The Transistor Bench

“It’s just two diodes, right?”

12 minutes · the myth that has to die first

Recall From chapter 1: what does forward bias actually do to a junction? show answer

A transistor has three legs. Before anything else, here is what each one is for.

Emitter

The source. It emits charge carriers into the device. It carries the largest current of the three, and it is the terminal the arrow sits on.

Base

The thin middle layer, and the control. A tiny current here decides how much gets through.

Collector

The far side. It collects whatever survives the crossing. This is where the useful current comes out.

Now the picture almost everyone is handed: an NPN transistor is n-type, p-type, n-type. Two junctions sharing a middle layer. Emitter diode, collector diode, back to back.

That picture is where most people stop, and it is why most people never build any intuition for what follows. So let’s take it seriously and find exactly where it breaks.

Commit before you touch anything

Two ordinary silicon diodes, soldered anode-to-anode, base wire taken from the joint. You feed the base 1 mA. What comes out of the collector?

Answer: C. Now find out why by dragging the base wider yourself.

What the bench is showing you

The picture is a slice through the silicon. Electrons are pouring out of the emitter into the base — that is just a forward-biased junction, exactly as in chapter 1, lesson 6.

What happens next is a race.

An electron enters the base. The base is p-type, so it is full of holes.
The electron has a limited distance it can travel before it meets a hole and disappears. That distance is the diffusion length — around 100 µm in this silicon.
If the base is thinner than that distance, most electrons make it across. The collector junction is reverse-biased, so its field grabs them and sweeps them out. That is collector current.
Any electron that doesn’t make it has to be replaced through the base wire. That is base current.

So β is just the score of that race: survivors divided by casualties. Drag the base wider and watch the survival rate on the bench collapse.

Gain is not a property of the junctions. It is a property of the gap between them.

Why two real diodes can never work: between them sits a metal contact and a bond wire. An electron injected into the first diode’s p region hits that metal within nanometres and recombines instantly. Zero carriers reach the second junction. You have two independent diodes, one of them permanently reverse-biased, and no coupling whatsoever.

In the wild

Why chips got faster. A thinner base means fewer casualties — and also less distance to cross, so the electron gets there sooner. Higher gain and higher speed come from the same change. Sixty years of the semiconductor industry is largely the story of making that middle layer thinner: from 25 µm in a 1950s alloy transistor to well under a micrometre today.

The trick that does work. Tie a transistor’s base to its collector and it behaves as a very good diode. You will see this everywhere in real circuits — it is the standard way to build a current mirror, and it is how a chip makes a reference voltage that barely drifts with temperature. Transistor-as-diode: fine. Diodes-as-transistor: impossible.

The germanium radios your grandparents had. Early alloy-junction transistors had thick bases and β around 20–30, and they leaked badly when warm. That is not history for its own sake — it is the same slider on this bench, parked further to the right.

One sentence to keep

A transistor is not two diodes. It is two junctions close enough together to talk to each other.

Before you move on

A 1950s alloy transistor had a base about 25 µm thick. A modern one is under 1 µm. Which statement follows?

Right. Thinner base → shorter race → a larger fraction survives → less base current for the same collector current → higher β. Set the slider to 25 µm and then to 0.5 µm and watch β on the readout.

Bench 01 · inside the silicon ACTIVE

● Locked until you commit a prediction above.

Base region width 0.5 µm

0.2 µm10 µm2 mm

I base—

I collector—

β = Ic / Ib—

Survive the base—

Where the collector current actually comes from

14 minutes · the valve, and who pays for the water

Recall From lesson 1: what physically sets β? show answer

Here is the picture that survives contact with real circuits. A transistor is a valve.

The base is the handle

Turn it a little and the pipe opens a lot.

The collector–emitter path is the pipe

This is where the useful current flows.

The supply is the tank

Every drop that comes through the pipe was already in the tank. The handle did not supply it.

That last line is where most people quietly go wrong. The collector current does not come from the base. It comes from the supply. The base only decides how much of it is allowed through.

Commit before you touch anything

The transistor is happily amplifying, drawing 4 mA at the collector from a 12 V supply. You leave everything else alone and drop the supply to 8 V. What happens to the collector current?

Answer: B, barely moves. Drag Supply on the bench and watch Ic sit almost perfectly still until it hits the wall.

