BJT structure (NPN and PNP): Worked examples — Hardware deep dive

This is a Deep Dive child of BJT structure (NPN and PNP). The parent built the structure and the three core relations. Here we drill every case the current relations can throw at you — every combination of what's given and what's asked, plus the weird edge cases (limits, zero inputs, temperature leakage, reverse-active) and a couple of exam twists.

Before we start, here is a map of how the five quantities feed into each other. It is not a circuit — it is a "which arrow do I follow?" chart. Read it like this: the three current boxes on the left ( $I_{E}$ , $I_{C}$ , $I_{B}$ ) are tied by the red charge-conservation double-arrow ( $I_{E} = I_{B} + I_{C}$ ); the two gain boxes on the right ( $α$ , $β$ ) are each fed by a pair of currents (coloured arrows) and are joined to each other by the grey bridge formulas. Whenever a problem hands you two of these boxes, trace the arrows to reach any third box. Every example below is just one such trace.

Figure s01 — Dependency map of the five BJT quantities. Left column: the three currents, joined by charge conservation (red). Right column: the two gains, joined by the bridge formulas (grey). Coloured arrows show which two currents define each gain. The take-away printed at the bottom: give any two independent values and all five are fixed.

The scenario matrix

Every problem this topic can ask is one row of this table. The last column names the example that clears that cell.

Cell	Given → Find	Edge / twist	Cleared by
A	$I_{C}, I_{B}$ → $β, α, I_{E}$	none	Ex 1
B	$α$ → $β$	edge: $α$ near 1 (limit)	Ex 2
C	$β$ → $α$	edge: $β$ large (limit)	Ex 3
D	$I_{B}, β$ → $I_{C}, I_{E}$	twist: design/forecast a current	Ex 4
E	$I_{E}, α$ → $I_{C}, I_{B}$	none (small- $I_{B}$ sanity)	Ex 5
F	Degenerate: $I_{B} = 0$	edge: cutoff / open base	Ex 6
G	Degenerate: $α \to 1$ , $I_{B} \to 0$	edge: limiting behaviour of $β$	Ex 6
H	Leakage $I_{C B O}$ included	edge: zero base current still leaks	Ex 7
I	Real-world word problem	twist: sensor drives an LED	Ex 8
J	Reverse-active	twist: emitter/collector swapped, low gain	Ex 9
K	Percent-change / sensitivity	twist: how much does $β$ move?	Ex 10

We'll do them in order. Every number is checked in the verify block. Unit convention: unless a step says otherwise, every current is written in milliamps (mA); gains $α, β$ are unitless ratios. Where a step switches to microamps ( $μ A$ ) or nanoamps ( $nA$ ) it says so explicitly and converts back to mA before combining.

Ex 1 — Cell A: two currents given

Worked example Find every quantity from

$I_{C}$ and $I_{B}$ Given $I_{C} = 4.95 mA$ and $I_{B} = 0.05 mA$ . Find $β$ , $I_{E}$ , and $α$ .

Forecast: $I_{B}$ is tiny next to $I_{C}$ (one-hundredth the size). So $β = I_{C} / I_{B}$ will be big (near 100), and $α$ will sit just under 1. Guess before reading on.

Step 1. $β = \frac{I _{C}}{I _{B}} = \frac{4.95 mA}{0.05 mA} = 99$ (the mA cancel → unitless). Why this step? $β$ is defined as collector-over-base current — it's a direct ratio of the two things we were handed, no other fact needed.

Step 2. $I_{E} = I_{B} + I_{C} = 0.05 mA + 4.95 mA = 5.00 mA$ . Why this step? Charge conservation ( $I_{E} = I_{B} + I_{C}$ ) is the one law that always holds; it turns the two given currents into the third. Both terms already in mA, so no conversion.

