There's a very good chance your 4-20 ma is really a 1-5 VDC signal developed across a 250 ohm resistor. )OK, maybe 2-10VDC, check and see.)
If it is, then just use a potentiometer and pick a voltage off the wiper arm for the input. Free if you have one in the junk box, 25 cents at radio shack. Any value will work, just use a pot and not a rheostat.
Use a Battery and a potantiometer.
If the battery provides 9V, you'll need 450 Ohms for 20 mA, 2250 Ohms for 4 mA.
Take a 2.5 or 2.2 kOhms Poti in series with a 390 Ohms resistor, and you'll be able to go savely through the full range.
The simplest design I have is a 470 ohm 2 watt resistor in series with a 10K ohm 1 watt 10 turn pot. The fixed resistor limits your maximum current to 51mA on a 24vdc supply. If you need to read the actual current, put your meter in series with this.
Now, if you do this a lot, get a bunch of resistors and pots, mount them in some kind of panel (we just bent and painted 16ga steel), and add terminals and decently long twisted pair cables.
There is one missing piece of information in your question - resistance of
PLC input (I assume 250 ohm).
If so, a voltage source higher than 5 VDC (say four AA batteris) and a
potentiomeneter (wired as rheostat) can do.
I leave the calculations to you.
Meir.
I have a very good simple design based on a 317 Voltage regulator
It uses 5 switches and gives 4-20 mA in 1 mA increments.
It's very accurate and independent of loop resistance.
If you're only trying to light up the PLC inputs to verify wiring from a field terminal, you could hook up a 10k potentiometer to the existing 24VDC bus. It will max out at 5mA and will trigger the PLC inputs. This would <b><i>NOT</b></i> be a good way to calibrate something, but i use this cheap potentiometer in lieu of a very expensive process meter to ensure the field terminals are on the right PLC inputs. =)
A pot can go all the way down to zero Ohms, so the current in the loop would only be limited by the input resistance of the analog card. A fair number of analog input cards have 50 ohm dropping resistors, so a zero ohm external resistance to the power supply positive on a 2 wire circuit would give you almost 500 mA of current, not 5 mA.
I have run across a few AIN cards that use a 250 Ohm dropping resistor. Even so, the max that would give you would be something like almost 100 mA.