Circuit Response
every model has an edge, and the figure carries it
Where the models stop

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.
16 min read 8 figures Every model has an edgeThe constants decideMeasured, not tabulated

Assumes: The one current a constant is right at · Two millivolts a kelvin, and the wrong sign

The constant that is a window found that a diode’s two constants are the output of a fit over a window of current, and that the second of them, the saturation current, travels worse than the first: it is an intercept extrapolated back to zero volts from a window that may be six decades away. The same essay attached a temperature law to it in a caption — that it roughly doubles every ten kelvin — which is one of the most quoted rules about a junction, and which four earlier essays had printed, two of them attached to that constant.

The rule is correct. It is about a different current.

A junction has two currents that go by the name. One is the constant in the forward law, the current the exponential is written in multiples of. The other is what a reverse-biased junction actually conducts, the current a data sheet lists as leakage and a textbook introduces as the current the junction saturates at. They come from different mechanisms, they differ on this model junction by five orders of magnitude, and they have different temperature laws — so a number measured as one and used as the other is wrong by a factor, and a rule learned for one and applied to the other can reverse a sign.

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.
Fig. 1 The two current scales of one model junction against temperature, on a logarithmic axis. The forward law’s saturation current doubles every 4.489 K at 300 K; the generation current a reverse-biased junction conducts 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, the current the rule belongs to. At 300 K the scales are 10 fA and 2 nA — this junction’s parameters, not silicon’s — and they meet at 616.8 K.

The current the forward law is written in

The saturation current of the exponential law is a diffusion current. It is set by how many minority carriers the junction’s neutral regions hold in equilibrium, and that is proportional to the square of the intrinsic carrier density. The intrinsic density itself goes as the temperature to the three halves times a Boltzmann factor in half the band gap, so its square carries a cube of the temperature and the whole band gap.

On this junction, with a band gap of 1.12 electron-volts, that current doubles every 4.489 kelvin at room temperature, which is a factor of 4.46 for every ten kelvin. Two exponentials, and where they meet uses exactly that factor for an oscillator’s limiting diodes and is right to.

The current a reverse-biased junction conducts is not a diffusion current. It is generated in the depletion region, where the carrier densities are far below equilibrium and pairs are created at a rate proportional to the intrinsic density itself rather than its square. Its temperature law therefore has half the band gap in its Boltzmann factor, and it doubles every 8.978 kelvin — exactly twice the interval, at every temperature, because halving the exponent halves the logarithmic derivative whatever the temperature is.

So the ten-kelvin rule is a rounded version of the second interval. Nine kelvin, at room temperature, on a junction with this band gap: close enough to ten that the rule is a fair summary of leakage, and more than twice the interval of the current the forward law needs.

What a junction conducts backwards

The single exponential makes a precise prediction about reverse bias. Put a few thermal voltages across the junction the wrong way and the exponential term vanishes, leaving the constant: the current saturates at the saturation current, and stays there at any reverse voltage. That flat line is where the name comes from.

At 300 K the junction conducts 3.1 × 10⁵ times its saturation current backwards. computed by solving, not by drawing. The current a reverse-biased junction conducts, from ten millivolts to a hundred volts, at 300 K. The dashed line is what the single exponential says: the saturation current, 10 fA, reached within a few thermal voltages and flat from there. The solid line is the same model junction with its second mechanism run backwards — generation in the depletion region, over a width that grows as the square root of the bias — and it conducts 3.06 nA at one volt, 3.06 × 10⁵ times as much, and 6.58 times more again by 75 V, so it never saturates. Both scales are this junction's parameters. The same second mechanism sets the resistance at zero bias: 25.9 MΩ, against the 2.59 TΩ the exponential alone gives.
Fig. 2 The current a reverse-biased junction conducts at 300 K, from 10 mV to 100 V. Dashed, the single exponential: 10 fA, reached within a few thermal voltages and flat from there. Solid, the same model junction with its second mechanism run backwards — generation over a depletion width that grows as the square root of the bias — conducting 3.06 nA at one volt, 3.06 × 10⁵ times as much, and 6.58 times more again by 75 V. The zero-bias resistance is 25.9 MΩ, against 2.59 TΩ from the exponential alone.

