NPN bipolar transistor

A small current into the base lets a current about $\beta$ times larger flow from collector to emitter.

Controlled by: Current into the base
With no drive: Off: it conducts only when driven
Fully on: A small fixed drop, $V_{CE(sat)}$ of 0.05 to 0.4 V
Terminals: B base, C collector, E emitter
Used for: Low-side switches for LEDs, relays, small motors; Small-signal amplifiers; Current sources, mirrors, differential pairs

How it works

An NPN is three layers of silicon: an N-type emitter, a very thin P-type base, and an N-type collector. Forward-biasing the base-emitter junction, about 0.65 V, pushes electrons from the emitter into the base. The base is so thin that almost all of them cross it, and the reverse-biased collector-base junction (collector above base) sweeps them into the collector; only a small fraction leave through the base wire.

So the base current sets a collector current $\beta$ times larger:

I_C = \beta\, I_B \qquad I_E = I_C + I_B = (\beta + 1)\, I_B

$\beta$ , written $h_{FE}$ on a datasheet, is between 100 and 300 for a 2N3904 at 10 mA, with only 30 guaranteed at 100 mA. It also changes with temperature and from one part to the next. A circuit that works only for one value of $\beta$ is a fragile circuit.

Seen from the base-emitter voltage instead, the collector current is exponential, like a diode:

I_C = I_S\, e^{V_{BE}/V_T} \qquad V_T = \frac{kT}{q} \approx 25.9\ \text{mV at } 300\ \text{K}

$I_S$ , the saturation current, is a tiny scale factor (6.7 fA in the 2N3904 SPICE model). $V_T$ is the thermal voltage: $k$ is Boltzmann's constant, $T$ the absolute temperature and $q$ the electron charge. Ten times more current takes only $V_T \ln 10 \approx 60$ mV more on the base. That is why $V_{BE}$ hardly moves, about 0.6 to 0.72 V from 0.1 mA to 10 mA, and why a design can treat it as a fixed 0.65 V. At a fixed current it also falls by about 2 mV for every degree of warming.

Regions of operation

Region	Base-emitter	Base-collector	What happens	Used as
Cutoff	under about 0.5 V	reverse	only leakage flows	open switch
Active	about 0.65 V, forward	reverse	$I_C = \beta I_B$ , nearly independent of $V_{CE}$	amplifier, current source
Saturation	forward	forward	$V_{CE}$ down to 0.05 to 0.4 V, $I_C$ set by the load, $I_C < \beta I_B$	closed switch
Reverse active	reverse	forward	collector and emitter swap roles, gain near 1	rarely

"Saturation" means opposite things for a BJT and a MOSFET. In a BJT it is the fully-on switch, with almost no voltage across it. A MOSFET's "saturation" is its flat, current-source region, the equivalent of the BJT's active region.

Small-signal model

Around an operating point in the active region the transistor behaves, for small signals, like a controlled current source. Four numbers describe it: the transconductance $g_m$ , the resistance $r_e$ seen looking into the emitter, the resistance $r_\pi$ seen looking into the base, and the output resistance $r_o$ . The first two follow from the current alone, $r_\pi$ also needs $\beta$ , and $r_o$ needs the Early voltage $V_A$ :

g_m = \frac{I_C}{V_T} \qquad r_e = \frac{V_T}{I_E} \approx \frac{1}{g_m} \qquad r_\pi = \frac{\beta}{g_m} \qquad r_o = \frac{V_A}{I_C}

At $I_C = 1$ mA: $g_m = 38.7$ mS, $r_e \approx 26\ \Omega$ , $r_\pi = 2.6$ kΩ for $\beta = 100$ , and with $V_A \approx 74$ V (the 2N3904 SPICE model) $r_o \approx 74$ kΩ. The gain of a common-emitter stage or a follower is then a ratio of resistors to $r_e$ .

Relatives

A resistor-equipped transistor (also sold as a digital or pre-biased transistor) has its base resistor, and often a base-emitter resistor, built in: a logic pin drives it directly. Its resistors are only ±30 %, so it is used as a switch, never to bias an amplifier.

Matched pairs and arrays put two or more transistors in one package, often on one die, so their $V_{BE}$ match (2 mV at most for a BCM847BS at 2 mA, 0.45 mV typical and 5 mV at most for an LM3046 at 1 mA) and track in temperature: the parts for current mirrors and differential pairs that have to be accurate.

