For cascading a system that have an output impedance of \$R_1\$ and an op amp that has an input impedance of \$R_2\$ then in order to use a isolated model we need a value for \$R_2\$ that is minimal 10 times higher than \$R_1\$. But why is that?

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    \$\begingroup\$ At one point you had accepted my answer. Is there something missing from my answer that prevents you from accepting it after all? If so let me know and I will try to explain further. \$\endgroup\$ – Null Mar 29 '18 at 17:39
  • \$\begingroup\$ @Null I'm sorry, you were editing so I waited \$\endgroup\$ – Looper Mar 29 '18 at 17:47
  • \$\begingroup\$ Ah, no worries. I was just concerned my answer didn't explain sufficiently for you. Thanks for accepting! \$\endgroup\$ – Null Mar 29 '18 at 18:13

\$R_1\$ and \$R_2\$ form a voltage divider which reduces the voltage at the input of the second stage. The Thevenin equivalent of the first stage combined with the second stage modeled as an input impedance with a voltage controlled voltage source looks like this:


simulate this circuit – Schematic created using CircuitLab

The requirement that \$R_2\$ is "minimal 10 times higher than \$R_1\$" is a rule of thumb, but it ensures that the voltage drop is (probably) negligible (i.e. you can assume \$v_1 \approx v_2\$, as if the stages were isolated). If \$v_1\$ is the output of the first stage, \$v_2\$ is the input of the second, and \$R_2 = 10\times R_1\$:

$$v_2 = \frac{R_2}{R_1 + R_2}v_1 = \frac{10R_1}{R_1 + 10R_1}v_1 = \frac{10}{11}v_1 \approx 90.9\% \times v_1$$

Depending on your application this may not be good enough, in which case you'd make sure \$R_2\$ is even higher with respect to \$R_1\$.

Note that this only applies if the signal is a voltage. If the signal is a current then you want the output impedance of the first stage to be as high as possible and the input impedance of the second stage to be as low as possible. This is because the output impedance is in parallel with the current source in the Norton equivalent, and you want as much of the current to flow into the second stage as possible rather than through the first stage's output impedance.


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