Hi, I have been trying to build a 4-20 milliamp tester but I need soom help. I,m an Electrician and when it comes to semiconductors i,m a little lost. I work on equipment that uses 4-20 feed back. I want to build a simple tester i can plug into when i,m fault finding, calibrating. I have a 10K pot and a supply of resistors, but dont have the brains to work out where and what value resistors are required. Does anybody have a simple circuit i could use.

Thanks
Doug

 

We need a bit of help here, Doug. Are you trying to generate a 4 -20 ma signal, or simply sense its presence?

If it's a sender, then this link might be interesting - http://www.edn.com/article/CA419566.html.

 

Basically, all you would need is a 100 Ohm resistor wired in series with the circuit, and a voltmeter to measure the voltage across it.

If there is 0v across the resistor, the circuit is open. (problem)
If there is 0.4v/-0.4v across the resistor, that is the 4mA signal, or a logic low.
If there is 2.0v/-2.0v across the resistor, that is the 20mA signal, or a logic high.
This works because I = E/R, or Current = Voltage / Resistance.

I suggest that you don't go higher than 100 Ohms with the resistor's value, as 4-20 circuits are usually limited to around 24v to 28v; using a large resistor may introduce faults in an otherwise working system.

You could simply use an ammeter directly. However, if there is a problem in the circuit, it's easy to accidently subject your meter to overcurrent, thus blowing the internal fuse. This can be very inconvenient, particularly if you don't have spare fuses for your meter on hand.

 

Doug,

If you have a schematic of your circuit, it would help us help you if you could post it.

hgmjr

 

Thanks for the replys. Yes i want to generate a 4-20 milliamp signal. I need it adjustable thru out 4 to 20 milliamps so i can calabrate the monitoring software. I understand your theroy regarding the 100 ohm resistor, but i thought 4-20 was based on 1-5 volts. (ie) 1 volt = 4 ma, 5 volt =20 ma.

Regards
Doug

 

Here is an example of a current loop tester.

hgmjr

 

PS. sorry i dont have a schematic of my project, thats what i was hoping you could help me with. I,m purely a plug and play guy when it comes to semiconductor component level.

Regards
Doug

 

If you wond to get up and running quickly there are a number of calibrator units that are commercially available. Such as this unit.

hgmjr

 

Thanks hgmjr, i was looking at those on the net. But i,m on a ship in the North sea so that is out of the question. As I said earlyer, i,m limited to a few pots (10K) and a selection of resistors and caps. I dont need this to be perfect, it will mainly be used for testing the circuit after the L.V.D.T.

Cheers
Doug

 

Here's a cheapie quickie I just threw together. See the attached schematic.

When you're verifying it's output current, your milliammeter should be as shown in the schematic.

When you close S1, S2 should give 20mA out in one position, and adjust R2 for 4mA in the other position. Make sure you adjust R2 to be about 312 Ohms before you install it in the circuit.

As shown, the 20mA output is not adjustable. You could replace the 62 Ohm resistor with a 100 Ohm pot, but make sure you set it to 60-65 Ohms before you wire it up. The lower the resistance, the more current will be passing through the resistor.

The red LED should light up brightly when it's cathode (the shorter lead) is connected to the battery -. It should still light up on the 4mA setting, but not nearly as bright.

A standard red LED has a forward voltage somewhere between 1.7 and 2.2v. Standard 9v "transistor" batteries start off at about 8.6v. The LM317 will drop about 3v across itself in current regulation mode.
That gives you 8.6v - (1.7v to 2.2v) - 3v = 3.4v to 3.9v for drops in your wiring.

A "transistor" 9v battery has a rating of about 150mAh, which isn't much. However, running it at 20mA should still get you about 7 hours of constant use.

 

Dang, Doug! That IS limiting!

OK, do you have some kind of portable power source, like batteries?

I'm afraid that capacitors aren't going to be much help here. The 10k pots may not be of much use either - but that remains to be seen.

In order to create a constant current source, you need some sort of DC voltage source such as a battery or batteries connected in series, and a means of limiting current.

I = E/R. This is a known fact.
So if you know the resistance of your load, you can apply a voltage across it to attain the desired current flow.

Your 10k pots are quite large in resistance value, and they would not likely survive if you attempted to use them for this project; you may also do grave damage to the systems in place. They would pass almost no current until they got down to a small percentage of their travel, and then they would likely overheat, or you would accidentally apply too much current to the circuit and burn it out.