The curves

interactive

The output curves a curve tracer draws, from the Ebers-Moll model of a small NPN ( $\beta = 150$ ). The red line is the load line of $R_C$ on $V_{CC}$ : the transistor can only sit where its curve crosses it.

Base current

I_B

20 µA Supply

V_{CC}

10.0 V Collector resistor

R_C

1.00 kΩ

active: V_CE = 6.73 V, I_C = 3.27 mA, V_BE = 0.683 V, I_C / I_B = 163, P = 22.0 mW

The base current picks a curve and the curve is nearly flat: $I_C$ is about $\beta I_B$ ( $\beta$ about 150 here) whatever $V_{CE}$ does. The resistor turns that current into a voltage, which is how the transistor amplifies.

Circuits to try

Falstad's simulator, one at a time

Each circuit runs in Paul Falstad's CircuitJS, loaded only on request. The sliders are in the panel on the right of the simulator, switches toggle with a click, and hovering a part shows its voltages, currents and state at the bottom right.

Schematic: NPN switch lighting an LED — Simulation
NPN switch lighting an LED

Schematic: Common-emitter amplifier — Simulation
Common-emitter amplifier

Schematic: Emitter follower — Simulation
Emitter follower

Design rules

A saturated switch: a base current of $I_C / 10$

To be sure the transistor saturates whatever its $\beta$ , the base gets about a tenth of the collector current: a "forced beta" of 10, which is also the ratio the 2N3904 and PN2222A datasheets test $V_{CE(sat)}$ at (the BC547 sheet uses 20).

R_B = \frac{V_{drive} - V_{BE(sat)}}{I_C / 10}

A 2N3904 switching 100 mA from 5 V logic: $I_B = 10$ mA and $V_{BE(sat)} \approx 0.9$ V, read off the typical curve at 100 mA (the table guarantees values only up to 50 mA), so $R_B = (5 - 0.9) / 10\ \text{mA} = 410\ \Omega$ , 390 Ω in the E12 series. 10 mA is a lot for a logic pin. A forced beta of 20 halves it (820 Ω) at the cost of margin when cold, and a Darlington, a ULN2003 or a logic-level MOSFET needs far less drive.

An inductive load needs a flyback diode

A relay coil, a solenoid or a motor keeps its current flowing when the transistor turns off, and the collector flies far above the supply until something breaks down. A diode across the coil, cathode to the supply, gives that current a path and clamps the collector one diode drop above the supply.

The coil's inductance $L$ sets the voltage $v$ it makes when its current $i$ changes, the energy $E$ it stores at a current $I$ , and the time constant $\tau$ of the current dying away through its own resistance $R_{coil}$ :

v = L\,\frac{di}{dt} \qquad E = \tfrac{1}{2} L I^2 \qquad \tau = \frac{L}{R_{coil}}

A 5 V relay coil of 100 Ω and 100 mH carries 50 mA. Cut off in 50 ns, the maximum fall time of a 2N3904, it would need $0.1 \times 0.05 / 50\ \text{ns} = 100$ kV to keep that current flowing. The transistor is rated 40 V ( $V_{CEO}$ ) and breaks down long before, and the 0.125 mJ stored in the coil ends up in its junction at every turn-off. With the diode the collector stops one diode drop above 5 V, and the current dies away with $\tau = 0.1\ \text{H} / 100\ \Omega = 1$ ms.

An amplifier is biased by resistors, not by beta

A divider sets the base voltage and an emitter resistor $R_E$ turns it into a current: $I_E \approx (V_B - 0.65) / R_E$ . With about 1 V across $R_E$ and a divider "stiff" enough, $R_1 \parallel R_2 \le (\beta_{min} + 1) R_E / 10$ , the collector current moves by a few percent over a threefold spread of $\beta$ .

On the +15 V lab rail, for 1 mA: $R_E = 1.5$ kΩ puts 1.5 V on the emitter and 2.15 V on the base. $R_1 = 56$ kΩ on top and $R_2 = 10$ kΩ below act as 2.27 V in series with 8.5 kΩ, under the 15.2 kΩ limit for $\beta_{min} = 100$ . $I_E$ is then 1.024 mA at $\beta = 100$ and 1.062 mA at $\beta = 300$ , 3.7 % apart.

