Electronic – Why does the differential amplifier not need capacitors at the input

Radio Frequency amplifiers quite often use common base circuits and I believe the main reason is because "miller capacitance" does not have the same detrimental effect.

On a normal common-emitter configuration, the input is fed to the base and the output is at the collector but, internally the capacitance between collector and base acts as negative feedback and can reduce gain.

To combat this, if the base is held at a fixed potential to ground (as per in its normally biased state), and this is supplemented by decent capacitance to ground, the miller capacitance is effectively shunted to ground. The down-side is that feeding an input to the emitter requires a stronger drive voltage because the emitter input impedance is lower.

It is lower because the emitter fed input has to also "handle" collector currents as well as the little bit of emitter-to-base current normally needed for amplification.

Consider also the differential amplifier: -

This is another circuit that uses the emitter as an input. On this occasion the input to the emitter comes from the emitter (outputting) of the other BJT. In fact, both emitters are simultaneously inputs and outputs.

Understanding transistors is a bit like peeling an onion- there are many layers. At the simplest large-signal level you can consider the transistor as a current sink that's controlled by the current through the base-emitter junction. The latter behaves like a forward-biased diode. Not much current until you get to some hundreds of mV, and way too much current if you put volts across the junction. As you say, the transistor will conduct excessive current and will be destroyed if you simply connect (say) 5V to the base with emitter grounded. This is in stark contrast to the behavior of a MOSFET.

At a more sophisticated level of understanding (which is required if you want to predict how most amplifiers work) and for small signals the base-emitter junction behaves like a resistor of Vt/Ib where Vt is the thermal voltage, about 26mV at room temperature. So if your base current is 2.5uA (say the beta is 300 and the transistor is biased with 0.75mA collector current), the base-emitter junction looks like about a 10K resistor for small signals. You can consider the transistor as a (somewhat imperfect because of r0) voltage controlled current source with an input resistance of Vt/Ib. This is the hybrid-\$\pi\$ model. Note that the transconductance gm (and thus the voltage gain in a common emitter configuration) is a function of the collector bias current and temperature and beta does not enter into it at all.

I must emphasize that this model is a linearized model about a bias point and is quite invalid if the (change in) input voltage is large (more than some millivolts). In other words we're talking about relatively small changes on top of a fixed base-emitter voltage of perhaps 600 or 700mV.