Circuit Response
every model has an edge, and the figure carries it
Concept

Saturation current — where it appears

The scale factor in a diode's exponential law, which fixes the current at which a stated forward drop appears and moves sharply with temperature. It carries the whole of the temperature dependence of a junction's forward drop, and it moves in the opposite direction to the thermal voltage.

Named by 13 essays across 4 fields — each of them below, with the objects they name alongside it.

What a 0.7 V constant costs, in the quantity it is used to predict. computed by solving, not by drawing by Newton's method on the exponential at 94 supplies through four resistors. The model is exact at 5.748 mA — the current at which the true drop is 0.7 V — and every curve crosses zero there, at four different supplies. Below it the model is low and above it high, and how much depends on the headroom rather than on the diode. Through the 87 Ω curve the drop is 49 mV out at 0.725 V and 147 mV out at 150.7 V — a factor of 3.0 — while the error in the current falls from -66% to 0.10%, a factor of 674, because the headroom underneath it has grown by 2017. On that curve the model is inside one per cent only above 1.12 V.

The one current a constant is right at

Seven-tenths of a volt is the true forward drop at 5.748 milliamperes and at no other current, and every circuit built on it crosses zero error there — four different resistors at four different supplies, all exact at the same current. What decides whether the model is any good is not the diode at all; it is how much of the supply the diode is taking.

limits · Diode model A diode's drop from 250 to 400 K, at 1.00 mA. computed by solving, not by drawing. Thirty-one operating points, each Newton's method on the exponential at its own temperature. The drop falls at 1.828 mV/K measured against 1.830 mV/K from the closed form — falls, although the thermal voltage in the exponent rises, because the saturation current rises faster. Over the same range the slope per decade of current goes the other way, from 49.6 mV to 79.4 mV, because that one is Vₜ ln 10 and nothing else.

Two millivolts a kelvin, and the wrong sign

Every number in the semiconductor field was computed at 300 K, and the model had no temperature in it at all. Putting it in moves a diode's drop by 1.828 mV/K — downwards, although the thermal voltage in the exponent is rising, because the saturation current rises by nine orders of magnitude across the same range. Two temperature dependences of one device, of opposite sign, from one solve.

semiconductors · Temperature One diode curve, eight one-decade fits, and eight different ideality factors. computed by solving, not by drawing. A junction with two conduction mechanisms — recombination near n = 2 at low current, diffusion near n = 1 above it — and a series resistance, which is what a real diode is. Fitting ln(i) against v over each decade in turn returns an ideality factor for each, and they run from 1.227 to 1.984 without being monotonic: the factor rises through the recombination region, falls through the diffusion region, and rises again where the series resistance takes over. Two of the windows are straight to a few parts in a thousand, so the residual gives no warning. The bars are what each fit predicts for the forward voltage at 1 mA: the worst is out by -186 millivolts, which is a current 0.01 times the truth.

The constant that is a window

A diode's ideality factor is quoted as a number and defined as a derivative, which means it has a value at every current and no value anywhere. Eight one-decade fits to one curve return factors from 1.23 to 1.98, two of them straight to a few parts in a thousand — so the residual gives no warning at all. Asked for the forward voltage at a milliamp, the window containing it is right to a third of a millivolt and the worst is out by 186, which is a current a hundredth of the truth.

limits · Diode model Two currents called saturation: one doubles every 4.49 K, the other every 8.98 K. computed by solving, not by drawing. The two current scales of one model junction against temperature, on a logarithmic axis. The saturation current of the exponential law goes as the square of the intrinsic carrier density — a cube of the temperature and the whole band gap in a Boltzmann factor — and doubles every 4.489 K at 300 K. The generation current a reverse-biased junction actually conducts goes as the density itself, with half the band gap, and doubles every 8.978 K: exactly twice as long, at every temperature. The dashed line is "doubles every ten kelvin" drawn through the generation scale, which is the current the rule belongs to. At 300 K this junction's two scales are 10 fA and 2 nA, which are its own parameters and not a property of silicon, and they become equal only at 616.8 K, or 343.6 °C.

