By the end of this chapter you should be able to:
- Describe the diffusion of majority carriers at the metallurgical junction and explain why it stops.
- Explain how uncovered dopant ions form the space-charge region and why the mobile carriers are absent from it.
- Derive \(V_{bi} = (kT/q)\ln(N_A N_D/n_i^2)\) from the equilibrium balance of drift and diffusion, and evaluate it.
- Draw the equilibrium band diagram and relate the band bending to \(qV_{bi}\) and to the single flat Fermi level.
- Sketch and justify the charge density, electric field and potential profiles through the depletion region.
- Compute the depletion width and its division between the two sides for unequal doping, and identify a one-sided junction.
- Derive the junction capacitance, explain its bias dependence, and use a \(1/C^2\) plot to extract \(V_{bi}\) and the doping.
Everything so far has concerned a uniform piece of semiconductor. Nothing built out of uniform material does anything interesting: a doped bar is a resistor, and the four chapters of Part 1 would amount to an elaborate account of a resistor if the story stopped here. What makes a device is a boundary — a place where the doping changes — and the simplest such boundary, one p-type region meeting one n-type region, is the p-n junction that underlies the diode, the bipolar transistor, the solar cell, the LED and the parasitic isolation of every integrated circuit.
This chapter treats that junction with no applied voltage. That may sound like the least interesting case, and it is the most important one: everything a biased junction does is a departure from the equilibrium state, and you cannot describe a departure without first describing what it departs from. Three quantities are established here and used constantly afterwards — the built-in potential \(V_{bi}\), which sets the forward voltage a diode needs and the turn-on of every transistor; the depletion width \(W\), which sets the breakdown voltage and the capacitance; and the peak field, which decides when avalanche occurs. The apparent paradox to be resolved is that a junction in equilibrium contains an electric field of tens of thousands of volts per centimetre and a potential difference of three-quarters of a volt, and yet carries no current and can light no lamp.
1 What Happens When the Two Halves Meet
Imagine, as a thought experiment, joining a bar of p-type silicon doped at \(N_A = 10^{17}\ \text{cm}^{-3}\) to a bar of n-type silicon doped at \(N_D = 10^{16}\ \text{cm}^{-3}\), so that the crystal lattice runs continuously through the plane where they meet. That plane is the metallurgical junction. In reality the two regions are made by diffusing or implanting one dopant into a wafer already doped with the other, and the transition takes place over a fraction of a micrometre; the idealisation of an abrupt step is close enough for everything in this chapter.
Consider the carrier concentrations either side of that plane, taken from Chapter 2 with \(n_i = 1.0\times10^{10}\ \text{cm}^{-3}\):
| p-side (\(N_A = 10^{17}\)) | n-side (\(N_D = 10^{16}\)) | Ratio across the junction | |
|---|---|---|---|
| Holes | \(1.0\times10^{17}\ \text{cm}^{-3}\) | \(1.0\times10^{4}\ \text{cm}^{-3}\) | \(10^{13}\) |
| Electrons | \(1.0\times10^{3}\ \text{cm}^{-3}\) | \(1.0\times10^{16}\ \text{cm}^{-3}\) | \(10^{13}\) |
A concentration step of thirteen orders of magnitude across a plane of zero thickness is the steepest gradient in this course, and Chapter 4 established what a gradient does: it drives a diffusion current. Holes pour from left to right and electrons from right to left, both of them majority carriers leaving the region where they are plentiful for the region where they are not. The initial diffusion current is enormous.
It does not last. Follow one electron that crosses from the n-side into the p-side. It leaves behind, in the n-material, the donor atom that supplied it — and that donor is now an uncompensated positive ion, fixed in the lattice and unable to follow. Meanwhile the electron arrives in p-material full of holes and recombines within a fraction of its lifetime, so it disappears as a carrier but its arrival has neutralised an acceptor, leaving that acceptor as an uncompensated negative ion. Exactly the same happens in mirror image to a diffusing hole.
The result is a thin layer either side of the junction from which the mobile carriers have gone, leaving only the fixed ionised dopants: negative acceptors on the p-side and positive donors on the n-side. This layer is the depletion region or space-charge region. It is depleted of mobile carriers, not of charge; on the contrary it holds a great deal of charge, and the whole behaviour of the junction follows from that charge being immobile.
