Howland Current Source

We assume $R_2/R_1=R_4/R_3$, and get

$$\frac{V^--V}{R_1}+\frac{V_0-V}{R_2}=0,\;\;\;\;\;\; \frac{V^+-V}{R_3}+\frac{V_0-V}{R_4}=\frac{V}{R_L};$$ (8)

Solving the first equation for $V_0-V$:

$$V_0-V=(V-V^-)\frac{R_2}{R_1}=(V-V^-)\frac{R_4}{R_3}$$ (9)

and substituting into the second equation, we get:

$$\frac{V^+-V}{R_3}+\frac{V-V^-}{R_3} =\frac{V^+-V^-}{R_3}=\frac{V}{R_L}=I_L$$ (10)

We see that the current through the load resistor $R_L$ is constant, independent of $R_L$, i.e., the circuit is a current source.