If you have a instrument loop consisting of a transmitter with a 4-20ma output going into a DCS (process control computer) if the transmitter has a 0% output and you place a VOLTMETER accross its terminals what will voltmeter read ? 1 volt or power supply voltage? Same loop with 100% output will the voltmeter read 5 volts or power supply voltage? THANKS

The voltmeter will read 0V.
It is a current source not a voltage source.

A 4-20mA process control loop transmitter will be scaled to output 4 mA at zero % of the process value, and 20mA at 100%.
If you are using a 250 ohm resistor at the input terminals of your receiving instrument, and you have 0 % transmitter output, then the 4mA current across your 250 load range resistor will result in a voltage of 1 volt, by ohms law.
Therefore, at 100% process value... 20mA across 250 ohms = 5 volts.

[This message has been edited by Redsy (edited 09-25-2005).]

hypress, if you're asking about the terminals on the DCS, the Redsy is correct.

If across the transmitter terminals (and it's loop powered) , then you'll get a different reading alltogether. And you'll have to know the voltage at the tramsmitter (with respect to P/S return) to estimate transmitter output.

Be aware that while many 4-20ma DCS (or PLC or process instruments) inputs do have a 250ohm resistor, there are a lot that use a lower value.

[edit for spelling]

[This message has been edited by Eng_Dave (edited 09-27-2005).]

I 'm just curious, .....why would you need to know the voltage in a current loop?

Good call Eng Dave.
I didn't notice that the question was about the reading at the transmitter terminals, in which case your reading would be different.
Interestingly though, from the negative transmitter terminal to ground you often see the same voltage that you would across receiving devices input terminals.

Togol,

Many instrumentation current loops actually drop the 4-20 mA current across a 250 ohm resistor to supply a 1-5 VDC input signal to the controller or recorder.

Ok, I just need to read better ... (pulling head out of ...) .. lol
Thank You Redsy

Waitaminnit...

The reading at the transmitter is probably not very different from the reading at the process control computer. The only difference is the wiring resistance.

If you've got 10 ohms of loop resistance (that's 250 feet of awg-20 cable) your 0% voltage at the transmitter is only 0.04 volts higher than at the computer.

As Eng_Dave pointed out, though, the computer's input resistance could be much lower than 250 ohms, and there's no reason why it couldn't even be zero ohms, which would make voltage readings pretty irrelevant.

That's why current-loop transmitters are so popular: they completely eliminate the influence of voltage drop. Current at the transmitter is equal to current at the receiver, no matter how much wire there is in between.

That also means that the right way to check them is by connecting your DMM as a milliammeter in series with the transmitter, instead of as a voltmeter across it.

An aside: When doing initial comminisoning / testing loops, I have the guy in the field jumper the transmitter terminals with a high value resistor (1K or so) and check the DCS (PLC, PC, whatever) for a sudden high reading (or Off Scale High) as most times I'll have a dozzen or two temperature / pressure / flow transmitters and they all will be reading ambient. A quick way to see that the field wiring guys hooked up the wires to the correct one.

Caution:

Using a too small resistor and the 250 ohm (or lower) resistor at the DSC might burn up if not rated for the wattage - some Siemens PLC analog input cards are fond of doing this.

Or if using I./S. barriers you can blow the internal fuse in them.

But heck, it is a whole lot easyer than removing a loop wire in the transmitter.

The reason I was checked the voltage on a current loop is we were doing some trouble shooting on a unrelated electrical problem that was showing up as a instrument problem.We had a indication that a soot blower was running and we knew different. We were actualy reading ground current and the grounded phase went through the motor running CT and I made a quick check with a voltmeter at the current transducer terminals.

Just wondering what the problem is.

Two different sorts of loops exist:

1. The transmitter actively drives the loop because it is externally powered.(4-wire)
Then the voltage over the transmitter terminal will be same as the voltage over the rest of the loop, but inverted polarity. If there is just one consumer in the loop and wiring is short, then voltages may appear to be identical on source and consumer but inverse.
For a 50 Ohm consumer you will need 0.2V at 0% and 1V at 100%, for a 250 Ohm resistor see above.


2. The loop (not the transmitter) is powered externally by f.i. 24VDC. Then the (2 wire)transmitter is one more consumer in the loop and must limit current to give correct reading at receiving consumer's terminals. Voltage is not inverted and equals just the rest that is not consumed by consumer and wiring:
example 24VDC/50 Ohm : 24V - 0.2V = 23.8 V (wiring not regarded)=> 0% or 24V-1V=23V => 100%

250 Ohm: 24V - 1V = 23 V(0%) and 24V-5V= 19V(100%)

Btw: A disadvantage of 2wire transmitters is that in case of failure of transmitter on certain models you might get a maximum signal like 22mA. So you have to have a routine to treat readings higher than 21mA as a failure in some applications.

A 4-wire transmitter usually will just give low or no voltage at all in case of failure, which is easier to treat.


[This message has been edited by Wolfgang (edited 10-03-2005).]

John,
That would be true for a four wire loop, but not for a two wire loop. On a two wire loop, you will read close to the supply voltage across the transmitter terminals.
Don