Bipolar Junction Transistor (BJT)

Semiconductor Devices · BJT · Learn

Figure 1. Three doped layers — N+/P/N (NPN) or P+/N/P (PNP). Emitter is heavy-doped, base is thin, collector is moderate.

Three layers, two junctions. Emitter–base (E–B) and base–collector (B–C). Each junction is a PN diode in its own right.
Base is thin and lightly doped so most carriers injected from the emitter sail straight through to the collector instead of recombining.

Figure 2. Schematic symbols. The arrow lives on the EMITTER and points OUT for NPN, IN for PNP — same direction as conventional current.

NPN ⇒ arrow OUT — the arrow points in the direction of conventional current. For an NPN in active mode, $I_E$ flows OUT of the emitter terminal externally (electrons flow IN — they are supplied by the external circuit and then injected from the emitter region into the base).
PNP ⇒ arrow IN — same rule, polarity flipped. Conventional current flows INTO the emitter externally (holes flow IN — supplied from the external circuit and injected into the base).

The three terminal currents obey KCL exactly:

I_E = I_B + I_C

Three operating regions

Figure 3. The three regions in (

V_{CE}

I_C

) state space. Cutoff is the bottom strip; saturation is the left slab; active is the bulk where

I_C

tracks

\beta \cdot I_B

Cutoff. $V_{BE} < V_{BE(\text{on})} \approx 0.5\,\text{V}$ — both junctions reverse-biased, $I_C \approx 0$ . The transistor looks like an open switch.
Active. BE forward-biased, BC reverse-biased. Here the transistor *amplifies*: $I_C = \beta \cdot I_B$ . $\beta$ (also written $h_{FE}$ ) is the DC current gain — typically 50–300.
Saturation. Both junctions forward-biased. The collector–emitter voltage clamps near $V_{CE(\text{sat})} \approx 0.2\,\text{V}$ ; $I_C$ is now limited by the load resistor, not by $\beta \cdot I_B$ . The transistor looks like a closed switch.

\beta \cdot I_B

is what the transistor would pass.

I_C

is what it actually passes — capped by the load as soon as

V_{CE}

hits

V_{CE(\text{sat})}

Figure 4. Common-emitter output curves. Each curve is one

I_B

value. In the active region (right of the dashed sat boundary) every curve is flat at

\beta \cdot I_B

Read these like contour lines: pick an $I_B$ , follow that curve from left to right. The flat plateau is the active region; you can move the load along that flat line by changing $V_{CE}$ with the load resistor without changing $I_C$ .

Why a base resistor

Figure 5. The base–emitter junction is a forward-biased diode that clamps

V_{BE}

near 0.7 V. Without

R_B

V_{in}

would drive an unbounded

I_B

and burn out the transistor.

The B–E junction clamps at $V_{BE} \approx 0.7\,\text{V}$ once it conducts (it's just a forward-biased diode).
Anything above 0.7 V on $V_{in}$ drops across $R_B$ . So $R_B$ sets $I_B$ :
$I_B = (V_{in} - V_{BE}) / R_B$
From $I_B$ we get $\beta \cdot I_B$ . If that exceeds $(V_{CC} - V_{CE(\text{sat})}) / R_C$ , the transistor saturates.