0..40mV to 4..20mA converter design - Page 1
Forum section: electronics / projects

author: joellinn
author_rank: Contributor
timestamp: December 17, 2016, 02:12:35 pm
content: Hi,I want to hook up a MPX 2100DP pressure sensor (potential free output?) to an 4..20mA PLC input (said to be <85 Ohms). I have a stupidly expensive multi purpose industrial converter lying around but what is the fun in using a 400 bucks converter for a 20 bucks sensorI tried to make a circuit that can operate using a cheap dual opamp, lm358 f.e.; but I think the min current output of the op amp (sink) is to low.Do you think my circuit is ok; or should I change to a instrumentation amp setup with 4 opamps or something else? As my personal challenge I thought it would be nice to keep the costs lowR16 is the PLC in this case

author: Marco
author_rank: Super Contributor
timestamp: December 17, 2016, 03:46:22 pm
content: A 3 decade old jellybean shouldn't be your goto for cheap general purpose opamp, we have CMOS RRIO today.My circuit-fu is weak, but I don't really see how your circuit can be a V to I converter.Usually a simple V to I circuit looks like this :

author: joellinn
author_rank: Contributor
timestamp: December 17, 2016, 06:00:29 pm
content: Yes, that indeed is a simple voltage to current converter.But with that design the load needs to be floating potential. PLC inputs usually have one side grounded.To have both input and output referenced to ground my search came up with the circuit on the right.In this case (R13/R12 = R15/R14 and R12 grounded) the Formula is I=U/R14.The problem with that thing however is:A) It loads the input.B) I feel like a resistor Value of about 2 Ohms is a fragile choiceSo i added a non inverting preamp to solve both problems.Please correct me if I'm wrong

author: IanB
author_rank: Super Contributor
timestamp: December 17, 2016, 06:09:16 pm
content: Quote from: joellinn on December 17, 2016, 06:00:29 pmI'm unsure what it means to say the PLC input has one side grounded. Grounded to what?

author: joellinn
author_rank: Contributor
timestamp: December 17, 2016, 08:05:40 pm
content: There is just one input pin which measures current flow to the PLC ground / 0V. And that's gonna be my GND as well.

author: Marco
author_rank: Super Contributor
timestamp: December 17, 2016, 08:32:56 pm
content: Quote from: joellinn on December 17, 2016, 06:00:29 pmDon't get too hung up about ground symbols in a circuit, it's not earth, just a reference point. In this case the negative side of a floating isolated power supply/battery. You can connect earth ground anywhere you wish, including at the bottom of the load.That said, the PLC input might not be simply a resistor. If it supplies voltage to run the current loop sensor you need a different circuit.

author: joellinn
author_rank: Contributor
timestamp: December 17, 2016, 10:46:07 pm
content: yeah sure but like I said, the plc measures current from the input to 0V. As i will be connected to the same power supply my 0V /GND is the same.https://www.beckhoff.com/english.asp?bus_terminal/kl3454.htmYes, it does supply 24V at one pin but I'm not required to use it and I think I can't because the static loss is above 4mA....

author: Zero999
author_rank: Super Contributor
timestamp: December 17, 2016, 10:52:42 pm
content: Does the power supply for the pressure sensor share the same 0V as the PLC?I'd probably use a Widlar or Howland current pump.http://www.circuit-fantasia.com/circuit_stories/understanding_circuits/current_source/widlar_op-amp_current_source/widlar_op-amp_current_source.htmwww.ti.com/lit/an/snoa474a/snoa474a.pdf

author: IanB
author_rank: Super Contributor
timestamp: December 17, 2016, 11:08:50 pm
content: Quote from: joellinn on December 17, 2016, 10:46:07 pmThe KL3454 says the inputs are electrically isolated from the PLC. So there is no concern about common grounds? (Especially if you don't use the supplied 24 V to drive the current loop.)

author: Benta
author_rank: Super Contributor
timestamp: December 17, 2016, 11:44:31 pm
content: There is some confusion here about how 4...20 mA loops work.So here goes:The KL3454 expects a variable-currentloadat the input.The modulesupplies+24 V (with an internal resistance of 80 ohms) from the 'active input', and expects your sensor to conduct 4...20 mA to the '0 input'.In other words, it's a passive sensor current loop input.If you are able to power your sensor and amp with less than 4 mA, it it feasible to make a loop-powered circuit which has an output that is a current sink (simple: opamp, plus open-collector NPN stage with current-sensing emitter resistor). You'll need to think about how to balance sensor supply current vs. total sink current, though. One solution is a 4 mA constant-current sink from which you bleed off power to the sensor/amp circuit, and let the amp sink 0...16 mA on top.Benta.

