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Termination Resistor - RS-485 line.

F

Franco

Jan 1, 1970
0
Hey guys, fast question: Is it possible to use as a termination
resistor (in a RS-485 bus) a combination of L and C in series instead?

I am just trying to find a way to set a termination resistor that does
not imply such a big load as a pure 120ohm resistor.

Cheers...

Franco.
 
J

John Larkin

Jan 1, 1970
0
Hey guys, fast question: Is it possible to use as a termination
resistor (in a RS-485 bus) a combination of L and C in series instead?

I am just trying to find a way to set a termination resistor that does
not imply such a big load as a pure 120ohm resistor.

Cheers...

Franco.

Sounds bad to me. If the data rate is low, just don't terminate it, or
use a series R-C. At high rates (relative to line length) any
non-resistive termination can cause pattern-dependent distortion,
"deterministic jitter" and possible errors.

John
 
R

Rene Tschaggelar

Jan 1, 1970
0
John said:
Sounds bad to me. If the data rate is low, just don't terminate it, or
use a series R-C. At high rates (relative to line length) any
non-resistive termination can cause pattern-dependent distortion,
"deterministic jitter" and possible errors.

John,
the idea of the poster was AC termination. Meaning a 100
Ohms in series with perhaps a 100nF.

Rene
 
J

John Larkin

Jan 1, 1970
0
John,
the idea of the poster was AC termination. Meaning a 100
Ohms in series with perhaps a 100nF.

Rene


I mentioned series R-C termination. Again, that can cause
pattern-dependent timing errors and may snarf the data. If the data
has no DC or low-frequency components (like NRZI or Manchester), the
series R-C will work, but saves no power!

John
 
F

Franco

Jan 1, 1970
0
I mentioned series R-C termination. Again, that can cause
pattern-dependent timing errors and may snarf the data. If the data
has no DC or low-frequency components (like NRZI or Manchester), the
series R-C will work, but saves no power!

John


Nice answers.

I think both are right, and I meant an RC termination implicitly. Now
I am
realizing is the most convenient termination for my network.

One more question to John, why you said "saves no power"?

Cheers...

Franco.
 
N

Noway2

Jan 1, 1970
0
Franco said:
Nice answers.

I think both are right, and I meant an RC termination implicitly. Now
I am
realizing is the most convenient termination for my network.

One more question to John, why you said "saves no power"?

Cheers...

Franco.
I guess I really don't see why you would want to use such a termination
in an RS485 system. Would you not want to use a terminating resistor
that closely matches the characteristic impedance of the cable?

The cable and terminating resistor would form a divider circuit. By
matching the terminating resistor to the impedance of the cable, you
would eliminate or at least greatly dampen any reflections on the lines.

You mention not wanting such a big load as a 120 ohm resistor. I am
curious as to what your line of thinking is here.
 
J

John Larkin

Jan 1, 1970
0
Nice answers.

I think both are right, and I meant an RC termination implicitly. Now
I am
realizing is the most convenient termination for my network.

One more question to John, why you said "saves no power"?

Cheers...

Franco.

If the data is balanced (NRZI, Manchester) the DC voltage across the
cap is zero, so it may as well be a short; so all the differential
signal swing appears across the resistor.

If the data is unbalanced, like async serial ASCII or anything that
pauses, the cap can charge up to almost the full driver swing in
either direction, which can severely skew the receive timing when
transitions start again.

Why not use a 120 ohm resistor?

John
 
F

Franco

Jan 1, 1970
0
If the data is balanced (NRZI, Manchester) the DC voltage across the
cap is zero, so it may as well be a short; so all the differential
signal swing appears across the resistor.

If the data is unbalanced, like async serial ASCII or anything that
pauses, the cap can charge up to almost the full driver swing in
either direction, which can severely skew the receive timing when
transitions start again.

Why not use a 120 ohm resistor?