Step 3. $α = \frac{I _{C}}{I _{E}} = \frac{4.95 mA}{5.00 mA} = 0.99$ (mA cancel → unitless). Why this step? Now that $I_{E}$ is known, $α$ is just its definition.

Verify: Cross-check with the $β$ – $α$ bridge: $α / (1 - α) = 0.99/0.01 = 99 = β$ ✓. Forecast (big $β$ , $α$ near 1) confirmed. Units: mA in, mA out — the gains are unitless ratios, as they should be. ✓

Ex 2 — Cell B: $α$ given, and the limit as $α \to 1$

Worked example Turn

$α$ into $β$ , then push $α$ toward 1 Given $α = 0.98$ . Find $β$ . Then predict what happens to $β$ if $α$ climbs to $0.995$ .

Forecast: $α$ near 1 means the denominator $1 - α$ is near zero, so $β$ should blow up. Higher $α$ → the denominator shrinks even more → $β$ jumps a lot. Guess the two numbers.

Step 1. $β = \frac{α}{1 - α} = \frac{0.98}{0.02} = 49$ (unitless in, unitless out). Why this step? This is the bridge formula derived in the parent by substituting $I_{C} = α I_{E}$ and $I_{B} = (1 - α) I_{E}$ . We use it because we're given $α$ and want $β$ — this is the exact tool that converts one to the other.

Step 2. Now $α = 0.995$ : $β = \frac{0.995}{0.005} = 199$ . Why this step? Same formula, new input — it shows the sensitivity: a jump of only $0.015$ in $α$ multiplied $β$ by roughly 4×.

Verify: Reverse both. $α = β / (β + 1) = 49/50 = 0.98$ ✓ and $199/200 = 0.995$ ✓. Forecast (blow-up) confirmed. This is exactly why real transistors quote $α$ to three or four decimals — the last digit swings $β$ wildly.

Ex 3 — Cell C: $β$ given (limit as $β \to \infty$ )

Worked example Recover

$α$ from $β$ , and read off the ceiling Given $β = 200$ . Find $α$ . Then say what $α$ approaches as $β \to \infty$ .

Forecast: $β$ is large, so $α$ must be just a hair below 1. And if $β$ could grow forever, $α$ would creep toward exactly 1 but never reach it.

Step 1. $α = \frac{β}{β + 1} = \frac{200}{201} \approx 0.99502$ (unitless). Why this step? This is the inverse of the bridge formula — we invert because we now know $β$ and want $α$ . Algebraically, solving $β = α / (1 - α)$ for $α$ gives $α = β / (β + 1)$ .

Step 2 (limit). As $β \to \infty$ , $\frac{β}{β + 1} = \frac{1}{1 + 1/ β} \to 1$ . Why this step? The term $1/ β$ shrinks to zero, so the fraction approaches $1/1 = 1$ . Physically: no base current ever recombines → every emitted carrier is collected → $α = 1$ . It's a ceiling you approach but can't cross (a real transistor always loses some carriers to recombination, so $α < 1$ always).

Verify: $200/201 = 0.995024 \dots$ ; feed back: $0.995024/ (1 - 0.995024) = 0.995024/0.004975 \approx 200$ ✓.

Ex 4 — Cell D: design a collector current from $I_{B}$ and $β$

Worked example Given the base drive and gain, predict the output

A transistor has $β = 120$ . You inject $I_{B} = 20 μ A$ . Find $I_{C}$ and $I_{E}$ .

Forecast: $I_{C}$ is $β$ times bigger than $I_{B}$ , so around $120 \times 20 μ A = 2.4 mA$ . $I_{E}$ is a whisker larger than $I_{C}$ .

Step 1. $I_{C} = β I_{B} = 120 \times 20 μ A = 2400 μ A = 2.4 mA$ (gain is unitless, so the unit of $I_{C}$ is the unit of $I_{B}$ , namely $μ A$ ; then convert $2400 μ A = 2.4 mA$ ). Why this step? Rearranging $β = I_{C} / I_{B}$ gives $I_{C} = β I_{B}$ — we go this direction because the cause (base current) and the amplification (gain) are given, and the effect (collector current) is asked.

