By the end of this chapter you should be able to:
- Explain how a metal-semiconductor contact forms a rectifying barrier, and relate the barrier height to the metal work function and the electron affinity.
- Account for the Schottky diode's lower forward drop using the Richardson equation, and show that the saving is independent of current.
- Explain why a Schottky diode has no reverse-recovery time, and quantify the switching loss this saves at 100 kHz.
- Use \(C = C_0/(1+V_R/V_{bi})^n\) to compute a varactor's capacitance ratio and the tuning range of a resonant circuit.
- Describe tunnelling conduction in a degenerate junction and identify the negative-resistance region of the tunnel-diode characteristic.
- State the condition under which a negative resistance sustains oscillation and estimate the resistive cut-off frequency.
- Explain the operation of PIN and step-recovery diodes and compute the PIN's RF resistance against bias current.
Every diode in Chapters 6 to 12 has been a p-n junction asked to rectify. Part 3 of this course turns to junctions engineered for something else entirely, and the three devices in this chapter make the point most sharply, because each one takes a property that an ordinary rectifier merely tolerates and makes it the whole purpose of the device.
The Schottky diode replaces one side of the junction with a metal. There is then no p-type region for electrons to be injected into, so there is no minority-carrier charge stored during forward conduction and nothing to be swept out when the diode is switched off. The consequence is a device that turns off in picoseconds rather than microseconds, and which drops 0.3 V rather than 0.7 V while doing it. The varactor takes the junction capacitance that Chapter 5 derived and that every other circuit tries to minimise, and sells it as a voltage-controlled capacitor — the component that tunes almost every radio receiver made since 1970. The tunnel diode dopes both sides so heavily that the junction conducts by tunnelling at forward biases too small for ordinary injection, producing a region in which current falls as voltage rises: a negative resistance, and therefore an amplifier and an oscillator made from a two-terminal device.
Two further devices complete the family. The PIN diode stretches the depletion region into a thick undoped layer to make a current-controlled resistor at radio frequencies, and the step-recovery diode engineers the stored charge that the Schottky abolishes, so that it can be released in a deliberately abrupt snap. What links all five is that each is a junction whose doping profile, and not its circuit, does the interesting work.
1 The Metal-Semiconductor Contact
Put a metal in contact with an n-type semiconductor and something has to happen to the electrons, because the two materials have different work functions. The work function \(\phi_M\) is the energy needed to remove an electron from the metal's Fermi level to vacuum; the semiconductor's equivalent quantity, measured from the conduction-band edge rather than from the Fermi level, is the electron affinity \(\chi\), which for silicon is 4.05 eV.
If \(\phi_M > \phi_S\), electrons in the semiconductor sit at a higher energy than those in the metal, so they flow into the metal until the two Fermi levels align. Exactly as in a p-n junction, the departing electrons leave behind uncompensated positive donor ions, forming a depletion region on the semiconductor side only — the metal, with \(10^{22}\) carriers per cm\(^3\), screens the field within an atomic layer. The bands in the semiconductor bend upward towards the interface, and an electron in the semiconductor now faces a barrier \(qV_{bi}\) if it is to reach the metal.
Figure 13.1 draws the result. The barrier is not the same seen from the two sides, and that asymmetry is the whole of the rectification. From the semiconductor the barrier is \(qV_{bi}\), and forward bias reduces it, so the electron flow from semiconductor to metal rises exponentially. From the metal the barrier is
the Schottky barrier height, and applying a bias does nothing to it at all, because the metal's Fermi level and the interface conduction-band edge both move together with the semiconductor's bulk. The flow from metal to semiconductor is therefore fixed, and it is what appears as the saturation current.
The Schottky-Mott relation \(\phi_B = \phi_M - \chi\) predicts a barrier that depends strongly on the metal. In practice it does not, because surface states at the interface pin the Fermi level; the measured barriers are more compressed than the relation suggests, and always somewhat larger than predicted for the low-work-function metals.
| Metal | \(\phi_M\) (eV) | Predicted \(\phi_B = \phi_M - 4.05\) | Typical measured \(\phi_B\) on n-Si |
|---|---|---|---|
| Titanium | 4.33 | 0.28 eV | 0.50 eV |
| Aluminium | 4.28 | 0.23 eV | 0.72 eV |
| Tungsten | 4.55 | 0.50 eV | 0.67 eV |
| Platinum silicide | 5.65 | 1.60 eV | 0.85 eV |
The depletion region behaves exactly as in a one-sided p-n junction, because all the charge is on the semiconductor side. With \(N_D = 10^{17}\ \text{cm}^{-3}\) and \(V_{bi} = 0.47\) V, Chapter 5's expression gives a zero-bias width of about 79 nm and a junction capacitance of the order of 130 pF for a 0.01 mm\(^2\) contact — larger than a small-signal p-n diode of the same area, which is worth remembering, because a Schottky's speed comes from having no stored charge and not from being a small capacitor.
