By the end of this chapter you should be able to:
- Describe the layer structure of npn and pnp transistors and justify the doping and width chosen for each of the three regions.
- Identify the four operating modes from the bias on the two junctions, and define the forward-active region.
- Trace the carrier transport from emitter injection through the base to collection, and sketch the minority-carrier profile.
- Derive the emitter injection efficiency \(\gamma\) and the base transport factor \(\alpha_T\), and compute both for a stated structure.
- Relate \(\alpha\) and \(\beta\) in both directions and explain why \(\beta\) is so sensitive to small changes in \(\alpha\).
- Apply \(I_C = \alpha I_E + I_{CBO}\) and \(I_C = \beta I_B + I_{CEO}\), and convert between \(I_{CBO}\) and \(I_{CEO}\).
- Estimate leakage at elevated temperature from the doubling rule and judge when it stops being negligible.
Part 2 got a great deal out of a single junction, but everything it produced was passive in the strict sense: a rectifier, a clipper, a regulator can shape power or throw it away, and none of them can make a small signal into a larger one. Amplification needs a device in which one terminal controls the current between two others, and the bipolar junction transistor achieves that by the simplest possible extension of Chapter 5 — put two junctions in the same crystal, and put them so close together that the first one decides what the second one carries.
That closeness is the whole invention. A p-n-p or n-p-n sandwich in which the middle layer is thick is not a transistor at all; it is two diodes back to back, and back-to-back diodes block current in both directions and do nothing useful. Make the middle layer thin compared with the distance a minority carrier diffuses before it recombines, and the carriers injected across the first junction arrive at the second one still alive, are swept across it by its reverse field, and appear in the output circuit. A large current then flows through a reverse-biased junction, controlled by the small current that maintains the injection. This chapter builds that picture in order: the structure and the deliberate asymmetry of its doping, the transport mechanism in the forward-active region, the two efficiencies that decide what fraction of the emitter current survives to the collector, the parameters \(\alpha\) and \(\beta\) that summarise them, and the leakage that spoils the summary as the device warms up.
1 Two Junctions in One Crystal
A bipolar transistor is three doped regions in one continuous crystal, alternating in type: n-p-n or p-n-p. The three are called the emitter, the base and the collector, and the two boundaries between them are two ordinary p-n junctions of exactly the kind Chapter 5 analysed — the emitter-base junction \(J_E\) and the collector-base junction \(J_C\). Each has its own depletion region, its own built-in potential of roughly 0.7 V, and its own capacitance.
What distinguishes a transistor from any other three-layer structure is that the three regions are deliberately made unequal. They differ in doping by four orders of magnitude and in thickness by a factor of a hundred, and every one of those choices is made for a reason that can be stated in one line.
| Region | Typical doping | Typical width | Why it is made that way |
|---|---|---|---|
| Emitter | \(N_E \approx 10^{19}\ \text{cm}^{-3}\) | a few µm | Doped as heavily as the crystal will tolerate, so that when \(J_E\) is forward biased almost all of the current across it is carried by the emitter's own majority carriers going into the base, and almost none by base majority carriers going the other way. |
| Base | \(N_B \approx 10^{17}\ \text{cm}^{-3}\) | 0.5–1 µm | Lightly doped so it cannot inject much back into the emitter, and made very thin so the carriers that enter it cross before they can recombine. |
| Collector | \(N_C \approx 10^{15}\ \text{cm}^{-3}\) | tens of µm | Lightly doped and physically large, so that \(J_C\) can support a wide depletion region and a high reverse voltage, and so that the heat generated there has somewhere to go. |
Chapter 5 established that a junction's depletion region extends mostly into the lightly doped side. Applied here, that tells you at once where the two depletion regions sit: \(J_E\) pushes almost entirely into the base rather than into the heavily doped emitter, and \(J_C\) pushes almost entirely into the collector rather than into the base. The base is squeezed from both sides, and the part of it that remains neutral — the effective base width \(W_B\) — is smaller than the metallurgical spacing of the two junctions and, as Chapter 17 will show, depends on the collector voltage.
