amplifier - Why BJT BE voltage allows to avoid distortions

In Common-Emitter amplifier the BE voltage is set near 0.7V but we apply both positive and negative swings to BJT and there is no distortion on output but the signal will be 0.7V+signal_value and 0.7V-signal_value. Why there is no distortion for 0.7V-signal_value? The BJT is not in conduting state now.

Answer

Your circuit is highly linearized by the 1.3 volts across the 500 ohm emitter resistor. Without that resistor (or if you boosted gain with 1,000uF in parallel with 500 ohms), you'd see serious distortion at 0.026 volts (major 2nd harmonic distortion), and about 10% distortion at 4 millivolts Peak-peak on the base.

The distortion is reduced by the ratio of 1.3/0.026 or 50:1.

Insert 4 volts peak-peak thru 1Kohm resistor (to protect the transistor) and you'll see distortion.

By the way, below 100Hz, you'll measure a dropoff in gain, because 1uF Cin implements HighPassFilter with Rin of your circuit.

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Regarding frequency respond, we'll consider the two nodes of the amplifier as having different frequency responses (thus different F3dB) because of the collector compliance (the stiffness of Icollector) does largely decouple the collector from the base.

Collector bandwidth will be 0.159 / (all resistors in parallel on collector * sum of all capacitors on the collector that are tied to GND). In this circuit, the resistors are 4K || 10K, or approx. 3K ohms. The capacitance is??? we'll assume 10pf (the 1uF does not tie to GND, so its not a parasitic capacitor). The time constant is 3K ohm * 10pF or 30 nanoseconds or 33MegaRadians/second. That divided by 2*pi == 5MHz F3dB (half power point, 40 degree phase shift).

Now for the input bandwidth. If you XFG1 has zero Zsource, the input bandwidth will be infinite (in reality, set by the base time constant: rbb' and Cemitter.) With finite Zsource, the amplifier's Cmiller will be a big bother. Assuming Cob is 10pF, and gain is 3Kohm / (500 + 26 ) ~~ 6x, the input capacitance will be 10pF * (1 + 6) = 70pF. If Zsource is 50 ohms, the 70pF * 50ohm = 3,500 nanoseconds or about 50MHz input bandwidth.