Answer 2—This is the “modified Howland current source,” and it has a number of important advantages over the original circuit. Its derivation is similar. Again, start by writing down the equations for the voltages at the inputs of the op-amp, using superposition to account for the voltage sources:

Set them equal to each other and isolate the quantity V_A – V_B, using the same trick as before:

Note that if R3 >> R5, the current through R5 and the load are essentially identical, and are: 

Combining the last two equations gives you: 

Very often, R2 = R3, so this reduces to the following: 

Contributor: David Tweed