4-20mA Current loop

[jondow]

Where do you get R5. On the schematic IN- goes to ground. not to R5. I am speaking about:  Basic, analog input, 2 wire, 4-20 mA loop powered transmitter.

Can you write your calculatings? I get other numbers.

[Seekonk]

Hero, look where the emitter is connected.

[Zero999]

Where do you get R5. On the schematic IN- goes to ground. not to R5. I am speaking about:  Basic, analog input, 2 wire, 4-20 mA loop powered transmitter.

Can you write your calculatings? I get other numbers.

R5 compensates for the op-amp bias currents. It's roughly equal to R3 & R4 in parallel.

The circuit works on the same principle as the other circuit I posted. The op-amp adjusts the voltage on its output to keep its inputs at 0V. Because the op-amp can't work when either of its inputs are below 0V, a potential divider consisting of R3 & R4 is used. When the voltage at the non-inverting input is 0V, then the voltage on the negative side of Rs will the same, but opposite polarity to  the input voltage. The current through Rs will be equal to -V/Rs. The current taken by the op-amp and your sensor both passes through Rs so this is compensated for. .

Hero, look where the emitter is connected.

Are you talking about my schematic? If so , what's your point?

[jondow]

R5 compensates for the op-amp bias currents. It's roughly equal to R3 & R4 in parallel.

The circuit works on the same principle as the other circuit I posted. The op-amp adjusts the voltage on its output to keep its inputs at 0V. Because the op-amp can't work when either of its inputs are below 0V, a potential divider consisting of R3 & R4 is used. When the voltage at the non-inverting input is 0V, then the voltage on the negative side of Rs will the same, but opposite polarity to  the input voltage. The current through Rs will be equal to -V/Rs. The current taken by the op-amp and your sensor both passes through Rs so this is compensated for. .

Ah I see. But at Vin=0V I get only 2.6mA.

[C]

The loop is a power source, a 4-20ma receiver, wire resistance, 4-20Ma transmitter, wire resistance and back to power source.
or
The loop is a power source, wire resistance, 4-20Ma transmitter, wire resistance, a 4-20ma receiver and back to power source.

Look at the extremes

1. 20Ma out what is min voltage needed across your 4-20ma transmitter.
2. 4Ma out what is max voltage allowed across your  4-20ma transmitter.
3. What Bandwidth of signals is supported.

With the limits set by #1 & #2 how linear is the change?
In real world all kinds of things happen.
V1 could be nice DC or could have all kinds of noise added. A long run could pick up a lot of AC power noise or other noise.

Think there is a need to look at more than just static DC

[Zero999]

Ah I see. But at Vin=0V I get only 2.6mA.

How did you calculate that?

Then Vin=0 the current taken from the power supply will be equal to the current through the LM317 circuit plus the quiescent current taken by the op-amp and the sensor.

[jondow]

From simulating with ltspcie. Because I need everytime 4mA avaliable. I actually have found an working schematic (in the Attachments), but I dont know how to calculate R3, R4, R5, R6. Thats the perfekt working circuit, but wihout explanating how to calculate.

[Zero999]

I don't have LTSpice on the PC I'm using at the moment so can't see that schematic.

Why do you need 4mA when Vin = 0, when the lowest input voltage you expect to see on the sensor is 0.2V? If the sensor becomes disconnected and the output voltage falls below 0.2V, causing the current to fall below 4mA, then this would trigger an error in the current loop monitor and error detection is one of the main advantages of using a current loop.

If you need 4mA all the time then why not just scale it from 4mA when Vin = 0V to 20mA when Vin = 1.2V? That would make it much easier to design.

[v8dave]

As long as your sensor draws far less than 4mA then simply use an XTR115 or XTR116 to convert the sensor output to current.

Texas Instruments have a version of Tina and a model of the XTR115/6 you can use to test your design with.

[Zero999]

As long as your sensor draws far less than 4mA then simply use an XTR115 or XTR116 to convert the sensor output to current.

Texas Instruments have a version of Tina and a model of the XTR115/6 you can use to test your design with.

Yes, that's a good idea.

The problem is, it will still be tricky to get 4mA, when the input voltage is between 0 and 0.2V, then increase it linearly from 4mA to 20mA, as the voltage rises to from 0.2V to 1.2V.

It could be possible to set the offset current to 4mA and the gain so the current will be 24mA, when Vin = 1.2V and subtract 0.2V from the input voltage using another op-amp. The input voltage would then be scaled to 0 to 1V but won't be able to go below zero, as the op-amp will have no negative rail. As Vin is between 0V to 0.2V, the output of the op-amp will be still 0V but as the input increases further, the output voltage of the op-amp will be equal Vin - 0.2V.

