# 4-20ma Current Source PLC Question

Discussion in 'General Electronics Chat' started by chevywaldo, Feb 25, 2009.

1. ### chevywaldo Thread Starter New Member

Feb 25, 2009
2
0
I have a question. The company I work for makes PLC's. On the PLC are analog outputs. These outputs put out 0-20 ma. Most devices we control require 0-10V input signals. Therefore we typically install a 500 ohm resistor across out analog output terminals to convert the 4-20ma in 0-10VDC. Simple - right?

Apparantly that is true only if the connected device has a high input resistance (impedance). Most devices we control have input impedance of 50K ohms or higher. We can therefore connect several of those in parallel and still be able to sustain our 0-10V signal across the terminals.

However, when a device with a low input impedance (10K ohms or less for example) is connected, our PLC output (with the 500 ohm resistor across it) cannot sustain it's required voltage level to drive the device. It appears that the low impedance load acts like a short circuit and the PLC does not have enough capacity to source that low impedance load.

Can somebody explain this please. Why this happens, how this happens, what can be done, etc.

We have been using higher resistance resistors in lieu of the 500 ohm and that sometimes helps - depending on the situation.

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
E = IR would strongly suggest you only have a 10 volt source available to drive the 4 - 20 ma current. A higher source voltage plus beefed-up circuitry (if needed) should let you work with lower input resistances. For instance, 20 volts would be adequate for driving 20 ma through 1000 ohms.

3. ### chevywaldo Thread Starter New Member

Feb 25, 2009
2
0
maybe you can help me understand how a 4-20 ma driver circuit works. I don't know. Are you saying there is a maximum voltage available to drive the 4-20 signal, and that maximum voltage on my PLC is likely 10V?

4. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
This is by using Ohm's law, where the voltage in a system is equal to the product of current times resistance (E = I*R). For 20 milliamps through a resistor of 500 ohms, it takes a source of 10 volts.

For more detail, here is a link to an application note on designing current loops - http://www.maxim-ic.com/appnotes.cfm/appnote_number/1064

Doesn't the company explain the theory of it's own equipment?