By the end of this chapter you should be able to:
- Explain forward and reverse bias as a change to the Chapter 5 barrier, and state what happens to the depletion width and the field in each case.
- Derive and interpret the Shockley equation \(I_D = I_S(e^{V_D/nV_T}-1)\), including the meaning and typical values of the ideality factor \(n\).
- Use the 60 mV/decade rule to move between diode currents and voltages without a calculator, and extract \(n\) from measured data.
- Account for the temperature behaviour of a diode: the doubling of reverse current per 10 °C and the \(-2\) mV/K drift of the forward drop.
- Choose between the ideal, constant-voltage-drop and piecewise-linear models and quantify the error each introduces.
- Distinguish static resistance from dynamic resistance and compute \(r_d = nV_T/I_D\) at a stated operating point.
- Explain junction and diffusion capacitance, relate diffusion capacitance to reverse recovery time, and distinguish Zener from avalanche breakdown.
Chapter 5 left the junction in equilibrium: a barrier of \(V_{bi} = 0.774\) V standing across a depletion region 0.302 µm wide, with a diffusion current and a drift current of hundreds of amperes per square centimetre cancelling each other so precisely that nothing at all comes out of the terminals. That is a complete description of a device that does nothing. This chapter connects a battery to it.
The whole of diode behaviour follows from one observation. An external voltage appears almost entirely across the depletion region, because that is the only part of the device with any appreciable resistance, and it therefore adds to or subtracts from the barrier. The drift current does not care — it is limited by how many minority carriers are generated, not by how hard they are swept. The diffusion current cares enormously, because it depends on the fraction of majority carriers with enough energy to climb the barrier, and that fraction is a Boltzmann exponential. Break the equilibrium by \(0.1\) V and one of the two cancelling currents changes by a factor of 48 while the other does not change at all. The result is a device that passes current in one direction and not the other, and the mathematics of it is the Shockley equation.
The second half of the chapter is about not using that equation. It is transcendental, it cannot be solved for \(V\) in a circuit by hand, and in most designs it is unnecessary. Three progressively cruder models replace it, and knowing which one is honest in a given circuit — and roughly how much error it costs — is a more useful skill than being able to quote the exponential.
1 Bias as a Change to the Barrier
Connect a battery across the junction, positive to the p-side and negative to the n-side. Where does that voltage appear? The neutral p and n regions are ordinary doped semiconductor, with a conductivity computed in Chapter 4; for a junction of \(10^{-4}\ \text{cm}^2\) area and a few hundred micrometres of neutral material the two of them together contribute perhaps a fraction of an ohm. The depletion region, by contrast, has essentially no mobile carriers in it at all, so its resistance is many orders of magnitude larger. Almost the whole applied voltage therefore falls across the depletion region, and the neutral regions stay very nearly equipotential. This is the assumption on which everything else rests, and it is the assumption that fails at high current — which is exactly where the bulk resistance of the third circuit model comes from.
With the applied voltage sitting across the depletion region, the barrier becomes \(V_{bi} - V_a\), where \(V_a\) is positive for forward bias. Everything follows mechanically:
- Forward bias (\(V_a > 0\)) opposes the built-in field, so the barrier falls. Fewer uncovered ions are needed to support the smaller potential drop, so the depletion region narrows, mobile carriers flood back towards the junction, and the peak field weakens.
- Reverse bias (\(V_a < 0\)) reinforces the built-in field, so the barrier rises, more ions must be uncovered, the depletion region widens and the peak field grows — which is why breakdown is a reverse-bias phenomenon.
Figure 6.1 draws all three cases on one set of axes, using the Chapter 5 junction with \(N_A = 10^{17}\) and \(N_D = 10^{16}\ \text{cm}^{-3}\). Since \(W \propto \sqrt{V_{bi}-V_a}\), half a volt of forward bias narrows the region from 0.302 to 0.180 µm and one volt of reverse bias widens it to 0.457 µm.
Now count carriers. The equilibrium diffusion current was suppressed by the Boltzmann factor \(e^{-qV_{bi}/kT}\). Lower the barrier by \(V_a\) and the suppression becomes \(e^{-q(V_{bi}-V_a)/kT}\), so the diffusion current is multiplied by \(e^{qV_a/kT}\). At room temperature \(kT/q = 25.85\) mV, so 0.1 V of forward bias multiplies the diffusion current by \(e^{0.1/0.02585} = 48\), and 0.6 V multiplies it by \(1.2\times10^{10}\).