The two structures behave identically once you translate the vocabulary. In an npn the injected carriers are electrons and the supply voltages are positive with respect to the emitter; in a pnp they are holes and every voltage and current direction reverses. Silicon npn devices dominate discrete practice because electron mobility is about three times hole mobility, so an npn of the same geometry is faster and carries more current, but pnp devices are indispensable as the upper half of a complementary pair — Chapter 20 builds an output stage out of exactly that pairing.
One asymmetry is easy to overlook and matters in the laboratory: the emitter and the collector are not interchangeable, even though the structure looks superficially symmetric. Swap them and you have a transistor whose "emitter" is lightly doped and whose "base" is heavily doped relative to it, which is precisely the arrangement Section 4 shows gives a poor injection efficiency. A 2N2222 wired backwards still works, with a current gain of about 2 to 5 instead of 200, and a reverse emitter-base breakdown of about 6 V instead of 60. Devices are asymmetric because the design deliberately made them so.
2 Four Modes, Set by Two Junctions
Each junction can be forward biased or reverse biased independently, so the device has four possible modes. Naming them and knowing what each is for takes a paragraph and saves a great deal of confusion later.
| Mode | \(J_E\) | \(J_C\) | Behaviour | Used for |
|---|---|---|---|---|
| Forward active | forward | reverse | \(I_C \approx \beta I_B\), controlled and nearly independent of \(V_{CE}\) | All linear amplification |
| Cut-off | reverse | reverse | Only leakage flows; the device is an open switch | The "off" state of a switch |
| Saturation | forward | forward | \(V_{CE}\) collapses to 0.1–0.3 V; \(I_C\) is set by the external circuit | The "on" state of a switch |
| Reverse active | reverse | forward | Works, badly: current gain of a few, low breakdown | Almost nothing, deliberately |
Only the first of these is the subject of this chapter and of Chapters 18 and 19; the other three are developed in Chapter 17, where the same four modes appear as four regions of the output characteristic. For an npn transistor, forward active means \(V_{BE} \approx +0.7\) V and \(V_{CB} > 0\), which together require \(V_{CE} = V_{CB} + V_{BE} > 0.7\) V. In practice the collector junction stops being reverse biased somewhere near \(V_{CE} = 0.2\) V, and the working rule is that a silicon npn is in the active region whenever \(V_{CE}\) exceeds about 0.3 V with \(V_{BE}\) held at 0.7 V.
Because the device is a single crystal, its three terminal currents are not independent. Kirchhoff's current law applied to the whole device gives
Written for an npn with all three currents taken in their conventional directions: \(I_C\) and \(I_B\) into the device, \(I_E\) out of it. For a pnp every arrow reverses and the same equation holds. This relation is exact under all conditions, in every mode, and is never an approximation.
The magnitudes, however, are wildly unequal. In a working amplifier \(I_C\) and \(I_E\) agree to within a per cent or so, while \(I_B\) is a hundred to three hundred times smaller than either. That is the entire commercial value of the device: a base current of 20 µA commanding a collector current of 2 mA, from a terminal that a preceding stage can drive with almost no effort.
Both carrier types take part. In an npn, electrons injected from the emitter are the current that crosses the base, while holes supplied through the base terminal are what recombine with them and what is injected backwards into the emitter. Remove either population and the device stops working. The field-effect transistors of Part 5 are unipolar: only the majority carrier of the channel is involved, which is why they have no minority-carrier storage and switch out of saturation faster.
3 What Actually Crosses the Base
Take an npn with \(V_{BE} = +0.70\) V and \(V_{CB} = +5\) V, and follow the carriers in order.
- Injection. Forward bias lowers the barrier at \(J_E\) by 0.70 V, so the electron concentration at the base edge of that junction rises above its equilibrium value by the Boltzmann factor \(e^{V_{BE}/V_T}\). With \(N_B = 10^{17}\) the equilibrium minority electron concentration is \(n_{B0} = n_i^2/N_B = 10^{3}\ \text{cm}^{-3}\), and at \(V_{BE} = 0.702\) V it becomes \(6.2\times10^{14}\ \text{cm}^{-3}\) — a rise of twelve orders of magnitude, and still only 0.6 per cent of the base doping, so the base remains quasi-neutral and low-level injection holds.
