Electronic – How to solve the transfer function for this OP-AMP circuit

operational-amplifier

I've came across the circuit below in EDN. This circuit is a "DC Offset Compensator" according to the article. It is used in an ultrasonic sensor circuit which requires self-adjustment to the level of an AC input signal. Also, this circuit accommodates the signal's unknown and variable DC bias voltage.

IC1 is a single supply, high input impedance, rail-to-rail input, rail-to-rail output dual operational amplifier.

I've tried to simulate this circuit using LTSpice, with LTC6241. However, the output stays the same as \$V_{REF}\$. Below is my spice netlist.

I have tried to solve the circuit using below equation, the general equation of operational amplifiers, however it gets too messy too soon.

\$V_{Out}=A_{OL}*(V^{+}-V^{-})\$

Can you, with details, explain me how to solve the transfer function of this circuit?

Schematic

SPICE Netlist:

XU1 N001 N004 0 Vcc N002 LTC6241
XU2 N002 N005 0 Vcc N003 LTC6241
R1 N005 ss 1.5k
R2 N003 N005 43k
R3 N004 N003 100k
R4 Vcc N001 1k
R5 N001 0 1k
C1 N002 N004 1ยต
V1 Vcc 0 5
V2 Signal 0 SINE(1.25 1 40k)
V3 ss 0 AC 1
.ac oct 100 1 1Meg
.lib LTC4.lib
.backanno
.end

Best Answer

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} \$