Meanwhile the drift current is unmoved. A minority electron generated in the p-side neutral region and wandering to the edge of the depletion layer is swept across it whatever the field, provided only that a field exists. Doubling the field does not double the number of electrons that arrive; the arrival rate is set by thermal generation. So the drift current is fixed at the value it had in equilibrium, and the net current is the difference between an exponential and a constant.
The barrier controls the majority-carrier diffusion current exponentially and the minority-carrier drift current not at all. Forward bias therefore lets the exponential run away; reverse bias kills the exponential and leaves the constant. A diode conducts one way because only one of the two cancelling currents can be steered.
2 The Shockley Diode Equation
Write the equilibrium drift current, per unit area, as \(I_S\). The diffusion current in equilibrium must equal it, so it too is \(I_S\), and under bias the diffusion current becomes \(I_S e^{V_D/V_T}\) where \(V_T = kT/q\). The net current is the difference:
The ideal diode equation, published by Shockley in 1949. Note that the \(-1\) is not a correction bolted on afterwards: it is the reverse drift current, and it is the whole of the reverse behaviour.
A fuller derivation, which Chapter 16 repeats for the transistor base, computes \(I_S\) from the minority-carrier diffusion profiles on either side of the depletion region and gives
\(L_p = \sqrt{D_p\tau_p}\) is the hole diffusion length in the n-material, the average distance an injected hole travels before recombining. For a small-signal silicon diode \(I_S\) is of order \(10^{-14}\) A; for a rectifier with a hundred times the area it is of order \(10^{-12}\) A.
Two features of that expression matter more than its exact form. It carries \(n_i^2\), which for silicon roughly doubles every five or six degrees, so \(I_S\) is violently temperature dependent. And it is inversely proportional to the doping, so a heavily doped diode has a small \(I_S\) and therefore a larger forward voltage at any given current — which is why a Schottky diode, whose current mechanism is quite different and whose effective \(I_S\) is a thousand times larger, drops only 0.3 V.
The equation as written is the ideal one, and real diodes do not obey it. Two mechanisms it ignores both matter at the extremes of the current range. At low current, carriers recombine inside the depletion region rather than diffusing across it; that recombination current also grows exponentially, but with \(e^{V_D/2V_T}\) rather than \(e^{V_D/V_T}\). At high current the injected minority concentration approaches the majority concentration, the "low-level injection" assumption fails, and the exponent is again halved. Both are absorbed into one empirical fudge, the ideality factor or emission coefficient \(n\):
\(n \to 1\) where diffusion dominates — the mid-current region of a good silicon diode; \(n \to 2\) where depletion-region recombination dominates, at low currents and in wide-gap materials. A silicon signal diode measured over its useful range typically shows \(n\) between 1.0 and 1.8, and a light-emitting diode of Chapter 14 often shows \(n\) near 2.
One simplification is worth doing at once. For \(V_D\) more than about four thermal volts positive — that is, above \(4 \times 25.85 = 103\) mV — the exponential is at least 55, so the \(-1\) contributes under 2 % and is dropped: \(I_D \approx I_S e^{V_D/nV_T}\). For \(V_D\) more than four thermal volts negative the exponential is under \(0.02\) and it is the exponential that is dropped: \(I_D \approx -I_S\). That is where the name reverse saturation current comes from — the reverse current saturates at \(-I_S\) and then, ideally, does not depend on reverse voltage at all.
3 The Exponential in Practice: 60 mV per Decade
An exponential is an awkward thing to reason about until it is turned into a rule about decades. Take the ratio of two forward currents on the same diode:
Set the current ratio to ten and evaluate at 300 K, where \(V_T = 25.852\) mV:
The 60 mV per decade rule for \(n = 1\); 119 mV per decade for \(n = 2\). It is the single most useful number in device electronics, and it reappears unchanged as the subthreshold slope of a MOSFET in Chapter 23.
Used the other way round, the rule is a measurement. Plot measured \(\ln I_D\) against \(V_D\); the slope gives \(n\) directly, and the intercept extrapolated back to \(V_D = 0\) gives \(I_S\). A device that shows 60 mV/decade at milliamperes and 90 mV/decade at nanoamperes is telling you that recombination has taken over at the bottom of the range, exactly as the two-mechanism picture predicts.