Two consequences of that table are worth noting. Barrier height is a design variable — a low-barrier device has a low forward drop and a large leakage, a high-barrier device the reverse, and manufacturers offer both. And if \(\phi_M < \phi_S\), electrons flow the other way, the bands bend downward, there is no barrier at all, and the contact is ohmic. Every integrated circuit relies on that case for its metal contacts, which is why the same physics that gives a Schottky diode also explains why the aluminium tracks on a chip do not rectify.
2 No Stored Charge, No Recovery
In a forward-biased p-n junction, electrons are injected into the p region and holes into the n region, where each is a minority carrier and must wait a lifetime \(\tau\) — typically microseconds — before recombining. At any instant the neutral regions therefore hold a stored charge \(Q = I\tau\). When the diode is suddenly reverse biased that charge must be removed before the junction can support a reverse voltage, and until it is, the diode conducts backwards freely. That is the reverse recovery of Chapter 6, and its duration \(t_{rr}\) is 30 µs for a 1N4007, 75 ns for a fast-recovery UF4007 and 4 ns for a 1N4148.
A Schottky diode has no p region. Current is carried entirely by electrons crossing the barrier by thermionic emission, and an electron arriving in the metal is instantly a majority carrier among \(10^{22}\) others, indistinguishable and unstored. There is no minority-carrier charge anywhere in the device, so there is nothing to sweep out.
Figure 13.2 puts the two turn-offs side by side, and the loss this saves is easy to quantify. Each turn-off transfers a charge \(Q_{rr}\approx\tfrac12 I_{RM}t_{rr}\) through the diode while the reverse voltage \(V_R\) stands across it, so the power dissipated is \(P = Q_{rr}V_Rf\). At 50 Hz a 1N4007 with \(t_{rr} = 30\) µs and \(I_{RM} = 1\) A loses 75 mW at \(V_R = 100\) V, which is negligible. At 100 kHz the same figure is 150 W, which is absurd — the diode never turns off at all, since 30 µs is three times the whole period. This is why mains rectifiers cannot be used in switching supplies, and it is the quantitative version of the warning at the end of Chapter 11.
The second advantage follows from the same physics. Thermionic emission over a barrier gives a saturation current density governed by the Richardson equation:
with \(A^{*} = 110\ \text{A/cm}^2\text{K}^2\) for n-silicon. At 300 K and \(\phi_B = 0.67\) eV this gives \(J_S = 5.50\times10^{-5}\ \text{A/cm}^2\), against about \(10^{-11}\ \text{A/cm}^2\) for a silicon p-n junction — larger by a factor of \(5.5\times10^{6}\).
Both devices obey \(J = J_S[\exp(V/V_T)-1]\), so the forward voltage needed for a given current density is \(V = V_T\ln(J/J_S)\), and the difference between them is
independent of the current, because the two exponentials are parallel on a semi-log plot. At 100 A/cm\(^2\) the Schottky needs 0.373 V and the p-n junction 0.774 V; at 1 A/cm\(^2\), 0.254 V and 0.655 V. The 0.4 V saving is a constant, and it is the reason Schottky rectifiers dominate low-voltage supplies: on a 5 V rail it recovers 8 % of the output that a silicon diode would waste.
\(J_S\) appears in the forward equation and in the reverse leakage alike, so the six-million-fold advantage in forward drop is a six-million-fold penalty in reverse current. A Schottky rectifier leaks microamperes where a p-n junction leaks nanoamperes, and because \(J_S\propto T^2\exp(-q\phi_B/kT)\) the leakage rises by a factor of 247 between 27 °C and 100 °C. In a hot rectifier this leakage dissipates power, which raises the temperature, which raises the leakage — thermal runaway is a real failure mode for Schottky diodes and is not for silicon ones. It is also why Schottky devices are rarely available above about 200 V: the low barrier and the lightly doped epitaxial layer needed for high voltage together make the leakage unmanageable.
