Electronic – ny symmetry in this opamp ckt

operational-amplifier

By thevenizing the at the inverting input, I notice that when the variable resistor equals \$R_2\$ the output \$v_{out}\$ becomes \$0\$. Can this result be "seen" with out doing the algebra?

I feel there is some symmetry but I couldn't quite figure it out. The top \$R_2\$ resistor is connected to \$v_{out}\$ but the bottom variable resistor is tied to ground. This is throwing me off. Any help?
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Best Answer

Yes. With negative feedback the circuit should stabilise at a state where \$ V_- = V_+ \$. The question then is simply what output voltage will achieve that for a given input signal.

To make the maths easy we'll assume that all resistors are equal values.

  • Since \$ R_1 = R_{var} \$ we know that \$ V_+ = \frac {V_{in}} 2 \$.
  • Therefore \$ V_- = \frac {V_{in}} 2 \$.

The only way this can happen is when \$ V_{out} = 0 \$.

I would then start to mentally adjust the variable resistor and consider what needs to happen to maintain balance.

  • If we adjust to half the value of R1 then \$ V_+ = \frac {V_{in}} 3 \$.
  • To achieve stability again the output needs to move to pull \$ V_- = \frac {V_{in}} 3 \$. This will happen when the voltage across R1, \$ V_{R1} = \frac 2 3 V_{in} \$ and the same across R2 meaning that \$ V_{out} = -\frac 1 3 V_{in} \$.
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