Why doesn't a normal npn transistor with common emitter configuration, in active mode, not behave as a DC amplifier?

Question

If a normal npn BJT transistor amplifies the base current (DC) to beta+1 times (the emitter current = (beta+1)*(the base current), why can't we call it a DC amplifier?

Answer 1

The "normal npn BJT transistor" is a DC amplifier; only its output quantity is a (collector) current. To make it a voltage amplifier, we have somehow to convert this current to a voltage. For this purpose, we connect a (collector) resistor between the collector and Vcc... and take the voltage drop VRc across it as an output quantity. The problem is that this voltage is not referenced to ground but to Vcc. So, we take its complement to Vcc (Vce = Vcc - VRc) that is grounded. That is why this circuit (aka "common emitter amplifying stage") is inverting.

Of course, to make a true voltage amplifier, we have to bias the transistor input (to add an initial input voltage about 0.7 V in series to Vin).

Answer 2

Speaking about a "DC amplifier" you certainly have an opamp with resistive fedback in your mind, which has - for example - a gain of (-10). Hence, an input voltage (DC) of 0.1V would give an ouput of (-1V).

In principle, the same applies to BJT (common emitter configuration). However, because the operating point is not at (ideally) zero V (as for the opamp) we have a 1V-decrease of the ouput voltage (across the coll. resistor Rc) below the DC operational point (which very often is app. at 50% of the supply voltage).

As another example, consider the well-known diff. amplifier (long-tailed pair). This amplifier is nothing else than a combination of an emitter follower (common-collector) and a common-base stage. For well-balanced supply voltages the DC operational point at the output is zero volts. In this case, a DC voltage at the input (again 0.1V) will give an output of (+-)1V at the ouput (for a gain of "10").