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For my opinion, it is not necessary to go down to KCL and KVL equations. Instead you should make use of basic available gain functions.

  • Between E and output node Vo there is a damped integrator (inverting lowpass)

  • Between node Vo and E there is a non-inverting active block with an inverting feedback loop (local opamp with R2, R3 and C3); the resistor R1 has no influence on the gain (ideal opamp); its only purpose is stability improvement because we have two opamps in one feedback loop.

  • A closer look into this block reveals that it is nothing else than a non-inverting integrator stage (integrating capacitor C3 multiplied by the driving inverting gain factor, which - in this case - is "-1")

  • These building blocks are arranged in one common feedback loop. Thus, we have one version of the classical two-integrator loop (one damped inverting integrator and one non-inverting integrator).

  • Following this approach, we have a system of two equations with two unknowns:

First equation: Vo=Vi * F1 + VE * F2 with F1=f(R5, C1, C2) and F2=f(R4, R5, C1)

(Comment: Both functions F1 and F2 are simple inverting gain functions)

2nd equation: VE=Vo * (1/sT) with T=R6C3

Unknowns: VE and Vo/Vi

(using this approach, I have found the transfer function - by hand calculation - within 8 minutes).

For my opinion, it is not necessary to go down to KCL and KVL equations. Instead you should make use of basic available gain functions.

  • Between E and output node Vo there is a damped integrator (inverting lowpass)

  • Between node Vo and E there is a non-inverting active block with an inverting feedback loop (local opamp with R2, R3 and C3); the resistor R1 has no influence on the gain (ideal opamp); its only purpose is stability improvement because we have two opamps in one feedback loop.

  • A closer look into this block reveals that it is nothing else than a non-inverting integrator stage (integrating capacitor C3 multiplied by the driving inverting gain factor, which - in this case - is "-1")

  • These building blocks are arranged in one common feedback loop. Thus, we have one version of the classical two-integrator loop (one damped inverting integrator and one non-inverting integrator).

  • Following this approach, we have a system of two equations with two unknowns:

First equation: Vo=Vi * F1 + VE * F2 with F1=f(R5, C1, C2) and F2=f(R4, R5, C1)

(Comment: Both functions F1 and F2 are simple inverting gain functions)

2nd equation: VE=Vo * (1/sT) with T=R6C3

Unknowns: VE and Vo/Vi

For my opinion, it is not necessary to go down to KCL and KVL equations. Instead you should make use of basic available gain functions.

  • Between E and output node Vo there is a damped integrator (inverting lowpass)

  • Between node Vo and E there is a non-inverting active block with an inverting feedback loop (local opamp with R2, R3 and C3); the resistor R1 has no influence on the gain (ideal opamp); its only purpose is stability improvement because we have two opamps in one feedback loop.

  • A closer look into this block reveals that it is nothing else than a non-inverting integrator stage (integrating capacitor C3 multiplied by the driving inverting gain factor, which - in this case - is "-1")

  • These building blocks are arranged in one common feedback loop. Thus, we have one version of the classical two-integrator loop (one damped inverting integrator and one non-inverting integrator).

  • Following this approach, we have a system of two equations with two unknowns:

First equation: Vo=Vi * F1 + VE * F2 with F1=f(R5, C1, C2) and F2=f(R4, R5, C1)

(Comment: Both functions F1 and F2 are simple inverting gain functions)

2nd equation: VE=Vo * (1/sT) with T=R6C3

Unknowns: VE and Vo/Vi

(using this approach, I have found the transfer function - by hand calculation - within 8 minutes).

7 deleted 61 characters in body
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For my opinion, it is not necessary to go down to KCL and KVL equations. Instead you should make use of basic available gain functions.

  • Between E and output node Vo there is a damped integrator (inverting lowpass)

  • Between node Vo and E there is a non-inverting active block with an inverting feedback loop (local opamp with R2, R3 and C3); the resistor R1 has no influence on the gain (ideal opamp); its only purpose is stability improvement because we have two opamps in one feedback loop.

  • A closer look into this block reveals that it is nothing else than a non-inverting integrator stage (integrating capacitor C3 multiplied by the driving inverting gain factor, which - in this case - is "-1")

  • These building blocks are arranged in one common feedback loop. Thus, we have one version of the classical two-integrator loop (one damped inverting integrator and one non-inverting integrator).