3 The Varactor: Capacitance Under Control
Chapter 5 showed that a reverse-biased junction has a depletion region of width \(W\propto\sqrt{V_{bi}+V_R}\) separating two conducting regions, and that this is a parallel-plate capacitor of value \(C = \varepsilon A/W\). Every diode has this capacitance and in most circuits it is a nuisance. A varactor (or varicap) is a diode optimised so that the capacitance is large, well controlled, low-loss, and above all strongly dependent on the applied voltage:
\(C_0\) is the zero-bias capacitance and \(n\) the grading coefficient, which is fixed by the doping profile: \(n = 1/2\) for an abrupt junction, \(n = 1/3\) for a linearly graded one, and \(n\) between about 1 and 2 for a hyperabrupt profile in which the doping increases towards the junction.
Figure 13.3 plots both cases on a logarithmic scale. The reason \(n\) matters is that a tuned circuit's frequency goes as \(1/\sqrt{LC}\), so a capacitance ratio of \(r\) gives a frequency ratio of only \(\sqrt r\). An abrupt varactor swinging 2.77:1 in capacitance tunes only 1.66:1 in frequency; the hyperabrupt device's 58.8:1 gives 7.67:1. Wideband tuners therefore use hyperabrupt devices even though they are noisier and less linear.
Design a varactor-tuned parallel resonant circuit to cover 88 to 108 MHz, using an abrupt varactor with \(C_0 = 100\) pF, \(V_{bi} = 0.7\) V and \(n = 0.5\). Take the lowest tuning voltage as 2 V, to keep the diode safely reverse biased on signal peaks.
Capacitance ratio required. \(f\propto C^{-1/2}\), so \(C_{88}/C_{108} = (108/88)^2 = 1.5062\).
Capacitance at 2 V. \(C = 100/(1+2/0.7)^{0.5} = 100/1.9640 = 50.92\) pF.
Voltage for the top of the band. Require \((1+V_2/0.7)^{0.5}/(1+2/0.7)^{0.5} = 1.5062\), so \((1+V_2/0.7) = 3.857\times1.5062^2 = 8.747\) and \(V_2 = 5.42\) V. The whole FM band is covered by a tuning voltage of 2.0 to 5.4 V, comfortably within a 9 V supply.
Inductance. \(L = 1/[C(2\pi f)^2] = 1/[50.92\times10^{-12}\times(2\pi\times88\times10^6)^2] = 64.24\) nH. Checking at the other end, \(C = 33.81\) pF gives exactly 108.00 MHz.
The correction nobody remembers. Stray capacitance — the coil's self-capacitance, the transistor's input capacitance, the layout — adds in parallel with the varactor and does not vary with the tuning voltage, so it compresses the tuning range. With just 5 pF of stray, the same design covers only 83.97 to 100.80 MHz: a span of 16.8 MHz instead of 20.0 MHz, and the whole band shifted down. Recovering the top of the band needs the varactor pulled to 28.8 pF, which requires 7.74 V rather than 5.42 V. Every practical varactor tuner therefore has a trimmer, and the design must start by measuring the strays.
A varactor's other figure of merit is its \(Q\). The depletion capacitance is in series with the resistance of the undepleted semiconductor, so
which falls as frequency rises. A device with \(R_S = 1\ \Omega\) tuned to 30 pF has \(Q = 53\) at 100 MHz. Note that \(Q\) improves as the diode is reverse biased further, because \(C_j\) falls, so the top of a tuning range is always the low-loss end.
Two applications beyond tuning are worth naming. Because \(C\) depends on voltage, a varactor is a non-linear reactance, and a non-linear reactance can be used as a parametric amplifier that transfers energy from a high-frequency pump into a signal with almost no added noise — the lowest-noise amplifiers available before cryogenic transistors. And because a varactor in a phase-locked loop's oscillator lets a control voltage set the frequency, it is the tuning element in every voltage-controlled oscillator, a circuit Chapter 29 returns to.
4 The Tunnel Diode and Negative Resistance
Dope both sides of a junction so heavily that the material becomes degenerate — the Fermi level lies inside the conduction band on the n side and inside the valence band on the p side — and the depletion region shrinks to about 10 nm. Chapter 12 showed that such a junction breaks down by tunnelling at a very small reverse voltage. What is new here is what happens in forward bias.
