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.

  • \$\begingroup\$ you made a mistake when replacing the leftmost occurrence of \$V_1\$. \$\endgroup\$
    – FrancoVS
    Commented Nov 14, 2015 at 22:17
  • \$\begingroup\$ In the left side you have a mistake is Vout * R1 and not Vout + R1. \$\endgroup\$
    – joe billy
    Commented Nov 14, 2015 at 23:15
  • \$\begingroup\$ FYI, you can use qsapecng to solve circuits like this symbolically. \$\endgroup\$ Commented Nov 14, 2015 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
    Commented Nov 14, 2015 at 23:55
  • \$\begingroup\$ It should come negative..then why you are worried? \$\endgroup\$
    – Virange
    Commented Nov 15, 2015 at 5:28

1 Answer 1


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


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:


The denominator is thus equal to


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


and all these examples solved in the book



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