A silicon diode has \(I_S = 10\ \text{fA} = 1.0\times10^{-14}\) A and \(n = 1\) at 300 K. Find \(V_D\) at 1 µA, 100 µA, 1 mA, 10 mA and 100 mA.
| \(I_D\) | 1 µA | 10 µA | 100 µA | 1 mA | 10 mA | 100 mA |
|---|---|---|---|---|---|---|
| \(V_D\) | 0.476 V | 0.536 V | 0.595 V | 0.655 V | 0.714 V | 0.774 V |
Five decades of current span 0.476 V to 0.774 V — a total of 298 mV, or 59.5 mV per decade, exactly as the rule says. Read the table the other way and its message is blunter: a hundred thousand-fold change in current requires a 300 mV change in voltage. That is why the diode is such a good clamp and such a bad resistor.
Notice also where 0.7 V falls. This diode reaches 0.7 V somewhere between 1 and 10 mA — interpolating, at \(I_D = 5.75\) mA. If your circuit runs the diode at 100 µA, the honest constant-drop value is 0.6 V, not 0.7 V, and using 0.7 V will make the predicted current wrong.
The same exponential controls how the diode responds to a small change about an operating point, and this is where the distinction between two quite different resistances arises. The static or DC resistance is simply the ratio of the coordinates of the operating point:
The dynamic or AC resistance is the reciprocal slope of the characteristic there, obtained by differentiating the Shockley equation:
The famous \(r_d = 26/I_D\ (\text{mA})\ \Omega\). It contains no device parameter whatever — not \(I_S\), not the area, not the doping. Every silicon diode in the world has \(r_d = 26\ \Omega\) at 1 mA, which is why the result reappears as \(r_e\) in the transistor models of Chapter 19.
For the diode of Example 6.1 at \(I_D = 1\) mA: \(V_D = 0.655\) V, so \(R_{DC} = 0.655/0.001 = 655\ \Omega\), while \(r_d = 0.025852/0.001 = 25.9\ \Omega\). At \(I_D = 10\) mA: \(R_{DC} = 0.714/0.01 = 71.4\ \Omega\) and \(r_d = 2.59\ \Omega\). Increasing the current tenfold divides \(r_d\) by exactly ten but divides \(R_{DC}\) by only 9.2, because the numerator moved as well.
The two numbers answer different questions. \(R_{DC}\) tells you the power dissipated and the DC voltage you will lose; \(r_d\) tells you what a 5 mV signal riding on the bias will see. Never substitute one for the other. Note too that \(r_d\) is a small-signal quantity only: it is meaningful for excursions of a few millivolts, because over \(2V_T\) the characteristic curves visibly.
4 Reverse Current, Temperature and Breakdown
The ideal equation predicts a reverse current that saturates at \(-I_S\) and stays there. For germanium that is roughly what happens. For silicon it is not, for a reason the ideal derivation omits: carriers are also generated thermally inside the depletion region, and every one of them is immediately swept out. That generation current is proportional to the depletion volume, and since \(W \propto \sqrt{V_{bi}+V_R}\) it grows slowly with reverse voltage instead of saturating. In a silicon diode this generation current is typically far larger than the ideal \(I_S\): the diode of Example 6.1 has \(I_S = 10\) fA, but its datasheet reverse leakage might be 5 nA, five orders of magnitude more. Add surface leakage across the package and the discrepancy grows further. When a datasheet quotes \(I_R\), it is quoting the sum, measured at a stated voltage and temperature.
Temperature is where reverse current becomes a design problem rather than a curiosity. Both mechanisms are driven by thermal generation, and the useful engineering rule is:
\(I_R(T) \approx I_R(T_0)\times 2^{(T-T_0)/10}\), which is 7.2 % per degree. A diode leaking 5 nA at 25 °C leaks 40 nA at 55 °C, 905 nA at 100 °C and 5.1 µA at 125 °C — a factor of 1024. Silicon's advantage over germanium is entirely here: germanium's smaller gap makes \(n_i\) about 2400 times larger at room temperature, and its leakage becomes intolerable above 75 °C.
The ten-degree rule is an empirical description of the depletion-region generation current, which follows \(n_i\) rather than \(n_i^2\). The ideal \(I_S\), being proportional to \(n_i^2\), rises much faster: evaluating \(n_i^2 T^3 e^{-E_g/kT}\) for silicon gives a doubling interval of about 4.6 °C near room temperature, widening to 6.2 °C at 77 °C. Quote the ten-degree rule for datasheet leakage and be aware that the underlying \(I_S\) moves faster still.
