Components (#7–11)

BJT — bipolar transistor

Current controlhFE2N3904wiki/embedded-bjt

TL;DR

A BJT is a current-controlled device where a small base current steers a large collector current; its two uses are current amplification and on/off switching. Internal junctions split it into NPN/PNP, and it turns on past the base-emitter threshold (~0.7V). Three regions — cut-off (OFF), active (amplify, IC=IB·hFE), saturation (switch, VCE(sat)≈0.2V). That switching happens in *saturation* is the opposite of a MOSFET (switch = linear/ohmic) — the key trap. Design back-calculates base current from hFE: for a 50mA switch on a 2N3904, size RL and RB from the datasheet VCE(sat)/VBE(sat).

What a BJT is — current controls current

In an NPN, a small base current (IB) lets an hFE-times-larger current (IC) flow collector→emitter. It conducts once the base-emitter voltage passes the ~0.7V threshold — the PN-junction threshold itself (PNP is the opposite polarity).
Two uses — current amplification (boosting a small signal) and on/off switching.

Base current IBVBE > 0.7V

📶×hFE

Collector current ICIC = IB·hFE

Amplify / switch

Small IB → ×hFE → large IC

Three regions — switch in saturation

Cut-off (IB≈0 → IC≈0, OFF), active/linear (IC=IB·hFE, amplify), and saturation (VCE(sat)≈0.2V, fully-on switch).
Trap: the 'fully-on switch' region is called saturation for a BJT but linear (ohmic) for a MOSFET — opposite names. BJT saturation = ON; MOSFET saturation = the lossy transition.

Region	Condition	Behavior / use
Cut-off	IB ≈ 0	IC≈0, OFF (open)
Active	IB flows	IC=IB·hFE — amplify
Saturation	IB large enough	VCE(sat)≈0.2V, fully ON — switch

Three regions — amplify in active, switch in saturation

hFE gain + 2N3904 switching design

In active region IC = IB×hFE. hFE comes from the datasheet and drops by half or more in saturation, so switching design assumes it conservatively (e.g., ~10).
A 50mA switch on a 2N3904 (VCC=12V, VIN=5V): after checking Max Ratings (VCEO 80V, IC 0.2A), RL sets collector current and RB sets base current.

R_{L} = \frac{V _{C C} - V _{C E (s a t)}}{I _{C}}, R_{B} = \frac{V _{I N} - V _{B E (s a t)}}{I _{B}}

$I_{B}$: = IC/hFE = 50mA/10 = 5mA
$V_{C E (s a t)}$: 0.3V (datasheet)
$V_{B E (s a t)}$: 0.95V (datasheet)

RL=(12−0.3)/50mA≈234Ω→245Ω · RB=(5−0.95)/5mA=810Ω→820Ω

Pitfalls & gotchas

Biggest trap — the fully-on switch region is 'saturation' for a BJT but 'linear (ohmic)' for a MOSFET; learn them together and you'll mix them up, so nail down that the 'ON region' names are reversed. The inverter's main switch is the MOSFET (≈0 gate current, easy drive, low loss); the BJT is its current-control archetype, still used for small-signal amplification. (The lecture also rounded RL from the computed 234Ω to the standard 245Ω.)