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below is a sample of simple conditioner converting a 4-20mA (two wires) signal from process to a voltage from 0 to 10V. The circuit make use of LM358 as it guarantee better of 0.1V at zero output with single power rails.
The main voltage is derived from the same process voltage, usually 24V.
Nevertheless, LM358 positive power rail may be powered with any voltage from 15V to 30V. The 15V limit comes from minimum drop-out required by saturation of LM354 totem pole output, for better linearity.
In the schematic, the resistor R1 has been chosen smaller enough (100 ohms will suffice) do not penalize output compliance of transmitter and allows process current over long wires distances.
You may have to replace the generator V1 with a band-gap 2.5 diode for better circuit stability.
the problem here is how to convert a current from 4 to 20 mA to a voltage from 0 to 10 V.
Using a simple resistor of 500 ohms (high precision or not), you will have a voltage accross resistor of 2V @ 4mA (500x4x10^-3)
and 10V @ 20mA (500x20x10^-3).
You will get an unwanted base voltage of 2V @ 4mA and a range of (10-2)V=8V , while the original need is 0V @ 4mA, and full range of 10V. So, in my circuit, U1:A cancels the base voltage, and U1:B make the full range for 0 to 10V.
Ther's other issues you've to consider. Current output sensors (called often transmitters) are very usuals in industrial process applications. The reason of their large employ is the need to transmitt the process signals at very long distances. In big process plants, long distances may be also 1km. Is almost unpossible send voltage signals over very long cables cause enviromental noise, electromagnetic fields and so on. Infact, voltage generators have very low impedence (ideally null impedence), so the induced noise current may be very high.
Transmitting current (current sensors output are current generators by fact) over long cable has the advantage that the noise current in the cable is very low, due to high impedence (theoretically infinite) of current generators, and the base current of 4mA is very usefull as the receiver equipment may understande if the cable in broken for example.
The drawback is, how much the cable is long, as much line resistor is high. For example, if line resistor is 500 Ohms (long wires resistance) and you include a further resistor of 500 Ohms, you will get total of 1000 Ohms. At 20mA output a voltage of 20V will drop over it, so only 4V (in case of 24V main voltage is used) will be available for our sensor transmitter. Often, 4V are not suffice, so the sensor does'n work in linear zone. (called not compliant zone)
All above just to say you've to load the current output with a resistor smaller enough do not add additional line resistance. Nevertheless, again, smaller resistors will not suffice to achieve the voltage range diegor747 need for.
0-10V is an usual voltage range for PC acquisition boards (like National and similar). May be he have to collects all signal from plant on his PC.
Let's say only a small thing, the James circuit puts around 100 ohm resistance in the current loop, that is a good value, if you want, you can get a slower value increasing the gain of the OP-Amp. (Be careful because how much great it's the gain less stability)
We understande your point of view. No reason to repeat it so many time.
The BurrBrown (formally Texas) RCV420 is a precision 4-20mA receiver converting to 0-5V. The average price is around 4$ for 1000+ off. Attached is datasheet.
If its tempco is OK for you - yes. Dont forget to supply it as schematics shows voltage source not bandgap diode .
Trimmer could be needed to set 0 volotage when 4 mA instead o R5 and R6 and optionally another trimmer in place of R9 to set 10 V output when 10 mA on input.
stojakovicv, do as follow:
- set I1 to 4 mA
- put variable resistor instead of R6 and adjust it to get 0 V on U1B output .
- set I1 to 20 mA
- Tune R9 to get 10 V
P.S. I assume Opamp are supplied with negative and positive voltage relative to ground (+ 15 V and -15 V)
using an Op-Amp with any gain will add gain error to the circuit.
you can use a simple buffer. and then a divider before your A/D(If you have any).
and the scale at your uP.
For fixing your circuit you need -Voltage for your Op-Amp.that should fix your problem.
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