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

Question

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?

Answer 1

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}\$

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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}\$

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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.

Answer 2

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\$