how do i create a current source
I don't really see the problem. Furthermore, your question is not clear. Your schematic has a R1 value of 330Ω, while the equation you discus has a R1 value of 500Ω

Anyways, the current/voltage ratio is determined by the value of R1. 

Basically, you can size R1 so that any arbitrary voltage on the op-amp input gives you any arbitrary current.



Effectively, the voltage at the negative summing node (e.g. pin 2) of the op-amp will always match the voltage at the positive summing node (e.g. pin 3). (Except when the op-amp is clipping)

As such, if you want a 5V input to result in 20 mA through your load, you simply need to do the math: \$R1Ω = \frac{5V}{0.020A}\$ or R1 = 250Ω. 

However, the output swing of the op-amp will need to be the drop over R1, plus the drop in your I->P converter.
The I->V converter in your device uses a 250Ω  resistor across the input to measure the 4-20 mA control signal. As such, at 20 mA, you will have 5V of drop in the device. 

So basically, with a 0-5V control input, and the device you have, you would need an op-amp able to swing 0-10V. 

However, you also need to account for the fact that there are very few op-amps that can drive a load of 20 mA anywhere near their rails (or even at all. The 741 may barely be able to drive 20 mA loads, but I doubt it'll do it anywhere close to the rails). 

Basically, if you have a 12V power rail, and a good, high-current rail-rail op-amp, it should work as drawn.



You do know that the electronic pressure regulator you link to in your question seems to have a plain-old 0-10V voltage input option, rather then a 4-20 mA current input option, right? Why not just use that.