At zero bias, filled states on the n side face filled states on the p side and no net current flows. Apply a small forward bias and the n-side bands rise: filled conduction-band states on the n side now face empty valence-band states on the p side at the same energy, and because the barrier is only nanometres thick, electrons tunnel straight across. The current rises quickly, reaching its peak \(I_P\) when the overlap of filled and empty states is greatest. Increase the bias further and the bands slide past each other: the filled n-side states now face the forbidden gap, where there are no states to tunnel into at all, so the tunnelling current falls. It reaches a minimum \(I_V\) at the valley, after which ordinary thermal injection over the barrier takes over and the characteristic becomes that of a normal diode.
The falling section of Figure 13.4 is a negative differential resistance. Taking the two marked points,
The peak-to-valley ratio \(I_P/I_V\) measures how much of the tunnelling current survives to the valley, and a large ratio means a strong negative resistance. Germanium devices reach about 8, gallium arsenide about 15, and the ratio sets how much gain the diode can provide.
A negative resistance is useful because it cancels a positive one. Put the diode across a parallel \(LC\) tank whose losses are represented by a shunt conductance \(G\); the diode contributes \(-1/R_n\), and if
the total conductance is negative, any disturbance grows rather than decays, and the circuit oscillates at \(f = 1/(2\pi\sqrt{LC})\). Amplitude is limited by the diode leaving the negative-resistance region, so the oscillation settles at whatever amplitude makes the average negative conductance equal to \(G\). With \(L = 5\) nH and \(C = 5\) pF the frequency is 1.01 GHz — from one two-terminal device, a coil and a capacitor.
The upper frequency limit is set by the device itself. The junction capacitance \(C\) shunts the negative resistance and the series resistance \(R_S\) of the leads and bulk material is in the path, so above a resistive cut-off frequency
the input resistance is no longer negative and the device cannot oscillate. With \(R_n = 32\ \Omega\), \(C = 5\) pF and \(R_S = 1.5\ \Omega\), \(f_r = 4.51\) GHz. Tunnel diodes worked to tens of gigahertz in the 1960s, long before any transistor could, which is why they mattered.
Two other uses follow from the same S-shaped curve. Because a load line can cross the characteristic at three points, two of which are stable, a tunnel diode is a bistable element and was used in early high-speed memories and logic. And biased in the negative-resistance region with a resonant load, it is a low-noise microwave amplifier, since the tunnelling process has no transit-time delay at all.
Tunnel diodes are now rare. They are two-terminal devices, so input and output share a port and isolation requires a circulator; the current swing is small, so the power output is a few milliwatts; and the transistors of Parts 4 and 5 caught up and passed them. The device survives in teaching, in a few microwave detectors, and as the direct ancestor of the resonant-tunnelling diodes used in research on terahertz sources.
5 PIN Diodes and Step-Recovery Diodes
Two more devices are built by changing the doping profile rather than the materials, and each is defined by what it does with stored charge — one exploiting it, the other releasing it abruptly.
A PIN diode inserts a thick, nearly undoped intrinsic layer between the p and n regions. Under reverse bias the whole of that layer is depleted, so the capacitance is small and almost independent of voltage, and the device is an excellent open circuit at radio frequencies. Under forward bias, carriers are injected into the I layer from both sides and, because they must live for a lifetime \(\tau\) before recombining, a large stored charge builds up. That charge makes the I layer conductive, and its resistance is
where \(W\) is the I-layer thickness. The device is therefore a current-controlled resistor, and the key point is that at radio frequencies it does not rectify: if the RF period is much shorter than \(\tau\), the stored charge cannot follow the RF cycle, and the diode presents the same resistance on both half-cycles.
With \(W = 50\) µm, \(\mu_n+\mu_p = 1830\ \text{cm}^2/\text{V}\cdot\text{s}\) and \(\tau = 1\) µs:
| Bias current | \(R_S\) | Use |
|---|---|---|
| 0.1 mA | 137 \(\Omega\) | Attenuator, high loss setting |
| 1 mA | 13.7 \(\Omega\) | Mid-range attenuation |
| 10 mA | 1.37 \(\Omega\) | Switch nearly closed |
| 100 mA | 0.14 \(\Omega\) | Switch fully on |
A thousand-to-one resistance range is 60 dB of controllable attenuation, set by a DC current and with no moving parts and no distortion of the RF signal. That is why PIN diodes are the switching element in every antenna changeover relay, transmit-receive switch and electronically variable attenuator in modern radio equipment. The lower frequency limit is \(f \gg 1/(2\pi\tau) = 159\) kHz for a 1 µs lifetime; below that the diode begins to rectify and distort.