The generated current does not saturate. It is proportional to the volume of the depletion region, and the depletion region widens as the reverse bias grows, as the square root of the bias for an abrupt junction; between one volt and seventy-five the current rises another 6.58 times. The current called the saturation current is the one that does not flow backwards, and the one that does flow backwards does not saturate.

The size of the gap is this junction’s rather than a law, and it is worth being exact about which part is which. The two scales, 10 femtoamps and 2 nanoamps, are the parameters the model junction was built with in the constant that is a window, chosen to give its forward curve two visible mechanisms. A different junction has different scales. What does not depend on them is the structure: the reverse current is set by the second scale, the forward law by the first, and nothing about measuring one says anything about the other.

The same second mechanism sets a quantity that matters in instruments: the junction’s resistance at zero bias. The exponential alone makes it the thermal voltage over the saturation current, 2.59 teraohms. With the generation current in the model it is 25.9 megohms, a hundred thousand times lower. A clamp diode across a high-impedance input presents that resistance to the source, and the current the instrument draws is what fifty nanoamps into a megohm costs.

Fifty kelvin warmer

At 350 K the junction conducts 1.1 × 10⁴ times its saturation current backwards. computed by solving, not by drawing. The current a reverse-biased junction conducts, from ten millivolts to a hundred volts, at 350 K. The dashed line is what the single exponential says: the saturation current, 7.74 pA, reached within a few thermal voltages and flat from there. The solid line is the same model junction with its second mechanism run backwards — generation in the depletion region, over a width that grows as the square root of the bias — and it conducts 85 nA at one volt, 1.10 × 10⁴ times as much, and 6.58 times more again by 75 V, so it never saturates. Both scales are this junction's parameters. The same second mechanism sets the resistance at zero bias: 1.08 MΩ, against the 3.9 GΩ the exponential alone gives.
Fig. 3 The same junction at 350 K. The single exponential’s saturation current is now 7.74 pA; the generation current is 85 nA at one volt, 1.10 × 10⁴ times as much, still rising 6.58 times by 75 V. The zero-bias resistance has fallen to 1.08 MΩ, against 3.9 GΩ from the exponential alone.

Fifty kelvin multiplies the generated current by 27.8 and the forward law’s constant by 774, so the gap between them closes from 3.06 × 10⁵ to 1.10 × 10⁴. The faster current is catching the slower one, and on this junction it would catch it at 616.8 kelvin, 343.6 degrees Celsius — above the temperature at which silicon is used, so at every temperature a circuit meets, the current named saturation is not the one that flows backwards.

The zero-bias resistance falls from 25.9 megohms to 1.08 over the same fifty kelvin. That is the number behind the familiar observation that a junction-input amplifier’s bias current becomes a problem on a hot day, and it is the leakage law, not the forward one, that sets how fast.

The floor a zero-bias junction sets

A resistance at zero bias is also a noise source, and the name decides that too. The junction that is a resistor at zero volts found that a junction in equilibrium makes exactly the Johnson noise of its zero-bias resistance, whatever it is made of, so the current noise across an unbiased junction is the square root of 4kT over that resistance.

Taken from the forward law’s constant, this junction’s zero-bias resistance is 2.59 teraohms and its current noise 0.080 femtoamps per root hertz at 300 kelvin. Taken from the generation current it is 25.9 megohms and 25.3 femtoamps per root hertz — 316 times more, the square root of the hundred thousand between the two resistances. At 350 kelvin the generation current’s 1.08 megohms makes 134 femtoamps per root hertz, 5.3 times the room-temperature floor for fifty kelvin of warming.