The gain of the common-emitter stage is then set by resistors too:

A_v \approx -\frac{R_C}{r_e + R_E}

Heat: power times thermal resistance

P = V_{CE}\, I_C + V_{BE}\, I_B \qquad T_J = T_A + P\, R_{\theta JA}

A TO-92 part has $R_{\theta JA} \approx 200$ °C/W: 0.5 W already lifts the junction 100 °C above the room, and 0.625 W takes it from 25 °C to its 150 °C limit. A datasheet's derating line (5 mW/°C for the 2N3904) is just $1/R_{\theta JA}$ .

The base-emitter junction breaks down at only 5 to 6 V in reverse

$V_{EBO}$ is 6 V on a 2N3904. An op-amp output swinging to -13 V on ±15 V rails, driving a base through its resistor, takes the junction into breakdown, which lowers $\beta$ at low current over time. A 1N4148 from emitter (anode) to base (cathode) clamps the reverse voltage near 0.7 V, and the base resistor limits the op-amp current through it.

Real parts and their datasheets

Pinouts as seen from the front, legs down. Every value comes with the condition it is measured at: a number from a datasheet means nothing without it.

flat face toward the viewer; a BC547 has the reverse order

2N3904

NPN, small signal

onsemi, TO-92. Datasheet (PDF), onsemi.com

Line	Value	Measured at
$V_{CEO}$	40 V max	base open
$I_C$	200 mA max	continuous
$h_{FE}$	100 to 300	$I_C = 10$ mA, $V_{CE} = 1$ V
$h_{FE}$	30 min	$I_C = 100$ mA, $V_{CE} = 1$ V
$V_{CE(sat)}$	0.2 V max	$I_C = 10$ mA, $I_B = 1$ mA
$V_{CE(sat)}$	0.3 V max	$I_C = 50$ mA, $I_B = 5$ mA
$V_{BE(sat)}$	0.65 to 0.85 V	$I_C = 10$ mA, $I_B = 1$ mA
$V_{EBO}$	6 V max	emitter-base, reverse
$f_T$	300 MHz min	$I_C = 10$ mA, $V_{CE} = 20$ V, $f = 100$ MHz
$P_D$ , $R_{\theta JA}$	625 mW, 200 °C/W	$T_A = 25$ °C, derate 5.0 mW/°C

What to read first. $h_{FE}$ at the current of the job: 100 minimum at 10 mA but only 30 at 100 mA, and the "typical 300" quoted online is the 10 mA maximum. Then $V_{CE(sat)}$ , measured at $I_C / I_B = 10$ , which is the forced beta to design a switch with.

flat face toward the viewer; the onsemi P2N2222A, also sold as "2N2222A", is C B E

PN2222A

NPN, small signal, higher current

onsemi (Fairchild), TO-92. Datasheet (PDF), onsemi.com

Line	Value	Measured at
$V_{CEO}$	40 V max	base open
$I_C$	1.0 A max	absolute maximum, steady state
$h_{FE}$	100 to 300	$I_C = 150$ mA, $V_{CE} = 10$ V, pulsed
$h_{FE}$	50 min	$I_C = 150$ mA, $V_{CE} = 1$ V
$h_{FE}$	40 min	$I_C = 500$ mA, $V_{CE} = 10$ V
$V_{CE(sat)}$	0.3 V max	$I_C = 150$ mA, $I_B = 15$ mA
$V_{CE(sat)}$	1.0 V max	$I_C = 500$ mA, $I_B = 50$ mA
$V_{BE(sat)}$	0.6 to 1.2 V	$I_C = 150$ mA, $I_B = 15$ mA
$f_T$	300 MHz min	$I_C = 20$ mA, $V_{CE} = 20$ V, $f = 100$ MHz
$P_D$ , $R_{\theta JA}$	625 mW, 200 °C/W	$T_A = 25$ °C, derate 5.0 mW/°C

What to read first. The 1 A on page 1 is a stress limit: at 500 mA the table guarantees an $h_{FE}$ of only 40 and allows 1.0 V of $V_{CE(sat)}$ , so a relay or LED switch is designed from the 150 mA rows. Then the body marking, since the PN2222A and the P2N2222A have opposite pinouts.