Separated positive and negative charge means an electric field, directed from the positive donor ions on the n-side towards the negative acceptor ions on the p-side — that is, from n to p. Now notice which way that field pushes the diffusing carriers. A hole diffusing from p to n moves into a region of field pointing back towards it, so the field opposes its motion. An electron diffusing from n to p is likewise pushed back. The field created by the diffusion opposes the diffusion. Diffusion therefore builds its own brake, and the process is self-limiting: as more carriers cross, the depletion region widens, the field grows, and the drift current it drives grows until it exactly cancels the diffusion current.
2 Equilibrium: Two Large Currents That Cancel
Equilibrium is reached in a time of order the dielectric relaxation time, which for silicon at \(10^{16}\ \text{cm}^{-3}\) is \(\varepsilon_s/\sigma = 1.04\times10^{-12}/2.0 = 5\times10^{-13}\) s. The junction settles in half a picosecond and then does nothing further until it is disturbed. The state it settles into is not one in which the carriers have stopped moving; it is one in which four large currents cancel in pairs:
The balance holds separately for each carrier type, not merely for the total. This is a stronger statement than "no net current flows", and it is what makes the derivation of \(V_{bi}\) in the next section possible.
The cancellation must be separate for each carrier because of the principle of detailed balance: in true thermal equilibrium every microscopic process is balanced by its exact reverse. If hole diffusion were balanced by electron drift, the junction would be steadily converting one carrier type into the other and generating heat, which an isolated system in equilibrium cannot do.
It is worth being clear about which carriers make up each of these currents, because it explains a great deal about how a diode behaves under bias. The diffusion current is carried by majority carriers — the enormous population of holes on the p-side, of which the small fraction with enough energy to climb the barrier crosses. The drift current is carried by minority carriers — the sparse electrons generated thermally on the p-side, which wander into the depletion region and are swept down the field to the n-side. Both currents are of the order of hundreds of amperes per square centimetre in a typical junction, and they cancel to better than one part in \(10^{20}\).
The drift current depends on the number of minority carriers arriving at the depletion edge and not at all on the height of the barrier — every minority carrier that arrives is swept across, whatever the field. The diffusion current depends exponentially on the barrier, through the Boltzmann factor \(e^{-qV_{bi}/kT}\). Lower the barrier by forward bias and the diffusion current rises exponentially while the drift current does not move: the result is the Shockley equation of Chapter 6. Raise the barrier by reverse bias and the diffusion current dies away, leaving only the small, bias-independent drift current: the reverse saturation current.
The last thing to establish is why no current flows in the external circuit, which students often find harder to accept than the internal balance. Connect a wire from one end of the device to the other and you have made two more junctions — metal to p-silicon at one end, metal to n-silicon at the other — and each has a contact potential of its own. Kirchhoff's voltage law applied round the loop requires the sum of all three to be zero, and it is: the contact potentials of the two metal junctions exactly cancel the junction's \(V_{bi}\). Any other outcome would be a device that circulated a current in a closed loop of one temperature for ever, which is forbidden by the second law of thermodynamics. A voltmeter across a diode reads zero, and it should.
3 Deriving the Built-In Potential
The potential difference across the depletion region — the n-side positive with respect to the p-side — is the built-in potential, or contact potential, \(V_{bi}\). It can be obtained in two ways, and both are worth having.
Route one: balance drift against diffusion. Take the hole current and set it to zero:
Write \(\mathcal{E} = -d\phi/dx\) and use the Einstein relation \(D_p/\mu_p = kT/q\):
Integrating from deep in the neutral p-region, where \(\phi = 0\) and \(p = p_p = N_A\), to deep in the neutral n-region, where \(\phi = V_{bi}\) and \(p = p_n = n_i^2/N_D\):
The central result of the chapter. Repeating the calculation with the electron current gives the identical answer, as it must; the two carrier balances are not independent.
