Exactly!



Gain calculation is as follows:





When V1=0V, the both non-inverting inputs of U2A and U3A are at 0.5V. Due to the action of OPamp (principle of negative feedback) at both terminals of R5 0.5V can be measured. So, R5 is without any current, which has the consequence, that R4 and R6 are without any current flowing through them either. So, at both the outputs of U2A and U3A a voltage of 0.5V can be measured, which U4A subtracts to 0V at its output.

When V1=10mV, on the other hand, the non-inverting input of U2A is at 0.505V, while the non-inverting input of U3A is still at 0.5V. Again, due to the action of negative feedback, a voltage of 0.505V - 0.5V = 5mV is dropping across R5, which will cause a current of 5mV / 1kOhm = 5µA through R5.

This current of 5µA causes a voltage drop of 5µA x 9kOhm = 0.045V across R4 and R6. So, at output of U2A a voltage of 0.505V + 0.045V = 0.55V will be measured, where at output of U3A it will be 0.5V - 0.045V = 0.455V. And as the output signal of U2A and U3A is the difference 0.55V - 0.455V = 0.095V, compared to the input signal of V1=10mV this gives a gain of 9.5.

Compared to what is actually present at the both non-inverting inputs of U2A and U3A, namely 0.505V - 0.5V = 5mV, this would means a gain of 19.

Kai