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Apparently the voltage of the inverting input to the op-amp in the 2nd order active LPF should be equal to \$v_2\$. Would anyone be able to explain why this would be the case?

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With negative feedback (see the two right-hand resistors feeding some fraction of the output potential back to the inverting input), the op-amp's behaviour is rather interesting; it adjusts its output in such a way as to minimise the difference between its two inputs. If the signal being fed back to the inverting input is too low, the output rises, and the feedback signal also rises. If feedback is too high, the output will fall, taking the feedback signal with it. The change in output is always in a direction that tends to reduce the difference between the two inputs.

The result is that the system always settles into a state of equilibrium, in which the two inputs have almost exactly the same potential.

From a more rigourous algebraic perspective, you need to understand the relationship between the op-amp's input and output potentials:

$$ V_{OUT} = A(V_{NONINV} - V_{INV}) $$

Here \$A\$ is some huge gain, like \$A=10^5\$ or \$A=10^6\$.

By inspection you can see that if \$ V_{NONINV} > V_{INV} \$ then the op-amp's output will have to rise to some large positive value, (constrained, of course, to its positive supply voltage), and if \$ V_{NONINV} < V_{INV} \$, then the output has to plummet towards some large negative potential. With negative feedback, as the output rises or falls, at some point it must pass through a condition in which the two are equal. If it overshoots that point, the difference changes sign, and the output swing changes direction. It eventually settles at whatever \$V_{OUT}\$ is necessary to obtain the condition \$V_{INV} \approx V_{NONINV}\$.

You can liken this behaviour to any system employing negative feedback. For example, while steering a car, you always turn the steering wheel in a direction that reduces the difference between where you are actually heading, and where you want to be heading. The result is that you don't drive into the ditch. With negative feedback, you are constantly adjusting the output (actual direction) so that it's always exactly equal to your desired direction, and you stay on the road.

Another example is room temperature. If you find the room too warm, you turn down the thermostat to reduce the temperature. When it's too cold, you turn it up. Again you are simply employing negative feedback to keep the room's actual temperature equal to your desired temperature. You are always acting to equalise these two values: actual temperature, and desired temperature.

The difference is called the "error" and negative feedback causes a system to settle at as close to zero error as possible. If you imagine the inverting input of an op-amp to be a measure of the actual condition, and the non-inverting input to be a signal representing the desired condition, then you have negative feedback, and the op-amp's output will change in such a way that causes the difference between them to tend towards zero.

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An ideal op-amp can be regarded as having infinite open-loop gain. This means one thing; if the op-amp output is somewhere mid-region between the power rails then, the voltage difference between the inverting and non-inverting inputs must be zero.

Apparently the voltage of the inverting input to the op-amp in the 2nd order active LPF should be equal to \$v_2\$

This concept applies to all op-amps using negative feedback (not just Sallen Key filters).

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This would be due to OpAmp action which strives, simply speaking, to make the inputs equal to each other. In your case it's not so much that \$V_{in-}\$ is equal to \$V_2\$, but \$V_{in+} = V_2\$ is "forced" equal to \$V_{in-}\$ by the OpAmp's internal circuitry (by changing the output until this condition is fulfilled).

This is, of course, a pretty narrowed down answer. There's a lot of reasons beneath this, but for the scope of this questions I think this is "good enough".

EDIT: Simon's answer goes into more detail, so please refer to his answer for clarification.

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    \$\begingroup\$ Slight correction: It is not the "OpAmps internal cicuitry" which causes the input differential voltage to be NEARLY to zero (in practice we ASSUME that it is zero). In fact, it is only the result of negative feedback (and NOT "a lot of reasons") which is responsible for this effect (as in all closed-loop systems with negative feedback). \$\endgroup\$
    – LvW
    Commented Jun 6, 2023 at 9:12
  • \$\begingroup\$ Sorry, isn't it the OpAmp that employs negative feedback by way of it's internal circuitry to make the input differential voltage (sort of) zero? You're right that it's mainly/only the negative feedback, I was only trying to convey that it's not as simple as the OpAmp performing some kind of internal subtraction and voltage control by saying "a lot of reasons". Well, now that I think about it that's pretty much exactly what it does, just not mathematically but with currents and voltages. I realize my explanation was somewhat basic but thats what I was trying to do. \$\endgroup\$ Commented Jun 6, 2023 at 14:10
  • \$\begingroup\$ My previous comment might unintentionally come off a bit challenging, I just wanted to clarify that I'm genuinely asking to further my understanding. \$\endgroup\$ Commented Jun 6, 2023 at 14:21

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