The ideal op-amp in open loop, has a frequency response with infinite bandwidth. What about an ideal op amp in closed loop (inverting, non-inverting etc.) configuration? Does it (still being an ideal op amp although now in closed loop) also have a frequency response with infinite bandwidth?
Electronic – Ideal Op Amp in Closed Loop
operational-amplifier
Related Solutions
Closed loop gain is the gain that results when we apply negative feedback to "tame" the open loop gain. The closed loop gain can be calculated if we know the open loop gain and the amount of feedback (what fraction of the output voltage is negatively fed back to the input).
The formula is this:
$$ A_{closed} = \frac{A_{open}}{1 + A_{open} \cdot Feedback} $$
The open-loop gain affects the performance generally like this. Firstly, look at the above formula. If the open loop is huge, like 100,000, then the 1 + does not matter. \$A_{open} \cdot Feedback\$ is a large number, and it doesn't matter whether or not we add 1 to this large number: it is like a drop in a bucket. Thus the formula reduces to:
$$ \begin{align}
A_{closed} &= \frac{A_{open}}{A_{open} \cdot Feedback} \\
&= \frac{1}{Feedback}\\
\end{align}
$$
So, with a huge open-loop gain, we can easily get the closed loop gain if all we know is the negative feedback: if it just the reciprocal. If the feedback is 100% (i.e. 1) then the gain is 1, or unity gain. If the negative feedback is 10%, then the gain is 10. With a huge open-loop gain, we can precisely set up gains: as precisely as we care to design and build our feedback circuit. With open-loop gain which is not that large, we may not be able to ignore that 1 +
. All the more so if \$Feedback\$ is small.
Okay, so far that's more of an issue of clean math and design convenience. Big open loop gain: closed loop gain is simple. But, practically speaking, small open-loop gains means that you must use less negative feedback to achieve a given gain. If the open loop-gain is a hundred thousand, then we can use 10% feedback to get a gain of 10. If the open loop gain is only 50, then we must use much less negative feedback to get a gain of 10. (You can work that out with the formula.)
We generally want to be able to use as much negative feedback as possible, because this stabilizes the amplifier: it makes the amplifier more linear, gives it a higher input impedance and lower output impedance and so on. From this perspective, amplifiers with huge open loop gains are good. It is usually better to achieve some necessary closed loop gain with an amplifier that has huge open loop gain, and lots of negative feedback, than to use a lower gain amplifier and less negative feedback (or even just an amplifier with no negative feedback which happens to have that gain open loop). The amp with the most negative feedback will be stable, more linear, and so on.
Also note that we don't even have to care how huge the open loop gain is. Is it 100,000 or is it 200,000? It doesn't matter: after a certain gain, the simplified approximate formula applies. Amplifiers based on high gain and negative feedback are therefore very gain-stable. The gain depends only on the feedback, not on the specific open-loop gain of the amplifier. The open loop gain can vary wildly (as long as it stays huge). For instance, suppose that the open loop gain is different at different temperatures. That does not matter. As long as the feedback circuit is not affected by temperature, the closed-loop gain will be the same.
Here's what the application hints say about Rload
I don't quite understand why there is feedback (the cap and the 10M resistor). Could someone please explain why?
Because the input offset voltage can be a few millivolt it's easy to apply a 10M feedback resistor to perfectly bias the op-amp DC wise so that ac measurements can be made. Becuase the input cap is 0.1uF and the feedback resistor is 10Mohm the circuit acts like a high pass filter at 0.159Hz so this is of no consequence when measuring ac gain. The real gain of the opamp can be inferred from the gain when using C and R.
how does one know what the maximum frequency that one can put into an op-amp is?
Here is an open loop gain (red) from an op-amp not disimilar to the LM358. Notice the blue line - this is the frequency response when negative feedback is applied. There is a constant 20 dB gain from DC to 1MHz (3dB down) and this would be the actual response with feedback. So if your square wave were 1kHz the square would be failry pure up to nearly the 1000th harmonic.
You can't break an op-amp with frequency - you can only break it with too much current or voltage.
Related Topic
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Best Answer
Yes, if the other components are ideal too and there is no parasitic capacitance, series inductance, etc. With these effects, every resistor has a non-flat frequency response. Since your feedback path will contain resistors, the frequency response will likewise be non-flat.
Then there are other considerations depending on how obsessive you want to get. Regardless of how ideal the opamp may be, there will be some finite propagation time of the feedback signal back to a opamp input. This represents a different phase shift at different frequencies, which will again make the frequency response non-flat and can make the whole system unstable.