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Powering isolated RS485 transceivers over cable

M

markp

Jan 1, 1970
0
Hi All,
I have an application that requires about 12 sensor devices over a RS485
cable, evenly spaced over about 20m with a controoler at the end.

I'd like to use galvanically isolated RS485 buffers such as IL3185 on each
sensor, but would also like to not have to put isolated DC-DC converters to
supply the power to the cable side of the buffers.

So can I supply +5V up the cable to all the devices to power their cable
side buffers? Since it is RS485 only one sensor will be transmitting at a
time so as long as the input buffers don't take much current it would seem
practicable (with some capacitance and possible ferrite filtering where the
power comes in to each sensor).

First, is this a pratical solution? If so, assuming CAT5 cable with a pair
dedicated to the 2 wire RS485 signals, a pair to +5V and a pair to ground,
how long a cable would be possible, or is it best to use a higher voltage
and use a cheap LDO on the receivers to generate +5V?

Thanks!
Mark.
 
M

markp

Jan 1, 1970
0
Sure. Typically this is done with a pair of wires (one each for
voltage and ground) Transformer couple the RS485 data and apply the
power to the center taps. In a similar way power is pulled off the
center taps at the other end. Power over Ethernet works the same way.
We power our remote widget over 1500' of CAT-5. We start out with 24V
and use a buck regulator at the far end. We could go further but the
drop-out on the regulator is pretty bad. 1500' is far enough for our
application so we haven't bothered fixing it.

I guess I'm concernred about the voltage drop across a 20m cable due to the
resistance of the cable when the worst case of all 12 sensor receivers are
powered and one sensor's transmitter is on. The caps would smooth things
locally so it's down to DC losses.

As a matter of interest, what drop-out voltages were you seeing (I'm
assuming that's down to cable losses), and how much current were you
supplying? At the moment I've got a 5V regulated supply available, but the
chips require 4.5V.

Also I guess since power has return currents I should use a twisted pair for
ground and power to reduce noise, maybe tripling up as I have 4 wire pairs
available and one is taken by the signal pair (?).

Mark.
 
P

Paul Keinanen

Jan 1, 1970
0
Hi All,
I have an application that requires about 12 sensor devices over a RS485
cable, evenly spaced over about 20m with a controoler at the end.

I'd like to use galvanically isolated RS485 buffers such as IL3185 on each
sensor, but would also like to not have to put isolated DC-DC converters to
supply the power to the cable side of the buffers.

So can I supply +5V up the cable to all the devices to power their cable
side buffers? Since it is RS485 only one sensor will be transmitting at a
time so as long as the input buffers don't take much current it would seem
practicable (with some capacitance and possible ferrite filtering where the
power comes in to each sensor).

First, is this a pratical solution? If so, assuming CAT5 cable with a pair
dedicated to the 2 wire RS485 signals, a pair to +5V and a pair to ground,
how long a cable would be possible, or is it best to use a higher voltage
and use a cheap LDO on the receivers to generate +5V?

While not exactly RS-485, the CAN bus based DeviceNet standard
transfers the actual data in one pair and the power for the
transceivers and small loads in an other pair. You might find some
interesting information about power arrangements etc.

If you have full control of both the master and slave side, have you
considered the good old current loop system, which is quite easy to
optoisolate (no local floating power needed at slaves) ?

Connect all devices, both master as well as slaves, both receivers
(photo transistors) and transmitters (optoisolator LEDs) into a single
20 mA current loop. When idle, all transmitter transistors are
conducting and the 20 mA current flows through the loop and when the
active station (master or selected slave) wants to send the "0" bit,
it will cut the loop current. Since the active station will hear its
own transmission on the receiver side, it must contain some echo
cancellation in software.

The problem with this arrangement, especially with optoisolators with
bipolar transistors, is the large voltage drop at each station, nearly
2 V for the receiver LED and 1 V for the Vce of the transmitter. With
12+1 stations the loop current loop would have to provide nearly 40 V,
which might be too much for some optoisolator transistors.

