Recently I learned about the functionalities of an Op Amp and how they are used to amplify voltages. Don’t they serve the same purpose as that of a boost converter but in a more cleaner fashion? Someone please highlight their subtle differences
In: Engineering
Those are completely different things:
**Boost converter**
They get an input voltage and turn it into a larger (or smaller if a buck converter) one, at the cost of correspondingly lower current. Very simplified they are the analogue of a transformer but for DC.
They are used when a different voltage is required, for example to drive a device; for example an Op-Amp!
**Op-Amp**
Those don’t actually amplify anything in the sense that their output is never higher than whatever you supplied it with. It typically has three connections (plus two for power supply): a positive input A, a negative one B, and an output C.
Given whatever voltages at A and B, an ideal Op-Amp outputs ∞·(A-B). This obviously is impossible and if nothing else this is bound above and below by the voltage the power supply provides. So it more or less outputs MAX when A>B and MIN when A<B. Here MAX and MIN are the maximal and minimal voltages the Op-Amp can provide, usually close to the supply voltage.
What purpose does it serve, apart from being a very basic check which of A and B is bigger? Well, for example put a resistor of 1k between B and C, and another from B to ground (“0V”). If you know voltage dividers or just do the calculations this now means that the voltage at B is half that at C by basic electrics.
But is that really correct? By what we have seen so far we would expect C to be at MAX (if A>B) or MIN (if A<B). But if C is at MAX; then B is also quite large and probably bigger than A, which makes C now MIN. And similarly C being at MIN makes it want to be at MAX. Peculiar…! Does it switch infinitely fast? Does the Op-Amp explode in flames? Does a black hole form and eat the universe?
But we missed one loophole: we never said what happens if A=B. And that is our way out. Only if A=B can we satisfy everything dictated by physics and our sanity. In this case we find C to be twice the voltage of B and therefore A; we _amplify_ whatever signal we feed into A, but still limited by MIN and MAX.(*)
So why do we really want this if it seems worse than the boost converter? Simply, it is not to make higher voltages, but to change signals, and very fast. Good Op-Amps switch millions, even billions, of times per second, a converter doing even a few switches per second quite possibly burns out. Op-Amps are entirely for analogue signalling, to control voltage (or current) in terms of another in a very specific, fast and reliable way.
You can also build circuits where C is smaller by a given factor, or even do integration and differentiation of whatever voltage-function A gets!
(*): We can even math this. C = ∞·(A-B) and also B = C/2, therefore C = ∞·(A-C/2) which rearranges to (1+∞/2)·C = ∞·A and well, that 1+ does not really change something as large as ∞/2, so we end with ∞/2 ·C = ∞·A, so again C = 2A.
No, they are completely different.
OpAmp doesn’t actually amplify voltage – it actually requires an external source of a greater voltage to work. OpAmp can only reduce that source voltage, it can never increase it.
OpAmps also have abysmal efficiency – they pretty much “burn away” the leftover voltage. They are supposed to manipulate only low power control signals – they are not supposed to be on the main power path.
It is pretty much the opposite for boost converters – they are designed for the power path. They can bump voltage, but even if they lower voltage – they don’t burn it, but convert into a higher current. They can have high efficiency. However, they can have slower response time and some other disadvantages, which make them bad for signal processing.
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