Two currents with one name

A junction's saturation current is two currents with one name. The one in the forward law doubles every 4.49 kelvin; the one a reverse-biased junction actually conducts is generated in its depletion region, doubles every 8.98, and on this model junction is 3.06 × 10⁵ times larger at a volt of reverse bias. Doubles every ten kelvin is the second current's rule, and applied to the first it turns the forward drop's −1.81 millivolts per kelvin into +0.39.

limits · Diode model At 300 K one junction holds a logarithm to ±1% over 2.4 decades as a diode and 8.5 at its collector. computed by solving, not by drawing. The voltage of one model junction against the logarithm of the current it carries, as a percentage error of that current from a straight line fitted over the widest range that stays within ±1%. Taken as a diode — both mechanisms and 0.6 Ω of series resistance — the range is 2.40 decades, from 50.1 nA to 12.6 µA, and its slope is an ideality of 1.982. Taken at the collector, where the recombination current is supplied from the base and 1.604 Ω remains, it is 8.50 decades, from the axis's own end at 1 pA to 316 µA, at an ideality of 1.0001. Nothing arrives beside the collector current, so its lower end on this axis is the axis.

The logarithm is in the collector

A diode is the textbook logarithm, and a real junction holds one to within one per cent over only 2.40 decades — from 50 nanoamps to 12.6 microamps, at an ideality of 1.98 — because two mechanisms and a series resistance share its terminals. The same junction read at a transistor's collector, with its recombination current supplied from the base, holds 8.50 decades at an ideality of 1.0001. How far the logarithm reaches is decided by which terminal the current is taken from, and at the bottom of the range by a leakage current a millivolt is enough to switch on.

limits · Diode model The growth per cycle, and the form that is a fifth low at the top of the range. computed by solving, not by drawing. The factor the envelope is multiplied by each cycle, against the gain. The solid curve is exp(π(k−3)/√(1 − ((k−3)/2)²)), which is what the characteristic equation gives and what the netlist's own poles return to twelve digits; the dashed one is exp(π(k−3)), which drops the denominator. The circles are the marched envelope, fitted over the cycles that are still small — 80 of them at k = 3.01 and 4 at k = 3.2, and none at all above that. The two expressions differ by 3.9e-7 at k = 3.01 and by 20.462% at k = 3.8, so the approximation fails exactly where nothing is left to check it against.

Two exponentials, and where they meet

An oscillator's envelope grows by exp(π(k−3)/√(1 − ((k−3)/2)²)) a cycle, and the form usually quoted drops the denominator — exact to four parts in ten million at a hundredth above three, and 20.462 per cent low at 3.8, which is precisely where too few small cycles are left to measure it. Where the growth stops is the diode's own exponential: 108.5 millivolts of amplitude for every decade of saturation current, proportional to the ideality to four parts in a thousand. Above 60.121 nanoamperes the limiter is already conducting at zero signal and there is no oscillation at all.

applied · Oscillator A junction's noise against its own resistance's: exactly one at zero volts, and a half only far from it. computed by solving, not by drawing. A junction carries two currents at once, Is·e^(V/nVt) forwards and Is backwards, and each has its own shot noise. Their noise over the Johnson noise of the junction's own conductance is n(1 + e^−u)/2. At n = 1 it is 1.000000 at zero volts, 0.5676 at 50 mV, within one per cent of 0.50 only above 115.1 mV — where the forward current is ninety-nine saturation currents — and 2.978 at −40.00 mV of reverse bias. The half the forward-biased junction is known for is the limit of this curve, not its value.