- Diffusion across the base. The far edge of the base sits against the reverse-biased collector junction, whose field sweeps away any electron that reaches it. The electron concentration there is therefore pinned at essentially zero. A concentration of \(6.2\times10^{14}\) at one edge and zero at the other, across a gap of 1 µm, is an enormous gradient, and the electrons cross by diffusion.
- Collection. At the collector edge the electrons enter the depletion field of \(J_C\), which points from collector to base and therefore accelerates electrons towards the collector. They are swept across and become collector current. Note what has happened: current is flowing freely through a reverse-biased junction. There is no contradiction — a reverse-biased junction has never had trouble passing minority carriers, only in finding any. The emitter supplies them.
Because the base is far thinner than the electron diffusion length, recombination inside it removes only a tiny fraction of what enters, and the electron profile across the base is very nearly a straight line rather than the decaying exponential of a long diode. The diffusion current is then simply proportional to the slope of that line:
Two consequences follow immediately. The collector current depends exponentially on \(V_{BE}\) and, to first order, not at all on \(V_{CE}\) — which is why the output characteristics of Chapter 17 are almost horizontal. And it varies as \(1/W_B\), which is why anything that changes the base width changes the gain.
The transit across the base is fast because it is short. For diffusive transport the mean crossing time is \(\tau_B = W_B^2/2D_{nB}\), which for \(W_B = 1\ \mu\text{m}\) and \(D_{nB} = 20\ \text{cm}^2/\text{s}\) gives 250 ps. The quadratic dependence is the reason every high-frequency process pushes the base thinner: halve \(W_B\) and the transit time falls by four, while the current gain rises. Chapter 25 returns to this when it defines the transition frequency \(f_T\).
4 Injection Efficiency and Base Transport
Not all of the emitter current reaches the collector, and the losses happen at two distinct places. Separating them is the most useful piece of bookkeeping in the whole subject, because the two losses are cured by two different design decisions.
Loss one: the wrong carriers cross \(J_E\). A forward-biased junction injects in both directions. Electrons go from emitter into base, which is what we want; holes go from base into emitter, which is pure waste, since those holes must be supplied through the base terminal and never reach the collector. The emitter injection efficiency \(\gamma\) is the useful fraction:
Each factor of the ratio is the corresponding term of the two diffusion currents, written out from the expression of the previous section applied on each side of \(J_E\). The dominant factor is \(N_B/N_E\), and it is entirely under the designer's control.
This single ratio explains the doping asymmetry of Section 1. Making the emitter a hundred times more heavily doped than the base makes the unwanted back-injection a hundred times smaller. Nothing else in the structure gives so large a return for so little effort, which is why every bipolar process dopes the emitter to degeneracy.
Loss two: some carriers never get across the base. Of the electrons that do enter the base, a fraction recombine with the holes waiting there. The surviving fraction is the base transport factor \(\alpha_T\), obtained by solving the diffusion equation in the base with the two boundary conditions of Figure 16.2:
The approximation is excellent whenever \(W_B \ll L_{nB}\), which is the design condition for a transistor. Note the quadratic: the loss is not proportional to the base width but to its square, so halving \(W_B\) cuts recombination by four.
An npn silicon transistor has \(N_E = 10^{19}\), \(N_B = 10^{17}\ \text{cm}^{-3}\), effective base width \(W_B = 1.0\ \mu\text{m}\), \(D_{nB} = 20\ \text{cm}^2/\text{s}\), \(L_{nB} = 10\ \mu\text{m}\), and in the heavily doped emitter \(D_{pE} = 1.0\ \text{cm}^2/\text{s}\), \(L_{pE} = 0.5\ \mu\text{m}\).