The forward drop moves too, and in the opposite direction. Hold \(I_D\) constant and solve for \(V_D\): as \(T\) rises, \(I_S\) rises steeply, so a smaller \(V_D\) suffices to pass the same current. Working the algebra through with \(I_S \propto T^3 e^{-E_g/kT}\) for a diode biased at 9.3 mA gives \(V_D = 0.712\) V at 300 K and 0.631 V at 350 K, a slope of \(-1.63\) mV/K. Different bias currents and different diodes give values between about \(-1.5\) and \(-2.5\) mV/K, and the universal rule of thumb is:
The same coefficient as the \(V_{BE}\) of a transistor, because it is the same junction. It is the physical basis of the bandgap reference, of diode temperature sensors, and of the thermal-runaway problem in the power amplifiers of Chapter 20.
Breakdown. Push the reverse voltage far enough and the reverse current stops being small. Two entirely distinct mechanisms produce this, and they are routinely confused because both are called "Zener" in casual speech.
Zener breakdown is quantum-mechanical tunnelling. In a heavily doped junction the depletion region is very thin — tens of nanometres — so a modest reverse voltage produces a field above about \(10^6\) V/cm, and valence electrons on the p-side tunnel straight through the forbidden gap into empty conduction states on the n-side. No collisions are involved. It dominates below about 5 V.
Avalanche breakdown is impact ionisation. In a lightly doped junction the depletion region is wide, so a carrier crossing it can be accelerated to more than the bandgap energy between collisions; it then knocks a bound electron out of a lattice bond, and the two carriers repeat the process. The multiplication is a chain reaction. It dominates above about 7 V.
They are told apart by their temperature coefficients, which is the practical test. Heating narrows the bandgap slightly, making tunnelling easier, so a Zener breakdown voltage falls with temperature. Heating also increases lattice vibration, shortening the mean free path so carriers gain less energy between collisions, so an avalanche breakdown voltage rises with temperature. Between about 5 and 6 V the two coefficients cancel, which is why 5.6 V reference diodes are prized for their stability. Chapter 12 uses all of this to design a regulator; the point here is that neither mechanism is destructive in itself. What destroys a diode is the power \(V_ZI_Z\), and an ordinary rectifier has no mechanism for spreading that power evenly, so it fails.
5 Three Circuit Models and When Each Is Honest
The Shockley equation cannot be solved analytically in a circuit. Put a diode in series with a resistor and a supply and Kirchhoff gives \(V_S = IR + nV_T\ln(I/I_S)\), which has no closed-form solution for \(I\). Chapter 7 shows how to solve it graphically and iteratively; here we take the other route, which is to replace the diode with something linear and accept a known error.
The ideal diode model. The diode is a switch: zero volts and any current when forward biased, zero current and any voltage when reverse biased. Its characteristic is the two half-axes. This is not as crude as it looks. In a rectifier fed from a 240 V peak transformer, the 0.7 V you are neglecting is 0.3 % of the peak, and the model gets the answer right to well inside the tolerance of the transformer. It is also the right model when the question is topological — which diodes conduct, and when — as in the clippers and multipliers of Chapter 11.
The constant-voltage-drop model. The diode is a switch in series with a 0.7 V battery: zero current below 0.7 V, and 0.7 V across it at any current above that. This is the everyday model, and it is worth being clear about what makes it work. It is not that the diode has a 0.7 V threshold; it is that the diode's voltage moves by only 60 mV per decade of current, so over the two decades that most circuits span the drop really is nearly constant. The correct value of the constant is the drop at the current your circuit actually runs, which for a signal diode at a few milliamperes is 0.7 V, for a power rectifier at an ampere is 0.9–1.0 V, and for a Schottky is 0.3 V.
The piecewise-linear model. The diode is a switch, a battery \(V_{D0}\) and a series bulk resistance \(r_B\): \(V_D = V_{D0} + I_D r_B\) when conducting. This is a chord drawn across the real characteristic between two chosen currents, and it captures the fact that at high current the drop really does climb. The resistance is genuinely resistive — it is the ohmic resistance of the neutral p and n regions and the contacts, the thing the first section assumed away — and it is typically 0.1–1 \(\Omega\) for a power rectifier and 5–20 \(\Omega\) for a signal diode. Below a few tens of milliamperes \(r_B\) contributes a millivolt or two and is not worth carrying; at an ampere it contributes hundreds of millivolts and dominates.
