BJT bias creates room for a small signal

Bias first, signal second

A bipolar junction transistor can amplify a small change around a DC operating point. For an NPN in forward-active operation, use $I_C\approx\beta I_B$ and a chosen $V_{BE}$ approximation. Neither β nor VBE is an exact constant. Saturation and cutoff are different operating regions, so check voltages after using the active-region model.

Consider a deliberately idealized 5 V circuit with base held by an ideal 1.7 V source, emitter resistor 1 kΩ to ground, and collector resistor 2 kΩ to 5 V. Assume VBE = 0.7 V and β = 100. Emitter voltage is 1 V and emitter current 1 mA. Collector current is β/(β+1) times emitter current, or 0.990099 mA. Collector voltage is 3.019802 V and VCE is 2.019802 V.

Linearize near that operating point

At room-temperature approximation VT = 25.85 mV, transconductance $g_m=I_C/V_T$ is 38.3017 mS. Its inverse is approximately 26.1085 Ω. Ignoring transistor output resistance and taking β as large for a simple small-signal estimate, an unbypassed emitter resistor gives gain $-R_C/(R_E+1/g_m)$, approximately −1.9491.

A 10 mV peak base signal therefore produces approximately 19.49 mV peak inverted collector change. This small-signal estimate is local; increasing amplitude until the transistor reaches cutoff or saturation breaks the linear model. The emitter resistor reduces gain but makes behavior less dependent on β and the raw junction characteristic.

The base source matters

Our ideal base source can supply the roughly 9.90 μA base current without droop. A real resistor divider cannot be assumed ideal. Replace it with its Thevenin voltage and resistance, include base current, and solve the DC point again. Coupling and bypass capacitors also have frequency-dependent effects.

Record the bias point separately from signal gain, and specify which approximations produced each number. A textbook transistor estimate is not a qualified device model or a guaranteed physical amplifier design.