The junction that is a resistor at zero volts

A forward-biased junction makes half the noise power of a resistor of its own dynamic resistance, and that half is a limit rather than a value. Kept with the saturation current that flows backwards across it, the ratio is one exactly at zero volts, 0.5676 at 50 millivolts, and within one per cent of the half only above 115.1 — at ninety-nine saturation currents, which is a picoampere on a small silicon diode and a microampere on a leaky one. A photodiode held at zero volts has the Johnson noise of its shunt resistance and nothing else, and it becomes shot-noise-limited at 49.981 millivolts of photocurrent drop.

noise · Shot noise A diode thermometer measuring its own sense current, at 600 K/W. computed by solving, not by drawing. The same fixed point as the core and the thermistor, on a junction: the dissipation is I·V and V falls with temperature, so the loop gain is negative and the equation has one root at every current. What it costs is two errors. The junction sits above ambient by 0.29 millikelvin at a microamp and 4.26 kelvin at ten milliamperes, which a calibration removes; and a kelvin of ambient produces less than a kelvin of junction, by 1/(1 − R_th·dP/dT), which it does not. After a calibration at 25 degrees the reading at 85 is out by -583 millikelvin at ten milliamperes and -0.09 at a microamp. The coefficient itself moves too — -2.403 against -1.613 millivolts a kelvin — so a quoted tempco carries a sense current as well as a junction.

The sensor inside its own answer

A junction driven from a current source cannot run away, because its forward voltage falls with temperature and its loop gain is therefore negative. What that costs is a thermometer that is warmer than what it is measuring by 4.26 kelvin at ten milliamperes, and — the part a calibration cannot remove — under-reports every change in ambient by 9,584 parts per million, because the sense current's own dissipation falls as the reading rises. Calibrated at 25 degrees, it is out by 583 millikelvin at 85.

semiconductors · Thermal feedback Three mismatches, and only one of them reaches the output. computed by solving, not by drawing. Each of the three quantities that can differ between the two transistors is given a spread of its own, one at a time, and 200 pairs are solved at each. The saturation currents produce a spread that follows them exactly — exponent 0.998, so 1.910 per cent of copy error for two per cent of mismatch. The current gains produce a line of slope 2.003, which is second order rather than first, and land at 2.20e-4 per cent for the same two. There is no third line because there is no third component: with no emitter resistors there is nothing for a resistor tolerance to be a tolerance of, and matching a mirror is a statement about emitter area and about nothing else.

The mismatch that cancels itself

A current mirror's copy error is spread by three things the two transistors can differ in, and the population that measures it has always drawn all three at once. Turned on one at a time, a two per cent spread of saturation currents gives 1.910 per cent of copy error and a two per cent spread of current gains gives 0.00022 — because the gains enter only as a sum of reciprocals, which has no first derivative where they are equal. Then the standard cure un-cancels it, by a factor of 86.

semiconductors · Current mirror Two junctions at 8 to 1 in area, at four currents from 10 µA to 10 mA. computed by solving, not by drawing: each drop is the diode law solved at its own temperature, from 200 to 400 K, with a saturation current that rises 15.0 orders of magnitude over the range. The two drops themselves sit 179 mV apart across the four currents; their difference does not move at all. It is Vₜ·ln 8 to 0.40 parts in a million at 10 mA and to 0.040% at the worst point, 10 µA at 400 K, where the larger junction's saturation current is no longer negligible against the current — 53.758 mV at 300 K, rising 179.19 µV/K — and the straight line through it reaches 0.001 µV at absolute zero (dashed).

The voltage at which the slope is zero

Two junctions carrying one current, one of them eight times the other's area, differ by exactly the thermal voltage times ln 8 — 53.758 mV at 300 K, rising 179.19 µV/K along a line that reaches zero at absolute zero to a thousandth of a microvolt — while each drop alone moves 1.9 mV/K. Weighted by 10.72 and added to one drop, the difference cancels it, and the sum is then 1.171704 V. That voltage is not a coincidence of this design: every weighting of the sum leaves the band gap at absolute zero, so a reference's slope at 300 K is its distance from 1.1717 V divided by 300 K, and two hundred references built wrong three different ways sit on that one line to 0.16 nV/K. A reference is trimmed by its voltage, at one temperature, and the voltage it is trimmed to is a statement about how its current is made.

semiconductors · Temperature A reference flattened at 300 K, from −40 to 125 °C, with its core current as T to the 1. computed by solving, not by drawing every 1.25 K. The reference has zero slope at 300 K by construction and its output there is 1.17170 V. It is never flat anywhere else: it falls 1.391 mV by −40 °C and 2.526 mV by 125 °C, a bow of 2.526 mV that no choice of the weight can remove, and the residue matches −(xti − α)·Vₜ·[x ln x − (x − 1)] with x = T/300 K (dashed) to within the diode law's "+1".