author: IanB
author_rank: Super Contributor
timestamp: December 17, 2016, 11:56:17 pm
content: My reading of the documentation is that it is optional whether you use the supplied 24 V to drive the current loop. I understand it like this:Option 1: You don't use the supplied 24 V. In this case you connect the two current loop wires between the signal input and 0 V. The sensor must actively provide the 4-20 mA current.Option 2: You do use the supplied 24 V. In this case you connect the two current loop wires between 24 V and the signal input. The sensor is now passive and must simply conduct 4-20 mA back from the 24 V supply to the signal input.Do I not read it correctly?

author: IanJ
author_rank: Unknown
timestamp: December 18, 2016, 12:00:49 am
content: Back in the day MAXIM used to do a range of 4-20mA IC's for various types of hook up, active, passive including loop powered if I remember correctly. The one I used a lot was the XTR110, simple V to 4-20mA all in one package.With 4-20mA intrumentation you have to consider the load may change, i.e. a loop powered indicator could be inserted in the loop so it's important to be aware.Ian.

author: IanJ
author_rank: Unknown
timestamp: December 18, 2016, 12:07:42 am
content: Quote from: IanB on December 17, 2016, 11:56:17 pmImagine the PLC input as a 250ohm resistor on it's negative terminal (-) to 0vdc and you can't go too far wrong, but yes you don't need to use the PLC supplied +24v output (+)......depends on the complexity of the PLC system but it's usually better to do so. Most PLC inputs give you various ways of hookup, and there can be up to 4 terminals on each channel to give you the options.Ian.

author: Benta
author_rank: Super Contributor
timestamp: December 18, 2016, 12:09:51 am
content: Quote from: IanB on December 17, 2016, 11:56:17 pmNo. See above in my post #9.Quote from: IanJ on December 18, 2016, 12:00:49 amWhat??

author: Benta
author_rank: Super Contributor
timestamp: December 18, 2016, 12:11:38 am
content: Quote from: IanJ on December 18, 2016, 12:07:42 amGibberish.

author: IanB
author_rank: Super Contributor
timestamp: December 18, 2016, 12:15:07 am
content: Quote from: Benta on December 18, 2016, 12:09:51 amI disagree with your post #9. Can you provide some supporting evidence?At this point I agree with IanJ. Everything he wrote makes sense.

author: Benta
author_rank: Super Contributor
timestamp: December 18, 2016, 12:38:35 am
content: Quote from: IanB on December 18, 2016, 12:15:07 amDoesn't make sense to me, but 'when two like minds meet'. Here you go:The basics:https://en.wikipedia.org/wiki/Current_loopApplication information on KL3454:https://download.beckhoff.com/download/document/Application_Notes/DK9221-1111-0059.pdfPlease pay attention to the table on page 5. Sensors connected to KL3454 are "loop powered". If your German is up to scratch, the description supports what I wrote earlier and adds that the sensor is powered from 24 V through the loop.Benta.

author: Marco
author_rank: Super Contributor
timestamp: December 18, 2016, 01:23:09 am
content: His sensor has 6 mA quiescent current. So he needs to run 3 wires (24V, current input and ground) or use a separate isolated power supply or battery at the sensor. His opamp choice and talk of instrumentation amplifier does make sense now, but I still don't grok his circuit.

author: IanB
author_rank: Super Contributor
timestamp: December 18, 2016, 01:49:20 am
content: Quote from: Benta on December 18, 2016, 12:38:35 amOK, I see what you are saying. A 4-20 mA input in general can work with external power, but the KL3454 in particular does not provide that option. It only provides terminals to connect passive 2-wire sensors. When I now look closely at the application note and the documentation this becomes clear.If you wanted to use some other connection arrangement you would have to choose a different analog input card other than the KL3454.

author: Marco
author_rank: Super Contributor
timestamp: December 18, 2016, 01:55:50 am
content: There is an alternative to some kind of difference/instrumentation amplifier. You can feed the MPX2100D with an isolated floating power supply, that way the output also floats and you can use the standard single op amp current loop converter.

author: Benta
author_rank: Super Contributor
timestamp: December 18, 2016, 01:45:47 pm
content: QuoteYep, that would work.Or choose the KL3054, which is able to supply separate power to the sensor.