- Ideal — when the supply is more than about 20 times the drop, or when you only need to know which diode is on.
- Constant drop — the default. Error under about 1 % for supplies above 5 V, and it is the model implied whenever an examination question says "assume silicon".
- Piecewise linear — when the current is large enough that \(I_Dr_B\) is comparable with the accuracy you need: power rectifiers, low-voltage supplies, and anything where you must predict dissipation.
- Shockley, solved iteratively — when the supply is comparable with the drop, when the answer is exponentially sensitive (log amplifiers, current mirrors), or when you are simulating.
Chapter 7 puts numbers on the errors: for a 10 V supply and a 1 k\(\Omega\) resistor, the ideal model is 7.7 % high, the constant-drop model 0.13 % high and the piecewise-linear model 0.33 % low, against an iterative answer of 9.288 mA. For a 2 V supply the ideal model is 49.5 % high. That contrast is the whole argument for knowing more than one model.
6 Junction and Diffusion Capacitance
So far the diode has been treated as though it responded instantly. It does not, and the reason is that both bias conditions store charge. Two distinct capacitances describe that storage, one dominant in reverse bias and one in forward.
Junction (depletion, transition) capacitance. Chapter 5 derived it: the depletion region is an insulating layer of thickness \(W\) with conducting neutral material either side, so it is a parallel-plate capacitor whose plate separation depends on the voltage across it. With \(W \propto (V_{bi}-V_a)^{1/2}\),
\(m = 1/2\) for an abrupt junction and \(1/3\) for a linearly graded one; \(V_a\) is negative in reverse bias, so \(C_j\) falls as the diode is reverse biased further. A diode with \(C_{j0} = 4\) pF and \(V_{bi} = 0.774\) V has \(C_j = 2.64\) pF at 1 V reverse, 1.47 pF at 5 V and 1.07 pF at 10 V.
That voltage dependence is a nuisance in a detector, where it detunes the circuit, and it is the entire purpose of the varactor of Chapter 13, where the diode is used as a voltage-controlled capacitor to tune a radio.
Diffusion (storage) capacitance. Under forward bias the depletion region is thin and \(C_j\) is small, but a far larger effect takes over. Forward bias injects minority carriers into the neutral regions — holes into the n-side, electrons into the p-side — where they form an exponentially decaying excess concentration extending about one diffusion length in. That stored excess charge \(Q\) is proportional to the current, through the mean transit time \(\tau_T\):
Diffusion capacitance is proportional to current, unlike junction capacitance. For \(\tau_T = 100\) ns at \(I_D = 1\) mA, \(C_d = (10^{-7})(10^{-3})/0.025852 = 3.87\) nF; at 10 mA it is 38.7 nF. Three or four orders of magnitude larger than \(C_j\) — and this is why a forward-biased junction is slow.
Reverse recovery. The consequence is the most important dynamic limitation of a rectifier. Suppose a diode carrying a forward current \(I_F\) is abruptly reverse biased. The stored minority charge \(Q = \tau_T I_F\) does not vanish; it has to be removed, either by recombination (which takes a carrier lifetime) or by being swept back out as reverse current. Until it is gone, the junction is still flooded with carriers and the diode continues to conduct in the reverse direction, limited only by the external circuit. The waveform has two parts: a storage time \(t_s\) during which the reverse current sits at a constant \(I_R\) set by the external circuit, and a transition time \(t_t\) during which the current tails back to the leakage value as the depletion region re-forms. Their sum is the reverse recovery time \(t_{rr}\).
A diode carries \(I_F = 10\) mA with \(\tau_T = 100\) ns and is switched into a circuit that limits the reverse current to \(I_R = 20\) mA. The stored charge is \(Q = \tau_T I_F = (100\times10^{-9})(0.010) = 1.0\) nC. If none of it recombines, the storage time is \(t_s = Q/I_R = 1.0\times10^{-9}/0.020 = 50\) ns. In that time the diode is a short circuit in the wrong direction.