That is the floor a photodiode or a clamp diode puts across an electrometer’s input, and it is set by the current a reverse bias exposes rather than by the constant the forward law is written with. A design that estimated it from a forward extraction would be optimistic by more than two orders of magnitude in noise, and the error could not be calibrated out at one temperature, because the ratio of the two currents is itself a function of temperature.

The rule applied to the other current

The forward drop at a fixed current has a temperature coefficient that is easy to state and hard to get the sign of. The drop is nVₜ ln(I/Iₛ). The thermal voltage rises with temperature, which raises it; the saturation current rises too, which lowers it. The drop’s coefficient is the difference between the two, and the two are of similar size.

The ten-kelvin rule applied to Iₛ gives the drop the wrong sign at 1 mA. computed by solving, not by drawing. The forward drop at 1 mA from 250 K to 400 K, twice. The solid line is the junction's own law, whose saturation current doubles every 4.49 K: the drop falls at −1.809 mV/K at 300 K. The dashed line is the same law with a saturation current that doubles every 10 K instead, which is the rule of thumb usually quoted for it, and at 300 K it gives +0.391 mV/K. The rule's slope is positive at every current above 10.7 µA, and at this current the drop would stay flat only if the saturation current doubled every 8.21 K. The general part is the sign, which follows from the logarithm; the currents are for a saturation current of 10 fA at 300 K.
Fig. 4 The forward drop at 1 mA from 250 K to 400 K, twice. Solid, the junction’s own law, whose saturation current doubles every 4.49 K: the drop falls at −1.809 mV/K at 300 K. Dashed, the same law with a saturation current that doubles every 10 K instead: +0.391 mV/K. The rule’s slope is positive at every current above 10.7 µA, and at 1 mA the drop would stay flat only if the saturation current doubled every 8.21 K.

With the right doubling interval the drop falls at 1.809 millivolts per kelvin, which is the “two millivolts a kelvin” two millivolts a kelvin measures by solving the junction at thirty-one temperatures. With the rule’s interval the saturation current rises too slowly to beat the thermal voltage, and the drop rises, at 0.391 millivolts per kelvin. A diode used as a thermometer, as a reference or as the compensating element for another junction would be designed to the wrong sign.

The balance point is informative. At a milliamp the two terms cancel if the saturation current doubles every 8.21 kelvin, and the leakage interval, 8.98, is on the wrong side of that. So the rule does not merely overstate the interval by a little; it overstates it past the point at which the sign flips, and the leakage current’s own law would have flipped it too.

The ten-kelvin rule applied to Iₛ gives the drop a slope too shallow at 1 µA. computed by solving, not by drawing. The forward drop at 1 µA from 250 K to 400 K, twice. The solid line is the junction's own law, whose saturation current doubles every 4.49 K: the drop falls at −2.404 mV/K at 300 K. The dashed line is the same law with a saturation current that doubles every 10 K instead, which is the rule of thumb usually quoted for it, and at 300 K it gives −0.205 mV/K. The rule's slope is positive at every current above 10.7 µA, and at this current the drop would stay flat only if the saturation current doubled every 11.29 K. The general part is the sign, which follows from the logarithm; the currents are for a saturation current of 10 fA at 300 K.
Fig. 5 The same comparison at 1 µA. The junction’s law gives −2.404 mV/K and the ten-kelvin rule −0.205 mV/K: the right sign, twelve times too shallow. At this current the drop would stay flat only if the saturation current doubled every 11.29 K.

At a microamp the rule survives with the right sign and is off by a factor of twelve, because at a lower current the drop is smaller, the thermal voltage’s term is smaller with it, and less of the saturation current’s rise is needed to beat it. Below 10.7 microamps the rule is a factor; above it, a sign. Those two currents belong to a saturation current of ten femtoamps at 300 kelvin and move with it; what does not move is the structure, which follows from the logarithm — a drop that is the difference of two temperature terms of similar size inherits the sign of whichever one a rule misstates.

