Electronic – External offset voltage adjustment on NE5534

First, the DUT is in the positive feedback because this way you have an inverting and a non-inverting gain in the loop, so that multiplying them you obtain an overall negative loop gain.

Second, the gain is 1000 because Vin of DUT is Vos*50/(50K+50), so if you consider that V+ of the DUT should be 0, there is only the offset applied, so the feedback forces the output to be 1k times the offset voltage.

I think that you can look at the output this way: suppose that the situation is the one described, and you have a 500uV DUT offset and so 500mV output voltage.

Now, if you try to perturb the Servo Input, the feedback forces the output of the DUT to be almost the same of Servo Input, restoring the same Output voltage.

Note: Voff is the conventional name for input offset voltage, while Vos is the output voltage with Voff applied at the input pins.

I'll use the opamp's pin numbers to name voltages.

\$ \begin{cases} V_6 = \dfrac{R3}{R3+R4} (V_O - V_{IN}) + V_{IN} \\ \\ \\ V_O - V_2 = I_{R2} R2 \\ \\ \\ I_{R2} = j \omega \text{ } C1 (V_2 - V_5) \end{cases}\$

The opamps will regulate their output so that \$ V_6 = V_5\$ and \$ V_2 = V_{REF}\$. Then, solving for \$V_O\$:

\$ V_O = \dfrac{(R3 + R4)(1 + j \omega \text{ } C1 R2) \text{ } V_{REF} + j \omega \text{ } C1 R2 (2 R3 + R4) \text{ } V_{IN}}{(R3 + R4) + j \omega \text{ } C1 R2 R3} \$

or

\$ V_O = \dfrac{j \omega \text{ } C1 R2 (2 R3 + R4)}{(R3 + R4) + j \omega \text{ } C1 R2 R3} V_{IN} +\dfrac{(R3 + R4)(1 + j \omega \text{ } C1 R2)}{(R3 + R4) + j \omega \text{ } C1 R2 R3} V_{REF} \$

Filling in the component values:

\$ V_O = \dfrac{j \omega \text{ } 4600 \text{ } \Omega s}{44500 \text{ } \Omega + j \omega \text{ } 150 \text{ } \Omega s} V_{IN} +\dfrac{44500 \text{ } \Omega + j \omega \text{ } 4450 \text{ } \Omega s}{44500 \text{ } \Omega + j \omega \text{ } 150 \text{ } \Omega s} V_{REF} \$