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let's consider this simple example of current source:

enter image description here

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:

enter image description here

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?

enter image description here

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    \$\begingroup\$ You are not sensing current anywhere. And without Neg FB , the Op Amp is open loop comparator. Consider: Howland Current source or simply allaboutcircuits.com/uploads/articles/BMCS_diagram1.jpg \$\endgroup\$ Commented Dec 22, 2019 at 16:57
  • \$\begingroup\$ Your second schematic doesn't have to involve an opamp. Perhaps you think so? You could, for example, just use a high voltage and a large resistor value on the left side to get the job done. I think that you think that the 2nd schematic is supposed to somehow use your 1st schematic and that's how you came up with the 3rd schematic. But if so, you are way, way mixed up. \$\endgroup\$
    – jonk
    Commented Dec 22, 2019 at 19:09

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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 up 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

schematic

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.

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  • \$\begingroup\$ Interesting... It can be considered as a 'MOSFET current mirror with improved input part' - the op-amp makes it behave as an "ideal diode" with zero voltage drop across it. Is it your idea? \$\endgroup\$ Commented Dec 22, 2019 at 22:28
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    \$\begingroup\$ @Circuitfantasist I use the MOSFET inside the feedback loop to force to a high precision value its drain current \$I_{D_{M_1}}\$ and feed its gate to source voltage to the other MOSFET(s) in the mirror arrangement. The voltage across \$M_1\$ is not zero but is nevertheless constant and equal to \$+V_{CC}\$. \$\endgroup\$ Commented Dec 23, 2019 at 7:19
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The first circuit is really a perfect current source made by an op-amp. Its function is to compensate the voltage drop across the load by adding the same voltage in series to the load (figuratively speaking, the op-amp output acts as a negative resistor with resistance -RL that neutralizes the positive resistance RL). As a result, the simple input current source (composed by VBIAS and R) "sees" zero virtual resistance and voltage (virtual ground)... and its current I = VBIAS/R does not depend on the load.

The problem of this circuit is that the load is "floating" (not connected to ground). So, if you want to pass its output current through the input part of the current mirror (as you have drawn in the third picture), there is only one way - to supply the op-amp circuit (or the current mirror) by a separate floating voltage source.

Fortunately, there is no need to make such a weird connection since the input part of the current mirror behaves as a "diode" (aka active diode) with low and relatively constant voltage across it, i.e., voltage stabilizer or a kind of virtual ground. This behavior is achieved by connecting the gate to the drain (a voltage type negative feedback). So this "load" will not affect the current and you can drive it directly by the humble current source VBIAS+R.

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