4-20mA To 0-10V Conversion Question

Hey guys,
I have a level sensor, Flowline LU20, for a tank we have here. It is a simple 4-20mA output. However, the company that installed the system installed a Micro 810 PLC which only has 0-10V inputs.

I know you can use a resistor to help convert, but I need a little more help than that. On the original wiring directions from the manual (below on page 4) it shows that you need a 250 Ohm resistor across the 2 wires at the PLC. This of course is if you have a PLC designed for 4-20mA input.

http://www.flowline.com/pdf/ultrasonic-level-measurement/echotouch-lu20-m.pdf

Since I only have 0-10V input, what could I do to make this 4-20mA become 0-10V? Resolution is NOT a worry here. Really, a ballpark close number is OK. It is a VERY non critical level.

I read somewhere about putting a 500 resistor across the input, but I'm not sure if that applies being as the directions already recommend a 250 ohm with the proper setup.

Also, the OEM installed a resistor, I believe it was 170 ohm, in series with one of the leads. Not sure why they did this. It doesn't seem to make the voltage fluctuate hardly at all as the level rises in the tank.

FYI, the sensor is brand new and calibrated, and gives me 4-20mA when connected to another setup, so I know it works.

Any help would be awesome. I would love to know the math behind this trick too if any of you math buffs would like to share...

The 500 ohm resistor across the voltage inputs will give you 2-10 Volts. Can you mathematically make up for that in the program? 250 ohms give a 1-5 volt input.

The resistor converts the current to a voltage for 'voltage' type inputs. If you had a current type input there would be no resistor.

It doesn't say so, but the 250 ohm resistor shown in the manual is not for a 4-20mA input, but for a 1-5 V input.

As Bernie said, if you change the 250 ohm resistor to a 500 ohm resistor you'll get a 2-10 V input.

The mathematics is Ohm's law: V=IR

On a voltage input, the resistance provided by the PLC (input impedance) is considered infinite, so none of the current from the transmitter will flow into the PLC. Instead all of the current will flow through the resistor and provide a voltage drop according to Ohm's law above. The PLC will then measure the voltage drop.

4mA => 0.004 A * 500 ohm = 2 V
20mA => 0.020 A * 500 ohm = 10 V

or

4mA => 0.004 A * 250 ohm = 1 V
20mA => 0.020 A * 250 ohm = 5 V

Got it. Well, I know Ohm's law very well, so I kind of feel dumb for not understanding that. I guess the PLC "input impedance" is what confuses me.

I'll see if I can find a 250 ohm resistor (or combine what I have to make one) and see how that works. Thanks!

You'll want a 500 ohm for your 0-10 V input - not a 250. Unless you feel like tying them in series, but that's a mess.

Good point. Will do.

The 14.7 kohm input impedence will shift this slightly. 525 ohms will be a little closer to what you need. It's basically 14,700 ohms in parallel to whatever you put across the terminals. Otherwise 500 ohms will give just a little bit less than 10 volts when 20 ma is applied.

You could purchase an isolation transmitter to accurately convert the 4-20 mA signal to 0-10 VDC, if accuracy was your paramount concern. Since you have already stated that this measurement doesn't require a high level of accuracy, you should simply go with the resistor parallel to the analog input.

I wouldn't even worry about calculating the value of the 250 or 500 ohm resistor in parallel with the impedence of the analog input. Simply record the number that the analog input returns at 4 mA and at 20 mA. The numerical value at 4 mA is your zero offset and the difference between the number returned at 4 mA and at 20 mA is your span.

The percentage of full scale of your instrument will be ((the instantaneous value of the analog input minus the zero offset) divided by the span). For example, let's assume that you received a numerical value of 200 from your analog input at 4 mA and a value of 20,000 at 20 mA. If you are receiving a value of 10,000 from your instrument when it is in operation, then (10,000 minus 200) divided by 19,800 equals 49.49%.

Thanks guys...

And Bit_Bucket...I wasn't going to ask...but curiosity is really getting to me. What does your signature mean? Hopefully I'm not missing something too obvious...

shoelesscraig said:

Thanks guys...

And Bit_Bucket...I wasn't going to ask...but curiosity is really getting to me. What does your signature mean? Hopefully I'm not missing something too obvious...

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Would this approach (500 ohm resistor) also work for monitoring an existing 4 - 20ma signal, with an additional outboard device requiring a 0 - 10V (2-10V) signal? or would the resistor shunt or load the 4 - 20 signal, producing some degree of error?

Practical application: I've got a stand alone PID controller, delivering a 4 - 20ma signal to an I/P valve. I need to send that signal info (level) to another device (PLC Input). It would end up being a fairly long series loop to deliver the 4-20 signal to the additional PLC. Would this conversion to 2 - 10V potentially work for me?

Thanks for your help.

yes it would, as long as you keep the resistor shunt close to input because current is constant in circuit

I'm not sure I'm clear on your application. Which signal are you wanting to send to the PLC? The valve signal or the level signal? You need to be clear on this.

Theoretically, the short answer is yes - this could work for you.

But there are serious considerations you need to take into account. Whatever device is generating the 4-20 mA signal (either the PID controller or the level sensor) will have a limit to the total impedance allowed on the 4-20mA loop.

For the sake of discussion, I'm going to assume that you really meant that you wanted to send the level sensor signal to the PLC. You'll need to find the documentation for the level sensor and find the limitation for signal impedance there. You'll then need to find the PID controller documentation and find the controller's input impedance for it's analog input signal. If you use a 500 ohm resistor at the PLC, your total impedance seen at the level sensor will be the PID controller's input impedance plus 500 ohm. You just need to make sure that number is lower than that allowed by the level sensor.

If those numbers are close, you'll also need to consider the resistance that is imposed by the wiring itself. You say that you'll have "a fairly long series loop". Depending on how long that is, it may throw you over the limit.

So, if you've managed to do all that and still remain under the limit you should be fine. The nice thing about a current loop is that you don't lose resolution - at least not significantly - when you do multiple devices like this.

If the impedance limit becomes an issue I believe you can buy a booster or device to isolate the impedance, not sure of the name and I have no personal experience with them, but I have heard of them.
It is a similar concept to using a MOSFET or relay to drive a load when the controller cannot provide adequate power (or just for isolation purposes)

It occurs to me that you will want the resistor to be in series with the current load of the loop, and not across the terminals as a shunt. @cbuysse alluded to it in his explanation, but the application is not as straight forward as the OP.