Three consequences follow. The reverse current spike appears in the supply as a current pulse, and in a mains rectifier at 50 Hz it happens 100 times a second and radiates as conducted interference — the reason "soft-recovery" rectifiers and snubber networks exist. The energy \(\int v i\,dt\) during recovery is dissipated in the diode, so switching loss rises linearly with frequency and sets the upper frequency limit of the device. And a rectifier used above about 20 kHz must be a fast-recovery type (\(t_{rr}\) of 50–500 ns) or a Schottky (essentially zero, because a Schottky is a majority-carrier device with no stored minority charge at all — the subject of Chapter 13).
A 1N4007 mains rectifier has \(t_{rr} \approx 30\) µs and is perfectly adequate at 50 Hz, where the half-period is 10 ms; used in a 100 kHz switching converter with a 10 µs period it would spend its entire life in recovery and destroy itself.
Both capacitances appear in parallel with the small-signal resistance \(r_d\) in the diode's high-frequency model, giving a single-pole roll-off at \(f = 1/(2\pi r_d C)\). In reverse bias \(C_j\) dominates and \(r_d\) is enormous; in forward bias \(C_d\) dominates and \(r_d\) is small, and the product \(r_dC_d = \tau_T\) is independent of current, so the cutoff frequency of a forward-biased diode is set by its transit time alone.
7 Reading the Numbers Off a Real Device
It is worth closing by connecting all of this to the four or five lines of a datasheet that a designer actually uses, because the vocabulary is not always the vocabulary of the derivation.
| Datasheet entry | Typical value (1N4148 signal diode) | What it is, in this chapter's terms |
|---|---|---|
| \(V_F\) at stated \(I_F\) | 0.72 V at 5 mA, 1.0 V at 100 mA | Two points on the Shockley curve; the gap between them exposes \(n\) and \(r_B\) |
| \(I_R\) at stated \(V_R\), \(T\) | 25 nA at 20 V, 50 µA at 150 °C | Generation plus surface leakage, not the ideal \(I_S\) |
| \(V_{RRM}\) | 100 V | Repetitive peak reverse voltage — stay well below breakdown |
| \(C_D\) or \(C_T\) | 4 pF at \(V_R = 0\), f = 1 MHz | Junction capacitance \(C_{j0}\) |
| \(t_{rr}\) | 4 ns | Set by stored diffusion charge; the reason this part is called "fast" |
| \(I_{FSM}\) | 1 A for 1 s | Non-repetitive surge — the rating that rectifier inrush attacks, Chapter 9 |
The two forward-voltage entries are the most informative line on the sheet, because between them they pin down the model. Going from 5 mA to 100 mA is 1.301 decades, so an ideal \(n=1\) device would climb \(1.301\times59.5 = 77\) mV, from 0.72 V to 0.797 V. The datasheet says 1.0 V, some 200 mV higher. Attributing the excess to bulk resistance gives \(r_B \approx 0.203/0.095 = 2.1\ \Omega\); attributing it instead to ideality gives \(n = 280/77 = 3.6\), which is unphysical. So the extra drop is ohmic, and a piecewise-linear model with \(V_{D0} = 0.70\) V and \(r_B = 2\ \Omega\) will track this device usefully from a milliampere to a hundred.
Two habits are worth forming now. First, always ask at what current a quoted forward drop was measured, because a number without a current is not a specification. Second, treat every parameter in this chapter as having a temperature next to it: \(I_S\) doubles every ten degrees, \(V_F\) falls at 2 mV/K, \(V_{BR}\) moves in a direction that identifies the breakdown mechanism, and \(C_j\) is nearly the only thing that stays put. A circuit that works on the bench at 22 °C and fails in a sealed enclosure at 70 °C has almost always been designed with room-temperature numbers.
Chapter 7 now takes the models of this chapter and puts them to work solving circuits, and Chapters 8 to 10 build the one application — rectification — that made the diode a component rather than a curiosity.
