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In my assignment, I want to study this current transmitter circuit shown in the figure 1. The goal of this circuit is to output a constant load current \$ i_L\$ (denoted by \$ I_L\$ in the diagram) controlled by \$ v_S\$ (operating within a certain range of \$ v_S\$ and \$ R_L\$). I derived the expression for relating \$ v_S\$ and \$i_L\$; that expression is (see figure 2) $$\dfrac{i_L}{v_S}=-\dfrac{1}{R_2}$$ The circuit simulation can be found on Multisim online here.

In the assignment question, the circuit is specified to work for $$|v_S|<1 V, R_L < 1k\Omega$$ However, the circuit does not behave according to the relation \$ i_L/v_S=-1/R_2\$ at those extreme conditions (being \$v_S=1 V, R_L = 1k\Omega\$), and it's probably because the op-amp output voltage \$ V_o\$ saturates at \$\pm \$15 V. The expression that relates \$ V_o\$ with \$ v_S\$ and \$ R_L\$ is (see figure 3) $$V_o = -(2 v_S \times R_L/R_2 + v_S)$$ And this must stay within saturation limits. At \$v_S=1 V, R_L = 1k\Omega, V_o=-21\$ V, so that explains the odd behavior there. My concern is that, for a value of \$ V_o\$ within limits, specified by some \$ v_S\$ and \$ R_L\$ pair, the output is still anomalous. For example, when \$ v_S=0.8 V\$ and \$ R_L=300 \Omega\$, \$ V_o=-5.6 V\$, but simulated \$i_L=-5.66\$ mA as opposed to being -8 mA, and I can't think of any reason why we're deviating from the expected results. The current transmitter circuit schematic.

Derivation relating v_S and i_L.

Deriving relation between V_o, R_L and v_S.

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I did a simulation Dynamic DC Analysis with microcap v12 and a old 741.

Here is what I get ... EE&O, 10 < R5 < 10k (log stepping *2).

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For another configuration of resistors (1 k)

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And here is a Maple sheet ... for analytical results.

enter image description here

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Both the results you cited are correct. The 1k results show a voltage out that's below the negative rail, but that's the simulator's fault. I think your problem is that you aren't considering RL's effect on the positive feedback voltage. In both cases IL is -10*Vs ma.

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