Route two: use the Fermi level. Chapter 3 gave \(E_F - E_i = kT\ln(N_D/n_i)\) on the n-side and \(E_i - E_F = kT\ln(N_A/n_i)\) on the p-side. In equilibrium a single Fermi level runs through the whole device, so the intrinsic level — and with it the whole band structure — must shift between the two sides by the sum of those two displacements:
The same expression from a completely different starting point. Route one is the physics; route two is the bookkeeping, and it is the faster one to use in an examination.
Silicon at 300 K with \(N_A = 10^{17}\ \text{cm}^{-3}\), \(N_D = 10^{16}\ \text{cm}^{-3}\), \(n_i = 1.0\times10^{10}\ \text{cm}^{-3}\), \(kT/q = 0.025852\) V.
Three checks and comparisons are worth making.
Sensitivity to doping. Raising both dopings tenfold adds \(2kT\ln 10 = 0.119\) V, giving 0.893 V. The logarithm keeps \(V_{bi}\) between about 0.6 and 0.9 V for every practical silicon doping, which is why the "0.7 V diode drop" is such a durable rule of thumb.
Sensitivity to the material. Using the same dopings with germanium, where \(n_i = 2.4\times10^{13}\ \text{cm}^{-3}\), gives \(V_{bi} = 0.371\) V, and with gallium arsenide, where \(n_i = 2.1\times10^{6}\), gives 1.212 V. The observed forward drops of germanium, silicon and GaAs diodes — roughly 0.3, 0.7 and 1.2 V — are these numbers.
Sensitivity to \(n_i\). Using the older figure \(n_i = 1.5\times10^{10}\) instead gives 0.753 V, 21 mV lower. Since \(n_i\) enters squared, a factor of 1.5 in \(n_i\) costs \(2kT\ln1.5 = 21\) mV. Always state which \(n_i\) you used.
Temperature. Because \(n_i^2\) grows exponentially, \(V_{bi}\) falls as the junction is heated. Recomputing at 400 K, where \(n_i = 5.3\times10^{12}\ \text{cm}^{-3}\), gives \(V_{bi} = 0.600\) V, an average slope of \(-1.7\) mV/K; a symmetrically doped junction gives closer to \(-2.0\) mV/K over a wider range. This is the origin of the familiar \(-2\) mV/K coefficient of a forward-biased diode, used as a temperature sensor and treated as a nuisance in every bias circuit in Part 4.
4 The Junction in the Band Diagram
The single most useful picture of a junction is its band diagram, because one rule — the Fermi level is flat in equilibrium — forces everything else.
Before contact, the n-material has \(E_F\) 0.205 eV below its conduction band and the p-material has \(E_F\) 0.120 eV above its valence band. Those are different absolute energies. On contact, electrons flow from the material with the higher Fermi level to the one with the lower until the two are equal, which is exactly the diffusion described above seen from the energy side. Once equilibrium is reached, \(E_F\) is one horizontal line right across the device.
Since \(E_C - E_F\) is fixed at 0.205 eV wherever the material is neutral n-type, and \(E_F - E_V\) is fixed at 0.120 eV wherever it is neutral p-type, the only way to accommodate a flat \(E_F\) is for the band edges themselves to move. They are flat in each neutral region and bend through the depletion region, and the total bend is
For the example junction, \(1.12 - 0.205 - 0.120 = 0.795\) eV, against 0.774 eV from the doping formula. The 21 meV difference is the same \(n_i\) bookkeeping discrepancy noted in Chapter 3, and it is below the accuracy of anything else here.
Reading the diagram takes a little practice, and two conventions cause most of the trouble. The vertical axis is electron energy, so the bands are high where the potential is low: the p-side, being at the lower electrostatic potential, has its bands drawn higher. And the local electrostatic potential is \(\phi = -(E_i - E_{i,\text{ref}})/q\), so the potential profile of the next section is the band diagram turned upside down and scaled by \(q\). Once that is internalised, a band diagram becomes the fastest way to read off what a device is doing.
