal (the greater of the rise or fall
> times).
>
> Network Critical Length = tt = tprop(total)
>
> Therefore, a typical CAN driver may have a 50 ns transition time, and when
> considering a typical twisted-pair transmission line prop delay of 5 ns/m,
> the down-and-back delay for one meter becomes 10ns/m. The critical length
> becomes 5 m (50 ns / 10ns/m = 5 m), and the max un-terminated stub length
> for the network is 1/3rd of the critical length, or 5/3 m (1.67 m).
>
> When critical length is taken into consideration, driver slew-rate control
> becomes a valuable design asset. The Standard recommends a maximum
> un-terminated stub length of 0.3 m with a 1 Mbps signaling rate, but with
> slew rate control, reduced signaling rate, and careful design, longer stub
> lengths are easily obtained.
>
> For example, if a 10 kW resistor is applied for slope-control at the Rs pin
> (pin 8) of the  PCA82C251 CAN transceiver, an ~150 ns (as I recall) driver
> transition time increases the maximum stub length to 15/3 m or 5 meters.
>
> Cheers,
> Steve C.
>
>
> -----Original Message-----
> From: canlist-owner@...
> [mailto:canlist-owner@...] On Behalf Of Dimitri
> Karch
> Sent: Monday, September 14, 2009 9:49 AM
> To: canlist
> Subject: [CANLIST] Bit Timing Theory: the length of unterminated stubs
>
>
> Hello!
>
> In Philips application note for PCA82C250/251 CAN Transceiver (AN96116) the
> following formula is given to compute the "unterminated cable drop length"
> (see Chapter 6.5).
>
> The length of some unterminated stub should be:
>
> Lu < t_prop_seg / (50 * t_p)
>
> t_p is the specific line delay per length unit (e.g. 5 ns/m)
> t_prop_seg is the duration of the propagation segment
>
> Does anybody know how was the formula deduced? Does the 50 mean the signal
> reflections disappear in 25 periods?
> And why the length of propagation segment is used? I though the propagation
> segment should include only the round trip delay of signal edge. The
> reflections appear after signal edge and shell disappear during the phase
> buffer segment 1. Is it right?
>
> Thank you in advance!
>
> Dimitri Karch
>
> Technical University Hamburg Harburg
>
> P.S. Link to AN96116:
> http://www.icbase.com/pdf/PHI/PHI08860106.pdf
> --
> Archives and useful links: http://groups.yahoo.com/group/CANbus
> Subscribe and unsubscribe at www.vector-informatik.com/canlist/
> Report any problems to <canlist-owner@...>
>
>
> --
> Archives and useful links: http://groups.yahoo.com/group/CANbus
> Subscribe and unsubscribe at www.vector-informatik.com/canlist/
> Report any problems to <canlist-owner@...>
>

--
Archives and useful links: http://groups.yahoo.com/group/CANbus
Subscribe and unsubscribe at www.vector-informatik.com/canlist/
Report any problems to <canlist-owner@...>

[Michael Joshua]

Hi All,

We are using dsPIC30F6010A controller and transceiver used is TJA1041A.  Using an external crystal of 10MHz and system frequency is 20MHz.
We configured the CAN baud rate as 125kbps. We enabled strobe and chip enable signals by making it to HIGH. We are not getting signals out of (CAN transmit) C1TX/RF1 pin.

Please provide your valuable suggestions.

Thanks in advance
Michael Joshua.S




___________________________________________
Larsen & Toubro Limited
Embedded Systems and Software ( EmSyS)
SDB3, 1st Floor
Plot No 25-30,EPIP 1st Phase
Whitefield, Bangalore 560 066
T [phone]