Q:

Implementation of Current Source for a Current Mirror

let's consider this simple example of current source:

It is a current source because each load I connect to it, it will receive a value of current equal to Vbias/R, which does not depend on the value of the load. So, it is correct to represent it as a current source, like in the scheme at right.
Now let's consider this example of current mirror, whoose aim is that of generating a current for the block called "circuit" equal to that provided by the source current we have previously created:

In theory, to realize this last circuit, I would put the initial current source with op - amp inside this current mirror. But It seems strange to me: the op amp feedback results open, and the current flowing in R is not more Vbias/R. Where is my mistake?

A:

As said in the comments, and noticed by yourself, the main problem of your circuit is that there is no feedback so the OPAMP will go in negative saturation as soon as you power the circuit: this means that \$V_{GS}\approx 0\$ for both MOSFETs, so no \$I_D\$ will flow. You should change the circuit in order to avoid both the absence of feedback and the main problem of the inverting amplifier "current generator" which is the need of a floating load, as noted by Circuit fantasist in his answer. My proposal is the following one

simulate this circuit – Schematic created using CircuitLab
Circuit description.
While trying to keep its non inverting input node at the same potential of the inverting input (i.e. realizing the so called "virtual ground") the OPAMP forces the gate voltage of \$M_1\$ to make the drain current equal to the current flowing in the node, i.e.
$$
I_{D_{M_1}}\simeq \frac{V_{CC}}{R}
$$
And since \$V_{GS_{M_1}}=V_{GS_{M_2}}\$ if \$M_1\$ and \$M_2\$ have very similar characteristics, you get \$I_{D_{M_1}}\simeq I_{D_{M_2}}\$, as you want.
Edit: a note on the polarity of the comparison node. The comparison node must be the non inverting one, since the rise of \$V_+\$ must imply the rise of \$V_{GS_{m_1}}\$ and in turn the rise of \$I_{D_{M_1}}\$ which causes a decrease in the same voltage: simply stated, the overall feedback has to be negative, and this implies also the need of considering the phase inversion given by the common source linear amplifier made of the \$M_1\$ MOSFET.

