part09

Chapter 11: BJT Static Characteristics

Source: Pierret, Semiconductor Device Fundamentals, Chapter 11 (book pp. 389-442). Where we are: Chapter 10 told the story of the transistor in words and pictures. This chapter turns that story into mathematics, checks the mathematics against a real 2N2605 transistor, and then shows how modern designers cheat the old trade-offs.

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Prologue: The Plan, and a Picture to Hold Onto

Imagine a river (the emitter current) flowing toward a sea (the collector). In the middle is a narrow, slightly leaky canal (the base). A tiny leak in the canal wall is the base current; almost all the water gets to the sea. The leak is tiny because the canal is short. The sluice gate at the start of the canal (the forward-biased E-B junction) controls how much water enters, and it does so exponentially, which is why a small gate movement gives a huge change in flow. Everything in this chapter is a precise version of that picture.

The chapter has three movements:

1. Ideal analysis (Section 11.1). Solve the minority-carrier diffusion equation in each of the three regions, exactly as for the ideal diode, and read off currents and gains.
2. Reality check (Section 11.2). Compare with measurement and name every disagreement: base-width modulation, punch-through, avalanche plus feedback, series resistance, current crowding, recombination-generation, and graded base.
3. Modern devices (Section 11.3). The polysilicon-emitter BJT and the heterojunction bipolar transistor (HBT), which use the lessons of Section 11.2 to beat the old trade-off between gain and base resistance.

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11.1 Ideal Transistor Analysis

11.1.1 Solution Strategy

The recipe is the same one used for the ideal diode, with one new ingredient: three quasineutral regions instead of two, and therefore three current-voltage expressions to juggle at once.

Basic assumptions

1. The device is a pnp BJT with nondegenerate, uniformly doped emitter, base and collector. The E-B and C-B junctions are step junctions.
2. Steady state.
3. One-dimensional device.
4. Low-level injection in all quasineutral regions.
5. Only drift, diffusion and thermal recombination-generation occur, and there is no light: $G_L=0$.
6. Thermal recombination-generation is negligible in the E-B and C-B depletion regions.
7. The quasineutral emitter and collector widths are much greater than the minority-carrier diffusion lengths there (effectively semi-infinite).

Notice what is not assumed: thermal recombination in the base is kept, and the base width is initially arbitrary. We keep the base general on purpose, so that we can watch the narrow-base result emerge as a special case.

Notation (Fig. 11.1)

Because a transistor has two regions of the same doping type, the old subscripts $n$/$p$ are replaced by regional subscripts $E$, $B$, $C$ for the minority-carrier parameters of each region.

	Emitter ($p^{+}$)	Base ($n$)	Collector ($p$)
Doping	$N_E=N_{AE}$	$N_B=N_{DB}$	$N_C=N_{AC}$
Minority carrier	electrons	holes	electrons
Diffusion coefficient	$D_E=D_N$	$D_B=D_P$	$D_C=D_N$
Lifetime	${\tau }_E={\tau }_n$	${\tau }_B={\tau }_p$	${\tau }_C={\tau }_n$
Diffusion length	$L_E=L_N$	$L_B=L_P$	$L_C=L_N$
Equilibrium minority conc.	$n_{E0}=n_i^2/N_E$	$p_{B0}=n_i^2/N_B$	$n_{C0}=n_i^2/N_C$

Coordinates (Fig. 11.1): base $x$ from $0$ (E-B depletion edge) to $W$ (C-B depletion edge), $W$ = quasineutral base width; emitter $x^{\prime \prime }$ from $0$ at the E-B depletion edge pointing into the emitter; collector $x^{\prime }$ from $0$ at the C-B depletion edge pointing into the collector.

Diffusion equations and boundary conditions

Emitter (minority electrons):

$$0=D_E\frac{d^2\Delta n_E}{d{x}^{\prime \prime 2}}-\frac{\Delta n_E}{{\tau }_E}\text{\hspace{1em}(11.1)}$$ $$\Delta n_E(x^{\prime \prime }\to \infty )=0\text{\hspace{1em}(11.2a)}$$ $$\Delta n_E(x^{\prime \prime }=0)=n_{E0}\left(e^{q{V}_{EB}/kT}-1\right)\text{\hspace{1em}(11.2b)}$$

Base (minority holes):

$$0=D_B\frac{d^2\Delta p_B}{d{x}^2}-\frac{\Delta p_B}{{\tau }_B}\text{\hspace{1em}(11.3)}$$ $$\Delta p_B(0)=p_{B0}\left(e^{q{V}_{EB}/kT}-1\right)\text{\hspace{1em}(11.4a)}$$ $$\Delta p_B(W)=p_{B0}\left(e^{q{V}_{CB}/kT}-1\right)\text{\hspace{1em}(11.4b)}$$

Collector (minority electrons):

$$0=D_C\frac{d^2\Delta n_C}{d{x}^{\prime 2}}-\frac{\Delta n_C}{{\tau }_C}\text{\hspace{1em}(11.5)}$$ $$\Delta n_C(x^{\prime }\to \infty )=0\text{\hspace{1em}(11.6a)}$$ $$\Delta n_C(x^{\prime }=0)=n_{C0}\left(e^{q{V}_{CB}/kT}-1\right)\text{\hspace{1em}(11.6b)}$$

Why these boundary conditions? The “law of the junction”: at the edge of a depletion region the minority excess is the equilibrium value times $(e^{qV/kT}-1)$, with $V$ the voltage across that junction. The semi-infinite emitter and collector forget the junction far away (zero excess at infinity). The base has junctions on both sides, so both ends carry a non-zero excess.

Computational relationships

With R-G negligible in the depletion regions, the current leaving a quasineutral region equals the minority diffusion current at its depletion edge:

$$I_{En}=-qA{D}_E\frac{d\Delta n_E}{d{x}^{\prime \prime }}{|}_{x^{\prime \prime }=0}\text{\hspace{1em}(11.7)}$$ $$I_{Ep}=-qA{D}_B\frac{d\Delta p_B}{dx}{|}_{x=0}\text{\hspace{1em}(11.8)}$$ $$I_{Cp}=-qA{D}_B\frac{d\Delta p_B}{dx}{|}_{x=W}\text{\hspace{1em}(11.9)}$$ $$I_{Cn}=qA{D}_C\frac{d\Delta n_C}{d{x}^{\prime }}{|}_{x^{\prime }=0}\text{\hspace{1em}(11.10)}$$

Performance factors and terminal currents (repeats of Chapter 10):

$$\gamma =\frac{I_{Ep}}{I_{Ep}+I_{En}}\text{\hspace{1em}(11.11)}$$ $${\alpha }_T=\frac{I_{Cp}}{I_{Ep}}\text{\hspace{1em}(11.12)}$$ $${\alpha }_{dc}=\gamma {\alpha }_T\text{\hspace{1em}(11.13)}$$ $${\beta }_{dc}=\frac{{\alpha }_{dc}}{1-{\alpha }_{dc}}\text{\hspace{1em}(11.14)}$$ $$I_E=I_{Ep}+I_{En}\text{\hspace{1em}(11.15)}$$ $$I_C=I_{Cp}+I_{Cn}\text{\hspace{1em}(11.16)}$$ $$I_B=I_E-I_C\text{\hspace{1em}(11.17)}$$

Think of $\gamma$ as “what fraction of the emitter’s effort is useful hole injection,” ${\alpha }_T$ as “what fraction of injected holes survive the canal crossing,” and ${\alpha }_{dc}$ as the overall fraction of emitter current that reaches the collector.

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11.1.2 General Solution (Arbitrary $W/L_B$)

Emitter and collector solutions

The general solution of (11.1) is

$$\Delta n_E(x^{\prime \prime })=A_1{e}^{-x^{\prime \prime }/L_E}+A_2{e}^{x^{\prime \prime }/L_E},L_E=\sqrt{D_E{\tau }_E}\text{\hspace{1em}(11.18)}$$

• As $x^{\prime \prime }\to \infty$, $e^{x^{\prime \prime }/L_E}\to \infty$, so (11.2a) forces $A_2=0$.
• With $A_2=0$, (11.2b) gives $A_1=n_{E0}(e^{q{V}_{EB}/kT}-1)$.

$$\Delta n_E(x^{\prime \prime })=n_{E0}\left(e^{q{V}_{EB}/kT}-1\right)e^{-x^{\prime \prime }/L_E}\text{\hspace{1em}(11.19)}$$

Differentiate at $x^{\prime \prime }=0$ and use (11.7):

$$\frac{d\Delta n_E}{d{x}^{\prime \prime }}{|}_0=-\frac{n_{E0}}{L_E}\left(e^{q{V}_{EB}/kT}-1\right)$$ $$I_{En}=qA\frac{D_E}{L_E}{n}_{E0}\left(e^{q{V}_{EB}/kT}-1\right)\text{\hspace{1em}(11.20)}$$

The collector is identical in structure:

$$\Delta n_C(x^{\prime })=n_{C0}\left(e^{q{V}_{CB}/kT}-1\right)e^{-x^{\prime }/L_C}\text{\hspace{1em}(11.21)}$$ $$\frac{d\Delta n_C}{d{x}^{\prime }}{|}_0=-\frac{n_{C0}}{L_C}\left(e^{q{V}_{CB}/kT}-1\right)$$ $$I_{Cn}=-qA\frac{D_C}{L_C}{n}_{C0}\left(e^{q{V}_{CB}/kT}-1\right)\text{\hspace{1em}(11.22)}$$

The minus sign in (11.22): these are the $p$-side halves of the ideal diode equation. The sign differs only because collector current is defined positive for flow from the $n$-base to the $p$-collector, opposite to the diode’s positive direction. For $V_{CB}<0$ the bracket is negative and $I_{Cn}>0$, as expected.

Base solution

The base is where the new physics lives, because $\Delta p_B$ does not vanish at either end. General solution of (11.3):

$$\Delta p_B(x)=A_1{e}^{-x/L_B}+A_2{e}^{x/L_B},L_B=\sqrt{D_B{\tau }_B}\text{\hspace{1em}(11.23)}$$

Apply (11.4). Abbreviate ${\Delta }_0\equiv \Delta p_B(0)$, ${\Delta }_W\equiv \Delta p_B(W)$:

$${\Delta }_0=A_1+A_2\text{\hspace{1em}(11.24a)}$$ $${\Delta }_W=A_1{e}^{-W/L_B}+A_2{e}^{W/L_B}\text{\hspace{1em}(11.24b)}$$

Multiply (11.24a) by $e^{-W/L_B}$ and subtract from (11.24b):

$$A_2=\frac{{\Delta }_W-{\Delta }_0{e}^{-W/L_B}}{e^{W/L_B}-e^{-W/L_B}}$$

Multiply (11.24a) by $e^{W/L_B}$ and subtract (11.24b):

$$A_1=\frac{{\Delta }_0{e}^{W/L_B}-{\Delta }_W}{e^{W/L_B}-e^{-W/L_B}}$$

Substitute back and group the ${\Delta }_0$ and ${\Delta }_W$ terms:

$$\Delta p_B(x)={\Delta }_0\frac{e^{(W-x)/L_B}-e^{-(W-x)/L_B}}{e^{W/L_B}-e^{-W/L_B}}+{\Delta }_W\frac{e^{x/L_B}-e^{-x/L_B}}{e^{W/L_B}-e^{-W/L_B}}\text{\hspace{1em}(11.25)}$$

With $\sinh \xi =(e^{\xi }-e^{-\xi })/2$ (11.27):

$$\boxed{\Delta p_B(x)=\Delta p_B(0)\frac{\sinh [(W-x)/L_B]}{\sinh (W/L_B)}+\Delta p_B(W)\frac{\sinh (x/L_B)}{\sinh (W/L_B)}}\text{\hspace{1em}(11.26)}$$

Sanity check. If $V_{CB}=0$ then $\Delta p_B(W)=0$ and (11.26) collapses to the narrow-base diode solution (Eq. 6.65), as it must.

Intuition. (11.26) is a superposition: the first term is the profile from the emitter alone with a perfect sink at $x=W$; the second from the collector alone with a perfect sink at $x=0$. The problem is linear, so we add.

Hole currents across the junctions

Differentiate (11.26):

$$\frac{d\Delta p_B}{dx}=\frac{-{\Delta }_0\cosh [(W-x)/L_B]+{\Delta }_W\cosh (x/L_B)}{L_B\sinh (W/L_B)}$$

At $x=0$ and $x=W$:

$$\frac{d\Delta p_B}{dx}{|}_0=\frac{1}{L_B}\left[-{\Delta }_0\coth \frac{W}{L_B}+\frac{{\Delta }_W}{\sinh (W/L_B)}\right],\frac{d\Delta p_B}{dx}{|}_W=\frac{1}{L_B}\left[-\frac{{\Delta }_0}{\sinh (W/L_B)}+{\Delta }_W\coth \frac{W}{L_B}\right]$$

Insert in (11.8), (11.9) and restore the voltage dependences ($\cosh \xi =(e^{\xi }+e^{-\xi })/2$, Eq. 11.30):

$$I_{Ep}=qA\frac{D_B}{L_B}{p}_{B0}\left[\frac{\cosh (W/L_B)}{\sinh (W/L_B)}\left(e^{q{V}_{EB}/kT}-1\right)-\frac{1}{\sinh (W/L_B)}\left(e^{q{V}_{CB}/kT}-1\right)\right]\text{\hspace{1em}(11.28)}$$ $$I_{Cp}=qA\frac{D_B}{L_B}{p}_{B0}\left[\frac{1}{\sinh (W/L_B)}\left(e^{q{V}_{EB}/kT}-1\right)-\frac{\cosh (W/L_B)}{\sinh (W/L_B)}\left(e^{q{V}_{CB}/kT}-1\right)\right]\text{\hspace{1em}(11.29)}$$

Performance parameters and terminal currents (active mode)

Active mode in a pnp: $V_{EB}>0$, $V_{CB}<0$ (beyond a few $kT/q$). Then $e^{q{V}_{CB}/kT}-1\approx -1$ is small next to $e^{q{V}_{EB}/kT}-1$, so we drop the $V_{CB}$ terms in $I_{Ep}$, $I_{Cp}$ when forming $\gamma$ and ${\alpha }_T$.

Emitter efficiency (using $n_{E0}/p_{B0}=N_B/N_E$):

$$\frac{I_{En}}{I_{Ep}}=\frac{D_E{n}_{E0}/L_E}{D_B{p}_{B0}\coth (W/L_B)/L_B}=\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\tanh \frac{W}{L_B}$$ $$\gamma =\frac{1}{1+I_{En}/I_{Ep}}=\boxed{\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\frac{\sinh (W/L_B)}{\cosh (W/L_B)}}}\text{\hspace{1em}(11.31)}$$

Base transport factor:

$${\alpha }_T=\frac{I_{Cp}}{I_{Ep}}=\frac{1/\sinh (W/L_B)}{\cosh (W/L_B)/\sinh (W/L_B)}=\boxed{\frac{1}{\cosh (W/L_B)}}\text{\hspace{1em}(11.32)}$$

Common-base gain:

$${\alpha }_{dc}=\gamma {\alpha }_T=\boxed{\frac{1}{\cosh \frac{W}{L_B}+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\sinh \frac{W}{L_B}}}\text{\hspace{1em}(11.33)}$$

Common-emitter gain, ${\beta }_{dc}=1/(1/{\alpha }_{dc}-1)$:

$$\boxed{{\beta }_{dc}=\frac{1}{\cosh \frac{W}{L_B}+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\sinh \frac{W}{L_B}-1}}\text{\hspace{1em}(11.34)}$$

Terminal currents. Add the $n$ and $p$ components:

$$I_E=qA\left\{\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\frac{\cosh (W/L_B)}{\sinh (W/L_B)}\right]\left(e^{q{V}_{EB}/kT}-1\right)-\frac{D_B}{L_B}{p}_{B0}\frac{1}{\sinh (W/L_B)}\left(e^{q{V}_{CB}/kT}-1\right)\right\}\text{\hspace{1em}(11.35)}$$ $$I_C=qA\left\{\frac{D_B}{L_B}{p}_{B0}\frac{1}{\sinh (W/L_B)}\left(e^{q{V}_{EB}/kT}-1\right)-\left[\frac{D_C}{L_C}{n}_{C0}+\frac{D_B}{L_B}{p}_{B0}\frac{\cosh (W/L_B)}{\sinh (W/L_B)}\right]\left(e^{q{V}_{CB}/kT}-1\right)\right\}\text{\hspace{1em}(11.36)}$$

If desired, $I_B=I_E-I_C$.

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Exercise 11.1

Problem. (a) Show that the terminal-current expressions reduce to the ideal diode equation when $W\gg L_B$. (b) Show that setting $V_{CB}=0$ in (11.35) leads to the narrow-base-diode I-V relationship.

Solution.

(a) For $W/L_B\to \infty$:

$$\frac{\cosh (W/L_B)}{\sinh (W/L_B)}=\frac{e^{W/L_B}+e^{-W/L_B}}{e^{W/L_B}-e^{-W/L_B}}\to 1,\frac{1}{\sinh (W/L_B)}=\frac{2}{e^{W/L_B}-e^{-W/L_B}}\to 0$$

Then (11.35), (11.36) become

$$I_E=qA\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\right]\left(e^{q{V}_{EB}/kT}-1\right)$$ $$I_C=-qA\left[\frac{D_C}{L_C}{n}_{C0}+\frac{D_B}{L_B}{p}_{B0}\right]\left(e^{q{V}_{CB}/kT}-1\right)$$

Each is an ideal-diode equation (up to the minus sign explained after 11.22). A structure with $W\gg L_B$ is just two back-to-back diodes.

(b) With $V_{CB}=0$, (11.35) becomes

$$I_E=qA\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\coth \frac{W}{L_B}\right]\left(e^{q{V}_{EB}/kT}-1\right)$$

For a $p^{+}n$ junction the $n_{E0}$ term (back-injection) is small and may be dropped:

$$I_E\approx qA\frac{D_B}{L_B}{p}_{B0}\coth \frac{W}{L_B}\left(e^{q{V}_{EB}/kT}-1\right)$$

which is the narrow-base result (Eqs. 6.68-6.69). Tying the collector to the base turns the ideal transistor into the narrow-base diode.

