current loops

[jcole]

Hello guys and gals

I started a job last year that has a lot of equipment that uses current loops (4-20mA, 0-20mA,etc) for positioning control.

I am looking for sources to educate me more on the design and troubleshooting of these circuits. One of the questions I would like explained is "What is the deal with 250ohm resisters". Some circuits have them some do not. I am thinking it is to bring circuit resistance down because they are all hooked in parallel.

I also noticed that most of the devices have an input rating in ohms. Why is that important? Do all devices need to have the same input impedence rating?

Any suggetions on educating me on these circuits would be greatly appreciated.

 

[iaov]

The 250 ohm resistor is there because most current loops are 1-5 vt circuits. 5 volt out gives a 20 miliamp flow and 1 volt = 4 ma. I would suggest finding some stuff on constant current amplifiers and a book on instrumentation for process control.Good luck.

 

[jcole]

Thanks for the reply.

 

[gar]

090112-1028 EST

jcole:

The theory behind the idea of current loop signaling is to use current as the signal variable rather than voltage.

If you have a current signal source, then within limits, the variation of the wiring or the receiver load has no effect on the value of the current signal. So you can have long or short or different diameter wires from the source to the receiver and variation of the resistance due to temperature and not affect the accuracy of the current signal, and therefore the originating physical variable being measured is still replicated at the receiver independent of changes in the loop resistance.

This requires that the source produces a constant output current independent of load resistance over some reasonable range (0 to some maximum total resistance) for a specific signal level. The signal might be pressure, position, liquid level, voltage, power, or whatever. The output change of current is made proportional to that signal.

Internal to the source, or someplace in the current loop, will be a voltage source, assume 24 V, from which the constant current device must work. Suppose that the constant current device requires at least 4 V drop to work. That leaves a maximum possible voltage for resistance drop of 20 V. At 20 MA, full scale, this means the maximum loop resistance is 20/0.02 = 1000 ohms. If the receiver of this signal is 250 ohms, then the maximum wiring resistance is 1000 - 250 = 750 ohms.

At 68 deg F. and #22 copper wire the resistance is slightly over 16 ohms per 1000 ft. Thus, the source (sender) and receiver could be 750/(16*2) = 23 thousand feet apart. Actually you need to consider temperature and other margins so maybe your maximum range might be 15,000 ft, about 3 miles.

.

 

[mull982]

090112-1028 EST

Internal to the source, or someplace in the current loop, will be a voltage source, assume 24 V, from which the constant current device must work. Suppose that the constant current device requires at least 4 V drop to work. That leaves a maximum possible voltage for resistance drop of 20 V. At 20 MA, full scale, this means the maximum loop resistance is 20/0.02 = 1000 ohms. If the receiver of this signal is 250 ohms, then the maximum wiring resistance is 1000 - 250 = 750 ohms.

I belive also that each current device has a maximum voltage drop that it can tolerate. According to Kirikov's voltage law all voltage drops around a loop must equal the source voltage. So if the loop uses a 24V source and the device required drop is 4V and the remaining loop drops (reciever and resistance drops) are 16V then you have a total loop drop of 20V. Kirikov's law says that this remaining 4V must be accounted for, so it gets dropped across the current device. The current device will now have 8V dropped across it. You need to make sure that the current device can tolerate this total voltage drop.

Correct me if I'm wrong about this guys.

 

[gar]

090112-1158 EST

mull982:

Suppose the voltage source is in the sender and that it is 24 V, then if we short the sender output the power dissipation at 20 MA (full scale) in the current source is 24*0.02 = 4.8 W. Thus, if the sender has been properly designed it will have a sufficiently large heat sink for the purpose. Pick other parameters and/or set some restrictions and the dissipation requirements will be different.

If you put the voltage source somewhere else and do not add any series loop resistance, then there is still the same amount of power to dissipate.

.