Electronic – Working of JFET as amplifier

What you have is probably two strain-gauges in each package forming a half-bridge.

The wire colours could mean anything so what you have to do is take a multimeter and measure the resistance. You should be able to figure out what you measure from this: -

^{simulate this circuit – Schematic created using CircuitLab}

Between two of the three wires you should read a resistance that is twice as high as the resistance between the 3rd wire and either of the original wires. The wire in the middle (above circuit) is the input to your amplifier and the wires at the top and bottom is where you apply your excitation voltage or current.

As a start, two of the three-wire load-cells can be excited together and the individual outputs fed to an instrumentation amplifier but, there's one more thing; whatever colour are the excitation wires, make sure they are applied oppositely on one of the load-cells - this ensures that one output rises with load whilst the other one falls. If you don't do this you will not measure a signal.

That gets you two load-cells into one instrumentation amplifier and to get 4 load-cells either parallel them up or use another instro amp.

The image below is your circuit with the base clamp transistor Q2 flipped so it does not look like a "twister" session :-).

The circuit should work "well enough" IF you have connections correct and IF Q2 collector actually connects to Q1 base - there is some doubt fom the supplied diagram whether this connection exists (see circuit at end).

BUT the CC circuit and CV circuit are fighting each other. As Q1 current is limited by Q2 the Vout drops and the opamp drives Q1 harder to provide more voltage and current rises so Q2 .... . This is probably a race to the death. Maybe not.

If you move the Q1 Q2 block "upstream" and provide a new Q3 driven by the opamp as Q1 was then this separates the two functions. Ultimately you can only have CC or CV dominant at any one time so one has to "lose" but in simulation this may give you better investigative control. You do not say what "when I lower the resistance" means in Ohms or what current you then see or if this was bricks and silicon or simulation (presumably the latter).