Do these two things, in this order

Move the base knob. Watch three things at once: the dots speed up in all three legs, the two ammeters climb together, and the ratio between them stays pinned. That ratio is β.
Count the dots. The base wire carries a crawl. The collector wire carries a flood. That difference, seen rather than calculated, is the whole reason transistors exist.
Now drag the supply. From 12 V down to about 3 V, Ic hardly notices. Then it falls off a cliff.

The cliff is the transistor running out of room. The collector voltage has come down to meet the base, and the device stops being a valve at all. We will name that region properly in lesson 4.

Kirchhoff at the transistor: Ie = Ic + Ib, always, in every region, no exceptions. The emitter carries the most current of the three — not the collector. Over a quarter of engineering students get that ranking wrong on a test. Look at the three meters and let it become obvious instead of memorised.

In the wild

Your microcontroller driving a motor. An Arduino or ESP32 pin can supply about 20 mA before it is damaged. A small DC motor wants 500 mA. A transistor solves this because the 500 mA never touches the pin — it comes from the battery. The pin only turns the handle, with about 2 mA. This is the single most common transistor circuit in hobby electronics, and it is exactly this bench.

Why an amplifier has a power cord. A guitar amplifier does not create the extra energy in the sound. It uses a tiny signal to modulate a large current drawn from the mains. Unplug it and the “amplification” stops instantly, because there was never anything in the signal to amplify from. The tank was doing the work all along.

One honest warning about the model

“Base current controls collector current” is a useful fiction, and I am flagging it as a fiction now so you don’t find out later and feel cheated.

What the device actually obeys is a voltage law — the same exponential you met in chapter 1. But that law is brutally steep and drifts with temperature, so no sane engineer drives a transistor from a bare voltage. We drive current instead, and think in β, and it works. Lesson 3 is about what we are papering over.

Before you move on

In an amplifying transistor with Ib = 20 µA and β = 150, what is the emitter current?

Yes. 3.00 mA + 0.02 mA. In practice you round it to Ie ≈ Ic, and that is fine — but only once you know why the two are not identical.

Bench 02 · common emitter ACTIVE

● Locked until you commit a prediction above.

Base drive —

Supply 12.0 V

0 V7.5 V15 V

Ib—

Ic—

Ie—

Vbe—

Vce—

Ic / Ib—

Dot speed ∝ current. Rc = 1.5 kΩ, β_F = 150 — the Ic/Ib meter reads a little above 150 because collector current creeps up with Vce (the Early effect).

The 0.7 volts that isn’t a threshold

15 minutes · the most useful graph in analogue electronics

Recall From lesson 2: which terminal carries the largest current? show answer

You met this curve in chapter 1, on a diode. Here it is again, on the base–emitter junction of a transistor, where it decides everything.

Everyone is taught that a silicon junction “turns on at 0.7 V.” There is no switch in there. There is an exponential, and an exponential has no threshold anywhere along it.

Commit before you touch anything

You raise Vbe from 0.60 V to 0.66 V — sixty millivolts, about the thickness of a rounding error. What does the collector current do?

Answer: C — ten times. Sixty millivolts per decade. Drag the slider and watch the log plot climb in a dead straight line.

Read the two plots together

They are the same data. On the left, on ordinary axes, you see the famous knee and your brain says “threshold.” On the right, with current on a log scale, the knee vanishes and you get a straight line at 59.5 mV per decade.

The knee was never in the silicon. It was in the graph paper.

Rewrite the rule in your head: 0.65 V is not what the junction demands. It is what you happen to measure when the circuit is pushing a milliamp or so through it. Change the current a thousandfold and the voltage moves by 180 mV.

Now the part that actually bites

Because the curve is so steep, a small temperature change shifts it a lot. Two numbers:

−2 mV per °C

At a fixed current, Vbe falls by about two millivolts for every degree the junction warms.

+9% per °C

At a fixed voltage, the current rises by roughly nine percent for every degree. Ten degrees is a factor of two and a half.

Turn on self-heating and watch what happens when the transistor is fed a fixed base voltage. It warms, so it draws more. Drawing more makes it warm further. Ic climbs on its own, with nothing changed at the input.

That loop is not a fault. It is the honest behaviour of an exponential with a thermal feedback path. Now switch on emitter resistor and run it again: the loop stops dead. That one resistor is why real circuits survive, and you will meet it properly in lesson 5.