A step-recovery diode, or snap-off diode, takes the opposite view of stored charge. Its doping profile is graded so that the field pushes the injected carriers away from the junction and holds them in the bulk. On reverse drive the diode conducts backwards while that charge is withdrawn, exactly as an ordinary rectifier does — and then, when the last of the charge is gone, the current stops not gradually but in a snap of 50 to 200 ps.
An abrupt change of that kind is rich in harmonics. A transition of duration \(t_s\) carries significant energy up to about \(1/(\pi t_s)\), which for \(t_s = 100\) ps is 3.2 GHz, so driving a step-recovery diode at 200 MHz and filtering the output at the tenth harmonic gives a clean 2 GHz signal from a low-frequency source. Comb generators, frequency multipliers and impulse generators for time-domain reflectometry are all built this way, and until recently the step-recovery diode was the standard means of producing picosecond pulses.
All four devices in this chapter are the same junction with the doping profile changed. Degenerate doping on both sides gives the tunnel diode. Heavy doping on one side gives the Zener of Chapter 12. Light, graded or hyperabrupt doping gives the varactor. Replacing one side with metal gives the Schottky, and inserting an undoped layer gives the PIN. Nothing else about the device is different: the same silicon, the same processing steps, the same physics of drift and diffusion from Chapter 4. The lesson generalises to the whole of Parts 4 and 5 — in semiconductor engineering, the doping profile is the design.
6 Choosing Between Them
Because these devices are so different in purpose, the choice is rarely between two of them. What does arise constantly is the choice between a Schottky and an ordinary silicon diode, and it has a clean answer in three questions.
- What is the switching frequency? Above about 20 kHz the reverse-recovery loss of a standard rectifier becomes significant, and above 100 kHz it is disabling. Use a Schottky or a fast-recovery diode.
- What is the reverse voltage? Schottky devices are available to about 200 V, and their leakage worsens rapidly with rating. Above that, an ultrafast p-n diode with a 35 ns recovery is the practical choice.
- How hot will it get? Schottky leakage multiplies by about 250 between 27 °C and 100 °C, and the resulting dissipation can run away thermally. In a hot enclosure with a high reverse voltage, the silicon diode's higher forward drop may be the cheaper loss.
The comparison is worth setting out with numbers, for a 3 A device blocking 40 V.
| Property | Silicon p-n (1N5401) | Schottky (1N5822) |
|---|---|---|
| Forward drop at rated current | 0.95 V | 0.48 V |
| Reverse leakage at 25 °C | 5 µA | 500 µA |
| Reverse recovery \(t_{rr}\) | ≈ 3 µs | none measurable |
| Maximum reverse voltage | 100 V and above | 40 V |
| Conduction loss at 3 A | 2.85 W | 1.44 W |
| Suitable switching frequency | < 1 kHz | > 1 MHz |
The 1.4 W saved in conduction is worth having on its own; on a 5 V output it is the difference between 84 % and 91 % rectifier efficiency. It is also the reason the Schottky appears inside almost every integrated circuit that has to switch quickly: a Schottky clamp across a transistor's base-collector junction prevents the transistor from saturating and therefore from storing charge, which is the trick that made Schottky TTL three times faster than standard TTL.
One further practical point applies to all five devices: none of them is a general-purpose part. A Schottky put where a 400 V mains rectifier belongs will fail immediately; a varactor asked to pass current is simply a poor diode; a PIN diode used below its lifetime-set corner frequency rectifies and distorts; and a tunnel diode biased by a high-resistance source latches instead of oscillating. In every case the failure comes from ignoring the one parameter the device was optimised for. Reading the first page of the datasheet — barrier height and leakage for a Schottky, \(C_0\), \(n\) and \(Q\) for a varactor, \(W\) and \(\tau\) for a PIN — tells you what the device was built to do and therefore where it will fail.
The varactor, tunnel and PIN devices have no rivals within this course, because nothing else does what they do with two terminals. The varactor's competitor is a mechanically variable capacitor, which is larger, more expensive and cannot be tuned by a control voltage; the PIN's competitor is a relay, which is slower by four orders of magnitude and wears out; the tunnel diode's competitor is a transistor, which eventually won.
Part 4 begins the transition to three-terminal devices. Everything in Chapters 5 to 13 has been built on a single junction, and everything that follows is built on two of them placed close enough together that the current in one controls the current in the other. The physics does not change — drift, diffusion, injection, depletion and stored charge all reappear — but the possibility of gain does, and with it every circuit in the second half of this course.
