Level Transmitter

vikky_rathi

The question is of one without technical knowledge of instrumentation, or nothing to do with engineering at all.

Level transmitters do not have ports, they only have terminal connectors for electrical wires and cables. 2-wire system is most common nowadays, giving standard outputs of 4-20 mA DC. The transmitter is a current drain device (somewhat like a potentiometer), connected in series with the power supply and the rest of the circuit (recorder or controller inputs).

However, there are some 4-wire systems still running, they give Voltage output of 0 to 10 Volts DC or 0 to 5 Volts, they are rare these days, as the wire resistance hinders the transmitter to provide integrity signals over a long distance, the voltage drops over the wire resistance.

There are so many types and principles of level transmitters, some simple and has direct contact with the fluid, and some non invasive like those using ultrasonic method, and weight system, and some even with Gamma rays which is Radiation Class.

Go and do your homework first, then write down on a piece of paper, your question in technical terms and the gurus will be glad to help you.

anonymous

For a 4-20mA trnasmitter it depends on the supply voltage, the loop resistance and the input resistance of all the devices connected to it. Most instrument manufacturers give this data in their technical descriptions.

Many analog controllers have a 'retransmit' output, where the input is boosted and sent on to other instruments elsewhere, which can be handy. If stuck, several manufacturers do a thing called a 'current mirror' or a 'repeater', which supply the same mA output as is seen on the mA input, though at a boosted loop voltage, enabling more things to be hooked on downstream. Other techniques might involve connecting the instrument to an analog input to a PLC and either displaying the value with some sort of human-machine interface or retransmitting the input with analog outputs on a 1:1 basis.

The best bet is to have a chat with an experienced Controls & Instrumentation Engineer to determine solutions on an application-by-application basis.

If you need all over the place-even via Internet--

Then go Digital

National Instruments is just one familiar name.

As PWSlack says, it depends on your supply voltage.

Let's assume you have a 24 Vdc supply. Most current signal inputs have 250 ohms as a load. 20mA passing through a 250 ohm load drops 5 volts across it. If you put another 250 ohm load in series, that's another 5 volts (for a total of 10 volts).

Using the same type of current input, you'll find that four 250 ohm inputs equal 20 volts and that five 250 ohm inputs equal 25 volts. Obviously, you do not have 25 volts to give so four 250 ohm inputs are all you can put on one single current loop.

For an equation, you can use the formula R_t = V_s/0.02, where R_t is the maximum resistance you can have on your current loop, and V_s is your supply voltage. Remember, however, that R_t includes the total resistance of your signal cable.

To be on the safe side, instead of 0.02 as your divisor, substitute 0.024 (some instruments can generate as much as 21 to 23mA to signal over-range). Your equation now becomes R_t = V_s / 0.024.

From my experience, however, there are no more than two loads that are connected to a current loop. On rare occasions, I found three loads. I've never seen four loads on a current loop.