The barrier has a direct kinetic interpretation. An electron in the n-side conduction band that wants to reach the p-side must climb \(qV_{bi} = 0.774\) eV, and the fraction of the electron population with that much energy is, by Boltzmann, \(e^{-0.774/0.02585} = 1.0\times10^{-13}\). Multiply that fraction by the n-side electron concentration of \(10^{16}\ \text{cm}^{-3}\) and you get \(10^{3}\ \text{cm}^{-3}\) — precisely the equilibrium electron concentration on the p-side, as it must be. The barrier does not stop the diffusion; it reduces it by a factor of \(10^{13}\), to exactly the value the reverse drift current can cancel.
Finally, note the slope of the bands inside the depletion region. A sloping band edge means an electric field, since \(\mathcal{E} = (1/q)\,dE_C/dx\), and the slope is steepest at the metallurgical junction, which is where the field is largest. The band diagram already contains the field profile of Figure 5.3; the next section merely draws it separately.
5 Charge, Field and Potential Through the Region
The profiles follow from Poisson's equation and two integrations. The standard simplification is the depletion approximation: assume the space-charge region is completely free of mobile carriers and that the neutral regions are completely free of space charge, with an abrupt boundary between them. The real transition is smeared over a few Debye lengths, about 40 nm here, but the approximation gives depletion widths within a few per cent and it makes the problem solvable by hand.
Charge density. Inside the depletion region, \(\rho = -qN_A\) for \(-x_p < x < 0\) and \(\rho = +qN_D\) for \(0 < x < x_n\); everywhere else \(\rho = 0\). Because the device as a whole is neutral, the two blocks must contain equal and opposite charge:
The depletion region extends further into the more lightly doped side, in inverse proportion to the doping. This one line explains most of what is peculiar about real junctions.
Electric field. Poisson's equation in one dimension is \(d\mathcal{E}/dx = \rho/\varepsilon_s\), with \(\varepsilon_s = 11.7\varepsilon_0 = 1.036\times10^{-12}\) F/cm for silicon. Integrating from \(x = -x_p\), where the field must be zero because the neutral region beyond has none:
Both expressions give the same value at \(x = 0\) because \(N_A x_p = N_D x_n\), so the field is continuous, as it must be where the permittivity does not change. It is triangular, zero at both edges and peaking at the metallurgical junction:
Potential. Integrating the field once more, \(\phi(x) = -\int \mathcal{E}\,dx\), gives two parabolic arcs joined at the junction, rising from 0 in the neutral p-region to \(V_{bi}\) in the neutral n-region. Because \(V_{bi}\) is the area under a triangle of height \(|\mathcal{E}_{\max}|\) and base \(W = x_p + x_n\):
A relation worth remembering: it converts between the two quantities without any doping arithmetic, and it is how the breakdown voltage of Chapter 12 is estimated from the critical field.
Reading Figure 5.3 from the top down, three features deserve comment. The charge blocks look wildly different in shape but enclose equal areas, which is the neutrality condition drawn. The field triangle is not symmetric — its two sides have slopes \(qN_A/\varepsilon_s\) and \(qN_D/\varepsilon_s\), differing by a factor of ten — and yet it reaches the same peak from both directions. And the potential rise is grossly unequal: for this junction 0.070 V (9 per cent) falls across the heavily doped p-side and 0.704 V (91 per cent) across the lightly doped n-side. Almost all the voltage appears across the lightly doped material, which is why the lightly doped side of a rectifier determines its breakdown voltage and why a power diode has a thick, nearly intrinsic drift region.
6 Depletion Width and Doping Asymmetry
Combining the two results just obtained — \(V_{bi} = \tfrac{1}{2}|\mathcal{E}_{\max}|W\) and \(|\mathcal{E}_{\max}| = qN_Ax_p/\varepsilon_s\) — with the neutrality condition \(N_Ax_p = N_Dx_n\) and \(W = x_p + x_n\), and eliminating \(x_p\), \(x_n\) and \(\mathcal{E}_{\max}\), gives the depletion width:
and its division between the sides follows from neutrality, \(x_n = W\dfrac{N_A}{N_A+N_D}\), \(x_p = W\dfrac{N_D}{N_A+N_D}\). Under bias, \(V_{bi}\) is replaced by \(V_{bi} - V_a\), with \(V_a\) positive for forward bias.