One way around this problem is to use optoisolators with FET output
stage and hence a low voltage drop across the transmitter, reducing
the loop voltage requirement.

An other approach is to use two loops, one containing the master
transmitter and all the slave receiver LEDs in series, the other
containing the master receiver LED and the slave transmitter
transmitters (bipolar or FET) in series. 24 V should be sufficient for
both loops. In this configuration, echo cancellation is not needed.
 
C

Charlie E.

Jan 1, 1970
0
I guess I'm concernred about the voltage drop across a 20m cable due to the
resistance of the cable when the worst case of all 12 sensor receivers are
powered and one sensor's transmitter is on. The caps would smooth things
locally so it's down to DC losses.

As a matter of interest, what drop-out voltages were you seeing (I'm
assuming that's down to cable losses), and how much current were you
supplying? At the moment I've got a 5V regulated supply available, but the
chips require 4.5V.

Also I guess since power has return currents I should use a twisted pair for
ground and power to reduce noise, maybe tripling up as I have 4 wire pairs
available and one is taken by the signal pair (?).

Mark.

I had a simple three station design, but there were 120 feet between
stations. I used one pair for power, and individual bucks at the
stations. I had an old programmable laptop power supply on hand, so
just used it (I think I had it set to around 12 volts.) Depending on
the current draw for your individual stations, you shouldn't need too
high a voltage. Remember, voltage loss is a function of both voltage
and current!

Charlie
 
J

Jamie

Jan 1, 1970
0
Paul said:
While not exactly RS-485, the CAN bus based DeviceNet standard
transfers the actual data in one pair and the power for the
transceivers and small loads in an other pair. You might find some
interesting information about power arrangements etc.

If you have full control of both the master and slave side, have you
considered the good old current loop system, which is quite easy to
optoisolate (no local floating power needed at slaves) ?

Connect all devices, both master as well as slaves, both receivers
(photo transistors) and transmitters (optoisolator LEDs) into a single
20 mA current loop. When idle, all transmitter transistors are
conducting and the 20 mA current flows through the loop and when the
active station (master or selected slave) wants to send the "0" bit,
it will cut the loop current. Since the active station will hear its
own transmission on the receiver side, it must contain some echo
cancellation in software.

The problem with this arrangement, especially with optoisolators with
bipolar transistors, is the large voltage drop at each station, nearly
2 V for the receiver LED and 1 V for the Vce of the transmitter. With
12+1 stations the loop current loop would have to provide nearly 40 V,
which might be too much for some optoisolator transistors.

One way around this problem is to use optoisolators with FET output
stage and hence a low voltage drop across the transmitter, reducing
the loop voltage requirement.

An other approach is to use two loops, one containing the master
transmitter and all the slave receiver LEDs in series, the other
containing the master receiver LED and the slave transmitter
transmitters (bipolar or FET) in series. 24 V should be sufficient for
both loops. In this configuration, echo cancellation is not needed.
Funny you would talk about this, we have some Red Lion devices that use
the current loop and if you don't have more than one slave unit on the
loop, in hot operating conditions, the crappie current source circuit
used will bias itself more than 20 ma's and over time burn out the
optical input on the slave device!. So, what I have done to correct the
issue for now is to use a R in series. This really should be documented
in the Red Lion manual but I haven't see it..

The current source that is being used is your first year 101 basic
PNP transistor being biased to 20 ma. there is a diode on the base to
the (+) rail to help out but its not enough, it still gets out of spec..
I've see it reach up to 35 ma when thing get warm..

Jamie.
 
U

Uwe Hercksen

Jan 1, 1970
0
markp said:
First, is this a pratical solution? If so, assuming CAT5 cable with a pair
dedicated to the 2 wire RS485 signals, a pair to +5V and a pair to ground,
how long a cable would be possible, or is it best to use a higher voltage
and use a cheap LDO on the receivers to generate +5V?

Hello,

I would look for the cable data, especially the resistance per length of
one wire and calculate the voltage loss over a 20 m long cable with the
necessary current for the tranceivers.

Bye
 
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