In the wild

Every chip in your laptop measures its own temperature this way. Feed a transistor a steady current and its Vbe becomes a thermometer, falling 2 mV per degree, linear and repeatable and free. Your CPU’s die-temperature reading is this curve, read backwards.

The 1.25 V that every chip depends on. Take a voltage that falls with temperature (Vbe) and add one that rises with temperature (the difference between two transistors run at different currents), in the right proportion, and the sum barely moves at all. That is a bandgap reference, it lands near 1.25 V, and there is one inside almost every integrated circuit ever made. It exists only because this curve is so predictable.

Why your circuit works in the lab and fails in a car. A dashboard reaches 70 °C. That is 40 degrees above your bench, which is 80 mV of Vbe shift — more than a full decade of current if nothing is holding the bias in place. Automotive and industrial electronics are largely the art of designing so this does not matter.

Two numbers worth owning

60 mV per decade. 18 mV per doubling. If you can recall those on demand you can estimate a bias point in your head for the rest of your life.

Before you move on

You measure Vbe = 0.62 V on a transistor passing 1 mA. A second transistor in the same circuit reads 0.68 V. Roughly what is it passing?

60 mV more is one full decade more: 10 mA. And note that β never entered the calculation — this is a voltage law, not a current law. Which sets up lesson 5 rather neatly.

Bench 03 · the exponential ACTIVE

● Locked until you commit a prediction above.

Base drive voltage 0.650 V

0.40 V1.00 V1.60 V

Junction temperature 27 °C

−20 °C30 °C85 °C

Ic—

Vbe measured—

gm = Ic/VT—

re = VT/Ie—

mV per decade—

Making it a switch, and the word “saturated”

15 minutes · your first real circuit

Recall From lesson 3: how many millivolts of Vbe buy you ten times the current? show answer

An LED, a 220 Ω resistor, a transistor and a base resistor. This is the circuit a microcontroller pin uses to drive anything bigger than itself, and it is worth getting completely right before anything else.

Commit before you touch anything

5 V in on the base through a resistor. Which base resistor gives the brightest LED — 100 kΩ, 10 kΩ, or 1 kΩ?

Answer: D. Once the transistor is saturated, more base current buys you nothing at all. Sweep the slider and watch the current hit a ceiling.

Sweep the base resistor and watch two things

Drag from 1 MΩ down to 1 kΩ, keeping an eye on the region badge at the top of the bench and the moving dot on the load line.

Above about 30 kΩ: a genuine amplifier. The dot slides along the load line and the LED gets brighter as you turn the knob. Ic = β × Ib, just as in lesson 2.
Below about 4 kΩ: the dot has hit the left wall and stopped. More base current changes nothing. The LED is as bright as it will ever be.

That wall is saturation, and it is worth being precise about what it means.

Saturation means fully on, not maximum current. It is one of the worst-named ideas in electronics, and it gets worse: in a MOSFET, “saturation” means the opposite region. Here is the definition that actually helps — a transistor is saturated when both junctions are forward-biased. The collector has come down to within about 0.1 V of the emitter, there is no voltage left to control anything with, and the gain is gone. For a switch, that is exactly what you want.

Region	B–E junction	B–C junction	Behaves as
Cut-off	reverse / off	reverse	open switch
Active	forward	reverse	amplifier
Saturated	forward	forward	closed switch, Vce ≈ 0.1 V

The design rule to memorise

To use a transistor as a switch, deliberately over-drive the base by a factor of ten.

Work out the current the load needs. Here: (5 − 1.9 − 0.1) / 220 = 14 mA.
Pick Ib = Ic / 10 — about 1.4 mA.
Size the resistor: Rb = (Vdrive − 0.7) / Ib — about 3 kΩ. A 2.2 kΩ from the shelf is fine.

The forced β is then 10 instead of 150, which guarantees hard saturation even with the worst transistor in the batch on the coldest day of the year. Deliberately wasting gain is what buys you reliability.

Then press “remove base resistor.” There is nothing left to limit base current except the exponential from lesson 3, and you know how steep that is. The base junction is a forward diode straight across a 5 V supply. Watch the number, then never build it.

In the wild

Every relay, motor and LED strip driven from a dev board. BC547, 2N2222, 2N3904 — the same three or four part numbers turn up in every tutorial, and the base resistor is always somewhere between 1 kΩ and 10 kΩ for exactly the reason above. If someone hands you a schematic where the base goes straight to a pin with no resistor, they have made the mistake you just simulated.