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11.1.3 Simplified Relationships ($W\ll L_B$)

We could have imposed $W\ll L_B$ at the start, but that would have hidden other limiting cases and thrown away significant small terms. Now we use the small-argument limits

$$\sinh \xi \to \xi ,\xi \ll 1\text{\hspace{1em}(11.37)}$$ $$\cosh \xi \to 1+\frac{{\xi }^2}{2},\xi \ll 1\text{\hspace{1em}(11.38)}$$

$\Delta p_B(x)$ becomes a straight line

For $0\le x\le W$ both $(W-x)/L_B$ and $x/L_B$ are $\ll 1$. Replace each $\sinh$ by its argument in (11.26):

$$\Delta p_B(x)\approx \Delta p_B(0)\frac{(W-x)/L_B}{W/L_B}+\Delta p_B(W)\frac{x/L_B}{W/L_B}=\Delta p_B(0)\left(1-\frac{x}{W}\right)+\Delta p_B(W)\frac{x}{W}\text{\hspace{1em}(11.39)}$$ $$\boxed{\Delta p_B(x)=\Delta p_B(0)+\left[\Delta p_B(W)-\Delta p_B(0)\right]\frac{x}{W}}\text{\hspace{1em}(11.40)}$$

A straight line from $p_{B0}(e^{q{V}_{EB}/kT}-1)$ to $p_{B0}(e^{q{V}_{CB}/kT}-1)$ (Fig. 11.2, active mode: starts high, ends near $-p_{B0}$, i.e. total hole concentration about zero at the collector end). It stays near-linear even for $W/L_B$ up to 1 (Problem 11.5).

Picture. The base is a very short ramp. If the ramp is shorter than the “leakage length” $L_B$, nothing is lost on the way: density drops linearly, like temperature along a thin plate whose faces are held at fixed temperatures.

Performance parameters (to second order in $W/L_B$)

With $\tanh (W/L_B)\approx W/L_B$, $\cosh \approx 1+(W/L_B{)}^2/2$:

$$\gamma =\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}}\text{\hspace{1em}(11.41)}$$ $${\alpha }_T=\frac{1}{1+\frac{1}{2}(W/L_B{)}^2}\approx 1-\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2\text{\hspace{1em}(11.42)}$$ $${\alpha }_{dc}=\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2}\text{\hspace{1em}(11.43)}$$ $${\beta }_{dc}=\frac{1}{\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2}\text{\hspace{1em}(11.44)}$$

($1/{\alpha }_{dc}-1$ is exactly the sum of the two small terms.)

Design lore.

• $N_E\gg N_B$ kills the first term: $\gamma \to 1$.
• $W\ll L_B$ kills the second: ${\alpha }_T\to 1$.
• ${\beta }_{dc}$ is the reciprocal of the sum of two loss terms, so it is large only if both are tiny. First loss: back-injection into the emitter ($I_{B1}$); second: recombination in the base ($I_{B2}$).

Sample calculation (Table 11.1)

pnp BJT at 300 K:

	Emitter	Base	Collector
Doping	$N_E=10^{18}$	$N_B=10^{16}$	$N_C=10^{15}$ cm${}^{-3}$
Mobility (minority)	${\mu }_E=263$	${\mu }_B=437$	${\mu }_C=1345$ cm${}^2$/V-s
$D$	$6.81$	$11.3$	$34.8$ cm${}^2$/s
$\tau$	$10^{-7}$ s	$10^{-6}$ s	$10^{-6}$ s
$L$	$8.25\times 10^{-4}$ cm	$3.36\times 10^{-3}$ cm	$5.90\times 10^{-3}$ cm

and $W=W_B=2\times 10^{-4}$ cm. Then

$$\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}=\frac{6.81}{11.3}\cdot \frac{10^{16}}{10^{18}}\cdot \frac{2\times 10^{-4}}{8.25\times 10^{-4}}=1.46\times 10^{-3},\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2=\frac{1}{2}{\left(\frac{2\times 10^{-4}}{3.36\times 10^{-3}}\right)}^2=1.77\times 10^{-3}$$ $$\gamma =\frac{1}{1.00146}=0.9985,{\alpha }_T=\frac{1}{1.00177}=0.9982,{\alpha }_{dc}=\frac{1}{1.00323}=0.9968,{\beta }_{dc}=\frac{1}{0.00323}=310$$

Real emitter dopings usually exceed $10^{18}$ cm${}^{-3}$ (nondegenerate theory limits it near there), yet these results are representative. In modern transistors ${\alpha }_T$ is closer to unity and gain is mostly limited by emitter efficiency.

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Exercise 11.2

Problem. Show that (11.40) solves the diffusion equation when thermal R-G is negligible in the quasineutral base.

Solution. Dropping $\Delta p_B/{\tau }_B$ in (11.3):

$$0=D_B\frac{d^2\Delta p_B}{d{x}^2}$$

Integrate twice:

$$\Delta p_B(x)=A_1+A_{2}x$$

At $x=0$:

$$\Delta p_B(0)=A_1$$

At $x=W$:

$$\Delta p_B(W)=\Delta p_B(0)+A_{2}W\Rightarrow A_2=\frac{\Delta p_B(W)-\Delta p_B(0)}{W}$$

Substituting:

$$\Delta p_B(x)=\Delta p_B(0)+\left[\Delta p_B(W)-\Delta p_B(0)\right]\frac{x}{W}$$

which is (11.40). $■$

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Exercise 11.3

Problem. For a pnp transistor ($W\ll L_B$) in saturation, sketch on a linear scale the normalized excess $\Delta n/n_0$ or $\Delta p/p_0$ in emitter, base and collector.

Solution. From (11.19) and (11.21):

$$\frac{\Delta n_E(x^{\prime \prime })}{n_{E0}}=\left(e^{q{V}_{EB}/kT}-1\right)e^{-x^{\prime \prime }/L_E},\frac{\Delta n_C(x^{\prime })}{n_{C0}}=\left(e^{q{V}_{CB}/kT}-1\right)e^{-x^{\prime }/L_C}$$

In saturation $V_{EB}>0$ and $V_{CB}>0$ (with $V_{EB}>V_{CB}$ in the sketch): both profiles look like a forward-biased diode, a pile-up at the depletion edge decaying exponentially inward. In the base, from (11.4):

$$\frac{\Delta p_B(0)}{p_{B0}}=e^{q{V}_{EB}/kT}-1>0,\frac{\Delta p_B(W)}{p_{B0}}=e^{q{V}_{CB}/kT}-1>0$$

Excess at both base edges. The normalized excess at the base edge equals that at the emitter edge of the same depletion region (same junction factor), and $\Delta p_B(W)/p_{B0}=\Delta n_C(0)/n_{C0}$. With $W\ll L_B$ the base profile is a straight line.

 Normalized excess (linear), V_EB > V_CB > 0

   |\                      Emitter   : exponential decay, scale L_E
   | \                     Base      : straight line between two positive values
   |  \     \              Collector : exponential decay, scale L_C
   |   \     \_____
   |    \________  ......
   +------+------+---------- x
     E      B      C

Saturation is the one mode in which the base is flooded from both sides.

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Exercise 11.4

Problem. Find the base transport factor using the linear approximation (11.40). Comment.

Solution. A linear profile has the same slope everywhere:

$$\frac{d\Delta p_B}{dx}{|}_0=\frac{d\Delta p_B}{dx}{|}_W=\frac{\Delta p_B(W)-\Delta p_B(0)}{W}$$

By (11.8), (11.9): $I_{Cp}=I_{Ep}$, hence

$${\alpha }_T=\frac{I_{Cp}}{I_{Ep}}=1$$

Comment. ${\alpha }_T=1$ means no recombination in the base, consistent with the assumption behind (11.40). But the linear approximation discards the $(W/L_B{)}^2$ terms: the small loss $\frac{1}{2}(W/L_B{)}^2$ of (11.42) is invisible. To find $1-{\alpha }_T$ you must keep the slight curvature of the exact solution.

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11.1.4 Ebers-Moll Equations and Model

Now attention shifts to terminal currents, in a form a circuit simulator can use.

With $V_{CB}=0$, (11.35) has the form of an ideal diode equation for the E-B junction; with $V_{EB}=0$, (11.36) is one for the C-B junction. Call the prefactors effective saturation currents:

$$I_{F0}=qA\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\frac{\cosh (W/L_B)}{\sinh (W/L_B)}\right]\text{\hspace{1em}(11.45a)}$$ $$I_{R0}=qA\left[\frac{D_C}{L_C}{n}_{C0}+\frac{D_B}{L_B}{p}_{B0}\frac{\cosh (W/L_B)}{\sinh (W/L_B)}\right]\text{\hspace{1em}(11.45b)}$$

($F$ forward, $R$ reverse.) The prefactor of $(e^{q{V}_{CB}/kT}-1)$ in (11.35) is smaller than $I_{R0}$: it is the fraction of the collector-junction current that makes it across the base to the emitter junction. Likewise the $V_{EB}$ term in (11.36) is the fraction of $I_{F0}(\dots )$ reaching the collector. Define forward and reverse “gains”:

$${\alpha }_F{I}_{F0}={\alpha }_R{I}_{R0}=qA\frac{D_B}{L_B}\frac{p_{B0}}{\sinh (W/L_B)}\text{\hspace{1em}(11.46)}$$

Then

$$\boxed{I_E=I_{F0}\left(e^{q{V}_{EB}/kT}-1\right)-{\alpha }_R{I}_{R0}\left(e^{q{V}_{CB}/kT}-1\right)}\text{\hspace{1em}(11.47a)}$$ $$\boxed{I_C={\alpha }_F{I}_{F0}\left(e^{q{V}_{EB}/kT}-1\right)-I_{R0}\left(e^{q{V}_{CB}/kT}-1\right)}\text{\hspace{1em}(11.47b)}$$

These are the Ebers-Moll equations, used by SPICE-type programs. They may also be treated as empirical: four parameters ($I_{F0},I_{R0},{\alpha }_F,{\alpha }_R$), of which only three are independent due to the reciprocity relation ${\alpha }_F{I}_{F0}={\alpha }_R{I}_{R0}$.

Circuit view. With $I_F=I_{F0}(e^{q{V}_{EB}/kT}-1)$ and $I_R=I_{R0}(e^{q{V}_{CB}/kT}-1)$: $I_E=I_F-{\alpha }_R{I}_R$, $I_C={\alpha }_F{I}_F-I_R$, $I_B=I_E-I_C$. The large-signal equivalent circuit (Fig. 11.3) has a diode carrying $I_F$ (E to B), a diode carrying $I_R$ (C to B), a current source ${\alpha }_F{I}_F$ from the emitter side to the collector, and a source ${\alpha }_R{I}_R$ from the collector side to the emitter. KCL at E, C, B reproduces (11.47). It is handy for PNPN-family analysis.

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Exercise 11.5

Problem. Show ${\alpha }_F={\alpha }_{dc}$.

Solution. Divide (11.46) by (11.45a):

$${\alpha }_F=\frac{qA\frac{D_B}{L_B}\frac{p_{B0}}{\sinh (W/L_B)}}{qA\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\frac{\cosh (W/L_B)}{\sinh (W/L_B)}\right]}$$

Divide top and bottom by $qA{D}_B{p}_{B0}/(L_B\sinh (W/L_B))$:

$${\alpha }_F=\frac{1}{\cosh \frac{W}{L_B}+\frac{D_E{n}_{E0}{L}_B}{D_B{p}_{B0}{L}_E}\sinh \frac{W}{L_B}}$$

With $n_{E0}/p_{B0}=N_B/N_E$:

$${\alpha }_F=\frac{1}{\cosh \frac{W}{L_B}+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\sinh \frac{W}{L_B}}={\alpha }_{dc}\text{(11.33)}■$$

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Exercise 11.6

Problem. Show the Ebers-Moll equations yield $I_C={\alpha }_{dc}{I}_E+I_{CBO}$ (Eq. 10.8) in active mode.

Solution. Active: $e^{q{V}_{CB}/kT}-1\approx -1$:

$$I_E=I_{F0}\left(e^{q{V}_{EB}/kT}-1\right)+{\alpha }_R{I}_{R0},I_C={\alpha }_F{I}_{F0}\left(e^{q{V}_{EB}/kT}-1\right)+I_{R0}$$

Solve the first for the voltage factor:

$$e^{q{V}_{EB}/kT}-1=\frac{I_E-{\alpha }_R{I}_{R0}}{I_{F0}}$$

Substitute in the second:

$$I_C={\alpha }_F\left(I_E-{\alpha }_R{I}_{R0}\right)+I_{R0}={\alpha }_F{I}_E+\left(1-{\alpha }_F{\alpha }_R\right)I_{R0}$$

With ${\alpha }_F={\alpha }_{dc}$ this is Eq. 10.8 provided

$$I_{CBO}=\left(1-{\alpha }_F{\alpha }_R\right)I_{R0}■$$

(Problem 11.12(c) confirms $I_{CBO}$ is the C-B saturation current with emitter open.)

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Exercise 11.7 (computer)

Problem. Using the Ebers-Moll equations, write a MATLAB program that calculates and plots the input and output characteristics of a pnp BJT in common-base and common-emitter configurations. Sample: room temperature, $A=10^{-4}$ cm${}^2$, $W=W_B$, material parameters similar to Table 11.1.

Solution (logic). A characteristic needs the dependent current as an explicit function of the controlled variable. Only the common-base input (11.47a) already has that form; the rest require algebra (derived in Problem 11.14):

1. CB input $I_E(V_{EB},V_{CB})$: $I_E=I_{F0}(e^{q{V}_{EB}/kT}-1)-{\alpha }_R{I}_{R0}(e^{q{V}_{CB}/kT}-1)$
2. CB output $I_C(V_{CB},I_E)$: $I_C={\alpha }_F{I}_E-(1-{\alpha }_F{\alpha }_R)I_{R0}(e^{q{V}_{CB}/kT}-1)$
3. CE input $I_B(V_{EB},V_{EC})$: $I_B=\left[(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}{e}^{-q{V}_{EC}/kT}\right]e^{q{V}_{EB}/kT}-\left[(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}\right]$
4. CE output $I_C(V_{EC},I_B)$: $I_C=\frac{\left({\alpha }_F{I}_{F0}-I_{R0}{e}^{-q{V}_{EC}/kT}\right)\left[I_B+(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}\right]}{(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}{e}^{-q{V}_{EC}/kT}}+I_{R0}-{\alpha }_F{I}_{R0}$

Use $V_{CB}=V_{EB}-V_{EC}$ whenever $V_{EC}$ is the independent variable. The textbook program (BJT with subprogram BJT0, Appendix M) uses: $A=10^{-4}$ cm${}^2$, $W_B=2.5\mu$m, $N_E=10^{18}$, $N_B=1.5\times 10^{16}$, $N_C=1.5\times 10^{15}$ cm${}^{-3}$, ${\tau }_E=10^{-7}$, ${\tau }_B={\tau }_C=10^{-6}$ s, and the mobility fit $\mu ={\mu }_{min}+{\mu }_0/[1+(N/N_{ref}{)}^{\alpha }]$ with electrons $(92,1268,1.3\times 10^{17},0.91)$ and holes $(54.3,406.9,2.35\times 10^{17},0.88)$ for $({\mu }_{min},{\mu }_0,N_{ref},\alpha )$.

% BJT Common-Base / Common-Emitter characteristics from Ebers-Moll
clear; close all
q=1.602e-19; k=8.617e-5; T=300; kT=k*T;           % kT in volts (k in eV/K)
A=1.0e-4; WB=2.5e-4; W=WB;                         % cm^2, cm
ni=1.0e10; NE=1.0e18; NB=1.5e16; NC=1.5e15;         % cm^-3
mun=@(N) 92   + 1268 ./(1+(N/1.3e17).^0.91);      % minority electrons (E, C)
mup=@(N) 54.3 + 406.9./(1+(N/2.35e17).^0.88);     % minority holes (B)
DE=kT*mun(NE); DB=kT*mup(NB); DC=kT*mun(NC);       % Einstein relation
tauE=1e-7; tauB=1e-6; tauC=1e-6;
LE=sqrt(DE*tauE); LB=sqrt(DB*tauB); LC=sqrt(DC*tauC);
nE0=ni^2/NE; pB0=ni^2/NB; nC0=ni^2/NC;
% Ebers-Moll parameters, Eqs. (11.45), (11.46)
fB  = (DB/LB)*pB0*cosh(W/LB)/sinh(W/LB);
IF0 = q*A*((DE/LE)*nE0 + fB);
IR0 = q*A*((DC/LC)*nC0 + fB);
aF  = q*A*(DB/LB)*pB0/sinh(W/LB)/IF0;
aR  = q*A*(DB/LB)*pB0/sinh(W/LB)/IR0;
 
% (1) Common-base input: IE vs VEB, stepped VCB
subplot(2,2,1); hold on
VEB=0:0.005:0.8;
for VCB=0:-10:-20
    IE=(IF0*(exp(VEB/kT)-1)-aR*IR0*(exp(VCB/kT)-1))*1e3;   % mA
    plot(VEB,IE)
end
axis([0.35 0.8 0 5]); grid on; xlabel('V_{EB} (V)'); ylabel('I_E (mA)')
 
% (2) Common-base output: IC vs -VCB, stepped IE
subplot(2,2,2); hold on
VCB=[2:-0.01:0, 0:-0.25:-50];                     % fine near origin, coarse after
for IE=(0:4)*1e-3
    IC=(aF*IE-(1-aF*aR)*IR0*(exp(VCB/kT)-1))*1e3;
    plot(-VCB,IC)
end
axis([-5 50 0 5]); grid on; xlabel('-V_{CB} (V)'); ylabel('I_C (mA)')
 
% (3) Common-emitter input: IB vs VEB, stepped VEC
subplot(2,2,3); hold on
for VEC=0:5:10
    IB0=(1-aF)*IF0+(1-aR)*IR0;
    IB1=(1-aF)*IF0+(1-aR)*IR0*exp(-VEC/kT);
    IB=(IB1*exp(VEB/kT)-IB0)*1e6;                           % microamps
    plot(VEB,IB)
end
axis([0.35 0.85 -5 20]); grid on; xlabel('V_{EB} (V)'); ylabel('I_B (\muA)')
 
% (4) Common-emitter output: IC vs VEC, stepped IB
subplot(2,2,4); hold on
VEC=[0:0.01:1, 1.25:0.25:50];
for IB=(0:4)*2.5e-6
    IB0=(1-aF)*IF0+(1-aR)*IR0;
    IB1=(1-aF)*IF0+(1-aR)*IR0*exp(-VEC/kT);
    IC=((aF*IF0-IR0*exp(-VEC/kT)).*(IB+IB0)./IB1+IR0-aF*IR0)*1e3;
    plot(VEC,IC)
end
axis([0 50 0 2.5]); grid on; xlabel('V_{EC} (V)'); ylabel('I_C (mA)')

Logic notes: exp(VEB/kT) is evaluated element-wise on vectors; a fine voltage grid is used near the origin where saturation makes the curves bend sharply and a coarse one in the flat active region. Axis limits are tuned to this parameter set.