Injection efficiency.
so \(\gamma = 1/(1+0.001) = 0.999001\). One part in a thousand of the emitter current is wasted as back-injected holes.
Base transport factor. \(W_B/L_{nB} = 0.100\), so \(\alpha_T = 1/\cosh(0.1) = 0.995021\); the approximation \(1 - 0.01/2 = 0.995000\) agrees to five figures. Five electrons in a thousand recombine before reaching the collector — five times the injection loss.
Combining.
The ledger for \(I_E = 2.000\) mA. Of that emitter current, \(I_{nE} = \gamma I_E = 1.998002\) mA is electrons entering the base and \(I_{pE} = 1.998\ \mu\)A is holes going the wrong way. Of the electrons, \(I_{nC} = \alpha_T I_{nE} = 1.988053\) mA reaches the collector and \(9.949\ \mu\)A recombines in the base. The base terminal must supply both losses: \(I_B = 1.998 + 9.949 = 11.947\ \mu\)A. Check: \(I_C/I_B = 1.988053/0.011947 = 166.4\), as it must be.
What the design levers do. Recomputing with the same material but different geometry: \(W_B = 0.5\ \mu\text{m}\) gives \(\beta = 571\); \(W_B = 2\ \mu\text{m}\) gives \(\beta = 45\); \(W_B = 5\ \mu\text{m}\) gives \(\beta = 7.5\). Doping the base as heavily as the emitter, at \(N_B = 10^{19}\), collapses \(\gamma\) to 0.909 and \(\beta\) to 9.5. Both of the design rules stated in Section 1 — thin base, lightly doped base — are visible in those numbers.
5 Alpha, Beta and Why Beta Is Fragile
The two efficiencies are properties of the physics; the two parameters that appear on a data sheet are their consequences. The common-base current gain is defined as the fraction of emitter current that arrives at the collector,
and the common-emitter current gain as the ratio of collector current to base current,
Because \(I_E = I_C + I_B\), the two are not independent. Divide that relation through by \(I_C\) and rearrange:
The third form is worth memorising on its own: \((1+\beta)\) turns up whenever an emitter current is referred to the base, and it is exactly the reciprocal of the fractional loss.
The relation is violently non-linear near \(\alpha = 1\), which is precisely where transistors live. That has a practical consequence that catches out every student at least once.
| \(\alpha\) | \(1-\alpha\) | \(\beta\) | Comment |
|---|---|---|---|
| 0.950 | 0.050 | 19.0 | A poor transistor, or a good one wired backwards |
| 0.980 | 0.020 | 49.0 | A power device at high current |
| 0.990 | 0.010 | 99.0 | Typical small-signal |
| 0.994027 | 0.005973 | 166.4 | Worked Example 16.1 |
| 0.995 | 0.005 | 199.0 | A high-gain part |
| 0.999 | 0.001 | 999.0 | A superbeta device, with almost no breakdown voltage |
Read the table downwards. A change in \(\alpha\) from 0.990 to 0.995 — half of one per cent — doubles \(\beta\) from 99 to 199. Since \(\beta = \alpha/(1-\alpha)\), differentiating gives \(d\beta/\beta = [1/(1-\alpha)]\,d\alpha/\alpha \approx (1+\beta)\,d\alpha/\alpha\): a fractional change in \(\alpha\) is amplified by roughly \(\beta\) in \(\beta\). Every microscopic detail that shifts \(\alpha\) in the fourth decimal place — a slightly different base width across a wafer, a slightly different recombination centre density, the operating temperature, the collector voltage — moves \(\beta\) by tens of per cent.
A data sheet for a 2N3904 quotes \(h_{FE}\) between 100 and 300 at 10 mA — a factor of three, for parts out of the same box, at the same current and temperature. That spread is not sloppy manufacturing; it is the inevitable magnification of a tiny spread in \(\alpha\). Chapter 18 is entirely about designing bias circuits whose operating point is set by resistor ratios rather than by \(\beta\), and Chapter 19 shows that the mid-band voltage gain of a common-emitter stage happens not to contain \(\beta\) at all.