The diode across the relay coil. Switch off a coil and its collapsing magnetic field produces a large reverse voltage spike that will punch straight through the transistor. A single diode wired backwards across the coil gives that energy a harmless path. It is called a flyback diode, it costs one rupee, and leaving it out is the most common way beginners kill a transistor.

Why chips like the ULN2003 exist. Driving seven stepper-motor windings means seven transistors and seven base resistors. That whole array, with the flyback diodes built in, comes in one 16-pin package for a few rupees. Once you can size one of these by hand, you can read the datasheet of the chip that does it for you.

Before you move on

You measure Vce = 0.08 V across a transistor and Ic / Ib = 8, while the datasheet says β is 150. Is the transistor faulty?

Correct. In saturation the collector circuit sets the current, so Ic/Ib tells you about your base resistor, not about the transistor. β_F = 150 is still true; it is simply not in charge any more.

Bench 04 · LED switch ACTIVE

● Locked until you commit a prediction above.

Base resistor 10.0 kΩ

1 kΩ32 kΩ1 MΩ

Ib—

Ic (LED)—

Vce—

forced β—

β is not a design parameter

13 minutes · the habit that separates working circuits from lucky ones

Recall From lesson 4: what are the two junction states that define saturation? show answer

Pull ten BC547s out of the same bag and measure them. You will find β anywhere from 50 to 400.

The same transistor will shift its β by 30% between a cold morning and a warm enclosure, and again as you change its operating current. The datasheet gives you a range, not a value, and it is not being coy — there genuinely is no value.

So the question is never “what is β?” The question is: does my circuit care?

Commit before you touch anything

Two amplifier stages, both biased to sit at about 2 mA with a typical transistor. You swap both transistors for a batch with β = 400 instead of 100. What happens to their output voltages?

Answer: C. Same transistor, same β, two circuits — and only one of them survives. Drag the β slider and watch.

Circuit A — what beginners draw

One resistor from the supply to the base, sized so that Ib × β lands the collector current where you want it. It works beautifully on the bench with the transistor you tuned it with.

Drag β and the operating point walks straight off the end of the world: into saturation at high β, into a useless near-cut-off at low β. Across the range you will actually get from a bag of parts, the output moves nearly the entire supply.

Circuit B — what a working design looks like

Hold the base at a fixed voltage and put a resistor in the emitter. Now trace it through:

The base sits at 2.0 V, held there by the divider.
So the emitter sits at about 2.0 − 0.65 = 1.35 V. Lesson 3 says that 0.65 is soft, but it only ever moves by tens of millivolts.
The emitter current is then just Ohm’s law on the emitter resistor: 1.35 V across 1 kΩ is 1.35 mA.
And Ic ≈ Ie. β never appeared.

β only has to be large enough that base current is a rounding error in the divider. Drag it across the full 10:1 range and the operating point moves by a couple of percent.

The habit: design so that β appears only in inequalities, never in equations. “β > 50 is enough” is a robust design. “β = 150” is a prototype that will fail in production, in winter, or on the ninth unit.

This is also why the emitter resistor in circuit B turns up in almost every real transistor stage you will ever meet. It is the same resistor that stopped the thermal runaway in lesson 3. One component, two entire classes of failure removed — at the cost of some gain, which is a trade you will make happily.

In the wild

Why the same transistor has three part numbers. BC547A, BC547B, BC547C are one device sorted into β bins — roughly 110–220, 200–450, 420–800. The manufacturer measures every unit and puts it in a bin because it cannot control the spread. If a supplier ships you Cs when you designed around Bs and your circuit notices, the circuit was wrong, not the supplier.

The single most common production failure. One unit in five sitting at the wrong operating point almost always means a β-dependent bias network. Resistors vary by a few percent; β varies by a factor of eight. When a design is intermittently wrong across units, go looking for the parameter with the widest spread first.

Where you have already seen it. Open any audio amplifier or radio schematic and look for a small resistor from the emitter to ground with a capacitor across it. That resistor is doing this job; the capacitor removes its cost in gain for the signal while keeping its benefit for the bias. Chapter 4 is that trick.

Before you move on

A colleague’s prototype works perfectly. In production, one unit in five sits at the wrong operating point. What do you look at first?

Yes. Resistors vary by a few percent; β varies by a factor of eight. When a design is intermittently wrong across units, the device parameter with the widest spread is the first suspect — and there is only one candidate.