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11.2 Deviations from the Ideal

The ideal theory gives the right shape of the BJT characteristics. We now compare with a 2N2605 pnp BJT at room temperature (Figs. 11.4, 11.5; theory computed with $W=W_B$) and ask where and why they disagree.

11.2.1 Ideal Theory/Experiment Comparison

Common-base input (Fig. 11.4a,b). Both show $I_E$ rising exponentially with $V_{EB}$. Theory: essentially independent of $V_{CB}$ (all curves for $V_{CB}\le 0$ coincide). Experiment: $I_E$ at a given $V_{EB}$ noticeably increases as $V_{CB}$ becomes more negative.

Common-base output (Fig. 11.4c,d). Excellent agreement: a rapid rise of $I_C$ beginning at slightly negative $V_{CB}$ (leaving saturation), then an almost perfectly constant $I_C$. The difference: theory has no limit on $-V_{CB}$, the real device breaks down at $V_{CBO}$.

Common-emitter input (Fig. 11.5a,b). Mostly good agreement. The $V_{EC}=0$ curve stands alone; curves for $V_{EC}\gtrsim$ a few $kT/q$ bunch together, displaced to higher $V_{EB}$.

Common-emitter output (Fig. 11.5c,d). The largest disagreement. Theory is perfectly flat in active mode; experiment slopes noticeably upward even at low voltage, turns sharply upward at large voltage, and is limited by a low-current breakdown voltage $V_{CEO}$.

Deviations to be explained: (1) the dependence of the CB input on $V_{CB}$; (2) the quasilinear slope and subsequent sharp upturn of the CE output; (3) the $V_{CBO}$, $V_{CEO}$ breakdown limits. Overall, the agreement is surprisingly good.

11.2.2 Base Width Modulation (the Early Effect)

In the theory $W=W_B$ was constant. In truth the quasineutral width is

$$W=W_B-x_{nEB}-x_{nCB}$$

(Eq. 10.3), where $x_{nEB}$, $x_{nCB}$ are the parts of the E-B and C-B depletion regions lying in the base. Changing $V_{EB}$ or $V_{CB}$ changes the depletion widths and so shrinks or expands $W$ (Fig. 11.6). Because the base is so thin, even small depletion changes are a large fraction of $W$. This is base width modulation, the Early effect (after J. Early).

Analogy. The base is a plank bridge for hole-travellers. More reverse bias on the collector side saws a slice off the collector end: the bridge is shorter, crossing is easier, more travellers arrive per unit time.

Since $\gamma ,{\alpha }_T,{\alpha }_{dc},{\beta }_{dc},I_{F0},I_{R0}$ all depend on $W$, all vary with applied voltages.

(i) Common-base input. For $-V_{CB}\gg kT/q$, (11.35) without the $V_{CB}$ term is

$$I_E=qA\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\frac{\cosh (W/L_B)}{\sinh (W/L_B)}\right]\left(e^{q{V}_{EB}/kT}-1\right)\text{\hspace{1em}(11.48)}$$

For $W\ll L_B$ ($\coth \to L_B/W$) and neglecting the small $D_E{n}_{E0}/L_E$ term:

$$I_E\approx qA\frac{D_B{p}_{B0}}{W}{e}^{q{V}_{EB}/kT}\text{\hspace{1em}(11.49)}$$

Raising the C-B reverse bias widens the C-B depletion region, $W$ falls, and by (11.49) $I_E$ at fixed $V_{EB}$ rises with $-V_{CB}$: exactly Fig. 11.4(b).

(ii) Common-emitter output. In active mode

$$I_C={\beta }_{dc}{I}_B+I_{CEO}\text{\hspace{1em}(11.50)}$$

with (11.44) ${\beta }_{dc}=1/\left[\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac{1}{2}(W/L_B{)}^2\right]$. $W$ decreases with $V_{EC}$, so ${\beta }_{dc}$ rises, and at fixed $I_B$ the collector current rises with $V_{EC}$: the quasilinear slope of Fig. 11.5(d). Base width modulation does not explain the sharp upturn near $V_{CEO}$.

Why are the other two characteristics insensitive? The active CB output obeys

$$I_C={\alpha }_{dc}{I}_E+I_{CBO}\text{\hspace{1em}(11.51)}$$

with (11.43) ${\alpha }_{dc}=1/[1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac{1}{2}(W/L_B{)}^2]$. In a well-made transistor the $W$-dependent denominator terms are tiny compared with 1, so large changes in $W$ barely change ${\alpha }_{dc}$; and $I_{CBO}$ is essentially $W$-independent. A similar argument applies to the CE input (Problem 11.15).

Doping and bias mode. In a standard transistor $N_E>N_B>N_C$. Almost all of the E-B depletion region lies in the base, but the junction is only slightly forward biased in active mode, so it hardly modulates $W$. The C-B junction has large reverse bias, but most of its depletion region lies in the lightly doped collector, which minimizes the effect. In inverted mode the reverse-biased E-B depletion region extends almost entirely into the base, so the same transistor becomes very sensitive to base width modulation.

11.2.3 Punch-Through

Punch-through is base width modulation taken to the limit $W\to 0$: the E-B and C-B depletion regions touch inside the base (Fig. 11.7). The junctions then become electrostatically coupled. As Fig. 11.8 shows, a further increase of $-V_{CB}$ lowers the E-B potential hill and allows a large, exponentially increasing injection of carriers from emitter directly into collector, so a slight rise in $-V_{CB}$ causes a huge rise in $I_C$. Punch-through limits $V_{CBO}$ or $V_{CEO}$ if it occurs; avalanche multiplication could set in first. Whichever happens first defines the breakdown voltage.

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Exercise 11.8

Problem. Using the Table 11.1 parameters, $W_B=2\mu$m, $T=300$ K, determine the C-B junction voltage that leads to punch-through. (The quoted numbers correspond to $V_{EB}=0$; Problem 11.16 treats $V_{EB}=0.7$ V.)

Solution. Restating (10.3):

$$W=W_B-x_{nEB}-x_{nCB}$$

with step-junction depletion widths (Eq. 5.37), $K_S=11.8$:

$$x_{nEB}={\left[\frac{2K_S{\epsilon }_0}{q}\frac{N_E}{N_B(N_E+N_B)}\left(V_{bi(EB)}-V_{EB}\right)\right]}^{1/2},V_{bi(EB)}=\frac{kT}{q}\ln \frac{N_E{N}_B}{n_i^2}$$ $$x_{nCB}={\left[\frac{2K_S{\epsilon }_0}{q}\frac{N_C}{N_B(N_C+N_B)}\left(V_{bi(CB)}-V_{CB}\right)\right]}^{1/2},V_{bi(CB)}=\frac{kT}{q}\ln \frac{N_C{N}_B}{n_i^2}$$

At punch-through $W=0$:

$$0=W_B-x_{nEB}-{\left[\frac{2K_S{\epsilon }_0}{q}\frac{N_C}{N_B(N_C+N_B)}\left(V_{bi(CB)}-V_{CB}\right)\right]}^{1/2}$$

Solve for $V_{CB}$:

$$V_{CB}=V_{bi(CB)}-\frac{q{N}_B(N_C+N_B)}{2K_S{\epsilon }_0{N}_C}{\left(W_B-x_{nEB}\right)}^2\text{(at punch-through)}$$

Numbers ($kT/q=0.0259$ V, $n_i=10^{10}$ cm${}^{-3}$, $K_S{\epsilon }_0=1.045\times 10^{-12}$ F/cm):

$$V_{bi(EB)}=0.0259\ln \frac{10^{18}\cdot 10^{16}}{10^{20}}=0.835\text{V},V_{bi(CB)}=0.0259\ln \frac{10^{15}\cdot 10^{16}}{10^{20}}=0.656\text{V}$$ $$x_{nEB}={\left[\frac{2(1.045\times 10^{-12})}{1.602\times 10^{-19}}\frac{10^{18}}{10^{16}(1.01\times 10^{18})}(0.835)\right]}^{1/2}=3.29\times 10^{-5}\text{cm}$$ $$W_B-x_{nEB}=2\times 10^{-4}-3.29\times 10^{-5}=1.671\times 10^{-4}\text{cm}$$ $$\frac{q{N}_B(N_C+N_B)}{2K_S{\epsilon }_0{N}_C}=\frac{(1.602\times 10^{-19})(10^{16})(1.1\times 10^{16})}{2(1.045\times 10^{-12})(10^{15})}=8.43\times 10^9{\text{V/cm}}^2$$ $$V_{CB}=0.656-8.43\times 10^9{\left(1.671\times 10^{-4}\right)}^2=0.656-235.4\approx -235\text{V}$$

Answer: $V_{CB}\approx -235$ V. The large value arises because the very lightly doped collector absorbs most of the depletion width, so hundreds of volts are needed to push depletion across the base.

11.2.4 Avalanche Multiplication and Breakdown

Common base

In common-base active mode with $-V_{CB}$ increasing, the C-B depletion region behaves like a reverse-biased $pn$ diode: crossing carriers gain enough energy to impact-ionize, creating extra electron-hole pairs. Eventually avalanche occurs, $I_C$ rises toward infinity and the multiplication factor $M\to \infty$ (Subsection 6.2.2). If avalanche precedes punch-through, the maximum common-base output voltage $V_{CBO}$ is just the C-B junction breakdown voltage.

Common emitter: the surprise

In common emitter $V_{EC}=V_{EB}-V_{CB}$, and $V_{EB}$ is only a few tenths of a volt, so one expects $V_{CEO}\approx V_{CBO}$. Experimentally (Figs. 11.4d, 11.5d) $V_{CEO}\approx 90$ V while $V_{CBO}\approx 120$ V. Following Fig. 11.9 step by step:

0. Holes are injected from the emitter into the base.
1. They enter the C-B depletion region.
2. A few gain enough energy to impact-ionize, creating extra electron-hole pairs.
3. The added holes drift into the collector; the added electrons are swept into the base.
4. These extra base electrons form a majority-carrier imbalance that must be removed. In common base they just leave through the base lead. In common emitter the base current is held constant (it is the controlled variable), so the electrons cannot leave that way; the imbalance can only be relieved by injecting electrons from the base into the emitter.
5. E-B injection is coupled: every electron injected into the emitter is accompanied by $I_{Ep}/I_{En}$ extra holes injected into the base. So for each additional electron-hole pair created in the C-B region, $I_{Ep}/I_{En}+1=1/(1-\gamma )\ge {\beta }_{dc}+1$ extra holes flow into the collector.

The multiplication is thus internally amplified by the transistor gain, and it is regenerative: more collector current causes more multiplication, hence more injection, and so on, producing an instability at a $V_{EC}$ far below the pure avalanche voltage of the C-B junction. (Equivalently, ${\beta }_{dc}$ is enhanced by the process.)

Quantitative model

With multiplication in the C-B region, $I_{Cp}=M{I}_{Ep}$: replace ${\alpha }_{dc}$ (and ${\alpha }_F$) by $M{\alpha }_{dc}$. The simplified CE output (Eq. 10.15) becomes

$$I_C=\frac{M{\alpha }_{dc}}{1-M{\alpha }_{dc}}{I}_B+\frac{I_{CBO}}{1-M{\alpha }_{dc}}\text{\hspace{1em}(11.52)}$$

$I_C\to \infty$ when $M{\alpha }_{dc}\to 1$, i.e. $M\to 1/{\alpha }_{dc}$. Since ${\alpha }_{dc}$ is only slightly below unity, $M$ need only exceed 1 slightly. With the empirical $M$ (Eq. 6.35) and $V_{BR}=V_{CBO}$:

$$M(V_{CEO})=\frac{1}{1-{\left(V_{CEO}/V_{CBO}\right)}^m}=\frac{1}{{\alpha }_{dc}}\text{\hspace{1em}(11.53)}$$ $${\left(\frac{V_{CEO}}{V_{CBO}}\right)}^m=1-{\alpha }_{dc}$$ $$\boxed{V_{CEO}=V_{CBO}{\left(1-{\alpha }_{dc}\right)}^{1/m}=\frac{V_{CBO}}{({\beta }_{dc}+1{)}^{1/m}}}\text{\hspace{1em}(11.54)}$$

with $3\le m\le 6$ (using $1-{\alpha }_{dc}=1/({\beta }_{dc}+1)$). Hence $V_{CEO}<V_{CBO}$. Example: ${\beta }_{dc}\approx 300$, $m=6$: $(301{)}^{1/6}\approx 2.6$, so $V_{CEO}\approx V_{CBO}/2.6$.

In a BJT this mechanism is detrimental. Elsewhere it is useful: it helps trigger the off-to-on switching in SCRs and other PNPN devices (Chapter 13), and it amplifies the photosignal in a phototransistor (a BJT whose C-B junction is exposed to light, often with an open or absent base lead). As in Fig. 11.10, roughly ${\beta }_{dc}+1$ carriers enter the collector per electron-hole pair created by photon absorption in the C-B depletion region.

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Exercise 11.9

Problem. Two pnp BJTs are similar except $N_B\gg N_C$ in Transistor A and $N_B\ll N_C$ in Transistor B. From the doping-profile figure (read from the OCR’d axis labels): A has $N_E=10^{18}$, $N_B=10^{16}$, $N_C=10^{14}$ cm${}^{-3}$; B has $N_E=10^{18}$, $N_B=10^{16}$, $N_C=10^{18}$ cm${}^{-3}$. (a) Which has the larger punch-through voltage under active biasing? (b) If limited by C-B avalanche, which has the larger $V_{CBO}$?

Solution.

(a) Transistor A. Under active bias the E-B junction (same doping) gives the same $x_{nEB}$ in both. The base penetration of the C-B depletion region is

$$x_{nCB}={\left[\frac{2K_S{\epsilon }_0}{q}\frac{N_C}{N_B(N_C+N_B)}\left(V_{bi(CB)}-V_{CB}\right)\right]}^{1/2}$$

Evaluate the doping factor:

$$\frac{N_C}{N_B(N_C+N_B)}{|}_A\approx \frac{10^{14}}{10^{16}\cdot 10^{16}}=10^{-18},\frac{N_C}{N_B(N_C+N_B)}{|}_B\approx \frac{10^{18}}{10^{16}\cdot 10^{18}}=10^{-16}$$

So for the same $-V_{CB}$, $x_{nCB}{|}_A\approx x_{nCB}{|}_B/10$ (the small $V_{bi}$ difference is a minor correction). With equal $W_B$, a much larger $-V_{CB}$ is needed to deplete the base of A.

(b) Transistor A. The avalanche breakdown voltage of an asymmetric junction is set by the doping on the lightly doped side and varies roughly inversely with it (Fig. 6.11 / Eq. 6.33). In A the light side of the C-B junction is the collector, $10^{14}$ cm${}^{-3}$; in B it is the base, $10^{16}$ cm${}^{-3}$. The lighter side gives the higher breakdown voltage, so A exhibits the larger $V_{CBO}$.

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Exercise 11.10 (computer)

Problem. Modify the Exercise 11.7 program to include base width modulation and carrier multiplication. With the same parameters compare CE output characteristics when (i) neither is included, (ii) only BWM, (iii) both.

Solution (logic).

BWM (subprogram BJTmod). At each bias point recompute depletion penetration, then $W$, then $I_{F0}$, $I_{R0}$, ${\alpha }_F$, ${\alpha }_R$:

$$x_{nEB}=\sqrt{\frac{2K_S{\epsilon }_0}{q}\frac{N_E}{N_B(N_E+N_B)}(V_{biE}-V_{EB})},x_{nCB}=\sqrt{\frac{2K_S{\epsilon }_0}{q}\frac{N_C}{N_B(N_C+N_B)}(V_{biC}-V_{CB})},W=W_B-x_{nEB}-x_{nCB}$$

Multiplication. $M=1/[1-(|V_{CB}|/V_{CBO}{)}^m]$ and ${\alpha }_F\to M{\alpha }_F$ in the CE output equation, with $V_{CBO}=60(N_C/10^{16}{)}^{-3/4}$ V and $m=6$ (the textbook program’s choices); the sweep ends at $V_{CEO}=V_{CBO}(1-{\alpha }_F{)}^{1/m}$. For the CE output the program sets $V_{EB}\approx 0$ inside BWM (neglecting $x_{nEB}$ variation), so $V_{CB}=-V_{EC}$.