Step 2. Convert $I_{B} = 20 μ A = 0.02 mA$ , then $I_{E} = I_{B} + I_{C} = 0.02 mA + 2.4 mA = 2.42 mA$ . Why this step? Charge conservation again — the only way to reach $I_{E}$ . We convert $I_{B}$ to mA first so both terms share a unit before adding.

Verify: $α = I_{C} / I_{E} = 2.4/2.42 = 0.99174$ , and $β = α / (1 - α) = 0.99174/0.00826 = 120$ ✓. Units: $μ A \times$ (unitless) $= μ A$ ✓. This is the everyday amplifier design move — see Common-Emitter Amplifier.

Ex 5 — Cell E: $I_{E}$ and $α$ given

Ex 6 — Cells F & G: degenerate case, $I_{B} = 0$ (open base → cutoff)

Worked example What happens when no base current flows?

You leave the base terminal open, so $I_{B} = 0$ (ignore leakage for now — that's Ex 7). What are $I_{C}$ and $I_{E}$ ? What does $β = I_{C} / I_{B}$ mean here?

Forecast: No control current → the gate is shut → essentially no current anywhere. This is the cutoff region (see BJT Operating Regions (Active, Saturation, Cutoff)).

Step 1. With $I_{B} = 0 mA$ and no injection, $I_{C} = 0 mA$ and therefore $I_{E} = I_{B} + I_{C} = 0 mA$ . Why this step? $I_{C} = β I_{B} = β \times 0 = 0$ — a finite gain times zero drive is zero output. The transistor is off.

Step 2. The ratio $β = I_{C} / I_{B} = 0/0$ is undefined, not zero. Why this step? $β$ is a property of the device geometry, not something that vanishes when you stop driving it. Dividing $0/0$ tells you nothing — you cannot measure $β$ at zero current. It still equals whatever the structure fixes (say 99); you just can't extract it from these two numbers.

Step 3 (limit link to Cell G). As you make the base thinner and thinner in the limit $α \to 1$ , the fraction of $I_{E}$ that becomes base current, $(1 - α)$ , goes to 0. So for a fixed collector current $I_{C}$ , the base current $I_{B} \to 0$ and $β = α / (1 - α) \to \infty$ . An ideal, perfectly thin base needs zero base current for finite collector current. Why this step? This is the same $0/0$ tension resolved: the ideal device has infinite $β$ ; real devices lose a sliver to recombination so $β$ stays finite.

Verify: Plug $α = 1$ into $α / (1 - α)$ : denominator $= 0 \Rightarrow$ formula diverges ✓ (checked as a limit, not a plug-in, in the verify block). Cutoff logic: $I_{C} = β \cdot 0 = 0$ ✓.

Ex 7 — Cell H: the leakage twist ( $I_{B} = 0$ but current still trickles)

Worked example Even with the base open, a small current leaks

The collector–base junction is reverse biased, and reverse-biased PN junctions leak a tiny saturation current called $I_{C B O}$ (Collector–Base current with the third lead Open). Suppose $I_{C B O} = 10 nA$ and $β = 100$ . With the base left open ( $I_{B} = 0$ from outside), the collector current is $I_{C} = (β + 1) I_{C B O} .$ Find $I_{C}$ .

Forecast: The leakage itself is tiny (nanoamps), but it gets amplified by roughly $β$ , so expect around $100 \times 10 nA \approx 1 μ A$ — small, but no longer exactly zero.

