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When deriving the transfer functions of common OPAMP circuits, like an inverting amplifier we begin with:
VOUT=A×(V+ − V-)
A two stage operational amplifier for example, could be considered as a second order system. The equation above only signifies the DC gain of the amplifier.This equation does not account for the dependency of gain of opamp on frequency.
Shouldn't the transfer function of OPAMP's internal circuitry be considered for accurate analysis?
Shouldn't the transfer function of OPAMP's internal circuitry be considered for accurate analysis?
For DC (low frequencies) it is unlikely because, most engineers know that the open loop gain of the op-amp they will be familiar with is around 100 dB or greater. This is a gain of 100,000 and for 1V on the output, you need an input differential of 10 uV. This is usually far in excess of the input offset voltage error hence for DC (low frequencies) nobody really cares about the transfer function.
However, if you have to make a DC accurate amplifier with a gain of 1000 up to 20 kHz (flat to +/- 0.1 dB) then you get stuck into the data sheets to see what the open loop response looks like. Or you just use the gain bandwidth product value to give you a reasonable idea.
Or just start simulating in PSpice (or equivalent). However you still need to exercise care when using sims.
Sometimes yes, sometimes no. It depends what your application is.
Most people, even seasoned engineers, use the simplest equation they can use for their application.
If you want to amplify a DC signal, then that equation is more than enough, assuming you are using an amplifier that is already internally compensated to be stable with any gain, as most are.
If you want to amplify an audio signal, some will be too slow, most will be fast enough.
If you want to build an active filter, then you might be surprised how fast the amplifier needs to be to maintain the Q of your design.
Yes or no, as always: it depends when you apply the formula.
Electronics is not only about knowing which formula to apply when but also to know when you can use a simple formula (the one in your question, where the opamp's bandwidth is infinite) or when you need a more complex formula or even simpler when you need to consider the bandwidth of the opamp.
For a DC instrumentation amplifier (slow but very accurate), above formula is OK, I can ignore the bandwidth as my amplifier is slow.
For an audio amplifier I might get into trouble if I want too much gain and the opamp is not very fast.
For an opamp I'm using as a buffer for a 100 Mega samples/second for sure I need to consider the bandwidth of the opamp.
See, it depends.
Why are operational amplifiers equipped with such a high gain values? The only reason is that - in this case - the frequency-dependent amplifying characteristics of the active unit can be neglected for many applications.
The reason behind this effect is "negative feedback" because for very large open-loop gain values (Aol) the closed-loop behaviour is determined practically by the feedback network only. But note that this pre-condition (high Aol) is sufficienmtly fulfilled up to a certain frequency limit only (dependent on the specific unit and specified in the data sheet).
EDIT: But there are counter-examples: You can design active filters and oscillators without any external capacitors. In this case, it is only the frequency-dependent gain Aol which determines the pole frequencies (filter bandwidth or oscillation frequencies). However, these circuits have a very small practical relevance only (because of the large tolerances) and are of "academic" value only.