For the junction of Example 5.1 (\(N_A = 10^{17}\), \(N_D = 10^{16}\ \text{cm}^{-3}\), \(V_{bi} = 0.774\) V, \(\varepsilon_s = 1.036\times10^{-12}\) F/cm):
The division is \(x_n = 0.332 \times 10^{17}/1.1\times10^{17} = 0.302\ \mu\text{m}\) and \(x_p = 0.0302\ \mu\text{m}\) — ninety-one per cent of the depletion region lies in the lightly doped n-side, exactly the inverse of the doping ratio. The peak field is
Check against \(V_{bi} = \tfrac{1}{2}|\mathcal{E}_{\max}|W = 0.5 \times 4.66\times10^{4} \times 3.32\times10^{-5} = 0.774\) V. Forty-seven thousand volts per centimetre exists across an unbiased diode sitting in a drawer, in a region a third of a micrometre wide. It is nowhere near the \(3\times10^{5}\) V/cm at which silicon avalanches, which is why the junction is stable, but it is a reminder of how large the internal fields of a device are.
For an area of \(10^{-4}\ \text{cm}^{2}\) the uncovered charge on each side is \(Q = qN_Dx_nA = 1.602\times10^{-19}\times10^{16}\times3.02\times10^{-5}\times10^{-4} = 4.83\times10^{-12}\) C, or 4.83 pC. That charge is what has to be supplied or removed when the bias changes, and it is the subject of the last section.
When one side is doped very much more heavily than the other, the expression simplifies usefully. If \(N_A \gg N_D\) then \(1/N_A\) is negligible and
A one-sided or \(p^+n\) junction. The depletion region lies almost entirely in the lightly doped side, and the heavily doped side may be treated as a perfect conductor. Nearly every real junction is designed this way.
The reason is that it puts one parameter in charge. For \(N_A = 10^{19}\) and \(N_D = 10^{16}\ \text{cm}^{-3}\), \(x_p\) is 0.3 nm — about one lattice constant, which incidentally is a warning that the depletion approximation is being stretched — while \(x_n\) is 0.340 µm. The breakdown voltage, the capacitance and the series resistance are then all set by \(N_D\) alone, and the designer has one knob rather than two. Doping the lightly doped side more lightly widens the depletion region, raising the breakdown voltage and lowering the capacitance, at the cost of a higher series resistance; that single trade-off runs through the whole of Parts 2 and 3.
7 Junction Capacitance
A layer of fixed positive charge separated from a layer of fixed negative charge by a region containing no mobile carriers is a parallel-plate capacitor, with the neutral regions as plates and the depletion region as dielectric. Changing the bias changes \(W\), which changes how much charge is uncovered, so the junction stores charge and has a capacitance. This is the junction, transition or depletion capacitance, \(C_j\).
The parallel-plate formula applies directly, with \(W\) as the separation:
Because \(W \propto \sqrt{V_{bi}-V_a}\), the capacitance varies as \((V_{bi}-V_a)^{-1/2}\). Reverse bias makes \(V_a\) negative, widening \(W\) and reducing \(C_j\).
A subtlety of definition matters here. Capacitance is properly \(C = dQ/dV\), not \(Q/V\), because the relation between charge and voltage is not linear. Carrying out the differentiation on \(Q = qN_Dx_nA\) with \(x_n \propto \sqrt{V_{bi}-V_a}\) gives exactly \(\varepsilon_sA/W\), so the simple formula turns out to be right — but only because it was derived as a small-signal quantity. A junction is a non-linear capacitor, and quoting a single value for it without stating the bias is meaningless.