John

Ok. The point is than I am going to try to join (by using analog
switches) two or more
"little" RS-485 buses with a transceiver of 54ohms max drivers load.
If I use two
terminator resistors of 120ohm on each bus, I think I cannot drive
more than two
of these "little" buses. All of this, not considering the load of the
receivers (1/8).

Maybe is not possible, or maybe the way to do it is with bidirectional
buffer
kind of circuitry, but, so far, I am just trying to get just an
overall view.

Comments, please...

Best Regards...

Franco.
 
J

John Larkin

Jan 1, 1970
0
Ok. The point is than I am going to try to join (by using analog
switches) two or more
"little" RS-485 buses with a transceiver of 54ohms max drivers load.
If I use two
terminator resistors of 120ohm on each bus, I think I cannot drive
more than two
of these "little" buses. All of this, not considering the load of the
receivers (1/8).

Maybe is not possible, or maybe the way to do it is with bidirectional
buffer
kind of circuitry, but, so far, I am just trying to get just an
overall view.

Comments, please...

Best Regards...

Franco.

What sort of data? What rates? Are drivers gated?

John
 
M

mkaras

Jan 1, 1970
0
Ok. The point is than I am going to try to join (by using analog
switches) two or more
"little" RS-485 buses with a transceiver of 54ohms max drivers load.
If I use two
terminator resistors of 120ohm on each bus, I think I cannot drive
more than two
of these "little" buses. All of this, not considering the load of the
receivers (1/8).

Maybe is not possible, or maybe the way to do it is with bidirectional
buffer
kind of circuitry, but, so far, I am just trying to get just an
overall view.

Comments, please...

Best Regards...

Franco.


As you switch in different bus configurations you want to make sure
that the "little" busses do not join into non-linear structures. You
want two distinct ends (and those two places are where your two 120
ohm terminators go) and have a straight path to the other end going
past all intermediate nodes without side branches and stubs. If the
switching topology cannot be done with fixed end terminated nodes then
you may have to also put switches to turn the termiminators on and off
at some nodes that swap back and forth between end and intermediate
type nodes.

- mkaras
 
N

Noway2

Jan 1, 1970
0
Franco said:
Ok. The point is than I am going to try to join (by using analog
switches) two or more
"little" RS-485 buses with a transceiver of 54ohms max drivers load.
If I use two
terminator resistors of 120ohm on each bus, I think I cannot drive
more than two
of these "little" buses. All of this, not considering the load of the
receivers (1/8).

Maybe is not possible, or maybe the way to do it is with bidirectional
buffer
kind of circuitry, but, so far, I am just trying to get just an
overall view.

Comments, please...

Best Regards...

Franco.
Double check the data sheets. When you say that the drivers can handle
a max load of 54 ohms are you certain that this means that it can't
handle a larger impedance rather than a smaller one? The smaller (less
ohms) the load on the drivers, the more current they will have to pump
out to cause a signal to reach full height as it propagates on the
transmission line.
 
R

Rene Tschaggelar

Jan 1, 1970
0
I mentioned series R-C termination. Again, that can cause
pattern-dependent timing errors and may snarf the data. If the data
has no DC or low-frequency components (like NRZI or Manchester), the
series R-C will work, but saves no power!


The Texas Instruments application note SLLA070C discusses
the various terminations and list the 100 Ohms plus 1nF
betwwen the two lines as feasible alternative termination
for baudrates below 100kbit. It saves power as soon as the
bus is idle in either state.

Rene
 
J

John Larkin

Jan 1, 1970
0
The Texas Instruments application note SLLA070C discusses
the various terminations and list the 100 Ohms plus 1nF
betwwen the two lines as feasible alternative termination
for baudrates below 100kbit. It saves power as soon as the
bus is idle in either state.

Rene

That's a 100 ns tau for 10 microsecond bits. In that case, you may as
well not terminate at all. There might be some corner cases where such
a termination will reduce edge ringing.

But why not terminate with a resistor?

John
 
F

Franco

Jan 1, 1970
0
That's a 100 ns tau for 10 microsecond bits. In that case, you may as
well not terminate at all. There might be some corner cases where such
a termination will reduce edge ringing.