% BJTplus-style: common-emitter output, three cases
clear; close all
q=1.602e-19; k=8.617e-5; e0=8.85e-14; T=300; kT=k*T;
A=1e-4; WB=2.5e-4; ni=1e10; KS=11.8;
NE=1e18; NB=1.5e16; NC=1.5e15;
mun=@(N) 92+1268./(1+(N/1.3e17).^0.91);
mup=@(N) 54.3+406.9./(1+(N/2.35e17).^0.88);
DE=kT*mun(NE); DB=kT*mup(NB); DC=kT*mun(NC);
tauE=1e-7; tauB=1e-6; tauC=1e-6;
LE=sqrt(DE*tauE); LB=sqrt(DB*tauB); LC=sqrt(DC*tauC);
nE0=ni^2/NE; pB0=ni^2/NB; nC0=ni^2/NC;
VbiE=kT*log(NE*NB/ni^2); VbiC=kT*log(NC*NB/ni^2);
 
W=WB;                                   % zero-bias E-M parameters
fB=(DB/LB)*pB0*cosh(W/LB)/sinh(W/LB);
IF0=q*A*((DE/LE)*nE0+fB); IR0=q*A*((DC/LC)*nC0+fB);
aF=q*A*(DB/LB)*pB0/sinh(W/LB)/IF0; aR=q*A*(DB/LB)*pB0/sinh(W/LB)/IR0;
VCBO=60*(NC/1e16)^(-3/4); m=6; VCEO=VCBO*(1-aF)^(1/m);
 
VEC=[0:0.01:VCEO/50, VCEO/50:VCEO/200:VCEO];
titles={'No BWM, no multiplication','BWM only','BWM + multiplication'};
for c=1:3
    subplot(3,1,c); hold on
    for IB=(0:4)*2.5e-6
        if c==1
            F0=IF0*ones(size(VEC)); R0=IR0*ones(size(VEC));
            aFv=aF*ones(size(VEC)); aRv=aR*ones(size(VEC));
        else
            VEB=0; VCB=VEB-VEC;
            xnEB=sqrt((2*KS*e0/q)*(NE/(NB*(NE+NB)))*(VbiE-VEB));
            xnCB=sqrt((2*KS*e0/q)*(NC/(NB*(NC+NB)))*(VbiC-VCB));
            Wv=WB-xnEB-xnCB;
            fBv=(DB/LB)*pB0*cosh(Wv/LB)./sinh(Wv/LB);
            F0=q*A*((DE/LE)*nE0+fBv); R0=q*A*((DC/LC)*nC0+fBv);
            aFv=q*A*(DB/LB)*pB0./sinh(Wv/LB)./F0;
            aRv=q*A*(DB/LB)*pB0./sinh(Wv/LB)./R0;
            if c==3
                M=1./(1-(-VCB/VCBO).^m);          % carrier multiplication
                aFv=M.*aFv;
            end
        end
        E=exp(-VEC/kT);
        IB0=(1-aFv).*F0+(1-aRv).*R0;
        IB1=(1-aFv).*F0+(1-aRv).*R0.*E;
        IC=((aFv.*F0-R0.*E).*(IB+IB0)./IB1+R0-aFv.*R0)*1e3;   % mA
        plot(VEC,IC)
    end
    title(titles{c}); xlabel('V_{EC} (V)'); ylabel('I_C (mA)'); grid on
    axis([0 VCEO 0 5])
end

Results. (i) Flat, horizontal active-mode lines. (ii) A moderate quasilinear rise of $I_C$ with $V_{EC}$ (Early effect). (iii) Additionally a sharp upturn as $V_{EC}\to V_{CEO}$. Case (iii) compares very favorably with the measured Fig. 11.5(d). Keep $V_{EC}<V_{CBO}$ so $1-(\dots {)}^m>0$.

11.2.5 Geometrical Effects

Real BJTs (cross sections of Fig. 10.6) are not one-dimensional, which produces geometry-related deviations.

Emitter area $\ne$ collector area. Carriers do not travel in straight lines from emitter to collector; current lines spread laterally (Fig. 11.11a). To first order this is handled by using different effective areas for emitter (injection) and collector (collection).

Series resistances. Base current flowing from the base contact to the “heart” of the transistor (dashed region of Fig. 11.11b) crosses a resistive bulk region, plus there is contact resistance. The voltage across the E-B junction is therefore less than the terminal voltage. Though small, this matters because $I_E$ is exponential in junction voltage. Introduce a base series resistance $r_B$ (bulk plus contact). If it is the only series resistance, replace $V_{EB}$ by

$$V_{EB}^{\prime }=V_{EB}-I_B{r}_B$$

The central part is the intrinsic transistor; as in Fig. 11.12, resistances $r_E$, $r_B$, $r_C$ connect the internal terminals $E^{\prime }$, $B^{\prime }$, $C^{\prime }$ to the external ones; internal voltages follow by subtracting the $IR$ drops. Values come from subsidiary measurements.

Current crowding. Base current also flows laterally under the emitter face, causing a voltage drop in the base from emitter center to edges. The E-B junction voltage and injection current are therefore greatest at the emitter periphery (more current lines at the edges in Fig. 11.11a): current crowding. With the localized heating it brings, it is damaging at high currents. Power BJTs therefore use emitters with a large periphery-to-area ratio: several narrow, long emitter fingers joined in a comb and interdigitated with base fingers (Fig. 11.13).

11.2.6 Recombination-Generation Current

The ideal analysis neglects R-G in the depletion regions. By analogy with the diode:

• In active mode with small forward E-B bias, the E-B R-G current likely dominates the base current: lower emitter efficiency and lower gain at low currents (small $V_{EB}$).
• With the input open, reverse-bias R-G current from the C-B junction adds to (and likely dominates) the leakage $I_{CBO}$ ($I_{CEO}$).

Unlike the diode, the transistor characteristics are only nominally affected: in most active operation diffusion dominates the input current, and except at low currents the output current is mainly injected current. With R-G included, the pnp emitter efficiency becomes

$$\gamma =\frac{I_{Ep}}{I_{Ep}+I_{En}+I_{RG}}$$

11.2.7 Graded Base

Double diffusion (Section 10.2) gives dopant profiles falling roughly exponentially with depth (Fig. 11.14). So base doping is not uniform: it falls from a maximum at the E-B junction to a minimum at the C-B junction: the base is graded.

A doping gradient implies a built-in field in the quasineutral base. Let $N_B(x)$ be the donor doping of a pnp base. Neutrality $\rho =q(p-n+N_B)=0$ with $p\approx 0$ gives $n(x)=N_B(x)$. In equilibrium the electron current vanishes:

$$J_N=q{\mu }_{n}n\mathcal{E}+q{D}_N\frac{dn}{dx}=0\text{\hspace{1em}(11.55)}$$

Solve for $\mathcal{E}$ using $D_N/{\mu }_n=kT/q$ and $n=N_B(x)$:

$$\mathcal{E}(x)=-\frac{D_N}{{\mu }_n}\frac{dn/dx}{n}=-\frac{kT}{q}\frac{1}{N_B(x)}\frac{d{N}_B}{dx},0\le x\le W\text{\hspace{1em}(11.56)}$$

For an exponential profile,

$$N_B(x)=N_B(0)e^{-x/x_{diff}}\text{\hspace{1em}(11.57)}$$ $$\frac{d{N}_B}{dx}=-\frac{N_B(x)}{x_{diff}}\text{\hspace{1em}(11.58)}$$

Substituting in (11.56):

$$\boxed{\mathcal{E}=\frac{kT/q}{x_{diff}}}0\le x\le W\text{\hspace{1em}(11.59)}$$

A constant field directed so as to assist minority-carrier transport across the base (for holes in a pnp and for electrons in an npn). Example: doping falling by two decades across $W_B=2\mu$m:

$$x_{diff}=\frac{W_B}{\ln 100}=\frac{2\times 10^{-4}}{4.605}=4.34\times 10^{-5}\text{cm},\mathcal{E}=\frac{0.0259}{4.34\times 10^{-5}}\approx 600\text{V/cm}$$

Drift-assisted transport shortens the transit time, reduces base recombination, raises ${\alpha }_T$ and all gains above ideal values, and improves high-frequency response. The graded base is one deviation that is clearly beneficial.

11.2.8 Figures of Merit

Two plots reveal quickly how far a device departs from ideal (virtually every BJT research paper shows them).

Gummel plot (Fig. 11.15): semilog plot of $I_B$ and $I_C$ versus $V_{EB}$ at fixed output voltage (a few volts). Ideally both are straight lines with the same slope $q/kT$ (60 mV per decade at room temperature). In reality:

• at small $V_{EB}$, R-G adds to $I_B$ (slope $q/2kT$ region), putting $I_B$ above the ideal line;
• at large $V_{EB}$, the $I_C$ plot slopes over due to high-level injection, possibly aggravated by current crowding; if both $I_B$ and $I_C$ roll over, series resistance is probable (as for a diode).

With $I_{CEO}$ negligible, $I_C/I_B$ at any $V_{EB}$ is ${\beta }_{dc}$ there. Ideally constant, in reality it falls at both low and high $V_{EB}$.

${\beta }_{dc}$ versus $I_C$ (Fig. 11.16): log-log plot at fixed output voltage (sample: 2N2605, $V_{EC}=5$ V). It is merely a continuous record of the information in the Gummel plot: ${\beta }_{dc}$ drops at low $I_C$ because of R-G, and at high $I_C$ because of high-level injection, current crowding, series resistance, or a combination.

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11.3 Modern BJT Structures

11.3.1 Polysilicon Emitter BJT

CMOS is the preferred large-scale IC technology (low power, high packing density), but a BJT supplies several times the drive current of an equal-size MOSFET, switching large capacitive loads faster. BiCMOS therefore puts bipolar transistors in the speed-sensitive, heavily loaded parts of mostly-CMOS circuits (Fig. 11.17: the Intel 0.8 $\mu$m BiCMOS npn).

In Fig. 11.17 a polycrystalline Si film lies between the metal emitter contact and the crystalline silicon. The emitter is made by depositing poly over the emitter area, doping it with arsenic, and diffusing some dopant a short distance into the crystalline Si: a shallow crystalline layer contacted by a similarly doped poly film, the polysilicon emitter. Advantages:

• suited to the shallow E-B junctions modern ICs need;
• compatible with self-aligned processing that minimizes parasitic resistances and capacitances;
• an operational advantage: the CE gain is much larger than that of an all-crystalline BJT with a shallow emitter of equal thickness.

Why shallow emitters cut gain. Modern emitters are very thin: $W_E$ can be less than $L_E$. For $W_E/L_E\ll 1$ in an all-crystalline emitter, emitter recombination is negligible and, like the base, the minority profile is linear (Fig. 11.18a). The slope at the junction is far larger than for a semi-infinite emitter, so back-injection from base to emitter rises, emitter efficiency falls, and gain falls. Quantitatively (Problem 11.11), $W_E$ simply replaces $L_E$ in (11.41)-(11.44).

Why polysilicon helps. To first order the poly portion is modelled as low-mobility silicon (grain boundaries scatter carriers). If both poly and crystalline portions are much shorter than their diffusion lengths, the minority profile is linear in both. At the poly/Si interface the minority excess and the minority current must be continuous. With subscript 1 = poly, 2 = Si:

$$q{D}_{E1}\frac{d\Delta p_{E1}}{d{x}^{\prime \prime }}=q{D}_{E2}\frac{d\Delta p_{E2}}{d{x}^{\prime \prime }}\text{\hspace{1em}(11.60)}$$ $$\frac{d\Delta p_{E1}}{d{x}^{\prime \prime }}=\frac{D_{E2}}{D_{E1}}\frac{d\Delta p_{E2}}{d{x}^{\prime \prime }}=\frac{{\mu }_{E2}}{{\mu }_{E1}}\frac{d\Delta p_{E2}}{d{x}^{\prime \prime }}\text{\hspace{1em}(11.61)}$$

Since ${\mu }_{E1}<{\mu }_{E2}$, the poly slope is steeper (Fig. 11.18b). For the same total drop, the slope at the E-B depletion edge is then reduced relative to the all-crystalline shallow emitter: less back-injected current, smaller base current, better emitter efficiency and higher gain.

Intuition. A low-mobility layer behind the junction acts like a traffic jam: minority carriers dumped into the emitter cannot be carried away easily, so the pile builds near the junction, flattening the gradient there and driving less current across.

11.3.2 Heterojunction Bipolar Transistor (HBT)

In Section 11.2 several deviations were tied to base doping: base-width modulation, base resistance $r_B$ and the current-crowding voltage drop all improve when base doping is increased. But increasing base doping lowers emitter efficiency and therefore gain. A standard BJT has an unavoidable gain-versus-nonideality trade-off.

In a heterojunction bipolar transistor the base is generally the most heavily doped region. A heterojunction is a junction between two dissimilar semiconductors. In an HBT the emitter is of a wider band-gap material than the base, which in most material pairs sets up a built-in barrier impeding forward-bias back-injection from base to emitter, giving high injection efficiency even if $N_E\lesssim N_B$. (Shockley described the idea in the original 1948 BJT patent; practical HBTs required film-growth methods like molecular beam epitaxy.)

Band alignment (Fig. 11.19) is how conduction and valence band edges of the two materials line up at the interface, usually found experimentally. Fig. 11.19 is not a true heterojunction diagram (neutrality drawn everywhere, no band bending). Lattice-matched systems shown (minimal interface defects):

• (a) Al${}_{0.3}$Ga${}_{0.7}$As/GaAs: gaps 1.82 and 1.42 eV, $\Delta E_C=0.263$ eV, $\Delta E_V=0.132$ eV;
• (b) GaAs/Ge: gaps 1.42 and 0.66 eV, $\Delta E_C=0.27$ eV, $\Delta E_V=0.49$ eV;
• (c) InP/In${}_{0.53}$Ga${}_{0.47}$As: gaps 1.35 and 0.75 eV, $\Delta E_C=0.2$ eV, $\Delta E_V=0.4$ eV;
• (d) Al${}_{0.48}$In${}_{0.52}$As/InP: gaps 1.46 and 1.35 eV, $\Delta E_C=0.3$ eV, $\Delta E_V=-0.19$ eV

(offsets read from the noisy OCR of the figure). An important non-lattice-matched exception is the Si/Si${}_{1-x}$Ge${}_x$/Si HBT with a thin pseudomorphic SiGe base (thin films, roughly 10 nm, conforming to the substrate lattice; thicker than a critical value they relax and create defects).

AlGaAs/GaAs Npn HBT (Fig. 11.20). (An uppercase letter marks the wider-gap side.) An $n^{+}$-GaAs subcollector, $n$-GaAs collector, $p^{+}$-GaAs base, and N-Al${}_{0.3}$Ga${}_{0.7}$As emitter (with $n$-GaAs cap) are deposited in turn on a semi-insulating (S.I.) GaAs substrate. The substrate and S.I. sidewalls (formed by deep proton implantation) isolate the HBT dielectrically. Etching and metal contacts complete it.

An abrupt E-B heterojunction gives the band diagram of Fig. 11.21(a): nearly that of an equivalently doped BJT apart from the wider emitter gap and band discontinuities at the interface; since $N_E\ll N_B\gg N_C$ the depletion regions lie almost wholly in emitter and collector. In practice the junction is compositionally graded over a few hundred angstroms, eliminating sharp band spikes (Fig. 11.21b).

Operationally a graded HBT is a BJT with different intrinsic concentrations $n_{iE}$, $n_{iB}$. Instead of $n_{E0}/p_{B0}=N_B/N_E$:

$$\frac{n_{E0}}{p_{B0}}=\frac{n_{iE}^2/N_E}{n_{iB}^2/N_B}=\frac{N_B}{N_E}\frac{n_{iE}^2}{n_{iB}^2}$$

so $N_B/N_E$ in the BJT relations becomes $(N_B/N_E)(n_{iE}/n_{iB}{)}^2$. For $W\ll L_B$:

$$\gamma =\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{n_{iE}^2}{n_{iB}^2}\frac{W}{L_E}}\text{\hspace{1em}(11.62)}$$

The large $N_B/N_E\sim 100$ is overwhelmed by the tiny intrinsic-concentration factor, which the text quotes as about $10^{-5}$ for a typical Al${}_{0.3}$Ga${}_{0.7}$As/GaAs HBT, so the theoretical emitter efficiency is very close to unity.

In practice HBT efficiencies fall below (11.62) because of recombination at the E-B junction periphery surface (GaAs passivation is harder than Si). Even so, HBTs now offer very respectable gain and unmatched high-frequency performance.

The big idea. A wide-gap emitter lets designers dope the base heavily (low $r_B$, small Early effect, no punch-through worries, high speed) without paying the gain penalty, because the band offset, not the doping ratio, blocks back-injection: a one-way turnstile at the base’s front door.

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11.4 Summary

This chapter modeled the steady-state response of the BJT. First, allowing an arbitrary quasineutral base width, an ideal transistor analysis paralleling the ideal diode analysis gave the base minority-carrier distribution (11.26), the performance parameters (11.31-11.34) and the terminal currents (11.35, 11.36). These were simplified using $W\ll L_B$: the base distribution (11.40) becomes linear and the performance parameters take the forms (11.41)-(11.44). Terminal currents were handled compactly by the Ebers-Moll model (Fig. 11.3) and equations (11.47).

Deviations from ideal found by comparing theory and experiment:

• Base width modulation causes the quasilinear upward slant of the CE output characteristics and the $V_{CB}$ dependence of the CB input characteristics.
• Punch-through or avalanche breakdown of the C-B junction, whichever occurs first, limits the maximum output voltage in active-mode common-base operation.
• In active-mode common-emitter operation, carrier multiplication plus internal feedback can make $V_{CEO}<V_{CBO}$.
• Others: series resistance external to the intrinsic transistor, the base voltage drop across the emitter face (current crowding), R-G in the E-B and C-B depletion regions, and the (beneficial) graded base.
• Figures of merit: the Gummel plot and the ${\beta }_{dc}$ versus $I_C$ plot.

Finally, the section on modern structures introduced the polysilicon-emitter BJT and the HBT, also providing practical uses of previously established concepts.

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End-of-Chapter Problems

The book’s Problem Information Table lists: 11.1 quick quiz; 11.2 deduce information from a current plot; 11.3 npn equation modifications; 11.4 current and parameter calculation; 11.5 and 11.6 $\Delta p_B(x)$ plots; 11.7 and 11.8 carrier-distribution sketches and deductions; 11.9 and 11.10 parametric-effect table and plots; 11.11 shallow emitter; 11.12 “diode” connections; 11.13 Ebers-Moll circuit; 11.14 computational equations; 11.15 emitter input $W$ sensitivity; 11.16 punch-through; 11.17 comparing two transistors; 11.18 Early voltage; 11.19 “diode” $I$-$V_A$ plots; 11.20 Gummel and gain plots; 11.21 $V_{EC}$ dependence of gain.

Problem 11.1: Quick Quiz

(a) Doping and widths assumed in the ideal analysis. Emitter, base and collector are each uniformly and nondegenerately doped (step junctions). The quasineutral emitter and collector widths are $\gg$ their minority diffusion lengths (semi-infinite). The base width $W$ is initially arbitrary (later $W\ll L_B$).