Bench 05 · the same β, two circuits A: ACTIVE

● Locked until you commit a prediction above.

β of the transistor in both circuits 150

40220400

A · Vout—

A · Ic—

B · Vout—

B · Ic—

Drag β across the full range and watch which output holds still.

PNP is the same device in a mirror

12 minutes · and nothing more than that

Recall From lesson 5: how should β appear in a design — in equations or in inequalities? show answer

PNP transistors have a reputation for being confusing, and they have earned exactly none of it. There is one rule, it is complete, and it has no exceptions:

Negate every voltage. Reverse every current. Every equation is otherwise identical.

Commit before you touch anything

In a PNP transistor that is amplifying normally, which way does conventional current flow in the base wire, and where does the emitter connect?

Answer: B. Everything reverses, including which rail the emitter belongs to. Hit the mirror toggle and watch every sign flip.

Press the mirror button

The schematic flips top to bottom, the arrow on the emitter turns around, and every reading on the meters changes sign. The magnitudes do not change at all — it is the same circuit, the same resistors, the same operating point, described from the other side.

Two things to fix in your head while you look at it:

The emitter always goes to the more positive rail in a PNP, because the emitter is always the terminal that current flows into. In an NPN it is the terminal current flows out of, so it goes to the more negative rail.
Vbe is −0.65 V, meaning the base sits 0.65 V below the emitter. Some books write Veb = +0.65 V instead. Read the meter as “the base is 0.65 V below the emitter” and you will never get lost.

The arrow, once and for all

The arrow is always on the emitter and always points the way conventional current flows through it. NPN: Not Pointing iN — arrow points out, current leaves the emitter. PNP: arrow points in, current enters. Silly, but it survives exam pressure.

In the wild

Anything with a metal chassis. In a car, every load has its negative terminal bolted to the bodywork. There is no room to put a switch below it, so the switch has to sit between the battery and the load — which needs a PNP, because only a PNP has its emitter naturally at the positive rail. Nearly every automotive load is switched this way.

The chip that decides your laptop runs on battery or mains. A USB or laptop power path uses a pair of high-side switches to choose which supply feeds the machine and to cut it off if something is wrong. High side means the switch is above the load, which means a P-type device.

Making a motor go both ways. An H-bridge is two PNPs on top and two NPNs underneath. Turn on one diagonal and the motor spins one way; the other diagonal and it reverses. Every drone, printer and robot arm has one. Neither device could do it alone: an NPN is good at pulling a node down towards ground, a PNP is good at pushing it up towards the supply.

Where this is heading. Put one NPN and one PNP on the same node and you have a push-pull output stage — the front door to op-amps, audio amplifiers and motor drivers. That is chapter 6.

Before you move on

You are asked to switch a 12 V motor whose negative terminal is hard-wired to ground. Which device, and where does it go?

Correct — this is the everyday reason PNPs exist. The load is committed to ground, so the switch must sit between the supply and the load, and only a PNP has its emitter naturally at the positive rail. (Driving its base then needs a level shift from your 5 V logic, which is its own small puzzle.)

Bench 06 · NPN / PNP mirror ACTIVE

● Locked until you commit a prediction above.

Base drive —

Ib—

Ic—

Vbe—

Vce—

✓

Chapter checkpoint

8 minutes · deliberately shuffled, deliberately unlabelled

These are mixed up on purpose. Half the difficulty in electronics is not solving the problem — it is recognising which idea the problem is about before you start. Practising that recognition is the point.

Question 1

Vbe reads 0.66 V. You need ten times less collector current. What Vbe do you aim for?

One decade down is 60 mV down. The relationship is exponential, so current ratios map onto voltage differences, never voltage ratios.

Question 2

A transistor has Ic = 20 mA and Ib = 3 mA. What is it doing?

A forced β near 7 is the fingerprint of a deliberately over-driven switch. A real device would give you 100–300 if the collector circuit let it; here the collector circuit is the limit.

Question 3

Which of these can you build by soldering two 1N4148 diodes together?

The gain lives in the geometry. Two packaged diodes put a metal contact where the thin base needs to be, and every injected carrier dies there.

Question 4

An amplifier stage is biased with a single resistor from the supply to the base. What is the most likely failure across a production batch?

Ib is fixed by that resistor, so Ic = βIb is at the mercy of the device. High-β parts push Ic up until the collector resistor eats the whole supply, and the stage saturates.