The edges that move with the room found this collection quoting boundaries at one temperature and leaving the condition off. This is a sharper case of the same omission, because the condition was not left off: a temperature law was supplied, and it was the law of a neighbouring quantity.

Where the rule is right

The part chosen for picoamps, at the temperature it will run at. computed by solving, not by drawing. A bipolar input's bias current is a base current — a collector current divided by β — and both move slowly and in opposite directions, so it is nearly flat. A field-effect input's is a reverse-biased junction's leakage, which doubles every ten kelvin. At 300 K the field-effect part is 500 times better and the choice looks settled. They cross at 111.4 °C, which is inside the operating range of most things and well inside a self-heating die's. The instruction "use a FET-input part for high source impedance" is a statement about a temperature as much as about an impedance, and it is the one clause that is never in it.
Fig. 6 The input current of a bipolar part and of a field-effect part against temperature. The bipolar input’s bias current is a base current, a collector current over β, and moves slowly; the field-effect input’s is a reverse-biased junction’s leakage and doubles about every ten kelvin. At 300 K the field-effect part is 500 times better, and the two cross at 111.4 °C.

A field-effect input’s gate current is exactly the current the rule was written for: a gate junction held off by a small reverse bias, conducting what its depletion region generates. So the rule is the right one there, and the crossing it produces — a part five hundred times better at room temperature and worse above 111 degrees — is a genuine design fact rather than a rounding. The current that does not reach the input uses it for the same kind of input and is right to.

The distinction, then, is not between a correct rule and an incorrect one. It is between two currents a circuit meets in two different ways. A junction carrying forward current is described by the saturation current of the law, which doubles every four and a half kelvin. A junction held off is described by its generation current, which doubles every nine. The name is shared, the numbers are not, and the rule belongs to whichever current the circuit is actually drawing.

Neither of the currents a fit returns

The constant that is a window fitted the forward curve of this junction over eight one-decade windows and reported each window’s ideality factor. Each fit also returns a saturation current, as its intercept, and those eight intercepts can now be compared with the two scales the curve was built from.

Eight windows, eight saturation currents, and none of them the one the law was built with. computed by solving, not by drawing. The intercept of each of the eight one-decade fits to the two-mechanism curve — the saturation current each returns — against the two scales that generated the curve: 10 fA for diffusion and 2 nA for recombination. Every intercept is above the diffusion scale, by 322 times at the least, and the largest, 7.35 nA, is 3.67 times the recombination one. The window that predicts the forward voltage at 1 mA best (10⁻³ to 10⁻² A) returns 3.22 pA, which is 322 times the diffusion scale and 620 times short of the recombination one. The window whose intercept is nearest the recombination scale returns 1.91 nA and is 36.2 mV out at 1 mA.
Fig. 7 The intercept of each of the eight one-decade fits — the saturation current each returns — against the two scales that generated the curve, 10 fA and 2 nA. Every intercept is at least 322 times the diffusion scale, and the largest, 7.35 nA, is 3.67 times the recombination one. The window that predicts the drop at 1 mA best returns 3.22 pA, 322 times the first scale and 620 times short of the second; the window whose intercept is nearest the recombination scale returns 1.91 nA and is 36.2 mV out at 1 mA.

No window returns either current. The intercepts lie between the two physical scales, most of them far from both, and the one that makes the best forward prediction at a milliamp is three decades above the constant the law was written with. That is the forward-fit version of the confusion in the name: a quantity measured as “the saturation current” is a third number, belonging to a window, and it has no temperature law of its own that anything here establishes. Giving it the leakage current’s rule would be a second error stacked on the first.