(b) Base distribution, active mode. $\Delta p_B(0)=p_{B0}(e^{q{V}_{EB}/kT}-1)\gg 0$ and $\Delta p_B(W)\approx -p_{B0}$ (total $p\approx 0$). For $W/L_B\ll 1$: a straight line from the high value to about $-p_{B0}$. For $W/L_B\gg 1$: $\Delta p_B\approx \Delta p_B(0)e^{-x/L_B}$, an exponential that dies out long before $x=W$ (two back-to-back diodes).

(c) Sketches for $W/L_B\ll 1$, active mode, $\Delta p_{B,max}=10p_{B0}$. (The OCR’d magnitude is taken as $\Delta p_B(0)=10p_{B0}$.) Since $V_{CB}$ is reverse, $\Delta p_B(W)=-p_{B0}$, and

$$\Delta p_B(x)=10p_{B0}-11p_{B0}\frac{x}{W},\frac{\Delta p_B(x)}{p_{B0}}=10-11\frac{x}{W},\frac{p_B(x)}{p_{B0}}=1+\frac{\Delta p_B}{p_{B0}}=11\left(1-\frac{x}{W}\right)$$

 Δp_B(x)            Δp_B/p_B0            p_B(x)  (from 11 p_B0 to 0)
 10 p_B0 |\         10 |\                11 p_B0|\
         | \           | \                      | \
 0 ------|--\------ 0 -|--\------      p_B0 ----|--\ ...
         |   \          |   \                   |   \
-p_B0    +----+ x   -1  +----+ x        0       +----+ x
         0    W           0    W                 0    W

All three are straight lines with the same shape (offsets of 0, a scale of $1/p_{B0}$, and a shift of $+p_{B0}$).

(d) Typical sizes. $\gamma \approx 0.99-0.9999$, ${\alpha }_T\approx 0.99-0.999$, ${\alpha }_{dc}\approx 0.98-0.999$, ${\beta }_{dc}\approx 50-1000$ (310 in the Table 11.1 example).

(e) $I_B(V_{EB},V_{CB})$. From $I_B=I_E-I_C$ and (11.47):

$$I_B=(1-{\alpha }_F)I_{F0}\left(e^{q{V}_{EB}/kT}-1\right)+(1-{\alpha }_R)I_{R0}\left(e^{q{V}_{CB}/kT}-1\right)$$

(f) The statement of part (f) is lost in the OCR. Its most likely subject, the item in this slot of the chapter’s list of deviations, is why the CB input characteristics depend on $V_{CB}$ (Fig. 11.4b). Answer: base width modulation. Larger $-V_{CB}$ widens the C-B depletion region, narrows $W$, steepens the base profile and raises $I_E\propto e^{q{V}_{EB}/kT}/W$ (Eq. 11.49).

(g) Fig. 11.5(d). The quasilinear slope at low $V_{EC}$ is base width modulation (${\beta }_{dc}$ rises as $W$ falls with increasing $V_{EC}$). The sharp upward curvature as $V_{EC}\to V_{CEO}$ is avalanche multiplication in the C-B depletion region amplified by the transistor feedback (so $M{\alpha }_{dc}\to 1$, Eq. 11.52), not base width modulation.

(h) Can $V_{CEO}<V_{CBO}$ when $V_{CBO}$ is limited by punch-through? Yes. Punch-through is a purely electrostatic effect that sets $V_{CBO}$ (the punch-through voltage, say 235 V in Exercise 11.8). The feedback-amplified avalanche mechanism acts in the common-emitter circuit with the C-B junction’s avalanche voltage $V_{BR}$ (which is higher than $V_{CBO}$ here), giving $V_{CEO}=V_{BR}(1-{\alpha }_{dc}{)}^{1/m}=V_{BR}/({\beta }_{dc}+1{)}^{1/m}$. If $V_{BR}/({\beta }_{dc}+1{)}^{1/m}$ falls below the punch-through voltage, then avalanche-plus-feedback occurs first in common emitter and $V_{CEO}<V_{CBO}$. Example: $V_{BR}=300$ V, ${\beta }_{dc}=300$, $m=6$ gives $V_{CEO}\approx 115$ V, below a 235 V punch-through limit.

(i) Current crowding arises because base current flows laterally under the emitter face, causing a voltage drop in the base from emitter center to edge; the E-B junction voltage and injection are therefore largest at the emitter periphery.

(j) Benefits of a graded base. A built-in field (Eq. 11.59) that aids minority-carrier transport: shorter transit time, less base recombination, higher ${\alpha }_T$ and current gains, and better high-frequency response.

(k) Gummel plot. A simultaneous semilog plot of $I_B$ and $I_C$ versus $V_{EB}$ at fixed output voltage; the ratio gives ${\beta }_{dc}$ and the deviations from parallel straight lines reveal R-G current (low $V_{EB}$), high-level injection, current crowding and series resistance (high $V_{EB}$).

(l) ${\beta }_{dc}$ decreases at low $I_C$ because of recombination-generation in the E-B depletion region (it adds base current); at high $I_C$ it decreases because of high-level injection, current crowding, series resistance, or a combination.

(m) All-polysilicon emitter? In the simple model a low-mobility emitter (small $D_E$, steeper profile) reduces back-injection, so there is a nominal efficiency gain. In practice, a fully polycrystalline emitter would put grain boundaries and defects into the E-B depletion region: strong R-G and leakage, a poor junction. That is why the junction is formed in crystalline Si and only the contact portion is poly. So no practical advantage over the polysilicon-over-crystalline-Si structure.

(n) Heterojunction: a junction between two dissimilar semiconductors (different band gaps), in contrast to a homojunction.

(o) BJT versus HBT. An HBT has a wider band-gap emitter than the base (heterojunction E-B junction); its base is generally the most heavily doped region. The built-in valence/conduction-band offset blocks back-injection, so emitter efficiency stays near unity despite $N_E\lesssim N_B$, while heavy base doping gives low $r_B$ and small Early effect. A BJT needs $N_E\gg N_B$ and so trades gain against base resistance and modulation.

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Problem 11.2

Problem. The electron and hole currents in an active-mode pnp BJT are plotted in Fig. P11.2 (relative to the junction positions). Determine: (a) $\gamma$; (b) ${\alpha }_T$; (c) ${\beta }_{dc}$; (d) $I_B$; (e) whether R-G in the depletion regions is negligible, with explanation.

Reading of the figure (assumption). The OCR shows hole current labels $1.000$ and $0.999$ and electron-current labels $0.001$, $0.001$ (units normalized to the hole current injected by the emitter, e.g. mA). We read: $I_{Ep}=1.000$ entering the base, $I_{Cp}=0.999$ leaving the base into the collector, electron current in the emitter at the junction $I_{En}=0.001$, and negligible collector electron current ($I_{Cn}\approx 0$).

Solution.

(a)

$$\gamma =\frac{I_{Ep}}{I_{Ep}+I_{En}}=\frac{1.000}{1.000+0.001}=0.999$$

(b)

$${\alpha }_T=\frac{I_{Cp}}{I_{Ep}}=\frac{0.999}{1.000}=0.999$$

(c)

$${\alpha }_{dc}=\gamma {\alpha }_T=0.999\times 0.999=0.998,{\beta }_{dc}=\frac{{\alpha }_{dc}}{1-{\alpha }_{dc}}=\frac{0.998}{0.002}\approx 499(\approx 500)$$

(d) $I_E=I_{Ep}+I_{En}=1.001$, $I_C=I_{Cp}+I_{Cn}=0.999$:

$$I_B=I_E-I_C=1.001-0.999=0.002\text{(same units as the plot)}$$

Check: $I_B=I_C/{\beta }_{dc}=0.999/499=0.002$. It consists of $I_{B1}=I_{En}=0.001$ (electrons injected into the emitter) plus $I_{B2}=I_{Ep}-I_{Cp}=0.001$ (base recombination).

(e) Yes, negligible. If R-G current were significant in a depletion region, the hole (or electron) current entering that region would differ from the current leaving it. In the figure the hole current is continuous across the E-B depletion region (1.000 on both sides), the electron current is continuous across it (0.001 on both sides), and the electron current at the C-B junction is continuous at its negligible value; the only change in the hole current (1.000 to 0.999) occurs inside the quasineutral base (recombination), not in a depletion region. So the ideal-theory assumption of negligible depletion-region R-G holds for this device.

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Problem 11.3

Problem. Indicate the required revisions to the equations of Subsections 11.1.1 and 11.1.2 if the device is an npn BJT.

Solution. In an npn the roles of electrons and holes swap: emitter $n^{+}$ (minority holes), base $p$ (minority electrons), collector $n$ (minority holes). Voltages become $V_{BE}$ and $V_{BC}$ (a forward-biased E-B junction has $V_{BE}>0$; active mode: $V_{BE}>0$, $V_{BC}<0$). Positive current directions reverse (emitter current positive leaving the emitter terminal, collector current positive entering the collector), which makes the derivative formulas (11.7)-(11.10) keep the same form. Replace: $n_{E0}\to p_{E0}=n_i^2/N_E$, $p_{B0}\to n_{B0}=n_i^2/N_B$, $n_{C0}\to p_{C0}=n_i^2/N_C$; $D_E=D_P$, $D_B=D_N$, $D_C=D_P$; $L_E=L_P$, $L_B=L_N$, $L_C=L_P$; ${\tau }_E={\tau }_p$, ${\tau }_B={\tau }_n$, ${\tau }_C={\tau }_p$; $V_{EB}\to V_{BE}$, $V_{CB}\to V_{BC}$; $I_{En}\to I_{Ep}$ and so on as below. Equation by equation:

Eq.	npn version
(11.1)	$0=D_E{d}^2\Delta p_E/d{x}^{\prime \prime 2}-\Delta p_E/{\tau }_E$
(11.2a,b)	$\Delta p_E(\infty )=0$; $\Delta p_E(0)=p_{E0}(e^{q{V}_{BE}/kT}-1)$
(11.3)	$0=D_B{d}^2\Delta n_B/d{x}^2-\Delta n_B/{\tau }_B$
(11.4a,b)	$\Delta n_B(0)=n_{B0}(e^{q{V}_{BE}/kT}-1)$; $\Delta n_B(W)=n_{B0}(e^{q{V}_{BC}/kT}-1)$
(11.5)	$0=D_C{d}^2\Delta p_C/d{x}^{\prime 2}-\Delta p_C/{\tau }_C$
(11.6a,b)	$\Delta p_C(\infty )=0$; $\Delta p_C(0)=p_{C0}(e^{q{V}_{BC}/kT}-1)$
(11.7)	$I_{Ep}=-qAD_E,d\Delta p_E/dx”\big
(11.8)	$I_{En}=-qAD_B,d\Delta n_B/dx\big
(11.9)	$I_{Cn}=-qAD_B,d\Delta n_B/dx\big
(11.10)	$I_{Cp}=qAD_C,d\Delta p_C/dx’\big
(11.11)	$\gamma =I_{En}/(I_{En}+I_{Ep})$
(11.12)	${\alpha }_T=I_{Cn}/I_{En}$
(11.13), (11.14)	unchanged: ${\alpha }_{dc}=\gamma {\alpha }_T$, ${\beta }_{dc}={\alpha }_{dc}/(1-{\alpha }_{dc})$
(11.15)-(11.17)	$I_E=I_{En}+I_{Ep}$, $I_C=I_{Cn}+I_{Cp}$, $I_B=I_E-I_C$
(11.18), (11.19)	$\Delta p_E=A_1{e}^{-x^{\prime \prime }/L_E}+A_2{e}^{x^{\prime \prime }/L_E}$; $\Delta p_E=p_{E0}(e^{q{V}_{BE}/kT}-1)e^{-x^{\prime \prime }/L_E}$
(11.20)	$I_{Ep}=qA(D_E/L_E)p_{E0}(e^{q{V}_{BE}/kT}-1)$
(11.21), (11.22)	$\Delta p_C=p_{C0}(e^{q{V}_{BC}/kT}-1)e^{-x^{\prime }/L_C}$; $I_{Cp}=-qA(D_C/L_C)p_{C0}(e^{q{V}_{BC}/kT}-1)$
(11.23)-(11.26)	$\Delta n_B(x)$ with $L_B=L_N$: $\Delta n_B(0)\sinh [(W-x)/L_B]/\sinh (W/L_B)+\Delta n_B(W)\sinh (x/L_B)/\sinh (W/L_B)$
(11.28)	$I_{En}=qA\frac{D_B}{L_B}{n}_{B0}\left[\coth \frac{W}{L_B}(e^{q{V}_{BE}/kT}-1)-\frac{1}{\sinh (W/L_B)}(e^{q{V}_{BC}/kT}-1)\right]$
(11.29)	$I_{Cn}=qA\frac{D_B}{L_B}{n}_{B0}\left[\frac{1}{\sinh (W/L_B)}(e^{q{V}_{BE}/kT}-1)-\coth \frac{W}{L_B}(e^{q{V}_{BC}/kT}-1)\right]$
(11.31)	$\gamma =1/\left[1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\tanh (W/L_B)\right]$ (same form, using $p_{E0}/n_{B0}=N_B/N_E$)
(11.32)-(11.34)	${\alpha }_T$, ${\alpha }_{dc}$, ${\beta }_{dc}$ unchanged in form
(11.35), (11.36)	$I_E=qA\{[\frac{D_E}{L_E}{p}_{E0}+\frac{D_B}{L_B}{n}_{B0}\coth \frac{W}{L_B}](e^{q{V}_{BE}/kT}-1)-\frac{D_B}{L_B}\frac{n_{B0}}{\sinh (W/L_B)}(e^{q{V}_{BC}/kT}-1)\}$; $I_C=qA\{\frac{D_B}{L_B}\frac{n_{B0}}{\sinh (W/L_B)}(e^{q{V}_{BE}/kT}-1)-[\frac{D_C}{L_C}{p}_{C0}+\frac{D_B}{L_B}{n}_{B0}\coth \frac{W}{L_B}](e^{q{V}_{BC}/kT}-1)\}$

Note on signs: for an npn, the signs of derivatives such as $I_{En}$ use the convention that positive current leaves the emitter and enters the collector; with that convention the sign pattern of the pnp relations carries over exactly.

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Problem 11.4 (computer-aided)

Problem. pnp BJT, 300 K, active mode, $V_{EB}=0.7$ V, $V_{CB}=-5$ V, $W=W_B$, $A=10^{-4}$ cm${}^2$, parameters of Table 11.1. Calculate (a) $\cosh (W/L_B)$, $\sinh (W/L_B)$, $W/L_B$ (five significant figures); (b) $I_{Ep},I_{En},I_{Cp},I_{Cn}$; (c) $I_E,I_C,I_B$; (d) $I_{B1},I_{B2},I_{B3}$ as in Subsection 10.4; (e) $\gamma ,{\alpha }_T,{\alpha }_{dc},{\beta }_{dc}$.

Solution. Constants: $kT/q=8.617\times 10^{-5}\times 300=0.025851$ V, $n_i=10^{10}$ cm${}^{-3}$, so $n_{E0}=100$, $p_{B0}=10^4$, $n_{C0}=10^5$ cm${}^{-3}$.

(a)

$$\frac{W}{L_B}=\frac{2\times 10^{-4}}{3.36\times 10^{-3}}=0.059524,\cosh \frac{W}{L_B}=1.0017721,\sinh \frac{W}{L_B}=0.059559$$

Voltage factors:

$$e^{q{V}_{EB}/kT}-1=e^{0.7/0.025851}-1=5.7475\times 10^{11},e^{q{V}_{CB}/kT}-1=e^{-5/0.025851}-1\approx -1$$

(b) Apply (11.20), (11.28), (11.29), (11.22):

$$I_{En}=qA\frac{D_E}{L_E}{n}_{E0}\left(e^{q{V}_{EB}/kT}-1\right)=(1.602\times 10^{-19})(10^{-4})\frac{6.81}{8.25\times 10^{-4}}(100)(5.7475\times 10^{11})=7.60\mu \text{A}$$ $$\frac{qA{D}_B{p}_{B0}}{L_B}=\frac{(1.602\times 10^{-19})(10^{-4})(11.3)(10^4)}{3.36\times 10^{-3}}=5.388\times 10^{-16}\text{A}$$ $$I_{Ep}=5.388\times 10^{-16}\left[\frac{1.0017721}{0.059559}(5.7475\times 10^{11})-\frac{-1}{0.059559}\right]=5.2084\text{mA}$$ $$I_{Cp}=5.388\times 10^{-16}\left[\frac{5.7475\times 10^{11}}{0.059559}-\frac{1.0017721}{0.059559}(-1)\right]\approx 5.1992\text{mA}$$

(the $V_{CB}$ terms are of order $10^{-14}$ A, utterly negligible next to the mA-level $V_{EB}$ terms);

$$I_{Cn}=-qA\frac{D_C}{L_C}{n}_{C0}\left(e^{q{V}_{CB}/kT}-1\right)=(1.602\times 10^{-19})(10^{-4})\frac{34.8}{5.90\times 10^{-3}}(10^5)(1)=9.45\times 10^{-15}\text{A}$$

Results: $I_{Ep}=5.2084$ mA, $I_{En}=7.600\mu$A, $I_{Cp}=5.1992$ mA, $I_{Cn}=9.45\times 10^{-15}$ A.

(c)

$$I_E=I_{Ep}+I_{En}=5.2084+0.0076=5.2160\text{mA}$$ $$I_C=I_{Cp}+I_{Cn}=5.1992\text{mA}$$ $$I_B=I_E-I_C=16.8\mu \text{A}$$

(d) The base-current components of Section 10.4 are $I_{B1}$ (electron back-injection into the emitter), $I_{B2}$ (hole recombination in the base), $I_{B3}$ (collector thermal-generation electrons), with $I_B=I_{B1}+I_{B2}-I_{B3}$:

$$I_{B1}=I_{En}=7.60\mu \text{A},I_{B2}=I_{Ep}-I_{Cp}=5.2084-5.1992=9.21\mu \text{A},I_{B3}=I_{Cn}=9.45\times 10^{-15}\text{A}$$

Check: $7.60+9.21-0=16.8\mu$A.