The bow a trim leaves

A band-gap reference trimmed to zero slope at 300 K is flat at exactly one temperature. Everywhere else it falls: 1.39 mV by −40 °C and 2.53 mV by 125 °C for a core biased proportional to absolute temperature, 3.79 mV for one run at a constant current, and nothing at all for one whose current rises as the cube of temperature. The residue is (xti − α) thermal voltages times x ln x − (x − 1), with x the temperature over the trim temperature, and six references of different bias, exponent and current fall on that one curve. The least box, 1.9 mV, comes from making the slope zero at 39 °C rather than at room temperature; quoted per kelvin it is 9.8 ppm/K over the automotive range and 4.2 over the commercial one. A third junction supplies the bow's own shape, and subtracting it leaves 18 µV — at an output that is then the band gap itself.

semiconductors · Temperature A field-effect input against a bipolar one biased for each source, at 25 °C. computed by solving, not by drawing. The field-effect input (dark) has 1 mA of drain current and a gate leakage that is 1 pA at 25 °C, from the junction's own law; the bipolar input (pale) is at its best current for each source. It wins between 2.83 kΩ and 5.14 GΩ. At 25 °C the ceiling was 5.14 GΩ; it falls in proportion to the leakage, which the slider's temperature multiplies.

The window that closes from the top

A field-effect input beats a bipolar one biased for each source between 2.83 kΩ and 5.14 GΩ at 25 °C, and the top of that window is twice the thermal voltage over √β times the transistor's gate leakage: it falls as fast as the input junction leaks. Carried through the junction's own law, the ceiling is 39.5 GΩ at 0 °C, 913 MΩ at 50 and 5.79 MΩ at 150 — while the floor does not move from 2.83 kΩ until 200 °C. A 1 GΩ sensor leaves the window at 48.6 °C, a 10 GΩ one at 16.4, about 33 kelvin a decade. The leakage's doubling interval is not ten kelvin but 7.5 at 0 °C and 15.0 at 150, so the rule of thumb is right at one temperature only.

noise · Shot noise An eighteen-segment diode table built at 300 K, carried to other temperatures four ways: the worst error over six decades of current. computed by solving, not by drawing: the two-mechanism junction's law moved with temperature, eighteen equal-ratio minimax segments from 100 nA to 100 mA fitted at 300 K (worst 1.886 mV). Unchanged, the table is out by 163.0 mV at 250 K and 265.5 mV at 400 K. One shift of every drop leaves 47.14 mV and 38.32 mV at 250 and 350 K. A shift per segment leaves 3.768 mV and 3.839 mV — and 14.67 mV at 400 K. Refitting every segment on the same breakpoints holds 1.573 mV to 2.379 mV at every temperature: the breakpoints never need to move.

The table that has to be refitted

Eighteen straight segments model a real diode from 100 nA to 100 mA to within 1.886 mV at 300 K. Carried to 350 K unchanged they are 149 mV out; moved by one common shift, 38.3 mV, because the drops at the bottom of the table fall 3.31 mV a kelvin and those at the top 1.54; moved by a shift each, 3.84 mV. That last correction holds within twice the table's own error over fifty kelvin either way and fails at 400 K, where the bottom of the table has dropped below the junction's own recombination current and is no longer an exponential at all. Refitted on the same breakpoints the table holds 1.57 to 2.38 mV at every temperature: the breakpoints never need to move, and the lines always do.

limits · Diode model

Named alongside it

The objects these essays reach for when they reach for this one.

Model rangeThermal voltageTemperature coefficientIdeality factorOperating pointBand gapMeasurement conditionBulk resistanceDesign tradeoffLeakage currentLinearisationClosed form

All concepts