  • Following this approach, we have a system of two equations with two unknowns:

First equation: Vo=Vi * F1 + VE * F2 with F1=f(R5, C1, C2) and F2=f(R4, R5, C1)

(Comment: Both functions F1 and F2 are simple inverting gain functions)

2nd equation: VE=Vo * (1/sT) with T=R6C3

Unknowns: VE and Vo/Vi

For my opinion, it is not necessary to go down to KCL and KVL equations. Instead you should make use of basic available gain functions.

  • Between E and output node Vo there is a damped integrator (inverting lowpass)

  • Between node Vo and E there is a non-inverting active block with an inverting feedback loop (local opamp with R2, R3 and C3); the resistor R1 has no influence on the gain (ideal opamp); its only purpose is stability improvement because we have two opamps in one feedback loop.

  • A closer look into this block reveals that it is nothing else than a non-inverting integrator stage (integrating capacitor C3 multiplied by the driving inverting gain factor, which - in this case - is "-1")

  • These building blocks are arranged in one common feedback loop. Thus, we have one version of the classical two-integrator loop (one damped inverting integrator and one non-inverting integrator).

For my opinion, it is not necessary to go down to KCL and KVL equations. Instead you should make use of basic available gain functions.

  • Between E and output node Vo there is a damped integrator (inverting lowpass)

  • Between node Vo and E there is a non-inverting active block with an inverting feedback loop (local opamp with R2, R3 and C3); the resistor R1 has no influence on the gain (ideal opamp); its only purpose is stability improvement because we have two opamps in one feedback loop.

  • A closer look into this block reveals that it is nothing else than a non-inverting integrator stage (integrating capacitor C3 multiplied by the driving inverting gain factor, which - in this case - is "-1")

  • These building blocks are arranged in one common feedback loop. Thus, we have one version of the classical two-integrator loop (one damped inverting integrator and one non-inverting integrator).

  • Following this approach, we have a system of two equations with two unknowns:

First equation: Vo=Vi * F1 + VE * F2 with F1=f(R5, C1, C2) and F2=f(R4, R5, C1)

(Comment: Both functions F1 and F2 are simple inverting gain functions)

2nd equation: VE=Vo * (1/sT) with T=R6C3

Unknowns: VE and Vo/Vi

6 deleted 61 characters in body
source | link

For my opinion, it is not necessary to go down to KCL and KVL equations. Instead you should make use of basic available gain functions.

  • Between E and output node Vo there is a damped integrator (inverting lowpass)

  • Between node Vo and E there is a non-inverting active block with an inverting feedback loop (local opamp with R2, R3 and C3); the resistor R1 has no influence on the gain (ideal opamp); its only purpose is stability improvement because we have two opamps in one feedback loop.

  • A closer look into this block reveals that it is nothing else than a non-inverting integrator stage (integrating capacitor C3 multiplied by the driving inverting gain factor, which - in this case - is "-1")

  • These building blocks are arranged in one common feedback loop. Thus, we have one version of the classical two-integrator loop (one damped inverting integrator and one non-inverting integrator).

  • Question: Are you sure about the signal input node via C2)?

For my opinion, it is not necessary to go down to KCL and KVL equations. Instead you should make use of basic available gain functions.

  • Between E and output node Vo there is a damped integrator (inverting lowpass)

  • Between node Vo and E there is a non-inverting active block with an inverting feedback loop (local opamp with R2, R3 and C3); the resistor R1 has no influence on the gain (ideal opamp); its only purpose is stability improvement because we have two opamps in one feedback loop.

  • A closer look into this block reveals that it is nothing else than a non-inverting integrator stage (integrating capacitor C3 multiplied by the driving inverting gain factor, which - in this case - is "-1")

  • These building blocks are arranged in one common feedback loop. Thus, we have one version of the classical two-integrator loop (one damped inverting integrator and one non-inverting integrator).

  • Question: Are you sure about the signal input node via C2)?

For my opinion, it is not necessary to go down to KCL and KVL equations. Instead you should make use of basic available gain functions.

  • Between E and output node Vo there is a damped integrator (inverting lowpass)

  • Between node Vo and E there is a non-inverting active block with an inverting feedback loop (local opamp with R2, R3 and C3); the resistor R1 has no influence on the gain (ideal opamp); its only purpose is stability improvement because we have two opamps in one feedback loop.

  • A closer look into this block reveals that it is nothing else than a non-inverting integrator stage (integrating capacitor C3 multiplied by the driving inverting gain factor, which - in this case - is "-1")

  • These building blocks are arranged in one common feedback loop. Thus, we have one version of the classical two-integrator loop (one damped inverting integrator and one non-inverting integrator).

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