Step 1. $I_{C} = (β + 1) I_{C B O} = 101 \times 10 nA = 1010 nA = 1.01 μ A$ ( $β$ unitless, so $I_{C}$ keeps the unit of $I_{C B O}$ , namely nA; then $1010 nA = 1.01 μ A$ ). Why this step? The leakage acts like a built-in base current the transistor can't switch off. The transistor amplifies it just like a real base current, hence the $(β + 1)$ factor (the $+ 1$ because the leakage also passes through as part of $I_{E}$ ). Why $(β + 1)$ and not $β$ ? The leakage current itself flows through the device to the emitter, adding one extra copy on top of the $β$ amplified copies.

Step 2. So "cutoff" from Ex 6 is really "as-off-as-possible," not literally zero — this leftover $I_{C}$ is called $I_{C E O}$ (Collector–Emitter current with the base Open). Why this step? It corrects the idealization of Ex 6: real transistors have a floor set by temperature-driven leakage, which roughly doubles every $1 0^{\circ}$ C.

Verify: $101 \times 10 = 1010 nA$ ✓. Sanity: $1.01 μ A ≫ 0$ but $≪$ the mA-scale active currents, so cutoff is still "effectively off." ✓

Ex 8 — Cell I: real-world word problem

Worked example A light sensor switches on an LED

A photodiode feeds a base current that rises from $0$ to $30 μ A$ as a room brightens. It drives an NPN transistor with $β = 150$ whose collector runs an LED that needs $4 mA$ to glow visibly. Does the LED light up at full brightness? What is $I_{E}$ when it does?

Forecast: $150 \times 30 μ A = 4.5 mA$ , which is more than the $4 mA$ the LED needs — so yes, it should glow. (In practice the transistor would then sit in saturation, capped by the LED's resistor, but for the current relation we compute the demanded $I_{C}$ .)

Step 1. Max collector current available: $I_{C} = β I_{B} = 150 \times 30 μ A = 4500 μ A = 4.5 mA$ ( $β$ unitless, so $I_{C}$ inherits $μ A$ ; convert $4500 μ A = 4.5 mA$ ). Why this step? The transistor tries to supply $β I_{B}$ ; since $4.5 mA > 4 mA$ , the LED gets what it needs.

Step 2. Convert $I_{B} = 30 μ A = 0.03 mA$ , then $I_{E} = I_{B} + I_{C} = 0.03 mA + 4.5 mA = 4.53 mA$ . Why this step? Charge conservation gives the total current the transistor pulls from the supply; convert $I_{B}$ to mA first so both terms share a unit.

Verify: $4.5 mA \geq 4 mA$ → LED lights ✓. $α = 4.5/4.53 = 0.99338$ , and $β = α / (1 - α) = 150$ ✓. This is a real switching/amplifier circuit — the tiny sensor current controlling a big LED current is the parent's water-gate story made concrete.

Ex 9 — Cell J: exam twist, reverse-active (emitter and collector swapped)

First, why running the transistor backwards forces a poor gain. The figure below shows the same NPN two ways: forward (heavily-doped emitter injects) and reverse (moderately-doped collector is forced to inject). Injection efficiency depends on how much more heavily the injector is doped than the base; the emitter wins big, the collector barely wins — so the reverse fraction $α_{R}$ that survives the base is far from 1.

Figure s02 — Why $α_{R}$ is small. Left (forward): the heavily-doped emitter (n⁺, dense dots) floods carriers into the thin base; almost all reach the collector, so $α_{F} \to 1$ . Right (reverse): the moderately-doped collector (n, sparse dots) is a weak injector into a base that now looks relatively more doped, so many carriers recombine and $α_{R} \approx 0.5$ . Structure, not symbols, sets the gain.

Worked example Why swapping E and C ruins the gain

A student wires an NPN "backwards": the collector plays the role of emitter. Now the moderately-doped collector does the injecting into the base and the heavily-doped emitter does the collecting. This reverse-active mode has a reverse gain $α_{R} = 0.5$ . Find the reverse $β_{R}$ and compare with a normal $β$ of, say, 99.