8 Summary and Key Results
| Quantity | Expression | Value or rule |
|---|---|---|
| Barrier under bias | \(V_{bi} - V_a\), \(V_a > 0\) forward | 0.774 V → 0.274 V at 0.5 V forward; 1.774 V at 1 V reverse |
| Depletion width | \(W \propto \sqrt{V_{bi}-V_a}\) | 0.302 µm → 0.180 µm forward, 0.457 µm at 1 V reverse |
| Shockley equation | \(I_D = I_S\!\left(e^{V_D/nV_T}-1\right)\) | \(-1\) negligible above \(4V_T\) = 103 mV forward |
| Ideality factor | \(1 \le n \le 2\) | 1 for diffusion, 2 for depletion-region recombination |
| Decade rule | \(\Delta V = nV_T\ln 10\) | 59.5 mV/decade at \(n=1\); 119 mV/decade at \(n=2\) |
| Forward drop | \(V_D = nV_T\ln(I_D/I_S)\) | 0.595 V at 100 µA, 0.655 V at 1 mA, 0.714 V at 10 mA |
| Static resistance | \(R_{DC} = V_D/I_D\) | 655 \(\Omega\) at 1 mA, 71.4 \(\Omega\) at 10 mA |
| Dynamic resistance | \(r_d = nV_T/I_D\) | 25.9 \(\Omega\) at 1 mA, 2.59 \(\Omega\) at 10 mA — device-independent |
| Reverse current | \(I_R(T) = I_R(T_0)2^{(T-T_0)/10}\) | 5 nA at 25 °C → 905 nA at 100 °C, 5.1 µA at 125 °C |
| Forward temperature drift | \(dV_D/dT|_{I_D}\) | −1.6 to −2.5 mV/K; use −2 mV/K |
| Junction capacitance | \(C_j = C_{j0}(1-V_a/V_{bi})^{-1/2}\) | 4 pF at 0 V → 1.47 pF at 5 V reverse |
| Diffusion capacitance | \(C_d = \tau_T I_D/nV_T = \tau_T/r_d\) | 3.87 nF at 1 mA for \(\tau_T\) = 100 ns |
| Reverse recovery | \(t_s = \tau_T I_F/I_R\) | 50 ns for \(I_F\) = 10 mA, \(I_R\) = 20 mA, \(\tau_T\) = 100 ns |
| Breakdown mechanism | Zener below 5 V, avalanche above 7 V | Zener \(V_{BR}\) falls with \(T\); avalanche \(V_{BR}\) rises; cancel near 5.6 V |
| Model error (10 V, 1 k\(\Omega\)) | ideal / 0.7 V / PWL | +7.7 % / +0.13 % / −0.33 % against 9.288 mA |
9 Common Mistakes
The characteristic is a smooth exponential with no threshold anywhere in it. The diode of Example 6.1 conducts 2.5 µA at 0.5 V and 0.12 mA at 0.6 V — small, but not zero, and quite enough to matter in a high-impedance circuit or a leakage-sensitive one. The apparent knee is an artefact of the current scale you plot on: replot with a 100 µA full scale and the knee sits at 0.6 V. Use 0.7 V as a modelling convenience for milliampere circuits, not as a physical property of silicon.
At 1 mA the same diode has \(R_{DC} = 655\ \Omega\) and \(r_d = 25.9\ \Omega\), a factor of 25 apart. Using \(R_{DC}\) in a small-signal calculation understates the signal by that factor; using \(r_d\) in a DC calculation understates the voltage drop and the dissipation by it. \(R_{DC}\) is a ratio of coordinates, \(r_d\) is a slope, and the two agree only for a genuine resistor.
Tunnelling and avalanche are different physics with opposite temperature coefficients, and a device marketed as a "Zener diode" above about 6 V is actually an avalanche diode. The distinction is not pedantry: a 3.3 V reference drifts down with temperature and a 15 V one drifts up, so a circuit compensated for one will be doubly wrong with the other. If you need a stable reference, use 5.6 V, where the two coefficients cancel.
10 Chapter Review
1. A silicon diode has \(I_S = 2\times10^{-14}\) A and \(n = 1.2\) at 300 K. Find \(V_D\) at 2 mA, the dynamic resistance there, and the voltage change needed to reach 20 mA.
\(V_T = 25.852\) mV, so \(nV_T = 31.02\) mV. \(V_D = 0.031022\ln(2\times10^{-3}/2\times10^{-14}) = 0.031022\ln(10^{11}) = 0.031022\times25.328 = 0.786\) V. The dynamic resistance is \(r_d = nV_T/I_D = 0.031022/0.002 = 15.5\ \Omega\). One decade of current costs \(nV_T\ln10 = 0.031022\times2.3026 = 71.4\) mV, so 20 mA is reached at \(0.786+0.071 = 0.857\) V, where \(r_d\) has fallen to \(1.55\ \Omega\). Notice that the larger ideality factor has raised both the forward drop and the decade step by 20 % relative to an \(n = 1\) device — which is exactly how you would measure \(n\) in the laboratory.