Two smaller points complete the picture. First, \(\alpha_{dc}\) and \(\beta_{dc}\) are ratios of total currents, whereas the small-signal parameters \(\alpha_{ac} = \partial I_C/\partial I_E\) and \(\beta_{ac} = \partial I_C/\partial I_B\) are ratios of changes. Data sheets call them \(h_{FE}\) and \(h_{fe}\) respectively, with the case of the subscript carrying the distinction. They differ by ten or twenty per cent in the middle of the current range and much more at the ends, and for most hand calculations they are treated as equal. Second, \(\beta\) is not constant with current: it falls at low \(I_C\) because recombination in the emitter-base depletion region takes a larger share of a small current, and it falls at high \(I_C\) because the injected carrier density approaches the base doping and high-level injection sets in. The useful, flat plateau typically spans one or two decades of collector current, and a bias design that sits in the middle of it is a design that will behave.
6 Leakage and Its Temperature Dependence
Everything so far has assumed that the collector current vanishes when the emitter current does. It does not. The collector junction is a reverse-biased junction, and Chapter 5 showed that such a junction passes a small, bias-independent current of thermally generated minority carriers. In a transistor this reverse saturation current has its own symbol and its own subscript convention.
\(I_{CBO}\) is the current from collector to base with the emitter open — the leakage of the collector junction on its own. Adding it to the transported current gives the exact common-base relation:
The equation every common-base calculation starts from. When \(I_E = 0\) it correctly gives \(I_C = I_{CBO}\).
\(I_{CEO}\) is the current from collector to emitter with the base open. Setting \(I_B = 0\) in \(I_E = I_C + I_B\) gives \(I_E = I_C\), and substituting into the relation above yields a result that surprises people the first time they see it:
Leakage is amplified by the transistor's own gain. With \(\beta = 100\), the collector-to-emitter leakage is 101 times the collector-junction leakage.
The physical reason is worth stating, because the algebra hides it. With the base terminal open, the leakage holes arriving in the base have nowhere to go: they cannot leave through the base lead, so they accumulate, raise the base potential, and forward-bias the emitter junction slightly. The emitter then injects, and the injected current is transported to the collector in the usual way. A floating base is not an off base — it is a base biased by its own leakage.
Substituting \(I_E = I_C + I_B\) into the common-base relation and rearranging gives the corresponding common-emitter equation:
Both leakage terms rise steeply with temperature, because they are generated currents and generation depends on \(n_i\). The engineering rule, accurate enough for design work over the ordinary range, is that \(I_{CBO}\) doubles for every 10 °C rise:
A silicon npn has \(I_{CBO} = 10\) nA at 25 °C and \(\beta = 100\). It is biased with \(I_B = 20\ \mu\)A. Find the collector current at 25 °C and at 100 °C.
At 25 °C, \(I_{CEO} = 101 \times 10\ \text{nA} = 1.010\ \mu\text{A}\), so
Leakage contributes 0.051 per cent — entirely negligible, which is why it never appears in a room-temperature design.
At 100 °C the temperature has risen by 75 °C, or 7.5 doublings, so \(I_{CBO} = 10\ \text{nA}\times 2^{7.5} = 1.810\ \mu\text{A}\) and \(I_{CEO} = 101 \times 1.810 = 182.8\ \mu\text{A}\). Then
a shift of 9.1 per cent in the operating point from leakage alone, before any account is taken of \(V_{BE}\) falling by 2 mV/°C or \(\beta\) rising with temperature. Both of those push in the same direction. This is the quantitative reason Chapter 18 exists.
Germanium for comparison. A germanium device of the same size starts from \(I_{CBO} \approx 2\ \mu\)A at 25 °C, because its smaller band gap makes \(n_i\) some \(10^{3}\) times larger. That gives \(I_{CEO} = 202\ \mu\)A at room temperature and 6.46 mA at 75 °C — larger than the intended signal current. Germanium transistors were unusable above about 70 °C, and silicon displaced them for that reason as much as any other.