Question 5

A PNP stage reads Vbe = −0.67 V and Ic = −4 mA. Is anything wrong?

Perfectly healthy. The base is 0.67 V below the emitter and 4 mA flows out of the collector. Same magnitudes as the NPN, opposite signs.

Question 6

You raise a transistor’s temperature by 30 °C while holding Vbe fixed. What happens to Ic?

The curve shifts about −2 mV/°C, so 30 °C is a 60 mV shift — one full decade. This is exactly why holding Vbe fixed is a bad way to bias anything, and why an emitter resistor is standard practice.

Checkpoint score

0 / 6

Answer all six. Four or more and the chapter has landed.

Every word this chapter introduced

If any of these is still a word rather than a picture, go back to the lesson in brackets before starting chapter 3.

Emitter, base, collector L1

Source, control, and destination. The emitter carries the largest current and holds the arrow; the base is the thin middle layer.

Transistor action L1

Carriers injected from the emitter surviving the crossing of a very thin base. It needs the thinness; it is not a property of the two junctions.

β (hFE) L1

Ic / Ib. The score of the race across the base. A range, never a value.

α L1

The fraction that survives the base, Ic / Ie. Always just under 1. β = α/(1−α).

Ie = Ic + Ib L2

True in every region, always. The emitter carries the most current of the three.

Common emitter L2

The configuration on every bench here: emitter to the common rail, signal in at the base, output at the collector.

V_T and the 60 mV rule L3

V_T = kT/q ≈ 26 mV. Sixty millivolts of Vbe multiplies the current by ten; eighteen doubles it.

g_m and r_e L3

Transconductance g_m = Ic/V_T, and its reciprocal r_e = 26 mV / Ie — the transistor’s own internal emitter resistance. Chapter 5 is built on these.

Thermal runaway L3

Fixed Vbe plus self-heating: warmer means more current, which means warmer. An emitter resistor breaks the loop.

Cut-off / active / saturated L4

Both junctions reverse; B–E forward and B–C reverse; both forward. Open switch, amplifier, closed switch.

Load line L4

What the collector circuit will allow, drawn on the same axes as what the transistor wants. The operating point is where they meet.

Forced β L4

Ic / Ib in saturation. It reflects your base resistor, not the device. Aim for 10 in a switch.

Emitter degeneration L5

A resistor in the emitter. Sets the current by Ohm’s law instead of by β, and kills thermal runaway. Costs gain; worth it.

PNP L6

The same device mirrored. Negate every voltage, reverse every current, emitter to the more positive rail.

Where this goes next

This chapter deliberately never asked you to analyse a circuit with algebra. That comes, and it comes easily, once the behaviour is already in your hands. The order of the remaining chapters is the one Horowitz and Hill use in The Art of Electronics, for the same reason: it puts every useful circuit before the theory that explains it.

CH 03

The emitter follower

Gain of exactly one, which sounds useless and is the most-used transistor circuit in existence. Where impedance stops being an abstraction.

CH 04

Bias, and DC versus signal

The single most documented source of confusion in transistor circuits: what sits still and what wiggles. Two views of the same circuit, side by side.

CH 05

Gain, and the wall you hit

−Rc/Re, then shrink Re to nothing and meet re = 25/Ic. Where the gain finally stops, and why the ceiling is the same for every transistor.

CH 06

The output stage

One NPN and one PNP on the same node, taking turns: push-pull, crossover distortion, and Class AB.

CH 07

FETs

The other kind of transistor, with a gate that draws no current at all — and a very different set of compromises.

CH 08–12

Op-amps, then integrated circuits

The black box that made most of this optional — but only for people who know what is inside it — then current mirrors, the differential pair, CMOS logic and stability.

About the simulations. Every circuit on this page is solved with a damped Newton–Raphson iteration on the Ebers–Moll transport equations — the same model SPICE uses, with the Early effect included and no region-switching shortcuts. That matters: hard-coded “if saturated then” branches lie precisely at the boundaries where learners are most confused. Default device parameters are 2N3904-class: I_S = 10⁻¹⁴ A, β_F = 150, β_R = 2, V_AF = 75 V. Two simplifications are worth naming: β is held flat with current here, where a real device humps and rolls off at both ends; and the base-width slider in lesson 1 uses a simple diffusion model, α = 0.995 e^−W/L, which has the right shape but is not a fabrication-grade number.