Forecast: $α_{R}$ is nowhere near 1 (see the figure — the weak injector loses half its carriers), so $β_{R}$ will be small — the "amplifier" barely amplifies. This is the parent's "E and C are not interchangeable" mistake, quantified.

Step 1. $β_{R} = \frac{α _{R}}{1 - α _{R}} = \frac{0.5}{0.5} = 1$ (unitless). Why this step? The same bridge formula still holds — the physics of counting carriers doesn't change direction — but with a poor $α_{R}$ it delivers a gain of only 1, i.e. no amplification at all.

Step 2. Compare: normal $β = 99$ vs reverse $β_{R} = 1$ , a ratio of $99/1 = 99$ , i.e. 99× worse. Why this step? It shows why structure ≠ symmetry: the heavy emitter doping and thin-base geometry only work in the forward direction, so $α_{R} ≪ α_{F}$ and the gain collapses.

Verify: $0.5/ (1 - 0.5) = 1$ ✓, and $99/1 = 99$ ✓. The device still conducts, but as an amplifier it's useless backwards — exactly the parent's steel-manned mistake, now with a number attached.

Ex 10 — Cell K: sensitivity / percent-change

Worked example How much does

$β$ move when $α$ drifts? Temperature nudges $α$ from $0.990$ up to $0.992$ . By what percent does $β$ change?

Forecast: $α$ barely moved (about $0.2%$ ), but since $β$ lives on the steep $1/ (1 - α)$ curve near 1, the percent change in $β$ should be much bigger than $0.2%$ . Guess the direction (up) and roughly how big.

Step 1. $β_{1} = \frac{0.990}{1 - 0.990} = \frac{0.990}{0.010} = 99$ (unitless). Why this step? Baseline gain from the bridge formula, using the starting $α$ .

Step 2. $β_{2} = \frac{0.992}{1 - 0.992} = \frac{0.992}{0.008} = 124$ (unitless). Why this step? New gain after the drift — the denominator shrank from $0.010$ to $0.008$ (a 20% shrink), which is what balloons $β$ .

Step 3. Percent change: $\frac{β _{2} - β _{1}}{β _{1}} \times 100% = \frac{124 - 99}{99} \times 100% \approx 25.3%$ . Why this step? $%$ change $= \frac{Δ β}{β _{1}} \times 100%$ — the standard sensitivity measure; the $%$ makes the answer a pure ratio (no current units survive).

Verify: $(124 - 99) /99 = 0.2525 \dots \Rightarrow 25.3%$ ✓. A $0.2%$ move in $α$ caused a $\approx 25%$ move in $β$ — that roughly 125× amplification of the fractional change is why circuits are designed to depend on $α$ /geometry rather than on the twitchy $β$ .

Recall Which relation do I reach for?

Given the two currents I have, which formula gets the target? ::: If given two currents use $I_{E} = I_{B} + I_{C}$ + a ratio; if given a gain, use $β = α / (1 - α)$ or its inverse $α = β / (β + 1)$ . Why is $β$ undefined (not zero) at $I_{B} = 0$ ? ::: $β = I_{C} / I_{B} = 0/0$ ; $β$ is set by geometry, not by drive current, so it can't be read off at zero. As $α \to 1$ for fixed $I_{C}$ , what happens to $I_{B}$ and $β$ ? ::: $I_{B} \to 0$ and $β \to \infty$ — the ideal, perfectly-thin base needs no base current. Why does reverse-active mode have tiny gain? ::: The heavy emitter doping only injects well forward; swapped, the weak collector injector gives $α_{R}$ far from 1 so $β_{R}$ is tiny. Why quote $α$ to many decimals? ::: $β$ is hypersensitive near $α = 1$ ; a $0.2%$ shift in $α$ can move $β$ by ~25%.

Prerequisites & neighbours: PN Junction Diode · Doping and Semiconductors · BJT Operating Regions (Active, Saturation, Cutoff) · Common-Emitter Amplifier · FET Structure · ↩ back to parent