(e)

$$\gamma =\frac{I_{Ep}}{I_E}=\frac{5.2084}{5.2160}=0.99854,{\alpha }_T=\frac{I_{Cp}}{I_{Ep}}=\frac{5.1992}{5.2084}=0.99823$$ $${\alpha }_{dc}=\gamma {\alpha }_T=0.99678,{\beta }_{dc}=\frac{{\alpha }_{dc}}{1-{\alpha }_{dc}}=309$$

(consistent with $I_C/I_B=5.1992\text{mA}/16.8\mu \text{A}=309$ and with the approximate formulas’ 310).

% Problem 11.4: BJT current components and performance parameters
q=1.602e-19; kT=8.617e-5*300; ni=1e10; A=1e-4;
NE=1e18; NB=1e16; NC=1e15;
DE=6.81; DB=11.3; DC=34.8;
LE=8.25e-4; LB=3.36e-3; LC=5.90e-3; W=2e-4;
nE0=ni^2/NE; pB0=ni^2/NB; nC0=ni^2/NC;
VEB=0.7; VCB=-5;
x=W/LB; c=cosh(x); s=sinh(x);
fprintf('%.5g  %.5g  %.5g\n', c, s, x)
fe=exp(VEB/kT)-1; fc=exp(VCB/kT)-1;
IEn=q*A*(DE/LE)*nE0*fe;
IEp=q*A*(DB/LB)*pB0*((c/s)*fe-fc/s);
ICp=q*A*(DB/LB)*pB0*(fe/s-(c/s)*fc);
ICn=-q*A*(DC/LC)*nC0*fc;
IE=IEn+IEp; IC=ICp+ICn; IB=IE-IC;
IB1=IEn; IB2=IEp-ICp; IB3=ICn;
gam=IEp/IE; aT=ICp/IEp; adc=gam*aT; bdc=adc/(1-adc);
disp([IEp IEn ICp ICn IE IC IB IB1 IB2 IB3 gam aT adc bdc])

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Problem 11.5 (computer)

Problem. For $V_{CB}=0$ ($\Delta p_B(W)=0$) plot $\Delta p_B(x)/\Delta p_B(0)$ versus $x/W$ for $W/L_B=10,5,1,0.5,0.1$ and comment.

Solution. From (11.26) with $\Delta p_B(W)=0$:

$$\frac{\Delta p_B(x)}{\Delta p_B(0)}=\frac{\sinh [(W/L_B)(1-x/W)]}{\sinh (W/L_B)}$$

xw=linspace(0,1,201);                 % x/W
r=[10 5 1 0.5 0.1];                   % W/LB
for i=1:length(r)
    y=sinh(r(i)*(1-xw))/sinh(r(i));
    plot(xw,y); hold on
end
hold off; grid on; axis([0 1 0 1])
xlabel('x/W'); ylabel('\Deltap_B(x)/\Deltap_B(0)')
legend('W/L_B=10','5','1','0.5','0.1')

Computed samples at $x/W=0,0.25,0.5,0.75,1$:

$W/L_B$	0	0.25	0.5	0.75	1
10	1	0.0821	0.0067	0.0005	0
5	1	0.2864	0.0815	0.0216	0
1	1	0.6997	0.4434	0.2150	0
0.5	1	0.7366	0.4848	0.2405	0
0.1	1	0.7495	0.4994	0.2496	0

Comment. For $W/L_B\gg 1$ (10, 5) the curve is essentially the decaying exponential $e^{-x/L_B}$ of a long diode: the excess vanishes well before $x=W$. For $W/L_B\le 0.5$ the curves are practically the straight line $1-x/W$ (e.g. 0.4994 vs 0.5 at midpoint for 0.1). Even at $W/L_B=1$ the curve departs from the line by only about 0.057 (at most roughly 11% at the midpoint), which confirms the claim that the linear approximation is good up to $W/L_B\approx 1$.

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Problem 11.6 (computer)

Problem. Construct four subplots of $\Delta p_B(x)/p_{B0}$ versus $x/W$ for the four biasing modes, with $W/L_B\ll 1$, $\Delta p_B(0)/p_{B0}$ and $\Delta p_B(W)/p_{B0}\le 10$; axes $x_{min}=0$, $x_{max}=1$, $y_{min}=-2$, $y_{max}=10$.

Solution. For $W\ll L_B$ the profile is linear between the end values $e^{q{V}_{EB}/kT}-1$ and $e^{q{V}_{CB}/kT}-1$. Choose illustrative end values (normalized):

Mode	$V_{EB}$	$V_{CB}$	$\Delta p_B(0)/p_{B0}$	$\Delta p_B(W)/p_{B0}$
Active	$>0$	$<0$	10	$-1$
Inverted	$<0$	$>0$	$-1$	10
Saturation	$>0$	$>0$	10	5
Cutoff	$<0$	$<0$	$-1$	$-1$

xw=linspace(0,1,100);
modes={'Active','Inverted','Saturation','Cutoff'};
d0=[10 -1 10 -1];     % dp(0)/pB0
dW=[-1 10  5 -1];     % dp(W)/pB0
for i=1:4
    subplot(2,2,i)
    y=d0(i)+(dW(i)-d0(i))*xw;          % Eq. (11.40), normalized
    plot(xw,y,'LineWidth',1.5); grid on
    axis([0 1 -2 10])
    title(modes{i}); xlabel('x/W'); ylabel('\Deltap_B/p_{B0}')
end

The plots show: active, a falling line from high to $-1$; inverted, a rising line from $-1$ to high (mirror image); saturation, a line between two positive values (base flooded); cutoff, a flat line at $-1$ (the base is depleted of minority carriers at both edges).

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Problem 11.7

Problem. For a pnp BJT with $W\ll L_B$, $N_E\gg N_B>N_C$, sketch (linear scale) majority and minority carrier distributions in the quasineutral regions under (a) active, (b) inverted, (c) saturation, (d) cutoff bias (Fig. P11.7 gives the equilibrium levels as dashed lines).

Solution. Rules used: minority excess at a depletion edge is $(e^{qV/kT}-1)$ times equilibrium; majority carriers follow the minority excess to preserve neutrality, but under low-level injection that change is negligible relative to the majority level, so the majority curves are essentially flat at the equilibrium values ($p_E\approx N_E$, $n_B\approx N_B$, $p_C\approx N_C$) with tiny bumps that mimic the minority excess. Minority profiles: exponential decay in emitter and collector, straight line in base.

(a) Active ($V_{EB}>0$, $V_{CB}<0$):

• Emitter minority $n_E$: peak at E-B edge $n_{E0}{e}^{q{V}_{EB}/kT}\gg n_{E0}$, decaying exponentially to $n_{E0}$.
• Base minority $p_B$: line from $p_{B0}{e}^{q{V}_{EB}/kT}$ (high) at $x=0$ to $\approx 0$ at $x=W$.
• Collector minority $n_C$: $\approx 0$ at the C-B edge, rising exponentially to $n_{C0}$ (depleted, since reverse bias).
• Majority: flat (emitter at $N_E$, base at $N_B$, collector at $N_C$), with the base majority bumping slightly with the same profile as $p_B$.

(b) Inverted ($V_{EB}<0$, $V_{CB}>0$): the roles of emitter and collector swap.

• $n_E\approx 0$ at the E-B edge rising to $n_{E0}$.
• $p_B$: from $\approx 0$ at $x=0$ rising linearly to $p_{B0}{e}^{q{V}_{CB}/kT}$ at $x=W$.
• $n_C$: peak $n_{C0}{e}^{q{V}_{CB}/kT}$ at the C-B edge, decaying to $n_{C0}$.
• Majority flat.

(c) Saturation ($V_{EB}>0$, $V_{CB}>0$): excess at all three depletion edges.

• $n_E$: peak $n_{E0}{e}^{q{V}_{EB}/kT}$, decaying to $n_{E0}$.
• $p_B$: line from $p_{B0}{e}^{q{V}_{EB}/kT}$ at $x=0$ to $p_{B0}{e}^{q{V}_{CB}/kT}$ at $x=W$ (both above $p_{B0}$).
• $n_C$: peak $n_{C0}{e}^{q{V}_{CB}/kT}$, decaying to $n_{C0}$.
• Majority flat (slight bumps).

(d) Cutoff ($V_{EB}<0$, $V_{CB}<0$): depletion of minority carriers at every edge.

• $n_E$: $\approx 0$ at the edge rising to $n_{E0}$.
• $p_B$: line from $\approx 0$ at both ends… i.e. flat near zero at the edges; with $W\ll L_B$ the whole base sits close to 0 (a line between two near-zero values).
• $n_C$: $\approx 0$ at the edge rising to $n_{C0}$.
• Majority flat.

 (a) active:      n_E  \         p_B \          n_C   ___
                       \___         \___          __/
                                        \_ 0 ____/
 (c) saturation:  n_E  \         p_B \___        n_C  \___
 (d) cutoff:      n_E ___/       p_B ~~~~ 0       n_C ___/

(Linear-scale sketches: minority curves in the heavily doped emitter are tiny on the scale of the base, as the problem’s figure indicates qualitatively.)

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Problem 11.8

Problem. The $\Delta n_E/n_{E0}$, $\Delta p_B/p_{B0}$, $\Delta n_C/n_{C0}$ distributions of a pnp BJT are as in Fig. P11.8. Determine: (a) polarity of $V_{EB}$; (b) polarity of $V_{CB}$; (c) magnitude of $V_{CB}$; (d) the bias mode; (e) repeat (a)-(d) for an npn.

Reading of the figure (assumption). The OCR retains the labels “10” and "$-1$" on the normalized axis. We read: at the emitter and at the base edge $x=0$ the normalized excess equals $+10$ (peak), the base line falls linearly to $-1$ at $x=W$, and the collector curve is at $-1$ at its edge, rising to 0.

Solution.

(a) $\Delta n_E(0)/n_{E0}=e^{q{V}_{EB}/kT}-1=10>0$, so $V_{EB}>0$ (forward bias). Its magnitude:

$$V_{EB}=\frac{kT}{q}\ln (1+10)=0.0259\ln 11=62\text{mV}$$

(b) $\Delta p_B(W)/p_{B0}=\Delta n_C(0)/n_{C0}=e^{q{V}_{CB}/kT}-1=-1<0$, so $V_{CB}<0$ (reverse bias).

(c) A value of exactly $-1$ requires $e^{q{V}_{CB}/kT}=0$, i.e. $|V_{CB}|\gg kT/q$. The plot only shows that $|V_{CB}|$ is several $kT/q$ ($\gtrsim 0.12$ V gives $e^{-4.6}-1=-0.99$); the actual magnitude cannot be determined from the data.

(d) Forward E-B and reverse C-B: active mode.

(e) npn. With the same normalized distributions (now $\Delta p_E/p_{E0}$, $\Delta n_B/n_{B0}$, $\Delta p_C/p_{C0}$): $V_{BE}=+62$ mV, $V_{BC}<0$ with $|V_{BC}|\gg kT/q$ (magnitude undetermined), active mode. In terms of the pnp-style definitions $V_{EB}=-V_{BE}$ and $V_{CB}=-V_{BC}$: $V_{EB}<0$ and $V_{CB}>0$, i.e. the terminal polarities are reversed relative to the pnp case, as expected.

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Problem 11.9

Problem. Fill in the table: effect of increasing $W_B$, ${\tau }_B$, $N_B$, ${\tau }_E$, $N_E$ on $\gamma$, ${\alpha }_T$, ${\beta }_{dc}$.

Solution. Use (11.41)-(11.44):

$$\gamma =\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}},{\alpha }_T\approx 1-\frac{W^2}{2L_B^2},{\beta }_{dc}=\frac{1}{\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac{W^2}{2L_B^2}}$$

with $L_B=\sqrt{D_B{\tau }_B}$, $L_E=\sqrt{D_E{\tau }_E}$ and mobilities (hence $D$‘s) decreasing with doping.

Change	$\gamma$	${\alpha }_T$	${\beta }_{dc}$	Reason
Increase $W_B$	decreases	decreases	decreases	$W$ appears in the numerator of both loss terms
Increase ${\tau }_B$	no effect	increases	increases	$L_B$ grows, so $W^2/2L_B^2$ shrinks; $\gamma$ has no ${\tau }_B$
Increase $N_B$	decreases	decreases (slightly)	decreases	$N_B/N_E$ grows; $D_B$ falls with doping so $L_B$ falls
Increase ${\tau }_E$	increases	no effect	increases	$L_E$ grows, shrinking $W/L_E$; ${\alpha }_T$ independent of emitter
Increase $N_E$	increases	no effect	increases	$N_B/N_E$ falls; also $D_E/L_E=\sqrt{D_E/{\tau }_E}$ falls as $D_E$ drops with doping, both lowering back-injection

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Problem 11.10 (computer)

Problem. With the Table 11.1 lifetimes and dopings as reference values (${\tau }_{B0}$, $N_{B0}$, ${\tau }_{E0}$, $N_{E0}$), and the mobility fits of Exercise 3.1, construct four log-log subplots of ${\beta }_{dc}$ versus ${\tau }_B/{\tau }_{B0}$, $N_B/N_{B0}$, ${\tau }_E/{\tau }_{E0}$, $N_E/N_{E0}$, each ratio from 0.1 to 10. $T=300$ K, $W=W_B=2\mu$m. Ignore emitter degeneracy. Do the results agree with 11.9?

Solution (logic). For each ratio, recompute the mobility-fit values $\mu$, then $D=(kT/q)\mu$, $L=\sqrt{D\tau }$, and use the general (11.34) (or the simplified (11.44)).

q=1.602e-19; kT=8.617e-5*300; W=2e-4;
mun=@(N) 92+1268./(1+(N/1.3e17).^0.91);
mup=@(N) 54.3+406.9./(1+(N/2.35e17).^0.88);
NE0=1e18; NB0=1e16; tE0=1e-7; tB0=1e-6;
r=logspace(-1,1,50);
beta=@(NE,NB,tE,tB) ...
   1./( cosh(W./sqrt(kT*mup(NB).*tB)) ...
      + (kT*mun(NE)./(kT*mup(NB))).*(NB./NE).* ...
        sqrt(kT*mup(NB).*tB)./sqrt(kT*mun(NE).*tE) ...
        .*sinh(W./sqrt(kT*mup(NB).*tB)) - 1 );
subplot(2,2,1); loglog(r, beta(NE0,NB0,tE0,tB0*r)); xlabel('\tau_B/\tau_{B0}'); ylabel('\beta_{dc}'); grid on
subplot(2,2,2); loglog(r, beta(NE0,NB0*r,tE0,tB0)); xlabel('N_B/N_{B0}');   ylabel('\beta_{dc}'); grid on
subplot(2,2,3); loglog(r, beta(NE0,NB0,tE0*r,tB0)); xlabel('\tau_E/\tau_{E0}'); ylabel('\beta_{dc}'); grid on
subplot(2,2,4); loglog(r, beta(NE0*r,NB0,tE0,tB0)); xlabel('N_E/N_{E0}');   ylabel('\beta_{dc}'); grid on

(Here beta is (11.34) written with $D_E/D_B$, $N_B/N_E$, $L_B/L_E$.)

Expected results and agreement. With the reference loss terms $a=1.46\times 10^{-3}$ (emitter) and $b=1.77\times 10^{-3}$ (base), ${\beta }_{dc}=1/(a+b)=310$ at the reference point.

• vs ${\tau }_B$: $b\propto 1/{\tau }_B$, so ${\beta }_{dc}$ increases with ${\tau }_B$ and saturates at $1/a\approx 685$ (the emitter-limited ceiling).
• vs $N_B$: $a\propto N_B$ (and $D_B$ shrinks, which raises $b$), so ${\beta }_{dc}$ decreases, roughly inversely with $N_B$ at large $N_B$.
• vs ${\tau }_E$: $a\propto 1/\sqrt{{\tau }_E}$, so ${\beta }_{dc}$ increases with ${\tau }_E$ and saturates at $1/b\approx 565$.
• vs $N_E$: $a\propto 1/N_E$ (times the mobility variation), so ${\beta }_{dc}$ increases, saturating at $1/b$ once emitter losses are negligible.

All four trends agree with the Problem 11.9 table (increase ${\tau }_B$: up; increase $N_B$: down; increase ${\tau }_E$: up; increase $N_E$: up).

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Problem 11.11: Shallow Emitter

Problem. Take a pnp BJT whose emitter has finite quasineutral width $W_E$ (Fig. P11.11), with $\Delta n_E=0$ at the metallic emitter contact ($x^{\prime \prime }=W_E$). (a) Paralleling the ideal analysis, obtain revised expressions for $\Delta n_E(x^{\prime \prime })$ and $I_{En}$. (b) Establish revised performance parameters analogous to (11.31)-(11.34). (c) Simplify for $W/L_B\ll 1$ and $W_E/L_E\ll 1$. (d) How are $\gamma$ and ${\beta }_{dc}$ affected as $W_E$ decreases? (e) Sketch the minority distribution in a shallow emitter, active mode.

Solution.

(a) The emitter equation (11.1) is unchanged, but the boundary conditions are now

$$\Delta n_E(0)=n_{E0}\left(e^{q{V}_{EB}/kT}-1\right),\Delta n_E(W_E)=0$$

This is the base-type problem with the second end value zero. Reusing (11.26) with $x\to x^{\prime \prime }$, $W\to W_E$, $L_B\to L_E$:

$$\Delta n_E(x^{\prime \prime })=n_{E0}\left(e^{q{V}_{EB}/kT}-1\right)\frac{\sinh [(W_E-x^{\prime \prime })/L_E]}{\sinh (W_E/L_E)}$$

Differentiate at $x^{\prime \prime }=0$:

$$\frac{d\Delta n_E}{d{x}^{\prime \prime }}{|}_0=-\frac{n_{E0}}{L_E}\left(e^{q{V}_{EB}/kT}-1\right)\frac{\cosh (W_E/L_E)}{\sinh (W_E/L_E)}$$

and by (11.7):

$$I_{En}=qA\frac{D_E}{L_E}{n}_{E0}\coth \left(\frac{W_E}{L_E}\right)\left(e^{q{V}_{EB}/kT}-1\right)$$

(As $W_E/L_E\to \infty$, $\coth \to 1$ recovers (11.20).)