7 Summary and Key Results
| Quantity | Expression | Value |
|---|---|---|
| Schottky barrier height | \(\phi_B = \phi_M - \chi\) (ideal) | 0.50 eV predicted for W on n-Si; 0.67 eV measured |
| Saturation current density | \(J_S = A^{*}T^2e^{-q\phi_B/kT}\) | \(5.50\times10^{-5}\) A/cm\(^2\), against \(10^{-11}\) for p-n |
| Forward-drop saving | \(\Delta V = V_T\ln(J_{S,\text{Sch}}/J_{S,pn})\) | 0.401 V at any current |
| Schottky forward drop | \(V = V_T\ln(J/J_S)\) | 0.373 V at 100 A/cm\(^2\) (p-n: 0.774 V) |
| Reverse recovery | \(t_{rr}\) | 30 µs (1N4007), 75 ns (UF4007), < 100 ps (Schottky) |
| Recovery loss | \(P = Q_{rr}V_Rf\) | 75 mW at 50 Hz; 150 W at 100 kHz for a 1N4007 |
| Schottky leakage vs temperature | \(\propto T^2e^{-q\phi_B/kT}\) | × 247 from 27 °C to 100 °C |
| Varactor capacitance | \(C_j = C_0/(1+V_R/V_{bi})^n\) | 50.92 pF at 2 V, 18.39 pF at 20 V for \(n\) = 0.5 |
| Capacitance ratio | \(n\) = 0.5 abrupt, \(n\) = 2 hyperabrupt | 2.77 and 58.8 over 2 V to 20 V |
| Frequency ratio | \(f\propto C^{-1/2}\) | 1.66 and 7.67 for the same two devices |
| FM tuner design | 88–108 MHz, \(C_0\) = 100 pF | \(L\) = 64.24 nH, \(V_R\) = 2.0 to 5.42 V |
| Varactor \(Q\) | \(Q = 1/(2\pi fC_jR_S)\) | 53 at 100 MHz, 30 pF, \(R_S\) = 1 \(\Omega\) |
| Tunnel-diode slope | \(-R_n = (V_V-V_P)/(I_V-I_P)\) | −32 \(\Omega\); PVR = 7.3 |
| Resistive cut-off | \(f_r = (1/2\pi R_nC)\sqrt{R_n/R_S-1}\) | 4.51 GHz at \(C\) = 5 pF, \(R_S\) = 1.5 \(\Omega\) |
| PIN resistance | \(R_S = W^2/[(\mu_n+\mu_p)\tau I]\) | 137 \(\Omega\) at 0.1 mA to 0.14 \(\Omega\) at 100 mA; 60 dB |
| Step-recovery harmonics | up to \(1/(\pi t_s)\) | 3.2 GHz for a 100 ps snap |
8 Common Mistakes
The speed has nothing to do with size, geometry or capacitance, and a Schottky's junction capacitance is in fact larger than a small-signal p-n diode's for the same area. The speed comes from the absence of a minority-carrier population: there is no p region into which electrons can be injected, so there is no charge \(Q = I\tau\) waiting to be swept out at turn-off. Recognising this makes the leakage penalty predictable rather than surprising, because the same large \(J_S\) that lowers the forward drop by 0.4 V raises the reverse leakage by the same factor of \(5.5\times10^{6}\).
A design that computes a 20 MHz tuning span from the varactor's capacitance ratio alone will not achieve it. Stray capacitance — coil self-capacitance, transistor input capacitance, layout — sits in parallel with the varactor and does not vary with the tuning voltage, so the achievable ratio is \((C_{\max}+C_s)/(C_{\min}+C_s)\), not \(C_{\max}/C_{\min}\). In Worked Example 13.1 just 5 pF of stray narrows the band from 20.0 MHz to 16.8 MHz and shifts it down by 4 MHz. Measure or estimate the strays first, then choose the varactor, then compute \(L\) — and fit a trimmer.
\(-R_n\) is a differential quantity, valid only for small changes about a bias point inside the falling region, and only between \(V_P\) and \(V_V\). The device does not deliver power at DC — its terminal current and voltage are both positive everywhere, so it absorbs DC power like any other diode. What the negative slope does is convert some of that DC power into signal power at the frequency of oscillation, which is exactly what an amplifier does. It also means the bias circuit must present a low enough resistance to intersect the characteristic at one point only; a high-resistance bias source gives three intersections and the circuit latches instead of oscillating.
