# Inverting op-amp gain 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.

• you made a mistake when replacing the leftmost occurrence of $V_1$. – FrancoVS Nov 14 '15 at 22:17
• In the left side you have a mistake is Vout * R1 and not Vout + R1. – joe billy Nov 14 '15 at 23:15
• FYI, you can use qsapecng to solve circuits like this symbolically. – Fizz Nov 14 '15 at 23:42
• 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? – Andy aka Nov 14 '15 at 23:55
• It should come negative..then why you are worried? – Virange Nov 15 '15 at 5:28

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