Electronic – Op amp outside resistive network

basicoperational-amplifier

Usually I don't care about algebra and just focus on the functionality. However, below circuit form appeared so many times in last few days in op amps that I thought of memorizing it by heart.

From my limited op amps study so far, it seems the entire op amp analysis boils down to figuring out the voltage at point \$\text{x}\$ shown in the diagram.

Working the equations using superposition gave a nice symmetry:
$$V(x) = \frac{V_1R_2 + V_2R_1}{R_1+R_2} = \frac{\left<V_1,V_2\right>\cdot \left<R_2,R_1\right>}{R_1+R_2}$$

If we think \$R_2, R_1\$ as weights, then \$V(x)\$ is the weighted average of \$V_1,V_2\$. It seems \$R_2\$ controls \$V_1\$ and \$R_1\$ controls \$V_2\$. This is beautiful and I'm dying to know if this symmetry has a nice physical explanation. Thanks!

schematic

simulate this circuit – Schematic created using CircuitLab


Few observations:

  1. When \$V_1=V_2=V\$ above formula gives \$V(x) = V\$. This means the entire branch including \$R_1, R_2\$ and every other point float to voltage \$V\$. Doesn't matter what \$R_1, R_2\$ are. This is used in instrumentation amplifiers to amplify AC while perfectly grounding common mode.
  2. \$x\$ will be virtual ground when the average value is \$0\$

Best Answer

I'm not sure where you're going with all the maths but the following may help. Often to get an intuitive feel for the various voltages in an op-amp circuit I tend to think of the circuit in proportional terms or, when looking at a virtual earth point, as a see-saw.

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Figure 1. With the values shown in the OP's circuit the voltage at x can be graphically or mentally calculated as 1/3 of the way between 1 V and 10 V. The graph shows this to be 4 V.

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Figure 2. The voltage see-saw about the virtual earth on an inverting op-amp.

When given the input voltage, V1, of the inverting op-amp circuit the output can be graphically or mentally calculated by using a see-saw or lever about the 0 V point with the lever arm lengths being in the same ratio as the resistor values.*

I hope that help.

Images: original.