3.1 BJT Structure, Operation & Current Relationships

3.1 BJT Structure, Operation & Current Relationships

The Bipolar Junction Transistor (BJT) is a three-terminal active semiconductor device fundamental to aircraft avionics, power distribution, and flight control instrumentation. Unlike unipolar Field-Effect Transistors (FETs) that depend on a single carrier type, the BJT is termed bipolar because its operation relies simultaneously on both majority and minority charge carriers (electrons and holes) crossing two closely spaced PN junctions.

In modern aviation Line Replaceable Units (LRUs)—such as autopilot servo amplifiers, cockpit audio management units, and Full Authority Digital Engine Controls (FADEC)—BJTs operate as linear small-signal amplifiers and saturation switches. Mastering BJT physical structure, terminal current equations, gain parameters, and biasing states is a core requirement for EASA Part-66 Module 04.

---

Physical Structure and Regional Doping Gradients

A BJT is formed on a single continuous semiconductor crystal (predominantly silicon in modern aviation; legacy units may employ germanium). It consists of three distinct, alternately doped regions forming either an NPN or PNP structure. Discrete diodes wired back-to-back cannot duplicate a transistor because the central base layer must be sub-microscopically thin to enable carrier diffusion without extensive recombination.

The three physical regions have distinct geometries and doping profiles:

1. Emitter ($E$):

   • Doping Level: Extremely heavily doped ($n^+$ in NPN, $p^+$ in PNP).
   • Function: Injects majority charge carriers into the base region. The high doping density ensures an abundant supply of carriers crossing into the base.
   • Geometry: Moderate physical size.

2. Base ($B$):

   • Doping Level: Very lightly doped ($p$ in NPN, $n$ in PNP).
   • Function: Serves as the control corridor through which injected carriers travel toward the collector.
   • Geometry: Ultra-thin, typically less than $1\ \mu\text{m}$. Light doping and minimal physical width ensure that minority carrier recombination remains minimal.

3. Collector ($C$):

   • Doping Level: Moderately doped ($n$ in NPN, $p$ in PNP), lighter than the emitter to support higher reverse breakdown voltages.
   • Function: Collects carriers surviving base transit and routes them to the external circuit.
   • Geometry: Physically the largest region. The reverse-biased collector-base junction dissipates substantial electrical power ($P = V_{CB} \cdot I_C$). Its large surface area conducts heat away to the transistor casing and aircraft chassis heat sinks.

---

Schematic Symbols and Terminal Polarities

On aircraft wiring schematics and LRU circuit diagrams, standard IEEE/IEC symbols distinguish NPN and PNP devices via the arrow located on the emitter lead:

• NPN Transistor: The arrow points outward from base to emitter (mnemonic: "Not Pointing iN").
• PNP Transistor: The arrow points inward from emitter to base (mnemonic: "Pointing iN Proudly").

[!WARNING] Conventional Current vs. Electron Flow: The schematic arrow strictly indicates conventional current direction (positive charge flow) when the base-emitter junction is forward-biased. In an NPN device, actual electron flow enters the emitter from the external circuit and flows inward toward the base and collector, directly opposite to the arrow.

---

Junction Biasing and Operating Regions

A BJT has two internal PN junctions: the Base-Emitter (BE) junction and the Collector-Base (CB) junction. Applying external DC bias voltages establishes one of four operational regions:

[!NOTE] Saturation Voltage Offset: Unlike an ideal mechanical switch that exhibits zero contact voltage, a saturated BJT maintains a residual voltage $V_{CE(sat)} \approx 0.1\text{ to }0.2\text{ V}$. In high-current aircraft actuator switching circuits (e.g., $5\text{ A}$ indicator lamps), this generates steady power dissipation ($P = 0.2\text{ V} \times 5\text{ A} = 1.0\text{ W}$), requiring proper thermal mounting.

---

Carrier Dynamics in Active Mode

In an active-biased NPN transistor:

1. Injection: Forward biasing the BE junction ($V_{BE} \approx 0.7\text{ V}$) lowers the barrier potential. The heavily doped $n^+$ emitter injects high concentrations of electrons into the $p$-type base.
2. Recombination: In the base, electrons are minority carriers. Because the base is lightly doped and extremely narrow, only $1%\text{ to }2%$ recombine with base holes, creating external base current ($I_B$).
3. Collection: The remaining $98%\text{ to }99%$ of electrons diffuse across to the reverse-biased CB junction depletion layer. The strong electric field sweeps them into the collector, forming collector current ($I_C$).

In PNP transistors, the mechanism is complementary: majority holes from the $p^+$ emitter are injected into an $n$-type base and swept into the $p$-type collector.

---

Fundamental Current Relationships and Gain Parameters

By Kirchhoff’s Current Law (KCL), total emitter current equals the sum of collector and base currents:

$I_E = I_C + I_B$

Because base current is very small, $I_C \approx I_E$.

Common-Emitter DC Current Gain ($\beta$ or $h_{FE}$)

The ratio of collector current to base current: $\beta = \frac{I_C}{I_B}$ In small-signal aircraft transistors, $\beta$ typically ranges from $50\text{ to }300$.

Common-Base DC Current Gain ($\alpha$)

The ratio of collector current to emitter current: $\alpha = \frac{I_C}{I_E}$ Because $I_C < I_E$, $\alpha$ is always slightly less than unity, typically between $0.95\text{ and }0.998$.

Mathematical Conversions Between $\alpha$ and $\beta$

Substituting $I_E = I_C + I_B$ yields the classic EASA exam identities:

$\alpha = \frac{\beta}{\beta + 1} \qquad \text{and} \qquad \beta = \frac{\alpha}{1 - \alpha}$

---

Worked Calculation: Terminal Currents and Current Gain

Problem

An avionics technician tests an NPN transistor in a radio receiver preamplifier. Measured parameters:

• DC Base Current: $I_B = 25\ \mu\text{A}$ ($0.025\text{ mA}$)
• Current Gain: $\beta = 120$

Calculate $I_C$, $I_E$, and $\alpha$.

Solution

1. Collector Current ($I_C$): $I_C = \beta \cdot I_B = 120 \times 25\ \mu\text{A} = 3000\ \mu\text{A} = 3.00\text{ mA}$

2. Emitter Current ($I_E$): $I_E = I_C + I_B = 3.00\text{ mA} + 0.025\text{ mA} = 3.025\text{ mA}$

3. Common-Base Current Gain ($\alpha$): $\alpha = \frac{I_C}{I_E} = \frac{3.00\text{ mA}}{3.025\text{ mA}} \approx 0.9917$ Alternatively: $\alpha = \frac{\beta}{\beta + 1} = \frac{120}{121} \approx 0.9917$

This confirms that $99.17%$ of injected emitter carriers reach the collector, with only $0.83%$ lost to base recombination.