For the junction of Example 5.2, with \(A = 10^{-4}\ \text{cm}^{2}\):
| \(V_a\) | \(V_{bi}-V_a\) | \(W\) | \(|\mathcal{E}_{\max}|\) | \(C_j\) |
|---|---|---|---|---|
| 0 | 0.774 V | 0.332 µm | \(4.66\times10^{4}\) V/cm | 3.12 pF |
| –1 V | 1.774 V | 0.502 µm | \(7.06\times10^{4}\) V/cm | 2.06 pF |
| –5 V | 5.774 V | 0.906 µm | \(1.27\times10^{5}\) V/cm | 1.14 pF |
| –10 V | 10.774 V | 1.238 µm | \(1.74\times10^{5}\) V/cm | 0.837 pF |
| –20 V | 20.774 V | 1.719 µm | \(2.42\times10^{5}\) V/cm | 0.603 pF |
Twenty volts of reverse bias reduces the capacitance by a factor of 5.2 — a 5:1 tuning range from one terminal. This is the varactor diode of Chapter 13, used to tune the oscillator of a radio receiver. Note also the last column of fields: at \(-20\) V the peak field is \(2.4\times10^{5}\) V/cm, approaching the \(3\times10^{5}\) V/cm at which avalanche multiplication begins, so this junction would break down at about 30 V. Capacitance and breakdown voltage are two readings of the same depletion width.
The most useful consequence is a measurement technique. Squaring and inverting the capacitance expression gives
A straight line in \(V_a\). Its intercept on the voltage axis is \(V_{bi}\) and its slope gives the effective doping \((1/N_A + 1/N_D)^{-1}\).
Figure 5.4 shows the plot for the example junction. The measured slope is \(1.33\times10^{23}\ \text{F}^{-2}\)/V, which gives an effective doping of \(9.1\times10^{15}\ \text{cm}^{-3}\); for this junction \((1/10^{17}+1/10^{16})^{-1} = 9.09\times10^{15}\), confirming that a one-sided junction returns the lightly doped side almost exactly. The extrapolated intercept is \(+0.774\) V, the built-in potential. Two quantities that cannot be probed directly — a potential inside a solid and a doping concentration — are recovered from a capacitance meter and a ruler.
Two warnings. Forward bias does not simply continue the curve: as \(V_a\) approaches \(V_{bi}\) the formula predicts an infinite capacitance, which is nonsense because the depletion approximation fails once the barrier is small enough for mobile carriers to flood back into the region. In practice \(C_j\) is taken as \(2C_{j0}\) beyond \(V_a = V_{bi}/2\). And in forward bias a second, much larger capacitance appears — the diffusion capacitance of the stored minority charge — which swamps \(C_j\) entirely and is the subject of Chapter 6.
8 Summary and Key Results
| Quantity | Expression | Value for the worked junction |
|---|---|---|
| Why diffusion stops | Uncovered ions create a field that opposes further diffusion | Self-limiting; equilibrium reached in about 0.5 ps |
| Equilibrium condition | \(J_{p,\text{diff}} + J_{p,\text{drift}} = 0\), and likewise for electrons | Two large currents cancel separately for each carrier type |
| Built-in potential | \(V_{bi} = \dfrac{kT}{q}\ln\dfrac{N_AN_D}{n_i^2}\) | 0.774 V; Ge 0.371 V, GaAs 1.212 V at the same doping |
| Temperature coefficient | \(V_{bi}\) falls as \(n_i^2\) rises with \(T\) | 0.600 V at 400 K, an average –1.7 mV/K; the source of the diode's –2 mV/K drift |
| Band bending | \(qV_{bi} = E_g - (E_C-E_F)_n - (E_F-E_V)_p\) | \(1.12 - 0.205 - 0.120 = 0.795\) eV; one flat \(E_F\) across the device |
| Charge neutrality | \(N_A x_p = N_D x_n\) | \(x_p = 0.030\) µm, \(x_n = 0.302\) µm — 91 % in the lighter side |
| Depletion width | \(W = \sqrt{\dfrac{2\varepsilon_s(V_{bi}-V_a)}{q}\left(\dfrac{1}{N_A}+\dfrac{1}{N_D}\right)}\) | 0.332 µm at zero bias; 0.906 µm at –5 V |
| Peak field | \(|\mathcal{E}_{\max}| = qN_Dx_n/\varepsilon_s = 2V_{bi}/W\) | \(4.66\times10^{4}\) V/cm at zero bias; \(2.42\times10^{5}\) V/cm at –20 V |
| One-sided junction | \(W \approx \sqrt{2\varepsilon_s(V_{bi}-V_a)/qN_D}\) when \(N_A \gg N_D\) | The lightly doped side alone sets width, capacitance and breakdown |
| Junction capacitance | \(C_j = \varepsilon_s A/W \propto (V_{bi}-V_a)^{-1/2}\) | 3.12 pF at 0 V falling to 0.60 pF at –20 V for \(A = 10^{-4}\) cm\(^2\) |
| \(1/C^2\) extraction | \(1/C_j^2\) is linear in \(V_a\) | Intercept gives \(V_{bi} = 0.774\) V; slope gives \(N_{\text{eff}} = 9.1\times10^{15}\) cm\(^{-3}\) |
9 Common Mistakes
Connecting probes to the diode creates two metal-semiconductor contacts, each with its own contact potential, and around the loop those two exactly cancel \(V_{bi}\). The reading is zero, and it must be: a device that maintained a voltage round a closed isothermal loop would be a perpetual motion machine of the second kind. \(V_{bi}\) is a barrier that carriers must climb, not a source that can drive a load. Energy comes out of a junction only when something — a bias supply, or light — disturbs the equilibrium.
