How does the voltage-current converter work?

In this op-amp circuit, one of the op-amp inputs is tied to ground and the other is then a virtual ground. Since the op-amp input terminals draw no current, all the current goes through the feedback resistor from the output to the negative terminal. The output is then $V_{\rm out} = -RI$, the drop across the resistor. So the output voltage is proportional to the input current. The device then acts like a voltage source, providing a voltage independent of load.