flat face toward the viewer; the reverse of a 2N3904

BC547B

NPN, small signal

onsemi (Fairchild), TO-92. Datasheet (PDF), onsemi.com

Line	Value	Measured at
$V_{CEO}$	45 V max	base open
$I_C$	100 mA max	continuous
$P_C$	500 mW max	absolute maximum; no thermal resistance in this document
$h_{FE}$	200 to 450	$I_C = 2$ mA, $V_{CE} = 5$ V, B grade (A: 110 to 220, C: 420 to 800)
$V_{BE(on)}$	0.58 to 0.70 V	$I_C = 2$ mA, $V_{CE} = 5$ V (0.66 V typ)
$V_{CE(sat)}$	0.25 V max (0.09 typ)	$I_C = 10$ mA, $I_B = 0.5$ mA
$V_{CE(sat)}$	0.6 V max (0.25 typ)	$I_C = 100$ mA, $I_B = 5$ mA
$f_T$	300 MHz typ	$I_C = 10$ mA, $V_{CE} = 5$ V, $f = 100$ MHz; no minimum given
$C_{ob}$	6.0 pF max (3.5 typ)	$V_{CB} = 10$ V, $I_E = 0$ , $f = 1$ MHz

What to read first. The suffix first: A, B and C are gain grades of the same die, and the pinout, which is the reverse of the 2N3904. The often-linked onsemi bc546-d.pdf is marked obsolete on every page; this bc550-d.pdf is the current document.

On the bench

Meter check. On the diode range, red lead on the base: 0.6 to 0.7 V to the emitter and to the collector, nothing with the leads swapped, nothing between collector and emitter either way. The base-emitter reading is usually a few millivolts higher than the base-collector one, which tells the emitter from the collector.

First build. The LED switch above on a breadboard: 2N3904, 4.7 kΩ to the base, 330 Ω and an LED from 5 V to the collector. Fully on, the meter reads about 0.75 V from base to emitter and under 0.1 V from collector to emitter.

Common traps

The base current flows on to the collector.

The base current leaves through the emitter. The collector current comes from the collector supply, enabled by the base current; the emitter carries both.
$\beta$ is a design value.

$\beta$ varies two- to threefold between parts of the same number and with current and temperature. A good design depends on resistors and uses $\beta$ only as a minimum.
$V_{BE}$ is exactly 0.7 V.

It runs from about 0.6 V at 0.1 mA to about 0.72 V at 10 mA and drops about 2 mV/°C at a fixed current. 0.65 V is a starting guess, never a constant to subtract precisely.
Saturation is the most amplification the transistor can give.

It is the opposite: in saturation the transistor no longer amplifies, $I_C$ is set by the load and $I_C / I_B$ falls below $\beta$ .
All TO-92 transistors have the same pinout.

A 2N3904 is E-B-C seen from the flat face, a BC547 C-B-E, and a TO-92 "2N2222A" can be either: the PN2222A is E-B-C, the P2N2222A C-B-E. The drawing in the datasheet of the exact part number, then a diode-test check, settle it.

Check yourself

answers fold out

A 2N3904 has to switch a 50 mA load from a 3.3 V logic pin. What base resistor?

Answer

Forced beta 10 gives $I_B = 5$ mA. The datasheet guarantees $V_{BE(sat)} \le 0.95$ V at exactly 50 mA and 5 mA, so $R_B = (3.3 - 0.95) / 5\ \text{mA} = 470\ \Omega$ , an E12 value. Check that the pin can source 5 mA.
A transistor with $\beta = 150$ carries $I_C = 12$ mA for $I_B = 1$ mA, with $V_{CE} = 0.1$ V. Which region?

Answer

Saturation: $I_C / I_B = 12$ is far below $\beta$ and the collector sits at 0.1 V. The load, not the transistor, sets the current.
Why does an emitter resistor keep the bias of an amplifier stable?

Answer

$I_E \approx (V_B - V_{BE}) / R_E$ depends on resistors and on $V_{BE}$ , not on $\beta$ . With $V_B = 2.15$ V and $R_E = 1.5$ kΩ, $I_E = 1$ mA. Warming by 50 °C takes 0.1 V off $V_{BE}$ , and $I_E$ only rises to 1.07 mA. With no $R_E$ and the base held at a fixed voltage, the same 0.1 V would multiply the current by about $e^{0.1/V_T} \approx 48$ . The mechanism is negative feedback: if the current rises, the emitter voltage rises and cuts $V_{BE}$ .

How it works

Regions of operation

Relatives

The curves

Circuits to try

NPN switch lighting an LED

Common-emitter amplifier

Emitter follower

Design rules

A saturated switch: a base current of IC/10I_C / 10IC​/10

An inductive load needs a flyback diode

An amplifier is biased by resistors, not by beta

Heat: power times thermal resistance

The base-emitter junction breaks down at only 5 to 6 V in reverse

Real parts and their datasheets

On the bench

Common traps

Check yourself

A saturated switch: a base current of $I_C / 10$