In all the textbooks the shown inputs to Op-Amps are AC.
Could one use Op-Amp to amplify a DC voltage? if the answer is no, why not? Why only transistors can be used to amplify DC, aren't Op-Amps made of transistors?
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Yes, you can amplify a DC voltage. Many signals in applications such as temperature, pressure, weighing, etc., change so slowly that they can be considered DC. The amplifiers that condition these signals will often use op-amps1 to buffer and boost the signal level.
1 "Op-amp" is an abbreviation of "operational amplifier". It is not an initialisation or acronym so we don't write it as OPAMP.
An opamp can certainly amplify a DC voltage.There is nothing inherent it opamp design that prevents DC amplifiction. All opamps are internally DC coupled. Obviously, coupling capacitors would not be used and the affect of the opamp offset voltage would have to be considered. Generally a dual supply opamp would have to be used to keep the input at close to 0 volts. Otherwise the opamp will amplify the bias voltages and probably saturate with any reasonable gain.
In principle, yes, you can amplify DC with an op-amp, subject to the constraints of the amplifier. Mainly, the amplified output voltage must be between the output limits of the op-amp. An amplifier powered with +12V and -12V cannot produce a 50V output, for example. A rail-to-rail op-amp can produce an output all the way to the supply voltages; others will not go all the way to the positive or negative supply voltage, but a few volts shy.
But why would you want to do this? A DC input never changes. So instead of, say, amplifying a 0.82V input signal by 10 to produce 8.2V out, just use an 8.2V power supply.
On the other hand, you can make a fixed power supply with a low-power accurate voltage reference as input to a fixed-gain amplifier, which feeds an external power transistor. (A fraction of the final output of the power supply is fed back to the op-amp's negative input to regulate the overall circuit.) So an op-amp is often part of a power supply.
simulate this circuit – Schematic created using CircuitLab The circuit has a 5.1V Zener diode as a reference, and the output is divided by 10/(6.2+10) = .62 to amplify by 1/.62 to produce 8.2V. Note you still need to power the amplifier with something greater than the output. Here's it's a 12V DC supply, but it could be a transformer and rectifier that runs from AC wall current.
If what you mean is an input that only slowly changes, for example when someone turns a knob, that's still AC. It's just very boring AC.