# How to calculate the gain of the OP-AMP in this configuration

Sorry for my ignorance in this field but I have failed miserably trying to figure the gain of this circuit:

simulate this circuit – Schematic created using CircuitLab

Simulation shows that the input impedance of the first stage is 509.4k I do not understand why not 504.7. Where this second 4k7 comes from?

• Show us your calculations so far. Start with the divider on the input. Hint: OA1 IN- is a virtual earth. Assume it is at 0 V. If that confuses you then omit R1 and R2 and calculate the gain of each stage but show your work (in the question - not in the comments). Nov 26 '17 at 22:59
• Looks like homework, and not too hard except for a “funny” trap. Nov 26 '17 at 23:59
• @user2233709 It looks like but I did my last homework about 30 years ago. I do programming - my analog electronics knowledge is a bit rusty. Nov 27 '17 at 0:08
• @PeterJ_01 Then I’d be curious to understand why you care about such a circuit… Anyway, I agree with you that the input impedance of the first stage should not be 509.4kΩ. I find 504.66kΩ. Perhaps you simulation takes into account the imperfections of the TL081 op-amp… Nov 27 '17 at 0:16
• @user2233709 I design the device and I cant get stable output when the OpAmp is in non-inverting configuration. Consulted dozen people but unfortunately all the solutions do not work so I have decided to have two OpAmps. Nov 28 '17 at 17:45

If we make use of the equations for inverting amplifier, then we can see that $V_{out}=-\frac{R_{f}}{R_{in}}V_{in}$.

Amplification = $\frac{V_{out}}{V_{in}}=-\frac{R_{f}}{R_{in}}$,
I will work in absolute, so amplification $= |-\frac{R_{f}}{R_{in}}|=\frac{R_{f}}{R_{in}}$

As @Transistor pointed out, OA1 IN- is a virtual earth, this means that $R_3$ is in parallel with $R_2$.

If we look back to the expression above and substitute we can understand that

• $R_{in}$ for OA1 is $R_1+R_2//R_3≃504.65\text{ kΩ}$
• $R_f$ for OA1 is $R_4=6.8\text{ kΩ}$
• $R_{in}$ for OA2 is $R_5=4.7\text{ kΩ}$
• $R_f$ for OA2 is $R_6=6.8\text{ kΩ}$

Amplification for OA1, is $0.0134≃-37.4\text{ dB}$, the gain is less than 1.

Amplification for OA2 is $1.44≃3.2\text{ dB}$.

Since everything is cascaded, we can just multiply the amplifications, logarithmic addition is the same as linear multiplication.

Amplification from $V_{in}$ to $V_{out}$ = $\frac{V_{out}}{V_{in}}=0.0134×1.44 = -37.4\text{ dB}+3.2\text{ dB}$

That's $-34.2\text{ dB}$, not that good. Or $0.019296$. This means that if $V_{in}=\frac{1}{0.019296}≃52\text{ V}$, then you will read $1\text{ V}$ at the output.

Another approach is to find the Thevenin equivalent at the input (right before the 4.7K resistor), the Thevenin voltage is the voltage of the voltage divider formed by the 500K resistors:

$\frac{500K}{500K+500K}V_{in}=1/2V_{in}$

The Thevenin resistance will be the parallel of the 500K resistors

$\frac{500K\cdot500K}{500K+500K}=250K$

The equivalent circuit at the input of the first stage is

simulate this circuit – Schematic created using CircuitLab

The output of the first stage is, according to the inverting amplifier equation $V_o=\frac{-R_f}{R_i}\cdot V_{i}$, in this case $V_i=1/2 V_{in}$

so

$V_{o1}=\frac{-6.8K}{254.7K}\frac{V_{in}}{2}$

The output of the second stage is

$V_{o2}=\frac{-6.8K}{4.7K}\cdot V_{o1}$

$V_{o2}=\frac{-6.8K}{4.7K}\cdot \frac{-6.8K}{254.7K}\frac{V_{in}}{2}$

The total gain is just $A=V_{o2}/V_{in}$

Thus $A=\frac{-6.8K}{4.7K}\cdot \frac{-6.8K}{254.7K}\frac{1}{2}=0.0193$