One further consequence deserves a sentence, because it is the seed of Chapter 18's most important idea. Leakage entering the base region behaves exactly like base current that the designer did not ask for. Anything that lets \(I_B\) rise freely lets \(\beta I_{CBO}\) rise with it; a bias circuit that holds the base voltage fixed instead, and lets \(I_B\) find its own level, denies leakage that amplification. The number that measures how well a circuit does this is the stability factor \(S(I_{CO}) = \partial I_C/\partial I_{CBO}\), and it is the subject of the chapter after next.
7 Summary and Key Results
| Quantity | Expression | Value for the worked device |
|---|---|---|
| Terminal currents | \(I_E = I_C + I_B\) | Exact in every mode; \(I_B\) is 100–300 times smaller than the other two |
| Forward active | \(J_E\) forward, \(J_C\) reverse | \(V_{BE} \approx 0.7\) V, \(V_{CE} > 0.3\) V for silicon |
| Injection efficiency | \(\gamma = \left[1 + \frac{D_{pE}}{D_{nB}}\frac{N_B}{N_E}\frac{W_B}{L_{pE}}\right]^{-1}\) | 0.999001 — set almost entirely by \(N_B/N_E = 10^{-2}\) |
| Base transport factor | \(\alpha_T = 1/\cosh(W_B/L_{nB}) \approx 1 - \tfrac{1}{2}(W_B/L_{nB})^2\) | 0.995021 for \(W_B/L_{nB} = 0.1\) |
| Common-base gain | \(\alpha = \gamma\alpha_T\) | 0.994027 |
| Common-emitter gain | \(\beta = \alpha/(1-\alpha)\), \(\alpha = \beta/(1+\beta)\) | 166.4; \(\alpha = 0.990 \to \beta = 99\), \(\alpha = 0.995 \to \beta = 199\) |
| Base transit time | \(\tau_B = W_B^2/2D_{nB}\) | 250 ps; falls as \(W_B^2\), which is why fast devices have thin bases |
| Common-base leakage | \(I_C = \alpha I_E + I_{CBO}\) | \(I_{CBO} = 10\) nA at 25 °C for a small silicon device |
| Common-emitter leakage | \(I_C = \beta I_B + I_{CEO}\), \(I_{CEO} = (1+\beta)I_{CBO}\) | 1.010 µA at 25 °C, 182.8 µA at 100 °C |
| Temperature rule | \(I_{CBO}\) doubles per 10 °C | 0.05 % of \(I_C\) at 25 °C rising to 9.1 % at 100 °C |
| Design levers | \(\beta\) rises with \(N_E/N_B\) and with \(1/W_B^2\) | \(W_B\): 0.5 µm → \(\beta = 571\); 2 µm → 45; 5 µm → 7.5 |
8 Common Mistakes
Wire two discrete diodes anode-to-anode and you have a structure that is p-n-p in the right order, and it will not amplify anything. The reason is geometric, not electrical: the two junctions are millimetres apart inside their packages, while the electron diffusion length in the base material is a few micrometres. Every carrier injected at the first junction recombines long before it reaches the second, so \(\alpha_T\) is effectively zero and \(\beta\) with it. A transistor is not two junctions; it is two junctions closer together than a diffusion length. The same reasoning tells you why base width dominates every performance figure in the chapter.
Because \(\beta = \alpha/(1-\alpha)\) and \(\alpha\) sits within a per cent of unity, a fractional change in \(\alpha\) is magnified roughly \(\beta\)-fold in \(\beta\). A data-sheet range of 100 to 300 is normal for one part number, and the same device will show a different \(\beta\) at 1 mA, at 100 mA, at 25 °C and at 100 °C. Any calculation whose answer is proportional to \(\beta\) has an uncertainty of a factor of three built into it. Designs that survive are those in which \(\beta\) appears only in a term that is deliberately made small — which is exactly what the voltage-divider bias of Chapter 18 and the swamped amplifier of Chapter 19 achieve.
