Checking Polarity

[Dale Hayes]

I have a temperature controller with a 4-20mA PV retransmit. I know the two terminals to connect the wires, but I am not for sure which terminal is for the positive wire and which is for negative wire. Is there a test procedure I can use to verify the (+) and the (-) connection? I do not want to end up reversing the two wires.

 

[gar]

131206-1455 EST

What is a "4-20 mA PV retransmit"? Is this the temperature controller?

You mention a temperature controller. Is the temperature controller a 4-20 mA output or input device? By that I mean is it the information (data) source (transmitter) or is it the receiver of information?

In this current loop I will assume one source and one receiver. There has to be an energy source somewhere in the loop. This could be in the transmitter, receiver, or external to either of these. Where is the energy source?

Is the energy source in this "4-20 mA PV retransmit" thing , or elsewhere?

A receiver of information will contain a resistor (call it a shunt) that is used to produce a voltage proportional to the current thru the resistor. The shunt voltage will be amplified and processed. Reverse polarity may be no problem here other than the system won't work correctly. Probably no damage to the receiver. If the shunt was at the center of a diode bridge circuit, then the input could be of either polarity. But this adds a little extra voltage drop in the current loop.

A transmitter (source) in the current loop with the energy source somewhere else might be damaged by reverse polarity. But there are ways that the circuit could be designed to prevent damage. If the circuit was protected, and connected backwards, then you would have a circuit that did not operate correctly, but was not damaged.

Wherever the energy source is you can determine polarity of that energy source with a voltmeter.

More accurately describe your circuit and at what point is your polarity question.

You may need to study your component manuals for more information.

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[Smart $]

Got a make and model number?

I can provide a procedure to verify which is which, but the best recourse is the documentation.

 

[gar]

131206-1620 EST

Dale:

Suppose the thing where polarity is your concern is a transmitter (signal source), then:

1. Put an autoranging DC VOM in voltage mode across the two terminals. If the voltage reading is 0, then go to step 2. Otherwise you have an energy source in the transmitter, and the voltmeter reading will tell you the polarity.

2. On the assumption the transmitter is passive (no energy source), then connect a 250 ohm 1/2 W resistor to the positive terminal of a 0-10 V DC adjustable power supply.

Set the voltage to 2.5 V and the power supply current limiting to 10 to 25 mA. The maximum current thru the resistor is 10 mA with 2.5 V applied. If I connect a 1N5060 diode forward biased with a nominal rating of 2 A in series with the 2.5 V and 250 ohms the drop across the diode is about 0.6 V. This is a reasonably safe test circuit because at this source voltage current will be less than 10 mA, and voltage does not exceed 2.5 V. Not likely to damage anything in the current loop transmitter.

If your transmitter contains a shunt reverse biased diode for protection, then with one polarity you should see something in the 0.6 V range. A 1N4148, a much smaller diode, has a drop of about 0.75 V with 250 ohms and a 2.5 V source. The other polarity would produce a higher voltage drop because the current controlling circuit requires some moderate voltage drop. Not likely as low as 1 V.

If the voltage drop across the transmitter terminals is much closer to 2.5 V with either polarity, then there is no reverse biased diode for protection. There could be a series diode, but this adds to transmitter voltage drop, and not likely.

If there is clearly reverse protection, then the polarity that does not cause this conduction should be the correct polarity for normal operation.

A transmitter almost certainly will require some minimum voltage across its terminals to provide current control.

If you have the correct polarity, then you can raise the power supply voltage and monitor the voltage (therefore current) across the 250 ohm resistor as you adjust the source signal and see the current loop current change. The source voltage in the loop has to be high enough to make it possible for the transmitter to control up to 20 mA.

10 V applied to a 250 ohm resistor will cause 0.4 W dissipation in the resistor.

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