Every intercept lies above the diffusion scale for a reason the fit cannot avoid. An intercept is where the fitted line reaches zero volts, and a line with an ideality above one descends towards zero volts more slowly than the diffusion law does, so it arrives higher. The window at a milliamp takes the curve’s 0.6393 volts back to zero at its own slope and covers 19.55 natural logarithms of current on the way, where the diffusion law would cover 25.33; the difference, 5.77, is the factor of 322 between 3.22 picoamps and 10 femtoamps. An ideality about a quarter above one, carried over six tenths of a volt, is three hundred times in the intercept — which is the whole of the reason an extracted saturation current means so little away from the window it came from.

Eight windows, eight saturation currents, and none of them the one the law was built with. computed by solving, not by drawing. The intercept of each of the eight one-decade fits to the two-mechanism curve — the saturation current each returns — against the two scales that generated the curve: 10 fA for diffusion and 2 nA for recombination. Every intercept is above the diffusion scale, by 322 times at the least, and the largest, 7.35 nA, is 3.67 times the recombination one. The window that predicts the forward voltage at 100 µA best (10⁻⁵ to 10⁻⁴ A) returns 1.08 nA, which is 107720 times the diffusion scale and 2 times short of the recombination one. The window whose intercept is nearest the recombination scale returns 1.91 nA and is 7.6 mV out at 100 µA.
Fig. 8 The same eight intercepts, asked about 100 µA instead. The window that predicts the drop there best (10⁻⁵ to 10⁻⁴ A) returns 1.08 nA, 107,720 times the diffusion scale and a factor of two short of the recombination one; the window nearest the recombination scale is 7.6 mV out at 100 µA.

Asked about a hundred microamps, the best window returns an intercept within a factor of two of the recombination scale, because recombination still carries most of the current there. So even the fitted intercept a careful extraction produces changes which of the two physical currents it resembles according to where the question is asked — near the generation scale at a hundred microamps, three decades below it at a milliamp.

The one measurement that tells them apart

At one temperature the two currents cannot be told apart on a forward curve: the forward law’s constant is never visible as a current in its own right, and at every reverse bias the generation current hides it by five orders of magnitude. What separates them is the property this page began with. The forward drop at a fixed current, measured at two temperatures, carries the diffusion current’s temperature law — provided the current is one where diffusion carries it, which on this junction means above a few milliamps, where recombination’s share has fallen below a fifth — and yields a doubling interval near four and a half kelvin. The reverse current at a fixed bias, measured at the same two temperatures, carries the generation current’s law and yields one near nine.

So a saturation current quoted with a doubling interval beside it says which of the two currents it is, and one quoted without one does not. The interval is the cheapest condition to attach to the number, and it is the one that decides whether the ten-kelvin rule is being used on the current it belongs to.

The number on a data sheet

A signal diode’s data sheet prints a reverse current at a stated reverse voltage, and the discussion above says precisely what that number is. It is the generation current, measured at a voltage, growing with that voltage, and doubling about every nine kelvin — and it is not the saturation current of the forward law, which no instrument measures directly and which is five orders of magnitude smaller on this junction.

Using the printed leakage as the forward law’s constant is a mistake that looks harmless because the units match. On this junction it would put the constant 3.06 × 10⁵ times too high at a volt of reverse bias, and the logarithm turns that into a forward drop about 330 millivolts too low at an ideality of one — the thermal voltage times the logarithm of the ratio. The one current a constant is right at found seven-tenths of a volt right at one current and no other; a constant borrowed from the reverse characteristic is right at none.

What this junction does not contain is also worth stating, because a real diode’s leakage has at least one more component. Current along the surface of the die, at the edge of the junction, is set by how the surface was passivated and is often the largest part of a small diode’s reverse current. It is not modelled here, it has no reason to follow either doubling interval, and nothing on this page says how large it is on any real part. The claim is the narrower one: of the two bulk mechanisms, the one the forward law uses and the one reverse bias exposes have different sizes and different temperature laws, and a rule for one is not a rule for the other.