The subscripts say which terminal is open: \(I_{CBO}\) is collector-to-base with the emitter open, \(I_{CEO}\) is collector-to-emitter with the base open, and they differ by the factor \((1+\beta)\), not by a little. Using the data-sheet value of \(I_{CBO}\) where \(I_{CEO}\) belongs understates the leakage by two orders of magnitude. The second half of the mistake is to note that 10 nA is negligible and stop thinking about it: at 100 °C that 10 nA has become 1.81 µA, and after multiplication by \((1+\beta)\) it is 183 µA, which is nine per cent of a 2 mA operating point. Leakage is negligible at room temperature and only at room temperature.
9 Chapter Review
1. An npn transistor has \(N_E = 5\times10^{18}\), \(N_B = 2\times10^{17}\ \text{cm}^{-3}\), \(W_B = 0.8\ \mu\text{m}\), \(L_{nB} = 12\ \mu\text{m}\), \(D_{nB} = 20\), \(D_{pE} = 1.2\ \text{cm}^2/\text{s}\) and \(L_{pE} = 0.6\ \mu\text{m}\). Find \(\gamma\), \(\alpha_T\), \(\alpha\) and \(\beta\), and say which loss dominates.
The injection ratio is \(I_{pE}/I_{nE} = (D_{pE}/D_{nB})(N_B/N_E)(W_B/L_{pE}) = (1.2/20)(2\times10^{17}/5\times10^{18})(0.8/0.6) = 0.06 \times 0.04 \times 1.3333 = 3.200\times10^{-3}\), so \(\gamma = 1/1.0032 = 0.996810\). For the base, \(W_B/L_{nB} = 0.8/12 = 0.066667\), so \(\alpha_T = 1/\cosh(0.066667) = 0.997781\), or from the quadratic approximation \(1 - (0.066667)^2/2 = 0.997778\). Hence \(\alpha = 0.996810 \times 0.997782 = 0.994599\) and \(\beta = 0.994599/0.005401 = 184.2\). The two losses are now comparable — 0.319 per cent from back-injection against 0.222 per cent from recombination — whereas in Worked Example 16.1 recombination was five times the larger. The change came from raising the base doping relative to the emitter (a ratio of 1/25 here against 1/100 there), which degrades \(\gamma\), while the thinner base improved \(\alpha_T\). Since injection now dominates, the way to raise \(\beta\) further in this device is to dope the emitter more heavily, not to thin the base.
2. A transistor is measured with \(I_E = 4.00\) mA and \(I_B = 32\ \mu\text{A}\). Find \(I_C\), \(\alpha\) and \(\beta\). If a second sample from the same batch has \(\alpha\) larger by only 0.3 per cent, what is its \(\beta\)?
From \(I_E = I_C + I_B\), \(I_C = 4.000 - 0.032 = 3.968\) mA. Then \(\alpha = I_C/I_E = 3.968/4.000 = 0.99200\) and \(\beta = I_C/I_B = 3.968/0.032 = 124.0\); the check \(\alpha/(1-\alpha) = 0.992/0.008 = 124.0\) agrees. For the second sample, \(\alpha' = 0.99200 \times 1.003 = 0.994976\), so \(1 - \alpha' = 0.005024\) and \(\beta' = 0.994976/0.005024 = 198.0\). A three-parts-in-a-thousand change in \(\alpha\) has raised \(\beta\) by 60 per cent. The magnification factor is \(1/(1-\alpha) = 1+\beta \approx 125\): multiply the fractional change in \(\alpha\) by that and you predict \(0.003 \times 125 = 0.375\), or 37 per cent, which is the right order and underestimates only because the derivative itself steepens over the interval. This is the arithmetic behind every data sheet's three-to-one \(h_{FE}\) range.
3. Explain why the base must be both thin and lightly doped, and state which of the two requirements is the more demanding to manufacture.