(b) $I_{Ep}$ and $I_{Cp}$ are unchanged (base unchanged), so in active mode $I_{Ep}\approx qA(D_B/L_B)p_{B0}\coth (W/L_B)(e^{q{V}_{EB}/kT}-1)$. Then

$$\frac{I_{En}}{I_{Ep}}=\frac{D_E{n}_{E0}\coth (W_E/L_E)/L_E}{D_B{p}_{B0}\coth (W/L_B)/L_B}=\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\coth \frac{W_E}{L_E}\tanh \frac{W}{L_B}$$ $$\gamma =\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\coth \frac{W_E}{L_E}\tanh \frac{W}{L_B}}$$ $${\alpha }_T=\frac{1}{\cosh (W/L_B)}\text{(unchanged)}$$ $${\alpha }_{dc}=\gamma {\alpha }_T=\frac{1}{\cosh \frac{W}{L_B}+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\coth \frac{W_E}{L_E}\sinh \frac{W}{L_B}}$$ $${\beta }_{dc}=\frac{1}{\cosh \frac{W}{L_B}+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\coth \frac{W_E}{L_E}\sinh \frac{W}{L_B}-1}$$

(c) For $W/L_B\ll 1$ ($\tanh \to W/L_B$) and $W_E/L_E\ll 1$ ($\coth \to L_E/W_E$):

$$\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{L_B}{L_E}\cdot \frac{L_E}{W_E}\cdot \frac{W}{L_B}=\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{W_E}$$ $$\gamma =\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{W_E}},{\alpha }_T\approx 1-\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2$$ $${\alpha }_{dc}=\frac{1}{1+\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{W_E}+\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2},{\beta }_{dc}=\frac{1}{\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{W_E}+\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2}$$

Thus $W_E$ simply replaces $L_E$ in (11.41)-(11.44).

(d) $W_E$ appears in the denominator of the emitter loss term, so decreasing $W_E$ increases that term: $\gamma$ decreases and ${\beta }_{dc}$ decreases (and ${\alpha }_T$ is unaffected). The thinner the emitter, the more steeply minority carriers are drawn from the junction toward the contact, so more back-injection current flows.

(e) Sketch: in the emitter ($\Delta n_E$ versus $x^{\prime \prime }$) a straight line from $n_{E0}(e^{q{V}_{EB}/kT}-1)$ at the junction to 0 at the contact; in the base a line from $p_{B0}(e^{q{V}_{EB}/kT}-1)$ to $-p_{B0}$; in the collector an exponential from $\approx -n_{C0}$ rising to 0.

 Δn_E | \                        Δp_B  \
      |   \  (straight line,           \  (straight line)
      |     \  hits 0 at W_E)           \
      +-------\------ x''               +------\----- x
         0     W_E                      0       W

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Problem 11.12: Diode-Like Connections

Problem. With one terminal floating or two terminals tied together, the transistor is a two-terminal “diode” (Fig. P11.12, six connections). Choose a connection and (a) derive $I$-$V_A$ from the Ebers-Moll equations; (b) find $\Delta p_B(0)/p_{B0}$ and $\Delta p_B(W)/p_{B0}$ in terms of $V_A$ and EM parameters; (c) simplify with ${\alpha }_F={\alpha }_R=\alpha$ and $I_{F0}=I_{R0}=I_0$ (identical emitter and collector parameters); (d) sketch the base distribution for $V_A>kT/q$ with $W\ll L_B$; (e) repeat for reverse bias $-V_A\gg kT/q$.

Interpretation of the figure (assumption). The six connections are read as: (a) E-B junction with collector open; (b) E-B with collector shorted to base; (c) C-B junction with emitter open; (d) C-B with emitter shorted to base; (e) E-C with base open; (f) the E-C counterpart with the base shorted. We solve (a)-(e) (including both open-circuit and short-circuit cases). Notation: $u\equiv e^{q{V}_A/kT}-1$, $a\equiv e^{q{V}_{EB}/kT}-1$, $b\equiv e^{q{V}_{CB}/kT}-1$. Useful: ${\alpha }_F{I}_{F0}={\alpha }_R{I}_{R0}$. Current $I$ is the current entering the terminal at $+V_A$ in the forward direction of that junction.

Solution.

Connection (a): E-B diode, collector open. $V_{EB}=V_A$ so $a=u$, and $I_C=0$:

$$0={\alpha }_F{I}_{F0}u-I_{R0}b\Rightarrow b=\frac{{\alpha }_F{I}_{F0}}{I_{R0}}u={\alpha }_{R}u$$

(using ${\alpha }_F{I}_{F0}={\alpha }_R{I}_{R0}$). Then

$$I=I_E=I_{F0}u-{\alpha }_R{I}_{R0}b=I_{F0}u-{\alpha }_R{I}_{R0}{\alpha }_{R}u$$

Using ${\alpha }_R{I}_{R0}={\alpha }_F{I}_{F0}$:

$$\boxed{I=I_{F0}\left(1-{\alpha }_F{\alpha }_R\right)\left(e^{q{V}_A/kT}-1\right)}$$

(b) $\frac{\Delta p_B(0)}{p_{B0}}=a=u,\frac{\Delta p_B(W)}{p_{B0}}=b={\alpha }_{R}u$.

(c) With ${\alpha }_F={\alpha }_R=\alpha$: $I=I_0(1-{\alpha }^2)u$; $\Delta p_B(0)/p_{B0}=u$, $\Delta p_B(W)/p_{B0}=\alpha u$.

Connection (b): E-B diode, collector shorted to base ($V_{CB}=0$, so $b=0$; $V_{EB}=V_A$):

$$\boxed{I=I_E=I_{F0}\left(e^{q{V}_A/kT}-1\right)}$$

(b) $\Delta p_B(0)/p_{B0}=u$, $\Delta p_B(W)/p_{B0}=0$. (c) Same with $I_{F0}=I_0$: $I=I_{0}u$, end values $u$ and 0.

Connection (c): C-B diode, emitter open. $V_{CB}=V_A$ so $b=u$, and $I_E=0$:

$$0=I_{F0}a-{\alpha }_R{I}_{R0}u\Rightarrow a=\frac{{\alpha }_R{I}_{R0}}{I_{F0}}u={\alpha }_{F}u$$

The current into the C-B junction in its forward direction is $-I_C$:

$$-I_C=-{\alpha }_F{I}_{F0}a+I_{R0}u=-{\alpha }_F^2{I}_{F0}u+I_{R0}u$$

With ${\alpha }_F^2{I}_{F0}={\alpha }_F{\alpha }_R{I}_{R0}$:

$$\boxed{I=-I_C=I_{R0}\left(1-{\alpha }_F{\alpha }_R\right)\left(e^{q{V}_A/kT}-1\right)}$$

For $-V_A\gg kT/q$ this tends to $-I_{R0}(1-{\alpha }_F{\alpha }_R)=-I_{CBO}$, confirming Exercise 11.6. (b) $\Delta p_B(0)/p_{B0}=a={\alpha }_{F}u$, $\Delta p_B(W)/p_{B0}=u$. (c) $I=I_0(1-{\alpha }^2)u$; end values $\alpha u$ and $u$.

Connection (d): C-B diode, emitter shorted to base ($V_{EB}=0$, $a=0$, $b=u$):

$$\boxed{I=-I_C=I_{R0}\left(e^{q{V}_A/kT}-1\right)}$$

(b) $\Delta p_B(0)/p_{B0}=0$, $\Delta p_B(W)/p_{B0}=u$. (c) $I=I_{0}u$; end values 0 and $u$.

Connection (e): E-C two-terminal, base open. $I_B=0$, so $I_E=I_C\equiv I$ and $V_A=V_{EC}=V_{EB}-V_{CB}$. From $I_B=I_E-I_C$:

$$0=(1-{\alpha }_F)I_{F0}a+(1-{\alpha }_R)I_{R0}b\Rightarrow b=-\frac{(1-{\alpha }_F)I_{F0}}{(1-{\alpha }_R)I_{R0}}a$$

Together with $V_A=V_{EB}-V_{CB}$ this fixes $a$ and $b$ numerically (transcendental). For $V_A\gg kT/q$: $V_{CB}\ll 0$, $b\approx -1$, so

$$a\approx \frac{(1-{\alpha }_R)I_{R0}}{(1-{\alpha }_F)I_{F0}},I=I_{F0}a+{\alpha }_R{I}_{R0}=I_{R0}\frac{1-{\alpha }_R}{1-{\alpha }_F}+{\alpha }_R{I}_{R0}=\frac{I_{R0}(1-{\alpha }_F{\alpha }_R)}{1-{\alpha }_F}=\frac{I_{CBO}}{1-{\alpha }_F}=I_{CEO}$$

(b) $\Delta p_B(0)/p_{B0}=a$, $\Delta p_B(W)/p_{B0}=-1$. (c) With the symmetric simplification (${\alpha }_F={\alpha }_R=\alpha$, $I_{F0}=I_{R0}=I_0$) the condition $I_B=0$ gives $b=-a$, i.e. $e^{q{V}_{CB}/kT}=2-e^{q{V}_{EB}/kT}$, and $I=I_E=I_0(a-\alpha b)=I_0(1+\alpha )a$, with $a$ fixed numerically by $V_A=V_{EB}-V_{CB}$. The base profile is antisymmetric: end values $a$ and $-a$.

(d) Sketch, forward bias $V_A>kT/q$, $W\ll L_B$, simplified case (c):

 Connection (a)/(c):   (a) high flat-ish line      (b)/(d) full ramp
 Δp_B/p_B0                                         Δp_B/p_B0
   u  |\                                             u |\
      | \__  αu  (almost flat, α≈1;                   | \
      |        both junctions forward biased,         |  \
 0 ---+----- x   base flooded: saturation-like)    0 --+---\--- x

(a) and (c): lines between $u$ and $\alpha u$ (nearly flat since $\alpha \approx 1$, both junctions forward biased: base flooded, like saturation). (b): line from $u$ to 0 (like the active or narrow-base diode); (d): from 0 to $u$.

(e) Sketch, reverse bias $-V_A\gg kT/q$ ($u\to -1$): (a): end values $-1$ and $-\alpha \approx -1$: a nearly flat line at $-1$ (base depleted of minority carriers, cutoff-like); (b): $-1$ to 0; (c): $-\alpha$ to $-1$ (flat, near $-1$); (d): 0 to $-1$.

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Problem 11.13

Problem. The CE output characteristics of a pnp BJT at small $V_{EC}$ are in Fig. P11.13; the d.c. operating point lies at the boundary between active mode and saturation (constant $I_B$). (a) Referring to Fig. 11.3, draw the simplified large-signal equivalent circuit at that point. (b) Obtain $V_{EC}$ at the operating point in terms of $I_B$ and the EM parameters.

Solution.

(a) At the active/saturation boundary $V_{CB}=0$. Then $I_R=I_{R0}(e^{q{V}_{CB}/kT}-1)=0$: the collector diode carries no current and the ${\alpha }_R{I}_R$ source is zero. What remains: the emitter diode carrying $I_F$ from E to B, and the current source ${\alpha }_F{I}_F$ from E to C:

        E o----+------------+
               |            |
             (diode: I_F)  (αF·I_F source)
               |            |
        B o----+            o C

with $I_E=I_F$, $I_C={\alpha }_F{I}_F$, $I_B=(1-{\alpha }_F)I_F$. (The collector terminal sees just the dependent source; the B-C diode is at zero bias.)

(b) From the circuit:

$$I_B=I_E-I_C=(1-{\alpha }_F)I_F\Rightarrow I_F=\frac{I_B}{1-{\alpha }_F}$$

and $I_F=I_{F0}(e^{q{V}_{EB}/kT}-1)$:

$$I_{F0}\left(e^{q{V}_{EB}/kT}-1\right)=\frac{I_B}{1-{\alpha }_F}\Rightarrow V_{EB}=\frac{kT}{q}\ln \left[1+\frac{I_B}{(1-{\alpha }_F)I_{F0}}\right]$$

With $V_{CB}=0$, $V_{EC}=V_{EB}-V_{CB}=V_{EB}$:

$$\boxed{V_{EC}=\frac{kT}{q}\ln \left[1+\frac{I_B}{(1-{\alpha }_F)I_{F0}}\right]}$$

(and $1-{\alpha }_F=1/({\beta }_{dc}+1)$.) This is the “knee” voltage separating saturation from the active region; it grows only logarithmically with $I_B$.

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Problem 11.14: Deriving the Computational Equations

Problem. Derive the quoted computational equations for (a) $I_C(V_{CB},I_E)$, (b) $I_B(V_{EB},V_{EC})$, (c) $I_C(V_{EC},I_B)$.

Solution. Let $a=e^{q{V}_{EB}/kT}-1$, $b=e^{q{V}_{CB}/kT}-1$.

(a) From (11.47a):

$$a=\frac{I_E+{\alpha }_R{I}_{R0}b}{I_{F0}}$$

Substitute in (11.47b):

$$I_C={\alpha }_F{I}_{F0}a-I_{R0}b={\alpha }_F\left(I_E+{\alpha }_R{I}_{R0}b\right)-I_{R0}b$$ $$\boxed{I_C={\alpha }_F{I}_E-\left(1-{\alpha }_F{\alpha }_R\right)I_{R0}\left(e^{q{V}_{CB}/kT}-1\right)}$$

(b) $I_B=I_E-I_C$ with (11.47):

$$I_B=(1-{\alpha }_F)I_{F0}a+(1-{\alpha }_R)I_{R0}b$$

With $V_{CB}=V_{EB}-V_{EC}$: $e^{q{V}_{CB}/kT}=e^{q{V}_{EB}/kT}{e}^{-q{V}_{EC}/kT}$, so $b=e^{q{V}_{EB}/kT}{e}^{-q{V}_{EC}/kT}-1$:

$$I_B=(1-{\alpha }_F)I_{F0}\left(e^{q{V}_{EB}/kT}-1\right)+(1-{\alpha }_R)I_{R0}\left(e^{q{V}_{EB}/kT}{e}^{-q{V}_{EC}/kT}-1\right)$$

Collect the terms multiplying $e^{q{V}_{EB}/kT}$ and the constants:

$$\boxed{I_B=\left[(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}{e}^{-q{V}_{EC}/kT}\right]e^{q{V}_{EB}/kT}-\left[(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}\right]}$$

(c) Write $X\equiv e^{q{V}_{EB}/kT}$, $E\equiv e^{-q{V}_{EC}/kT}$, $I_{B0}\equiv (1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}$, $I_{B1}\equiv (1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}E$. Part (b) reads $I_B=I_{B1}X-I_{B0}$, so

$$X=\frac{I_B+I_{B0}}{I_{B1}}$$

Now (11.47b) with $e^{q{V}_{CB}/kT}=XE$:

$$I_C={\alpha }_F{I}_{F0}(X-1)-I_{R0}(XE-1)=\left({\alpha }_F{I}_{F0}-I_{R0}E\right)X-{\alpha }_F{I}_{F0}+I_{R0}$$

Substitute $X$:

$$\boxed{I_C=\frac{\left({\alpha }_F{I}_{F0}-I_{R0}{e}^{-q{V}_{EC}/kT}\right)\left[I_B+(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}\right]}{(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}{e}^{-q{V}_{EC}/kT}}+I_{R0}-{\alpha }_F{I}_{F0}}$$

Note: the last term is $I_{R0}-{\alpha }_F{I}_{F0}$. (In the Appendix M listing it is typed IR0-aF*IR0, which is numerically different from $I_{R0}-{\alpha }_F{I}_{F0}$ unless ${\alpha }_F{I}_{F0}={\alpha }_R{I}_{R0}$ with ${\alpha }_R\approx {\alpha }_F$; both terms are tiny compared with the leading term, so the plotted curves are insensitive to the distinction, but the derivation above gives $I_{F0}$.) An equivalent form from direct elimination is $I_C=\{({\alpha }_F{I}_{F0}-I_{R0}E)I_B+(1-{\alpha }_F{\alpha }_R)I_{F0}{I}_{R0}(1-E)\}/I_{B1}$.

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Problem 11.15: Emitter Input and Base Width Modulation

Problem. In modern transistors gain is usually limited by emitter efficiency, so $\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}\gg \frac{1}{2}(W/L_B{)}^2$. (a) Show the CE input characteristics are then relatively insensitive to base width modulation. (b) Sketch the expected CE input characteristics if gain is not emitter-limited, i.e. $\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}\sim \frac{1}{2}(W/L_B{)}^2$.

Solution. (a) The CE input characteristic is (Exercise 11.7, item 3)

$$I_B=\left[(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}{e}^{-q{V}_{EC}/kT}\right]e^{q{V}_{EB}/kT}-\left[(1-{\alpha }_F)I_{F0}+(1-{\alpha }_R)I_{R0}\right]$$

For $V_{EC}$ more than a few $kT/q$ the $e^{-q{V}_{EC}/kT}$ term vanishes and, for $V_{EB}\gg kT/q$,

$$I_B\approx (1-{\alpha }_F)I_{F0}{e}^{q{V}_{EB}/kT}$$

Evaluate the prefactor using (11.45a), (11.46) and $\cosh -1=2{\sinh }^2(W/2L_B)$ (so $(\cosh -1)/\sinh =\tanh (W/2L_B)$):

$$(1-{\alpha }_F)I_{F0}=I_{F0}-{\alpha }_F{I}_{F0}=qA\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\frac{\cosh (W/L_B)-1}{\sinh (W/L_B)}\right]$$ $$=qA\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B}{L_B}{p}_{B0}\tanh \frac{W}{2L_B}\right]\approx qA\left[\frac{D_E}{L_E}{n}_{E0}+\frac{D_B{p}_{B0}}{2L_B^2}W\right]$$

The first term is $W$-independent, and the second is smaller by the ratio $\frac{1}{2}(W/L_B{)}^2/\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}$. (Check: dividing by the first term, the second/first ratio equals $\frac{D_B{p}_{B0}W/(2L_B^2)}{D_E{n}_{E0}/L_E}=\frac{\frac{1}{2}(W/L_B{)}^2}{\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}}\ll 1$ when emitter-limited.) Hence $I_B\approx qA(D_E/L_E)n_{E0}{e}^{q{V}_{EB}/kT}$ is essentially independent of $W$, and therefore of $V_{EC}$: the input characteristics for different $V_{EC}$ coincide. $■$

(b) When the two terms are comparable, $(1-{\alpha }_F)I_{F0}$ contains a significant part proportional to $W$. As $V_{EC}$ rises, $W$ falls, so $I_B$ at a given $V_{EB}$ decreases: the curves for increasing $V_{EC}$ shift toward higher $V_{EB}$ (to the right on a semilog plot, to lower current at fixed voltage), by a fixed ratio on a log scale. The expected sketch is a family of exponential curves, all with the same slope $q/kT$, spaced progressively to the right (lower current at fixed $V_{EB}$) as $V_{EC}$ goes from a few $kT/q$ up to tens of volts, with the $V_{EC}=0$ curve far to the left (higher current, from the $(1-{\alpha }_R)I_{R0}$ term).

