What is an example of current in voltage out amplifier? And if current
is in then some voltage also must be in why we call it current in?
A transimpedance amplifier (TIA) is "current in" and "voltage out" circuit. The input impedance is theoretically zero ohms so if you tried to put a milli-volt at the input the current taken (theoretically) would be infinite.
They are quite commonly used for amplifying photodiodes: -
Reason: the output from a photodiode (PD) is current for a given light-power-density on the device. It produces a current out and, if you connected the PD output to a resistor, it would still try to push the same current through that resistor for the same light input. The side effect of this is that there will be a voltage produced across that resistor to keep ohms law happy.
So why not just feed the PD output to a resistor and use a voltage amplifier? You can do this but what you will find is that if the light is modulated (i.e. a typical data transmission from say a fiber optic), the presense of the resistor (and the self-capacitance of the PD) will form a low pass filter and make the data look very non-ideal.
So a TIA is used because its input impedance is zero ohms i.e. it is a current-in device with zero volts produced. With zero volts produced at the input, the self-capacitance of the PD is defeated and you can get significantly higher bandwidths thru the amplifier.
For example is common-emitter BJT NPN amplifier voltage in voltage
out? But some current is also in and some current is also out. Why
Strictly speaking (with respect to the physics) a BJT is a voltage-input device and any current taken into the base is all part and parcel of the Shockley diode equation. Regarding the output, it is generally accepted that it is a current-out device because of this: -
For a given base voltage (that gives rise to a given base current), the collector current is largely constant (flat) for a vast change is colelctor-emitter voltage. This is normally where a linear amplifier uses a BJTs characteristic i.e. it's a current out device BUT that current out converts to a voltage out for a given collector resistance.
The Schockley equation for a diode tells you what current flows in the diode but a variant of this is the ebers-moll equation. This tells you the collector current for a given Vbe: -
The important thing to note is that the BJT is a voltage-in device and a current-out device. Look at the top equation - what drives collector current is Vbe and Vbc BUT Vbc is negative (collector and base reverse biased) so the dominant part of the equation is governed by Vbe and not Ib - base current is a by-product that is convenient to use because base current (as a side-effect) and collector current conveniently appear to be proportional to each other. As far as I know (and I'm no physics expert), this happy accident is just that and has "convenienced" the use of hFE as a measure of Iout/Iin.