20mA current loop

K

[email protected]

Jan 1, 1970
0
I have a temperature measuring instrument giving 0 - 10 Volts (30 mA)
corresponding to 0 - 100 celsius. This voltage signal needs to be
transfered 500 feet away on an ammeter. Is this possible by just using
a series resistor? What are the drawbacks if neglecting environmental
temperatue and noise pickup disturbances.

T

Tom Bruhns

Jan 1, 1970
0
I have a temperature measuring instrument giving 0 - 10 Volts (30 mA)
corresponding to 0 - 100 celsius. This voltage signal needs to be
transfered 500 feet away on an ammeter. Is this possible by just using
a series resistor? What are the drawbacks if neglecting environmental
temperatue and noise pickup disturbances.

You didn't say what sort of accuracy you need... Certainly it would
be easy to get decent accuracy if you keep the current in the wire
low, so that the voltage drop in the wire is low compared to the drop
in the fixed resistor you add (which of course should be chosen to be
stable). You need to account for changes in wire resistance and in
the resistance of the meter movement as temperature varies. A key
reason for using a 4-20mA current loop in industrial instrumentation
is that the sensors control the current, not the voltage, so voltage
drop in the connecting wire is irrelevant (up to a point). What you
propose is NOT a current loop, but should work fine anyway. Just pick
a meter with low enough full current range that the change in
resistance due to the wire and meter changing temperature doesn't
materially affect the current. The temperature coefficient of
resistance for copper is about 0.4%/C.

Cheers,
Tom

E

Eeyore

Jan 1, 1970
0
I have a temperature measuring instrument giving 0 - 10 Volts (30 mA)
corresponding to 0 - 100 celsius. This voltage signal needs to be
transfered 500 feet away on an ammeter. Is this possible by just using
a series resistor? What are the drawbacks if neglecting environmental
temperatue and noise pickup disturbances.

I've never used them but isn't 'zero' on a 20mA loop, 4mA ? That would
require a DC offset to be added as well as a series R. Since the
instrument is voltage source - not current source, you'll also need to
compensate for line resistance.

I'd use a highish impedance differential receiver at the destination and
send the signal as a voltage via screened twisted pair.

Graham

S

Spehro Pefhany

Jan 1, 1970
0
I have a temperature measuring instrument giving 0 - 10 Volts (30 mA)
corresponding to 0 - 100 celsius. This voltage signal needs to be
transfered 500 feet away on an ammeter. Is this possible by just using
a series resistor? What are the drawbacks if neglecting environmental
temperatue and noise pickup disturbances.

Suppose you use a 1mA meter, then the series resistance should be
10.0K. AWG18 wire would have a loop resistance of 6.5 ohms at 20'C, so
that would affect the reading by < 0.1 C (and the temperature effect
would be even less significant).

The meter will have a bit of resistance too. You can add that number
to the nominal loop resistance of the wire and adjust the 10.0K
downward a bit to compensate if it's important.

J

Jamie

Jan 1, 1970
0
I have a temperature measuring instrument giving 0 - 10 Volts (30 mA)
corresponding to 0 - 100 celsius. This voltage signal needs to be
transfered 500 feet away on an ammeter. Is this possible by just using
a series resistor? What are the drawbacks if neglecting environmental
temperatue and noise pickup disturbances.

You can use a R for that how ever, current transmitter devices are
designed to regulate to the desired current with regard to the distance
, receiving load and conductor size.

In other words, you need a current regulator that is biased via
the 0..10 volts.

In your case, you most likely can get away with just using a
R in series with a twisted pair and be ok.
R = 10/0.030 = 333 ohms..
etc..

I think you may find that the 0..10 source may not like a
30 ma load. You still may need to make you're self a voltage

It's your choice of what you want to do I guess.

http://webpages.charter.net/jamie_5"

A

Apostrophe Police

Jan 1, 1970
0
I think you may find that the 0..10 source may not like a
30 ma load. You still may need to make you're self a voltage
.. ^^^^^^^^^^^

yourself.

W

whit3rd

Jan 1, 1970
0
I have a temperature measuring instrument giving 0 - 10 Volts (30 mA)
corresponding to 0 - 100 celsius. This voltage signal needs to be
transfered 500 feet away on an ammeter. Is this possible by just using
a series resistor? What are the drawbacks if neglecting environmental
temperatue and noise pickup disturbances.

A simple circuit with one operational amplifier
can drive a current meter through a long wire
with no errors due to the wire resistance.
Since the output is 'an ammeter', that seems
appropriate.

The use of a 500 foot length of two-conductor cable
for a simple temperature measurement is a tad
expensive, though. Could you substitute a wireless
thermometer, or use a telescope to see a distant
meter?

Oregon Scientific makes wireless temperature sensors
at modest prices.

N

Nemo

Jan 1, 1970
0
I have a temperature measuring instrument giving 0 - 10 Volts (30 mA)
corresponding to 0 - 100 celsius. This voltage signal needs to be
transfered 500 feet away on an ammeter. Is this possible by just using
a series resistor? What are the drawbacks if neglecting environmental
temperatue and noise pickup disturbances.

Quite feasible, as your instrument can source 30mA. The usual problem
people hit with current loops is, it is easy to make a current *sink*
(ie variable resistor) but you sometimes find there's no power at either
end of the loop to drive it.

Next problem people hit is voltage drops: you will have some volts
dropped across the series resistor, but people tend to overlook the
cable. 500 feet each way = 1000 feet = several ohms. When we used to
drive 4-20mA over a kilometre or so, we needed to make sure the voltage
drops were OK *at max current*, ie 20mA. (Actually many instruments use
24mA to mean "fault", so you needed to be sure the supply could give
enough volts not to go non-linear at 24mA with a kilometre of cable,
doubled 'cos it's round a loop, plus the drop across the variable
resistor.) Include the tolerances of the supply and resistors and so on
and you can hit problems. It's just Ohm's Law, simple algebra, but it's
odd how many people just think of cable as a zero ohm resistor.

You can test this by looking up the specs of the cable you intend using.
Calculate the DC resistance of 1000 feet, put that on the output of your
instrument, add a current limiting resistor such that the total output
at 10V is say 10mA, and see if the output is indeed 10mA. If the
instrument only has a 10V supply within it it could have trouble
sourcing over 10mA.

Next problem - ground loops. If it's a simple unpowered ammeter at the
far end, no problem. If you end up connecting 2 earths together, you
could have an offset (that's one reason why 4-20mA loops start at 4mA)
and need to get an (expensive) galvanic isolator to sort the issue, or
discuss it further here.

We generally used unshielded twisted pair for current loops.

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