Common-emitter amplifier

The common-emitter amplifier is the workhorse single-transistor BJT voltage amplifier: input at the base, output at the collector, emitter held at AC ground. It’s the BJT analogue of the Common-source amplifier with the same qualitative properties: high gain, inverting, the default voltage-amplifier stage.

Why it amplifies

Bias the BJT in active mode at an operating point (see BJT DC analysis), then superimpose a small signal on the base–emitter voltage: $v_{BE}=V_{BE}+v_{be}$. In active mode the device is a voltage-controlled current source, so the small signal produces a collector-current change

$i_c=g_m{v}_{be}$

where $g_m=I_C/V_T$ is the BJT transconductance. That current flows through the collector resistor $R_C$, and the output is the collector voltage. Because $i_c$ flowing into the collector pulls the collector voltage down,

$v_o=-i_c{R}_C=-g_m{R}_C{v}_{be}\Rightarrow A_v=\frac{v_o}{v_{be}}=-g_m{R}_C$

The minus sign makes the stage inverting: a 180° phase shift between input and output. The transistor turns an input voltage into a current via $g_m$, and the resistor turns that current back into a much larger output voltage.

BJT voltage amplifier topology: $v_{BE}=V_{BE}+v_{be}$, where $V_{BE}$ is the DC bias and $v_{be}$ the small signal; the output $v_{CE}$ swings about its DC bias as $v_{be}$ varies.

$v_{CE}=V_{CC}-i_C{R}_C$ swings inversely with the input $v_{be}$, the source of the inversion.

Gain by differentiating the transfer characteristic

The same result drops out of the large-signal transfer characteristic. The output is $v_{CE}=V_{CC}-i_C{R}_C$ with $i_C=I_S{e}^{V_{BE}/V_T}$ (BJT collector current). The small-signal gain is the slope of $v_{CE}$ versus $v_{BE}$ at the bias point:

$A_v\equiv {\frac{d{V}_{CE}}{d{V}_{BE}}|}_{\text{bias}}=-R_C\frac{d{i}_C}{d{v}_{BE}}=-R_C\cdot \frac{I_S{e}^{V_{BE}/V_T}}{V_T}=-R_C\frac{I_C}{V_T}=-g_m{R}_C$

using $d{i}_C/d{v}_{BE}=I_C/V_T=g_m$. Same answer two ways: the linearised current model and the slope of the DC transfer curve agree, as they must.

Derivation: $A_v\equiv d{V}_{CE}/d{V}_{BE}=-R_C{I}_C/V_T=-g_m{R}_C$; the minus sign means the output is inverted (180° phase shift).

Detailed analysis of the BJT amplifier transconductance and output voltage.

Input/output resistance and the Early-effect ceiling

• Input resistance: $R_{in}=r_{\pi }=\beta /g_m$, the BJT input resistance, finite (a few kΩ). This is the main weakness versus the infinite-input-resistance MOSFET stage.
• Output resistance: $R_C\parallel r_o$, looking back into the collector.

Including the transistor’s own output resistance $r_o$ from the Early effect, the collector load is $R_C\parallel r_o$, so the more complete gain is

$A_v=-g_m(R_C\parallel r_o)$

This is why $r_o=V_A/I_C$ sets a hard ceiling on how much gain a single CE stage can produce. The input and output resistances matter once stages are cascaded.

The most useful BJT amplifier configuration: high gain, inverting.

CE amplifier characteristics: high voltage gain, moderate input resistance, moderate output resistance, signal inversion.

To trade some of this large gain for predictability and bandwidth, add an un-bypassed emitter resistor (Emitter degeneration). The full board-level realisation with biasing and coupling capacitors is the Discrete-circuit BJT amplifier.