But why not terminate with a resistor?

John

The system is battery powered. I think pure termination resistor
(parallel termination) is not an option.
Moreover, I want to join little buses, to expand a kind of global bus.
As far as I understand, if one driver
is going to send the signal towards two little buses, both terminated
at 120 ohm (multipoint buses), I am
over-clocking the transceivers (let said, min 54 load - not less). The
transcievers are: SP3078E.

I want high rates, as high as I can. The micro-controllers driving
each transceiver attached to the bus are fast
enough to all the span of rates (up to 10Mbs).

It is a distributed system.

I know I am going to break the guidelines of linear buses, when
joining two little buses. I cannot
do anything about that, more than just analyze the effects of that. In
this moment I am
trying to figure out what happen when placing AC termination at both
end of the buses.

Cheers...

Franco.
 
M

martin griffith

Jan 1, 1970
0
The system is battery powered. I think pure termination resistor
(parallel termination) is not an option.
Moreover, I want to join little buses, to expand a kind of global bus.
As far as I understand, if one driver
is going to send the signal towards two little buses, both terminated
at 120 ohm (multipoint buses), I am
over-clocking the transceivers (let said, min 54 load - not less). The
transcievers are: SP3078E.

I want high rates, as high as I can. The micro-controllers driving
each transceiver attached to the bus are fast
enough to all the span of rates (up to 10Mbs).

It is a distributed system.

I know I am going to break the guidelines of linear buses, when
joining two little buses. I cannot
do anything about that, more than just analyze the effects of that. In
this moment I am
trying to figure out what happen when placing AC termination at both
end of the buses.

Cheers...

Franco.
more stuff here
http://www.bb-elec.com/tech_articles/rs422_485_app_note/table_of_contents.asp
http://www.bb-elec.com/485successes/ about termination


martin
 
N

Nico Coesel

Jan 1, 1970
0
Noway2 said:
I guess I really don't see why you would want to use such a termination
in an RS485 system. Would you not want to use a terminating resistor
that closely matches the characteristic impedance of the cable?

The cable and terminating resistor would form a divider circuit. By
matching the terminating resistor to the impedance of the cable, you
would eliminate or at least greatly dampen any reflections on the lines.

This is true if the length of the cable is greater than (roughly) 1
fifth of the wavelength of the signal. It might be easier to use
slew-rate controlled (read: bandwidth limited) RS485 drivers because
they allow longer unterminated transmission lines.
 
R

Rene Tschaggelar

Jan 1, 1970
0
John said:
That's a 100 ns tau for 10 microsecond bits. In that case, you may as
well not terminate at all. There might be some corner cases where such
a termination will reduce edge ringing.

But why not terminate with a resistor?

Well, the swing is the differential voltage across
the resistor of 3V with either sign. With 100 Ohms,
this is a loss of 90mW. For a battery operated device,
these 30mA are a lot. The reflection of the signal
is a reflection of the slope, and it is the slope
that dissipates in the AC termination. I'd possibly
go for a tau in the order of a tenth of a bit.

Rene
 
J

jasen

Jan 1, 1970
0
Hey guys, fast question: Is it possible to use as a termination
resistor (in a RS-485 bus) a combination of L and C in series instead?

not really.
I am just trying to find a way to set a termination resistor that does
not imply such a big load as a pure 120ohm resistor.

if the 120 ohms is hurting go to a lower voltage or use a different cable.
with a higher characteristic impedance.

or go to a lower bitrate - low enough for one bit to flood the whole bus.
 
J

jasen

Jan 1, 1970
0
Well, the swing is the differential voltage across
the resistor of 3V with either sign. With 100 Ohms,
this is a loss of 90mW. For a battery operated device,
these 30mA are a lot. The reflection of the signal
is a reflection of the slope, and it is the slope
that dissipates in the AC termination. I'd possibly
go for a tau in the order of a tenth of a bit.

you ran data wires but forgot to run power wires?

Bye.
Jasen
 
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