9 Chapter Review
1. Explain why a Schottky diode has no reverse-recovery time, and compute the switching loss avoided when a 1 A, 100 V, 100 kHz rectifier is changed from a 1N4007 (\(t_{rr}\) = 30 µs) to a UF4007 (\(t_{rr}\) = 75 ns) and then to a Schottky.
In a forward-biased p-n junction, current is carried by minority carriers injected across the junction — electrons into the p side and holes into the n side — and each waits a lifetime \(\tau\) before recombining, so a charge \(Q = I\tau\) is stored in the neutral regions. Reversing the drive cannot establish a reverse voltage until that charge is removed, so the diode conducts backwards for \(t_{rr}\). A Schottky diode's forward current is carried entirely by electrons crossing a metal-semiconductor barrier; an electron arriving in the metal is a majority carrier among \(10^{22}\) others and is not stored anywhere, so there is no charge to remove and no recovery interval. The loss is \(P \approx Q_{rr}V_Rf\) with \(Q_{rr}\approx\tfrac12I_{RM}t_{rr}\). For the 1N4007, \(Q_{rr} = 0.5\times1\times30\times10^{-6} = 15\) µC, and at 100 V and 100 kHz that is 150 W — a meaningless number, because 30 µs is three times the 10 µs period, so the diode never turns off and the circuit simply does not work. For the UF4007, \(Q_{rr} = 0.5\times0.5\times75\times10^{-9} = 18.75\) nC, giving 0.188 W and a recovery occupying 0.75 % of each period — workable. For a Schottky the loss is essentially zero, leaving only the energy \(\tfrac12C_jV_R^2f\) to charge the junction capacitance, which for 200 pF at 100 V and 100 kHz is 0.1 W. Note that the Schottky also halves the conduction loss.
2. A varactor has \(C_0 = 47\) pF, \(V_{bi} = 0.7\) V and \(n = 0.5\). It tunes a circuit with \(L = 220\) nH and 8 pF of unavoidable stray capacitance. Find the frequency range for a tuning voltage of 1 V to 25 V.
First the varactor capacitance at each end. At 1 V, \(C_j = 47/(1+1/0.7)^{0.5} = 47/\sqrt{2.4286} = 47/1.5584 = 30.16\) pF. At 25 V, \(C_j = 47/(1+25/0.7)^{0.5} = 47/\sqrt{36.714} = 47/6.0592 = 7.757\) pF. Adding the 8 pF stray gives total tank capacitances of 38.16 pF and 15.76 pF. Then \(f = 1/(2\pi\sqrt{LC})\): at 38.16 pF, \(\sqrt{220\times10^{-9}\times38.16\times10^{-12}} = 2.898\times10^{-9}\) and \(f = 54.93\) MHz; at 15.76 pF, \(f = 85.48\) MHz. The range is therefore 54.9 to 85.5 MHz, a ratio of 1.556. Note what the stray costs: the varactor's own capacitance ratio is \(30.16/7.757 = 3.888\), which would give a frequency ratio of \(\sqrt{3.888} = 1.972\), but the fixed 8 pF reduces the effective ratio to \(38.16/15.76 = 2.421\) and the frequency ratio to 1.556 — a loss of more than a fifth of the tuning span. The lesson is that stray capacitance hurts most at the high-frequency end, where the varactor's own capacitance is smallest and the fixed stray is the largest fraction of the total.
3. Sketch and explain the tunnel-diode characteristic in terms of band alignment, and state the condition for a tunnel diode to oscillate in a parallel resonant circuit.