The two requirements attack two different losses. Light base doping raises the emitter injection efficiency, because the ratio of unwanted back-injection to wanted injection contains the factor \(N_B/N_E\) directly: dropping \(N_B\) by ten drops the wasted hole current by ten. A thin base raises the base transport factor, because the fraction of injected electrons lost to recombination is \(1 - \alpha_T \approx (W_B/L_{nB})^2/2\), quadratic in the width. A base that was thin but heavily doped would inject badly; one that was lightly doped but thick would transport badly; a transistor needs both. Manufacturing difficulty falls on the width. Doping is set by an implant dose, which is controlled to a fraction of a per cent routinely, whereas the effective base width is the difference between two diffusion depths minus two depletion edges, so a 1 µm base is the small difference of two much larger numbers and is the parameter that varies most across a wafer. That variation, magnified as in question 2, is the principal source of \(\beta\) spread. The light base doping also carries a penalty of its own: it raises the base spreading resistance, which limits high-frequency performance and forces real devices towards a compromise near \(10^{17}\ \text{cm}^{-3}\).
4. A silicon transistor with \(\beta = 120\) has \(I_{CBO} = 15\) nA at 25 °C. It is used in a circuit that biases it at \(I_B = 25\ \mu\text{A}\) and that will operate up to 85 °C. Find \(I_{CEO}\) and \(I_C\) at both temperatures, and comment.
At 25 °C, \(I_{CEO} = (1+\beta)I_{CBO} = 121 \times 15\ \text{nA} = 1.815\ \mu\text{A}\), and \(I_C = \beta I_B + I_{CEO} = 120(25\ \mu\text{A}) + 1.815\ \mu\text{A} = 3.000 + 0.001815 = 3.0018\) mA. At 85 °C the rise is 60 °C, which is six doublings, so \(I_{CBO} = 15\ \text{nA}\times 2^{6} = 0.960\ \mu\text{A}\) and \(I_{CEO} = 121 \times 0.960 = 116.2\ \mu\text{A}\). Then \(I_C = 3.000 + 0.1162 = 3.116\) mA, a rise of 3.9 per cent from leakage alone. Two comments are in order. First, the calculation deliberately held \(\beta\) and \(I_B\) fixed, and in a real circuit neither stays fixed: \(\beta\) typically rises by about half a per cent per degree, and if the bias network holds the base voltage rather than the base current then the fall of \(V_{BE}\) at 2 mV/°C raises \(I_B\) as well. The true drift is therefore several times the 3.9 per cent computed here. Second, the leakage enters multiplied by \((1+\beta)\) only because the base is free to float; the whole purpose of the bias circuits of Chapter 18 is to provide a low-resistance path that keeps that multiplication from happening.
5. A transistor is operated in the reverse-active mode by interchanging its emitter and collector leads. Predict qualitatively what happens to \(\gamma\), \(\alpha_T\) and \(\beta\), and estimate \(\beta_R\) if \(\alpha_R = 0.40\).
The base transport factor is almost unchanged, because \(\alpha_T\) depends only on \(W_B/L_{nB}\), and the base is the same base whichever way the carriers cross it; if anything it improves slightly, since the collector-side depletion region no longer eats into the base at the injecting end. The injection efficiency, on the other hand, collapses. It is now the collector that must inject into the base, and the collector is doped at \(10^{15}\) against a base at \(10^{17}\) — the ratio \(N_B/N_E\) that was \(10^{-2}\) in normal operation becomes \(10^{2}\) here, so back-injection from the base into the lightly doped region dominates and \(\gamma\) falls to a few tenths. There is also a geometric loss the forward direction does not suffer: the emitter junction is small in area, so most of what crosses the base misses it and is collected by the base contact instead. With \(\alpha_R = 0.40\), \(\beta_R = \alpha_R/(1-\alpha_R) = 0.40/0.60 = 0.667\) — less than unity, so the arrangement gives no current gain at all. The reverse mode is nonetheless useful in one place: because the reverse-biased emitter junction has both regions heavily doped, its breakdown is a sharp 6 to 7 V, and that is exactly how a low-voltage Zener reference is made on a bipolar integrated circuit.