Still open: heat, the collector, and a straight line

The resistance a warm curve hides. Every diode curve so far is at one temperature, and a forward curve taken on a bench is not: at a hundred milliamps the junction dissipates enough to warm, and each point is taken after it has. The drop’s temperature coefficient measured above is exactly the quantity that bends such a curve, and bending the top of a curve is what a fitted series resistance reads. The resistance a slow curve cannot see measures how much of the resistance disappears from the fit.

The terminal the logarithm lives at. The two forward mechanisms share a diode’s terminals, which is why no window fits either of them cleanly. In a transistor they do not share a terminal: diffusion is collected at the collector and recombination is supplied from the base. That separates the two currents this page and the one below it have been disentangling by fitting, and the logarithm is in the collector measures how much longer the exponential law holds once they are apart — and finds the leakage current of this page setting the bottom of the range.

A straight line against a constant. The one current a constant is right at compared a constant drop with the exponential. The model between them, a drop plus a resistance, has a best fit to a pure exponential over any decade of current that is out by ±8.00 millivolts whichever decade it is, where the best constant is out by ±29.76. Measuring that on the real two-mechanism curve would say where a piecewise-linear diode is better than a constant one and where it is worse, which the constant-drop comparison claimed without the numbers.

Part 3 on diode model

One argument about Diode model, and one of 6 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it:

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 Part 2 — 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. A fit to the held curve reads the series resistance falling to nothing at 532 K/W. computed by solving, not by drawing. Each point is a three-parameter fit — a constant, an ideality factor and a series resistance — to the held forward curve between 10 and 100 mA, for a junction built with 0.6 Ω and no temperature coefficient on it, mounted at the thermal resistance on the axis. With no thermal resistance the fit returns 0.580 Ω with a residual of 43.9 µV. At 350 K/W it returns 0.184 Ω, an ideality of 1.086 and a residual of 34.0 µV. The resistance it reports reaches zero at 531.7 K/W and is negative beyond. The resistance a slow curve cannot see Part 4 — A diode's series resistance is read off the top of its forward curve, and a bench curve is a slow one: each point is held until the junction has warmed to it. Through 350 kelvin per watt the held curve sits 39.7 millivolts below the pulsed one at 100 milliamps, and the three-parameter fit that reads 0.580 ohms from the pulsed curve reads 0.184 from the held one — with a smaller residual. The fitted resistance reaches zero at 531.7 kelvin per watt, and the resistance it hid is what keeps the junction from folding back: with 0.05 ohms instead of 0.6 the held curve turns over at 87.6 milliamps.

Where this goes next

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. 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. Where an amplifier's reading comes from, against the source it is reading. computed by solving, not by drawing. Three errors with three different dependences on the source, each measured by a solve with the other two set to zero. The offset voltage is flat — 50 microvolts wherever the source is. The bias current times the imbalance is linear in the source and is what balancing removes. The offset current times the source is linear too and is what balancing leaves. Unbalanced, the current overtakes the voltage at 1.77 kΩ; balanced, at 10.0 kΩ, which is the offset voltage divided by the OFFSET current and is the ratio of the two currents further along. Below about a kilohm, balancing makes the reading worse — the feedback network is already the larger resistance, and equalising means adding to the source. The current the instrument draws Every amplifier in this collection has had inputs that take no current, and that is not an idealisation of a small quantity — it is an idealisation of one whose size is decided by something outside the part. Fifty nanoamps is nothing until it flows in a megohm, and then it is fifty millivolts. The classical cure balances the two resistances and removes the bias current, leaving the offset current: worth a factor of ten, not a thousand, and it costs forty per cent of the noise density to get. 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.

What links here

Essays that reach for this one mid-argument — the half of a link its own author cannot write down.

The objects named here

The third axis, after the field and the idea: the things themselves, and every essay that touches each one.

Band gapDynamic resistanceIdeality factorInput bias currentLeakage currentMeasurement conditionModel rangeOperating pointSaturation currentTemperature coefficientThermal voltage