4-20mA - resistance converter

[Sheymalcolm]

Hi everyone, im after some help with integrating a PT100 sensor into my alarm panel (its a pretty smart alarm panel, just not quite smart enough to take a 4-20mA reading and convert it to temperature)

Basically I have a PT100 probe, and a 4-20mA converter for it, what I need to do is accurately convert the 4-20mA signal back to a known resistance change so the alarm panel can read it, preferably with the following increments (these resistances are what the panel recognises as the temperature)

10C - .229vdc - 99ohms
14C - .310vdc - 131.4ohms
18C - .399vdc - 170ohms
22C - .489vdc - 210ohms
24C - .523vdc - 226.9ohms
27C - .579vdc - 254ohms
30C - .653vdc - 287.5ohm
33C - .704vdc - 313.2ohm
36C - .755vdc - 338ohm
38C - .806vdc - 363.6ohm

This is the probe, and the transmitter -

http://www.kiatronics.com/temperature-s ... steel.html

http://www.kiatronics.com/temperature-s ... 70692.html


Thank you all for any help & ideas you might have.

Cheers


Shey

[pebe]

Reading from the values in your table, it appears that your alarm panel feeds its input point through a source resistance of about 5.7kohms from a 14V source. Inserting various resistance values will then cause a voltage drop across the resistors, and this is what the panel will be measuring.

So the appropriate voltages fed into the panel from a low impedance source can substitute for the range of resistances. The incoming 4-20mA from the current loop of the thermostat can be fed through a resistor to obtain a voltage proportional to the current. An op-amp can then be used to modify that voltage to suit the panel and provide an offset for the ‘lost’ 4mA of the current loop.

So here is a suitable circuit. Note that the design has only been calculated – not simulated or tested – but it is a simple circuit and there is not much that can go wrong.

As the input and output voltages vary in phase with each other (as distinct from being inverted), the opamp must be used in its non-inverting mode. In order to give isolation between the offset voltage and the feedback, it is necessary to have R2, so the amp has been given a fixed gain of x2. That means the range of input voltages must be 50% of the output range.

The output must swing from 0.229V to 0.806V; a range of 577mV. Half of that range (288.5mV) will be obtained when the load resistor for the loop equals 62.2ohms. The input voltage range will be from 0.3485V @ 10ºC to 0.637V @ 38ºC. These values can be matched to the required outputs when the amp has a gain of x2 and an offset at P2 wiper of 0.468V.

The alarm panel needs a current of 2.3mA. The amp chosen cannot sink that current so the 2N3704 is wired as an emitter follower. Because the ‘pull down’ current is now passive, the emitter load must be 91 ohms (less than the minimum 99ohms the panel needs). That arrangement is probably cheaper than finding an amp with rail-to rail outputs that can sink 3mA.

Finally, P1 and P2 should be multi-turn types for good setting accuracy. If the 5V supply is from a regulator then it will probably be smooth enough, but add extra smoothing if you like.

Best of luck.