Neutrality requires \(N_Ax_p = N_Dx_n\), so the region extends further into the lightly doped side, in inverse proportion to the doping. With \(N_A = 10^{17}\) and \(N_D = 10^{16}\), ninety-one per cent of the width and ninety-one per cent of the potential drop are on the n-side. A symmetric sketch will lead you to attribute breakdown to the wrong side of the device and to compute the wrong capacitance. The only junction with a symmetric depletion region is one with equal doping on both sides.
\(C_j\) varies as \((V_{bi}-V_a)^{-1/2}\), so a diode specified at 3 pF is 3 pF at one stated bias and nothing like it elsewhere — that variation is the whole point of a varactor. In forward bias the expression is worse than inaccurate: it predicts infinity at \(V_a = V_{bi}\), because the depletion approximation collapses once mobile carriers return to the region. Beyond \(V_a = V_{bi}/2\), take \(C_j \approx 2C_{j0}\) and remember that the diffusion capacitance of Chapter 6, which can reach nanofarads, dominates completely.
10 Chapter Review
1. A silicon junction at 300 K has \(N_A = 5\times10^{17}\) and \(N_D = 2\times10^{16}\) cm\(^{-3}\). Find \(V_{bi}\), \(W\), \(x_p\), \(x_n\) and \(|\mathcal{E}_{\max}|\). Take \(n_i = 10^{10}\) cm\(^{-3}\), \(\varepsilon_s = 1.036\times10^{-12}\) F/cm.
\(V_{bi} = 0.025852\ln[(5\times10^{17})(2\times10^{16})/10^{20}] = 0.025852\ln(10^{14}) = 0.025852 \times 32.24 = 0.834\) V. Next, \(1/N_A + 1/N_D = 2\times10^{-18} + 5\times10^{-17} = 5.2\times10^{-17}\) cm\(^3\), so \(W = \sqrt{2(1.036\times10^{-12})(0.834)(5.2\times10^{-17})/(1.602\times10^{-19})} = \sqrt{5.61\times10^{-10}} = 2.37\times10^{-5}\) cm = 0.237 µm. The split is \(x_n = W N_A/(N_A+N_D) = 0.237 \times 5\times10^{17}/5.2\times10^{17} = 0.228\) µm and \(x_p = 0.0091\) µm. The peak field is \(|\mathcal{E}_{\max}| = qN_Dx_n/\varepsilon_s = (1.602\times10^{-19})(2\times10^{16})(2.276\times10^{-5})/(1.036\times10^{-12}) = 7.04\times10^{4}\) V/cm; check with \(2V_{bi}/W = 2(0.8334)/2.367\times10^{-5} = 7.04\times10^{4}\) V/cm.
2. Explain why no current flows through a junction in equilibrium even though it contains a field of \(5\times10^{4}\) V/cm.
The field does drive a drift current, and a large one: minority carriers reaching either edge of the depletion region are swept across it. But the same field is the consequence of a diffusion that has already taken place, and the barrier it represents suppresses the majority-carrier diffusion current until the two are exactly equal and opposite. The balance holds separately for holes and for electrons, as detailed balance requires, so no net current of either species crosses the junction. The physical picture is of two currents of hundreds of amperes per square centimetre cancelling to better than one part in \(10^{20}\), not of stationary carriers. Nothing is different in kind from a beaker of water at rest, in which enormous numbers of molecules cross any plane every second in both directions.