As I said above, to avoid this complexity, the original poster should consider dropping the 4mA between 0V and 0.2V requirement.

[jondow]

I don't have LTSpice on the PC I'm using at the moment so can't see that schematic.

Okay have a picture attached. There ist the signal low=2.5V=4ma, high=5V=20ma and 0V=4mA.

As I said above, to avoid this complexity, the original poster should consider dropping the 4mA between 0V and 0.2V requirement.

But dont know how to calculate this 4 resistors.

I cant drop it, because its leading standard for industrial sensors.

The 4mA of current not required for information transfer serves two purposes: it can furnish power to a remote module, and it provides a distinction between zero (4mA) and no information (no current flow). In a 2-wire, 4-20mA current loop, supply current for the sensor electronics must not exceed the maximum available, which is 4mA (the remaining 16mA carries the signal).

[Zero999]

Okay have a picture attached. There ist the signal low=2.5V=4ma, high=5V=20ma and 0V=4mA.

That only works because the current in R1 gives the minimum current of just under 4mA, with the rest being taken by the op-amp. What about the current taken by your sensor? Any current taken by your sensor will be added on to the 4mA.

I cant drop it, because its leading standard for industrial sensors.

The 4mA of current not required for information transfer serves two purposes: it can furnish power to a remote module, and it provides a distinction between zero (4mA) and no information (no current flow). In a 2-wire, 4-20mA current loop, supply current for the sensor electronics must not exceed the maximum available, which is 4mA (the remaining 16mA carries the signal).

I think you've misunderstood.

I didn't mean you should scrap minimum current of 4mA. I meant the voltage should be scaled between 0V and 1.2V to 4mA and 20mA. It's starting at 0.2V, rather than 0V, which is creating the additional complexity.

[jondow]

That only works because the current in R1 gives the minimum current of just under 4mA, with the rest being taken by the op-amp. What about the current taken by your sensor? Any current taken by your sensor will be added on to the 4mA.

You are right! But this problem have your schematic too! If I change RL from 2k2 to a lower value then there is additional current added to the 4mA.


I can

I didn't mean you should scrap minimum current of 4mA. I meant the voltage should be scaled between 0V and 1.2V to 4mA and 20mA. It's starting at 0.2V, rather than 0V, which is creating the additional complexity.

I think thats not a problem, I can scale my Signal voltage from 0V (0%) to 1.2 or 4V or whatever (100%). But i have only one positive supply.

[Zero999]

You are right! But this problem have your schematic too! If I change RL from 2k2 to a lower value then there is additional current added to the 4mA.

Of course, changing the value to anything lower will cause it to draw more current. However increasing the value will no longer cause the current to fall.

I think thats not a problem, I can scale my Signal voltage from 0V (0%) to 1.2 or 4V or whatever (100%). But i have only one positive supply.

That makes it much easier then.

[v8dave]

The problem is, it will still be tricky to get 4mA, when the input voltage is between 0 and 0.2V, then increase it linearly from 4mA to 20mA, as the voltage rises to from 0.2V to 1.2V.

I see your point. I loaded up the simulation in Tina and I could not get any offset or range to work with with 0.2 offset.

[jondow]

Have found this Howland-source: http://www.elektronikpraxis.vogel.de/analogtechnik/articles/397228/
Have used the 2nd Equations. It works fine with input signal 1 to 5V I have calculated it that at 1V input I have 4mA and at 5V I have 20mA. But i forgot that I must have 4mA at 0V input too!

Do you think it is possible to mod this circuit that I get 4mA at input 0V too? I have R1=R3 2750 ohm, R2= 1k, R4 1k1, R5=100.

[Zero999]

You could scale the input voltage from 0 to 5V to 1 to 5V with a potential divider but you'll need a buffer to mimise the effects of loading.

[meeder]

You could use 3.8mA for 0V input. 3.8mA and 22mA are the default error currents for 4-20mA loops. When the instruments is in a fault state it will set the output to either 3.8 or 22mA so the PLC guys can trigger an alarm.

[jondow]

I have now found the best working circuit. It works from 2.5V to 5V V_signal. R3 and R4 is a voltage divider sets the voltage at opam minus to 0.5V because lt317_out is 5V. With R5 I set the initial current of 4mA. At v_signal 0V I get 4mA.

But how to calculate R1 and R2 to change v_signal maybe from 1V to 5V? How does it work?