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Problem 11.16

Problem. How is the Exercise 11.8 result modified if $V_{EB}=0.7$ V?

Solution. Only $x_{nEB}$ changes, since it contains $(V_{bi(EB)}-V_{EB})$:

$$x_{nEB}=3.29\times 10^{-5}\sqrt{\frac{0.835-0.7}{0.835}}=3.29\times 10^{-5}(0.402)=1.32\times 10^{-5}\text{cm}$$ $$W_B-x_{nEB}=2\times 10^{-4}-1.32\times 10^{-5}=1.868\times 10^{-4}\text{cm}$$ $$V_{CB}=V_{bi(CB)}-8.43\times 10^9{\left(W_B-x_{nEB}\right)}^2=0.656-8.43\times 10^9{\left(1.868\times 10^{-4}\right)}^2=0.656-294.1$$ $$V_{CB}\approx -293.5\text{V}\approx -294\text{V}$$

Forward bias shrinks the E-B depletion region, leaving more base width to be consumed by the C-B depletion region, so the punch-through voltage rises in magnitude from 235 V to 294 V.

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Problem 11.17

Problem. Two pnp BJTs are identical except that emitter and collector dopings are interchanged (Fig. P11.17). Reading the OCR’d figure: A has $N_E=10^{17}$, $N_B=10^{15}$, $N_C=10^{14}$ cm${}^{-3}$; B has $N_E=10^{14}$, $N_B=10^{15}$, $N_C=10^{17}$ cm${}^{-3}$. (a) Which has the greater emitter efficiency? (b) Which is more sensitive to base width modulation in active mode? (c) If limited by C-B avalanche, which has the larger $V_{CBO}$?

Solution.

(a) A. The emitter loss term in $\gamma$ is proportional to $N_B/N_E$ (the other factors are nearly the same):

$${\frac{N_B}{N_E}|}_A=\frac{10^{15}}{10^{17}}=10^{-2},{\frac{N_B}{N_E}|}_B=\frac{10^{15}}{10^{14}}=10$$

so ${\gamma }_A\approx 1/(1+0.01K)\approx 1$, while ${\gamma }_B\approx 1/(1+10K)$ is much smaller (B has a lightly doped emitter relative to the base, the opposite of good design).

(b) B. The base penetration of the C-B depletion region is

$$x_{nCB}={\left[\frac{2K_S{\epsilon }_0}{q}\frac{N_C}{N_B(N_C+N_B)}\left(V_{bi(CB)}-V_{CB}\right)\right]}^{1/2}$$ $$\frac{N_C}{N_B(N_C+N_B)}{|}_A=\frac{10^{14}}{10^{15}(1.1\times 10^{15})}=9.1\times 10^{-17},\frac{N_C}{N_B(N_C+N_B)}{|}_B=\frac{10^{17}}{10^{15}(1.01\times 10^{17})}=9.9\times 10^{-16}$$

The ratio is about 11, so $x_{nCB}{|}_B\approx 3.3x_{nCB}{|}_A$ at the same bias (ignoring the small difference in $V_{bi}$). In A most of the C-B depletion region lies in the lightly doped collector; in B it lies in the base. B’s base width therefore varies more with $V_{CB}$.

(c) A. The avalanche breakdown voltage is set by the lightly doped side of the C-B junction and varies roughly inversely with its doping. For A the light side is the collector ($10^{14}$); for B it is the base ($10^{15}$). The lower light-side doping gives A the larger $V_{CBO}$ (roughly an order of magnitude lower doping gives several times larger breakdown voltage, $V_{BR}\propto N^{-3/4}$ from the text’s fit).

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Problem 11.18 (Early voltage)

Problem. For $V_{EC}\ge 20$ V the characteristics in the middle figure of Exercise 11.10 (BWM included, no multiplication) are closely modelled by $I_C=B(V_{EC}-V)$ with $B$ and $V$ constants. Determine $B$ and $V$ for each of the four $I_B>0$ curves. $V$ is the Early voltage, the $V_{EC}$-axis intercept of the extrapolated linear region. Comment.

Solution (method). Use the BJTplus equations (Exercise 11.10, case ii) to compute $I_C(V_{EC})$ for $I_B=2.5,5,7.5,10\mu$A over $20\le V_{EC}\le 80$ V, then fit a straight line (polyfit(VEC,IC,1)): slope $=B$, intercept on the $V_{EC}$ axis $=V=-\text{intercept}/\text{slope}$.

% Early-voltage fit (run after defining parameters as in Exercise 11.10)
VEC=20:1:80; VEB=0; VCB=VEB-VEC;
xnEB=sqrt((2*KS*e0/q)*(NE/(NB*(NE+NB)))*(VbiE-VEB));
xnCB=sqrt((2*KS*e0/q)*(NC/(NB*(NC+NB)))*(VbiC-VCB));
Wv=WB-xnEB-xnCB;
fBv=(DB/LB)*pB0*cosh(Wv/LB)./sinh(Wv/LB);
F0=q*A*((DE/LE)*nE0+fBv); R0=q*A*((DC/LC)*nC0+fBv);
aFv=q*A*(DB/LB)*pB0./sinh(Wv/LB)./F0; aRv=q*A*(DB/LB)*pB0./sinh(Wv/LB)./R0;
E=exp(-VEC/kT);
for IB=(1:4)*2.5e-6
    IB0=(1-aFv).*F0+(1-aRv).*R0; IB1=(1-aFv).*F0+(1-aRv).*R0.*E;
    IC=(aFv.*F0-R0.*E).*(IB+IB0)./IB1+R0-aFv.*F0;
    p=polyfit(VEC,IC,1);
    fprintf('IB=%5.1f uA:  B=%7.3f uA/V,  V=%7.1f V\n',IB*1e6,p(1)*1e6,-p(2)/p(1));
end

Results (computed with the BJT0 parameters: $W_B=2.5\mu$m, $N_B=1.5\times 10^{16}$, $N_C=1.5\times 10^{15}$, $N_E=10^{18}$ cm${}^{-3}$):

$I_B$ ($\mu$A)	$B$ ($\mu$A/V)	$V$ (V)
2.5	4.69	$-131$
5.0	9.38	$-131$
7.5	14.07	$-131$
10.0	18.76	$-131$

(Over this range $W$ falls from about 1.83 $\mu$m at $V_{EC}=20$ V to about 1.34 $\mu$m at 100 V.)

Comment. The Early voltage is about $-131$ V, i.e. all four straight lines, when extrapolated backwards, meet the $V_{EC}$ axis at the same point ($|V_A|\approx 131$ V, a typical magnitude for small-signal transistors, large because the lightly doped collector keeps base modulation modest). $B$ is proportional to $I_B$. This follows from $I_C\approx {\beta }_{dc}(V_{EC})I_B$ (since $I_{CEO}$ is negligible): the shape of every curve is the shape of ${\beta }_{dc}(V_{EC})$, which is nearly linear with intercept $V$, and $B=I_{B}d{\beta }_{dc}/d{V}_{EC}\approx 1.88{\text{V}}^{-1}\times I_B$. The common intercept is a signature of base width modulation; the output conductance $d{I}_C/d{V}_{EC}=I_C/(V_{EC}-V)$ rises in proportion to $I_C$.

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Problem 11.19 (computer): Diode-Like $I$-$V_A$ Plots

Problem. Adapt the BJTplus programs to plot the $I$-$V_A$ characteristics of the Problem 11.12(a) relationships. (a) First assume $W=W_B$ and neglect BWM and multiplication. (b) Then include BWM and multiplication (including carrier multiplication in the E-B depletion region at large reverse bias on the E-B junction) and compare.

Solution (logic). Use the Problem 11.12 relations: connection (a) [C open] $I=I_{F0}(1-{\alpha }_F{\alpha }_R)u$; connection (b) [C shorted to B] $I=I_{F0}u$, $u=e^{q{V}_A/kT}-1$. For (a) with BWM the parameters depend on $V_{CB}$ which depends on the parameters ($b={\alpha }_{R}u$), so iterate to self-consistency. Multiplication: for reverse $V_A$ across E-B, $M_E=1/[1-(|V_A|/B{V}_{EBO}{)}^m]$ with $B{V}_{EBO}$ from the light-side (base) doping, $60(N_B/10^{16}{)}^{-3/4}$ V, multiplying the (reverse) current.

% Problem 11.19: diode-like I-VA characteristics (connections: C open and C shorted)
BJT0                                   % loads constants and parameters
m=6; VBRE=60*(NB/1e16)^(-3/4);         % E-B avalanche voltage (light side = base)
VAf=0:0.005:0.8;                       % forward bias
VAr=0:-0.5:-0.95*VBRE;                 % reverse bias, stay below breakdown
for part=1:2                           % 1: ideal (W=WB);  2: BWM + multiplication
  for k=1:2                            % k=1 collector open, k=2 collector shorted
    for VAset={VAf,VAr}
      VA=VAset{1}; I=zeros(size(VA));
      for i=1:length(VA)
        u=exp(VA(i)/kT)-1; VEB=VA(i); VCB=0;
        for it=1:15                    % self-consistency for W
          if part==1, F0=IF0; R0=IR0; aF_=aF; aR_=aR;
          else
            xE=sqrt((2*KS*e0/q)*(NE/(NB*(NE+NB)))*(VbiE-min(VEB,0.9*VbiE)));
            xC=sqrt((2*KS*e0/q)*(NC/(NB*(NC+NB)))*(VbiC-VCB));
            Wl=max(WB-xE-xC,0.05*WB);
            fBl=(DB/LB)*pBO*cosh(Wl/LB)/sinh(Wl/LB);
            F0=q*A*((DE/LE)*nEO+fBl); R0=q*A*((DC/LC)*nCO+fBl);
            aF_=q*A*(DB/LB)*pBO/sinh(Wl/LB)/F0; aR_=q*A*(DB/LB)*pBO/sinh(Wl/LB)/R0;
          end
          if k==1, VCB=kT*log(1+aR_*u); else, VCB=0; end   % b=aR*u when C open
        end
        if k==1, I(i)=F0*(1-aF_*aR_)*u; else, I(i)=F0*u; end
        if part==2 && VA(i)<0, I(i)=I(i)/(1-(abs(VA(i))/VBRE)^m); end
      end
      figure(part*10+k); hold on
      if VA(1)==0 && VA(end)>0, plot(VA,I*1e3); xlabel('V_A (V)'); ylabel('I (mA)');
      else, plot(VA,I*1e6); xlabel('V_A (V)'); ylabel('I (\muA), reverse'); end
      grid on
    end
  end
end

(Notes: pBO, nEO, nCO are the BJT0 names with letter O; the forward and reverse parts are plotted separately for clarity.)

Expected behaviour. (a) Without BWM or multiplication: forward $I\propto e^{q{V}_A/kT}$ (an exponential with the ideal factor $I_{F0}$, or $I_{F0}(1-{\alpha }_F{\alpha }_R)$ for collector open, lower by the small factor $1-{\alpha }_F{\alpha }_R\approx 0.4$ to 1 depending on ${\alpha }_R$); reverse: constant saturation current $-I_{F0}$ (shorted) or $-I_{F0}(1-{\alpha }_F{\alpha }_R)$ (open). (b) With BWM the parameters change slightly with bias and the curves tilt (forward: tiny change; reverse with collector shorted: no effect on $V_{EB}$-dependent $W$ other than through $x_{nEB}$, giving a mild increase in magnitude as $|V_A|$ grows); with multiplication the reverse current turns sharply up (avalanche) near $B{V}_{EBO}$ ($\approx 44$ V for these parameters), where the (a) result keeps a flat saturation current to arbitrarily large voltage.

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Problem 11.20 (computer): Gummel and Gain Plots

Problem. (a) Construct a theoretical Gummel plot like Fig. 11.15 with the BJT0 parameters; active mode with $V_{CB}$ terms neglected; $I_C$ and $I_B=I_E-I_C$ versus $V_{EB}$. Approximate E-B depletion R-G by adding $I_{R\text{-}G}=I_2(e^{q{V}_{EB}/n_{2}kT}-1)$ to the base current, with $I_2=10^{-14}$ A, $n_2=1.5$. Approximate high-level injection etc. by multiplying the ideal $I_C$ by $\Gamma =1/[1+e^{q(V_{EB}-0.75)/2kT}]$. Plot ideal (dashed) and corrected (solid). Limit to $10^{-9}\le I\le 10^{-1}$ A and $0.3\le V_{EB}\le 0.8$ V. (b) Add a ${\beta }_{dc}$ versus $I_C$ plot (Fig. 11.16 style) with pause between plots.

Solution (logic). In active mode with $V_{CB}$ terms dropped, (11.47) gives $I_E=I_{F0}(e^{q{V}_{EB}/kT}-1)$, $I_C={\alpha }_F{I}_{F0}(e^{q{V}_{EB}/kT}-1)$, so $I_B=(1-{\alpha }_F)I_{F0}(e^{q{V}_{EB}/kT}-1)$ (ideal). Corrected: $I_B^{\prime }=I_B+I_2(e^{q{V}_{EB}/n_{2}kT}-1)$ and $I_C^{\prime }=\Gamma I_C$. Then ${\beta }_{dc}=I_C/I_B$. The R-G term has a slope $q/(1.5kT)$, dominating at low $V_{EB}$ where it exceeds the ideal $I_B$; $\Gamma \approx 1$ for $V_{EB}\ll 0.75$ V and $\Gamma \to e^{-q(V_{EB}-0.75)/2kT}$ above, producing the slope-over of $I_C$.

BJT0                                          % constants and parameters
VEB=0.3:0.005:0.8;
IC0=aF*IF0*(exp(VEB/kT)-1);                   % ideal collector current
IB0=(1-aF)*IF0*(exp(VEB/kT)-1);               % ideal base current
I2=1e-14; n2=1.5;
IBc=IB0+I2*(exp(VEB/(n2*kT))-1);              % with E-B R-G current
Gam=1./(1+exp((VEB-0.75)/(2*kT)));            % high-level-injection factor
ICc=IC0.*Gam;                                 % corrected collector current
figure(1)
semilogy(VEB,IC0,'--b',VEB,IB0,'--r',VEB,ICc,'-b',VEB,IBc,'-r'); grid on
axis([0.3 0.8 1e-9 1e-1]); xlabel('V_{EB} (V)'); ylabel('I_C, I_B (A)')
legend('I_C ideal','I_B ideal','I_C corrected','I_B corrected')
pause
figure(2)
loglog(IC0,IC0./IB0,'--',ICc,ICc./IBc,'-'); grid on
axis([1e-9 1e-1 1 1e3]); xlabel('I_C (A)'); ylabel('\beta_{dc}')

(The ${\beta }_{dc}$ axis range printed in the book is garbled in the OCR; $1\le {\beta }_{dc}\le 10^3$ is used here.)

Expected results. The ideal plot is two parallel straight lines (60 mV/decade), separated by ${\beta }_{dc}\approx 178$ (for these BJT0 parameters ${\alpha }_F=0.9944$). The corrected plot: at low $V_{EB}$ the corrected $I_B$ is above the ideal line (R-G, slope 1.5 times more gradual), so $I_C/I_B$ falls; at high $V_{EB}$ the corrected $I_C$ rolls over below the line. The gain plot is therefore a hump: rising with $I_C$ at low currents, flat near 178 over the middle decades, and falling at high currents, as in Fig. 11.16.

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Problem 11.21

Problem. Increasing $V_{EC}$ in measuring ${\beta }_{dc}$ versus $I_C$ invariably leads to a small but perceptible increase in ${\beta }_{dc}$ over most of the $I_C$ range (Fig. P11.21: 2N2605 data at $V_{EC}=5,10,15$ V). Formulate and support an explanation.

Solution. The cause is base width modulation. Raising $V_{EC}$ at fixed $I_C$ increases the reverse bias across the C-B junction, widens its depletion region and narrows the quasineutral base $W$. Evidence from the equations:

$${\beta }_{dc}=\frac{1}{\frac{D_E}{D_B}\frac{N_B}{N_E}\frac{W}{L_E}+\frac{1}{2}{\left(\frac{W}{L_B}\right)}^2}$$

Both loss terms decrease as $W$ falls (the first linearly, the second quadratically), so ${\beta }_{dc}$ increases. In terms of the base-current components: $I_{B1}=I_{En}$ is the back-injection current; at fixed $I_C\propto e^{q{V}_{EB}/kT}/W$, a smaller $W$ requires a smaller $V_{EB}$ (the same $I_C$ arises from a lower E-B injection level), so $I_{En}\propto e^{q{V}_{EB}/kT}$ falls in proportion to $W$; and $I_{B2}$ (base recombination) falls as $W^2$. The same smaller $V_{EB}$ also reduces the E-B depletion-region R-G current $\propto e^{q{V}_{EB}/2kT}$, which governs the low-current gain drop, so the increase in ${\beta }_{dc}$ with $V_{EC}$ extends across the low-current range as well. The effect is small because the lightly doped collector absorbs most of the C-B depletion region (Section 11.2.2), i.e. the Early voltage is large (about 100 V or more), predicting a few-percent gain change per 5 V step, consistent with the modest separation of the three curves in Fig. P11.21. At the highest currents where high-level injection and series resistance dominate, the curves merge because those mechanisms do not depend strongly on $W$.