3. Two junctions are made with the same total doping: junction A has \(N_A = N_D = 10^{17}\) cm\(^{-3}\); junction B has \(N_A = 10^{19}\), \(N_D = 10^{15}\) cm\(^{-3}\). Compare their built-in potentials and depletion widths.
The built-in potentials depend on the product of the dopings, which is \(10^{34}\) in both cases, so both give \(V_{bi} = 0.025852\ln(10^{34}/10^{20}) = 0.025852 \times 32.24 = 0.834\) V — identical. The widths depend on the harmonic combination \((1/N_A + 1/N_D)\), which is very different: junction A gives \(2\times10^{-17}\) and junction B gives \(1.0001\times10^{-15}\), a factor of 50. Hence \(W_A = \sqrt{2(1.036\times10^{-12})(0.834)(2\times10^{-17})/1.602\times10^{-19}} = 0.147\) µm and \(W_B = 1.038\) µm, seven times wider. In A the region is shared equally; in B, 99.99 per cent of it lies in the \(10^{15}\) side. Junction B will have far the lower capacitance and far the higher breakdown voltage, and it is the design a rectifier uses; junction A would have lower series resistance. The product fixes the barrier, the harmonic mean fixes the geometry.
4. Sketch \(\rho(x)\), \(\mathcal{E}(x)\) and \(\phi(x)\) for a \(p^+n\) junction, and say where each of them is largest.
With \(N_A \gg N_D\), the charge density is a very tall, very narrow negative spike on the p-side and a low, wide positive block on the n-side, the two areas equal. The field is triangular with its apex at the metallurgical junction; on the p-side it rises almost vertically over the tiny \(x_p\), and on the n-side it falls linearly over the whole of \(x_n \approx W\), so the triangle is effectively right-angled. The potential is essentially a single parabola rising across the n-side, with a negligible step on the p-side. Each quantity is largest where the one above it changes fastest: \(|\rho|\) is largest on the heavily doped side, \(|\mathcal{E}|\) is largest at \(x = 0\) where the charge changes sign, and \(\phi\) reaches its maximum \(V_{bi}\) at the far edge \(x = x_n\) where the field has fallen back to zero. The practical readings are that avalanche breakdown starts at \(x = 0\), and that nearly all the applied reverse voltage is supported by the lightly doped side.
5. A capacitance-voltage measurement on a \(p^+n\) diode of area \(2\times10^{-4}\) cm\(^2\) gives a straight \(1/C^2\) plot with intercept \(+0.80\) V and slope \(4.0\times10^{22}\) F\(^{-2}\)/V. Find \(V_{bi}\), \(N_D\), and the zero-bias capacitance.
The intercept is \(V_{bi}\) directly, so \(V_{bi} = 0.80\) V. For a one-sided junction the slope is \(2/(q\varepsilon_sA^2N_D)\), giving \(N_D = 2/(q\varepsilon_sA^2 \times \text{slope}) = 2/[(1.602\times10^{-19})(1.036\times10^{-12})(4\times10^{-8})(4.0\times10^{22})] = 2/(2.656\times10^{-16}) = 7.5\times10^{15}\) cm\(^{-3}\). The zero-bias point on the line is \(1/C_{j0}^2 = 4.0\times10^{22} \times 0.80 = 3.2\times10^{22}\) F\(^{-2}\), so \(C_{j0} = 1/\sqrt{3.2\times10^{22}} = 5.6\times10^{-12}\) F = 5.6 pF. As a cross-check, \(W = \varepsilon_sA/C_{j0} = (1.036\times10^{-12})(2\times10^{-4})/5.6\times10^{-12} = 0.37\) µm, which is what \(\sqrt{2\varepsilon_sV_{bi}/qN_D}\) gives for these figures. The straightness of the plot is itself information: it confirms the junction is abrupt and the doping uniform, since a graded profile would curve it.