Both sides of the junction are degenerately doped, so the Fermi level lies inside the conduction band on the n side and inside the valence band on the p side, and the depletion region is only about 10 nm wide — thin enough to tunnel through. At zero bias the filled states on each side face filled states on the other and no net current flows. A small forward bias raises the n-side bands so that filled n-side conduction states face empty p-side valence states at the same energy; electrons tunnel across and the current rises steeply, reaching its peak \(I_P\) at \(V_P\approx65\) mV when the overlap is greatest. Increasing the bias further slides the bands past one another, so the filled n-side states come to face the forbidden gap where there are no states to receive them, and the tunnelling current falls to a valley \(I_V\) at \(V_V\approx340\) mV. Beyond the valley, ordinary thermal injection over the barrier takes over and the characteristic becomes that of a normal diode. The falling portion has a negative differential resistance, \(-R_n = (V_V-V_P)/(I_V-I_P) = 0.275/(-8.6\ \text{mA}) = -32\ \Omega\). For oscillation, the diode is biased at the middle of that region and connected across a parallel \(LC\) tank whose losses are a shunt conductance \(G\). The diode contributes \(-1/R_n\), so the net conductance is \(G - 1/R_n\); if \(1/R_n > G\), that is if the tank's loss resistance is greater than \(R_n\), the net damping is negative, any small disturbance grows, and the circuit oscillates at \(f = 1/(2\pi\sqrt{LC})\). The bias source must also have a low enough resistance that its load line crosses the characteristic only once, otherwise the circuit latches at one of the two stable states instead of oscillating.
4. A PIN diode has \(W = 100\) µm, \(\tau = 2\) µs and \(\mu_n + \mu_p = 1830\ \text{cm}^2/\text{V}\cdot\text{s}\). Find its RF resistance at 0.5 mA and at 50 mA, the attenuation range this gives in a series switch feeding a 50 \(\Omega\) load, and the lowest usable frequency.
Convert the thickness: \(W = 100\) µm \(= 100\times10^{-4}\) cm \(= 0.01\) cm, so \(W^2 = 10^{-4}\ \text{cm}^2\). Then \(R_S = W^2/[(\mu_n+\mu_p)\tau I] = 10^{-4}/(1830\times2\times10^{-6}\times I) = 0.02732/I\) with \(I\) in amperes. At 0.5 mA, \(R_S = 0.02732/0.0005 = 54.6\ \Omega\). At 50 mA, \(R_S = 0.02732/0.05 = 0.546\ \Omega\). In a series switch driving a 50 \(\Omega\) load from a 50 \(\Omega\) source, the insertion loss is \(20\log_{10}[1 + R_S/(2\times50)]\): at 0.546 \(\Omega\) that is \(20\log_{10}(1.00546) = 0.047\) dB, essentially transparent; at 54.6 \(\Omega\) it is \(20\log_{10}(1.546) = 3.79\) dB, which is a poor "off" state for a switch. This shows why a series PIN switch is normally run at a much lower bias for the off state or, better, why practical switches use a series-shunt pair: the shunt diode, turned on, short-circuits the line and adds another 20 to 30 dB. The lowest usable frequency is set by the requirement that the RF period be much shorter than the carrier lifetime, so that the stored charge cannot follow the RF cycle and the diode presents the same resistance on both half-cycles: \(f \gg 1/(2\pi\tau) = 1/(2\pi\times2\times10^{-6}) = 79.6\) kHz. In practice a factor of ten margin is used, so this diode is well behaved above about 800 kHz and distorts below it.
5. A 5 V, 3 A output is rectified by two diodes conducting alternately in a full-wave circuit. Compare the losses using silicon p-n rectifiers (\(V_F = 0.95\) V) and Schottky rectifiers (\(V_F = 0.48\) V), and say what other considerations decide the choice.
Each diode carries the full 3 A while it conducts, and the two together conduct for the whole cycle, so the average conduction loss is simply \(V_FI_o\) regardless of which is conducting. For the silicon diodes, \(P = 0.95\times3 = 2.85\) W; for the Schottkys, \(P = 0.48\times3 = 1.44\) W. The saving is 1.41 W. Against a useful output of \(5\times3 = 15\) W, the rectifier efficiency is \(15/(15+2.85) = 84.0\ \%\) with silicon and \(15/(15+1.44) = 91.2\ \%\) with Schottkys — a seven-point improvement, and a substantially smaller heatsink. Three other considerations decide the choice in practice. First, reverse voltage: the Schottky must block the same peak inverse voltage as the silicon device, and Schottkys are scarce above 200 V and expensive above 100 V; on a 5 V output the PIV is small and this is no obstacle. Second, leakage and thermal runaway: a 40 V Schottky may leak 500 µA at 25 °C and 250 times that at 100 °C, and the resulting dissipation heats the junction further, so the thermal design must be checked at the maximum ambient rather than at 25 °C. Third, switching frequency: if this is a mains-frequency supply the silicon diode's 3 µs recovery costs nothing, but if it is the output rectifier of a 100 kHz switching converter the silicon diode is unusable at any efficiency, and the Schottky is the only choice. For a 5 V rail the Schottky wins on every count that matters.