How a current buffers works, I want to make (1-5)V to (4-20)mA V-I converter, how can I make this circuit load independent?
Then you need to measure the current, you can either measure the current through a resistor on the low side, or on the high side.

High side means that neither of the ends of the resistor are connected to ground. 

Low side means that one of the ends resistor are connected to ground.

Low side is much easier to measure, but it also means that whatever current you are trying to measure through won't have a real ground. It will have a small resistor before its ground. 

If you're going to measure high side then you will most certainly need an instrumentation amplifier.



Here is a schematic that measures the high side. 



Link to circuit


The input is the "+4" on the rightmost op-amp


If you're going to make an instrumentation amplifier yourself out of four individual op-amps and four resistors, then don't. The precision needed for the resistor has to be within 0.1%, they must be matched. That's why you look here and find this one.



If whatever you are trying to control is okay with low side measurement, then you can use this circuit. 



Link to circuit


The input is the "+4" on the rightmost op-amp


In this setup there's no matching required, so you can easily do it with two individual op-amps. You don't need a special IC for low side measuring. 



Extra notes:

You don't have to use 1 Ω shunt resistor. You can use 5 Ω, 0.3 Ω or whatever. The problem with large resistances is that they will dissipate a lot of heat, or you need to drive your system with 12 V, or maybe even 24 V so you can force the current through your large resistance.

If you however use a small resistance for measuring the current, then you need to increase the gain. In the two schematics above, the gain is \$\frac{100k}{400}=250\$. If you would use a 0.1 Ω resistor instead the gain would have to be 2500. The bandwidth of the op-amps goes down as gain goes up. So if the unity gain of an op-amp is 1 MHz and you use a 0.1 Ω resistor and a gain of 2500. Then the bandwidth of this new system will be \$\frac{1\text{ MHz}}{2500}=400\text{ Hz}\$. 

Or simply change the voltage range, from 1-5 V => 4-20 mA to 0.1-0.5 V => 4-20 mA. But then the SNR will decrease. Noise will affect your op-amp more. So you will need more / better shielding than the two previous paragraph's.

In electronics it's all about compromises. You can't have it all. But you can cascade/cascode things in a clever way to reduce many problems.