How does this constant current circuit work?

I answered this SE EE question yesterday. Here is my answer.

I rely on the OP's claim that the circuit solution in question is an "improved Howland current source" and offer my explanation of the idea on which this famous circuit is built.

The idea

We can figuratively call it "dynamic voltage source". In some (usually transistor) current source implementations, to keep the current constant, we vary a resistance RI connected in series to the load RL while keeping the supply voltage V constant. With the same success, we can vary the voltage V while keeping the resistance RI constant... i.e., to supply the RI-RL network by a "dynamic voltage source" (see Fig. a on the left below). For example, if RL increases its resistance, the voltage drop VL = I.RL across it will increase... but the voltage source will increase its voltage V with the same value... and the current I = V/(RI + RL) will not change. Figuratively speaking, the voltage increase removes the load resistance increase... as though it acts as equivalent negative resistance that removes the positive resistance increase.

The implementation

An excellent implementation of this technique is the ingenious circuit solution of the so-called "improved Howland current pump" (Fig. b on the right above). Here the op-amp acts as the dynamic voltage source V which output voltage is "lifted" (by the inverting input) with the constant voltage VRI above the load voltage VL. The output voltage follows the variations of the load voltage VL by the mechanism of a positive feedback. The combination of the op-amp and the four resistors R can be thought as two cascaded circuits - the voltage divider on the right with ratio of R/(R + R) = 1/2 and the non-inverting amplifier on the left with gain of (R + R)/R = 2. Thus the total transfer ratio of the whole circuit is 1 and the load voltage variations (at the lower end of RI) appear at the op-amp output (at the upper end of RI). As a result, there is a constant voltage drop across the constant resistor RI; so the current through RI and RL is constant (does not depend on the load). The circuit can be thought as of a "shifted voltage follower" that behaves as a current source with extremely high differential internal resistance.

Is there a negative feedback here?

Note, although there are two kinds of feedback - negative and positive, this current source does not use any feedback to keep the current constant... it is a current source without negative feedback. The circuit blindly corrects the input voltage without monitoring the final result - the load current. That is why, the resistances should be precise.

Why two more op-amps?

Since I am mainly interested in circuit ideas, I work with conceptual circuit diagrams with symbolic notations of components. Because of this, the OP notations (with pin numbering) make it hard for me and I can only guess what it is. But still, my guess is that the two additional op-amps are buffer amplifiers that make the circuit more precise.

Maybe, the first op-amp is inserted between the input voltage source VREF (see my picture above) and the resistor of the inverting input; maybe, the second op-amp is connected between the load (RI-RL network) and the voltage divider of the non-inverting input.

See also

You can see more about the philosophy behind current source implementations in my Codidact paper How to create current sources. The improved Howland current source is considered in the section 5 figuratively named "Dynamic voltage source".

Also, in my circuit story Widlar Op-Amp Current Source for Grounded Load I have visualized the circuit operation by voltage bars and current loops. I have named it after Widlar because I first saw it described by him (as a "bilateral current source") in his paper IC Op Amp Beats FETs on Input Current in NSC's AN 29.