2. Why does a diode's reverse leakage double every 10 °C while its forward voltage falls by only about 2 mV/K? Are these two statements consistent?
They are the same fact seen from two directions. Reverse current is set by thermal generation and rises steeply with temperature — a factor of two per 10 K, or 7.2 % per degree. Forward voltage is measured at fixed current, so as \(I_S\) rises the exponential needs less help and \(V_D\) must fall. Quantitatively, \(V_D = nV_T\ln(I_D/I_S)\); differentiating, the rise in \(V_T\) pushes \(V_D\) up but the much faster rise in \(I_S\) pulls it down, and the second wins. Because \(V_D\) depends on the logarithm of \(I_S\), a huge multiplicative change in \(I_S\) becomes a small additive change in \(V_D\): a factor of 1024 in \(I_S\) is \(\ln 1024 = 6.93\) thermal volts, or about 180 mV over 100 K — the same order as \(-2\) mV/K × 100 K. Consistent, and the logarithm is what reconciles the two very different-looking numbers.
3. A 10 V supply feeds a 1 k\(\Omega\) resistor in series with a silicon diode. Compare the current predicted by the ideal, constant-drop and piecewise-linear models with the exact value of 9.288 mA, and say when each answer would be good enough.
The ideal model gives \(10/1000 = 10.00\) mA, high by 7.67 %. The constant-drop model gives \((10-0.7)/1000 = 9.300\) mA, high by 0.13 %. The piecewise-linear model with \(V_{D0} = 0.65\) V and \(r_B = 10\ \Omega\) gives \((10-0.65)/1010 = 9.257\) mA, low by 0.33 %. The constant-drop answer is the best here, and by a wide margin; the piecewise-linear model is not automatically better, because a chord fitted over a wide current range need not pass through your particular operating point. The ideal model's 7.7 % error would be acceptable for a first sanity check or for deciding which diodes conduct, but not for predicting a bias current. Repeat the exercise with a 2 V supply and the ideal error becomes 49.5 %, while the constant-drop error grows to 2.8 % — the errors scale with the ratio of the neglected drop to the supply.
4. Distinguish junction capacitance from diffusion capacitance. Which dominates in a reverse-biased varactor, which in a forward-biased rectifier, and why does the second one set the switching speed?
Junction capacitance is the charge stored on the fixed dopant ions bounding the depletion region: \(C_j = \varepsilon_sA/W\) with \(W \propto \sqrt{V_{bi}-V_a}\), so it falls as reverse bias increases, from 4 pF at zero bias to 1.47 pF at 5 V for a typical small diode. It dominates in reverse bias, where there is no injected charge at all, and its voltage dependence is what a varactor exploits. Diffusion capacitance is the charge stored as excess minority carriers injected into the neutral regions under forward bias: \(C_d = \tau_TI_D/nV_T\), proportional to current and typically nanofarads — three orders larger. It dominates in forward bias. It sets switching speed because that stored charge must be physically removed before the junction can block: with \(\tau_T = 100\) ns and \(I_F = 10\) mA the stored charge is 1 nC, and a circuit that can pull only 20 mA in reverse takes 50 ns to extract it, during which the diode conducts backwards. A Schottky diode stores no minority charge and so has no such delay.
5. A diode is measured at 0.72 V at 5 mA and 1.00 V at 100 mA. Fit a piecewise-linear model and state the current range over which you would trust it.
From 5 mA to 100 mA is \(\log_{10}(20) = 1.301\) decades. If the device were ideal with \(n = 1\), the drop would rise by \(1.301\times59.5 = 77\) mV, to 0.797 V. The measured 1.00 V is 203 mV higher, and attributing that excess to an ohmic term over a current increase of 95 mA gives \(r_B = 0.203/0.095 = 2.1\ \Omega\). Fitting the straight line through the two measured points instead gives slope \((1.00-0.72)/(0.100-0.005) = 2.95\ \Omega\) and intercept \(V_{D0} = 0.72 - 0.005\times2.95 = 0.705\) V, so \(V_D \approx 0.705 + 2.95I_D\). Trust it between roughly 5 and 100 mA, the range the chord was fitted over. Below 5 mA it will overestimate the drop badly — at 100 µA it predicts 0.705 V where the device delivers about 0.6 V — because a straight line cannot follow a logarithm down. Above 100 mA it will underestimate, as self-heating and high-level injection both take hold.