2
\$\begingroup\$

Inverting op-amp with inductor and capacitor

I have this amplifier as part of a larger system. To my understanding this circuit will always have a negative gain Vout/Vin, since we have ground on the non-inverting input and a non-zero signal on the inverting input. Is this correct? If so, would it become a non-inverting op-amp if the non-inverting input was Vin while the Vin from my original circuit is set to ground?

I tried finding the gain in the laplace domain like this:

Defining a node V1 and then using KCL I get the two expressions $$ \frac{V_{in}-V_1}{sL_1} = sC_1V_1+\frac{V_1}{R_1} $$ and $$ \frac{V_1}{R_1} = \frac{0-V_{out}}{R_2+sL_2} $$

Combining these two gives

$$ \frac{V_{in}}{sL_1}+\frac{V_{out}+R_1}{R_2+sL_2}=-\frac{sC_1R_1V{out}}{R_2+sL_2}-\frac{V_{out}}{R_2+sL_2} $$

Rearranging for the gain

$$ \frac{V_{out}}{V_{in}}=-\frac{R_2+sL_2}{R_1C_1L_1s^2+L_1(R_1+1)s} $$

I can't see where I might have gone wrong in the calculation of gain, but at the same time it doesn't really make sense to have a negative gain at this part of my system.

\$\endgroup\$
  • \$\begingroup\$ you made a mistake when replacing the leftmost occurrence of \$V_1\$. \$\endgroup\$ – FrancoVS Nov 14 '15 at 22:17
  • \$\begingroup\$ In the left side you have a mistake is Vout * R1 and not Vout + R1. \$\endgroup\$ – joe billy Nov 14 '15 at 23:15
  • \$\begingroup\$ FYI, you can use qsapecng to solve circuits like this symbolically. \$\endgroup\$ – Fizz Nov 14 '15 at 23:42
  • \$\begingroup\$ L1 and C1 will produce about a 90 degree phase shift at midband so why on earth are you worried whether the circuit is inverting or not? \$\endgroup\$ – Andy aka Nov 14 '15 at 23:55
  • \$\begingroup\$ It should come negative..then why you are worried? \$\endgroup\$ – Virange Nov 15 '15 at 5:28
1
\$\begingroup\$

You can determine the transfer function of this system using the fast analytical circuits techniques or FACTs. First, you start with \$s=0\$, shorting inductors and opening capacitors. The dc gain is simply

\$H_0=-\frac{R_2}{R_1}\$

Then, you look at the resistance offered by the energy-storing elements when temporarily removed from the circuit. You should find:

\$\tau_1=\frac{L_1}{R_1}\$ then \$\tau_2=C_1*0\$ and \$\tau_3=\frac{L_2}{R_{inf}}=0\$

Then, you determine the resistance seen from the energy-storing elements when one of them is set in its high-frequency state (inductors replaced by open circuit and capacitors replaced by short circuits). You should find:

\$\tau_{12}=C_1R_1\$ then \$\tau_{13}=\frac{L_2}{R_{inf}}=0\$ and \$\tau_{23}=\frac{L_2}{R_{inf}}=0\$

Finally, you determine the resistance seen from \$L_2\$ while \$L_1\$ and \$C_1\$ are set in their high-frequency state (inductors replaced by an open circuit and capacitors replaced by short circuits). You have:

\$\tau_{123}=\frac{L_3}{R_{inf}}=0\$

The denominator is thus equal to

\$D(s)=1+s(\tau_1+\tau_2+\tau_3)+s^2(\tau_1\tau_{12}+\tau_1\tau_{13}+\tau_2\tau_{23})+s^3(\tau_1\tau_{12}\tau_{123})\$

The zero exists when the impedance made of \$L_2\$ and \$R_2\$ becomes a transformed short circuit. This occurs when \$\omega_z=\frac{R_2}{L_2}\$. The complete transfer function is defined as

\$H(s)=H_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}\$ with \$H_0=-\frac{R_2}{R_1}\$, \$\omega_z=\frac{R_2}{L_2}\$, \$\omega_0=\frac{1}{\sqrt{L_1C_1}}\$ and \$Q=R_1\sqrt{\frac{C_1}{L_1}}\$

The complete Mathcad file appears below. I have purposely changed the labels so that time constant labels match that of the components but results are similar:

enter image description here

It looks a bit mysterious but FACTs are easy to learn and apply. Check out this APEC 2016 presentation

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

and all these examples solved in the book

http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.