The one current a constant is right at

The constant-drop diode model is the first model most engineers meet and the one they keep using longest. Its usual defence is that it is crude but adequate, and its usual criticism is that the drop is not a constant — it moves 59.5 millivolts for every factor of ten in current, so a model that picks one number is wrong by tens of millivolts over any real range.

Both of those are about the device. Neither is about the model, because the model is not used to predict a voltage. It is used to predict a current, and the error in the current is a different quantity with a different — and much better behaved — dependence.

The identity #

Write $V_s$ for the supply, $R$ for the series resistance, $V_t$ for the true forward drop that the exponential and Newton’s method produce together, and $V_m$ for the constant the model assumes.

The true current is $(V_s - V_t)/R$. The modelled current is $(V_s - V_m)/R$. So the fractional error in the current is

$$\frac{i_m - i_t}{i_t} = \frac{V_t - V_m}{V_s - V_t}$$

and the resistance cancels out of it entirely. That is exact algebra rather than an approximation, and it is checked here against the solved operating points at ninety-six supplies through four resistors — three hundred and eighty-four solves, worst departure below $10^{-9}$.

Everything else in this essay is a reading of that fraction. Its numerator is a property of the device and the constant; its denominator is the headroom the circuit leaves above the drop; and the resistance appears only through what current it happens to produce.

The one current #

Set the numerator to zero. The model is exact when the true drop equals the constant, which happens at one current: the one for which $V_T \ln(i/I_S) = V_m$.

For a saturation current of $10^{-14}$ A at room temperature and a constant of 0.7 V, that current is 5.748 mA.

It is a property of the diode and the constant, and of nothing else. In the figure four curves cross zero at four different supplies — 1.27 V through 8.7 Ω, 6.45 V through 870 Ω, and so on — and every one of those crossings is at 5.748 mA, to five figures. The resistor decides where on the supply axis the crossing happens and has no say in what current it happens at.

Move the constant and the current moves with it, exponentially:

Fifty millivolts is a factor of 6.92 in current, which is $e^{0.05/V_T}$ — the same exponential running backwards. So the choice of constant is a choice of the current at which the design is exact, and the sensible way to make it is to pick the operating current first and read the constant off the device.

The error changes sign at that current, which is the thing the usual phrasing cannot express. Below it the model predicts too little current; above it, too much. “The constant-drop model overestimates the current” is true of exactly half the axis.

Headroom, not current #

Now the denominator, which is what makes the model useful rather than crude.

Through 87 Ω the true drop is 49 mV away from 0.7 V at a supply of 0.725 V and 147 mV away at 150.7 V — a factor of three, and in opposite directions. The error in the current over the same range runs from −66% to +0.10%, a factor of 674, because the headroom underneath has grown by a factor of two thousand.

That is the sentence worth keeping: the accuracy of the constant-drop model is set by how much of the supply the diode is taking, and hardly at all by the diode.

A five-volt rail through a kilohm — the most ordinary bias arrangement in the subject — has the model right to two parts in a thousand, from a device whose drop is nine millivolts away from the number assumed. A one-volt rail through the same resistor has it 19% wrong.

So the model’s edge is a ratio, not a current: the error is one per cent when the drop’s own error is one per cent of the headroom. On the 87 Ω curve that is a supply of 1.12 V, which is 1.6 times the drop; on the 8.7 kΩ curve it is 6.5 V, which is ten times the drop, because at that resistance the operating current is low and the drop is further from 0.7 V to begin with.

What the model has no opinion about #

Three quantities the constant-drop model does not contain at all, and each of them is an ordinary design question.

The dynamic resistance. A constant-drop diode has zero incremental resistance: change the current and the drop does not move. The real one has $V_T/I$ — 2.6 Ω at 10 mA, 25.9 Ω at 1 mA and 259 Ω at 100 µA. In a reference or a bias string that resistance is the whole of what the diode is doing, and the model says it is zero.

The temperature coefficient. −2 mV/K, which over a hundred kelvins is 200 mV: three times the distance the constant is ever wrong by within a decade of current. The semiconductors field measures it, and it is the reason the exact-at-5.748-mA result above should be read as “at room temperature” throughout.

And any low-current behaviour. Below a microamp the exponential is still an exponential and the drop is around half a volt rather than 0.7. A one-volt supply through a megohm gives a true drop of 0.4603 V and a current of 539.7 nA; the constant-drop model predicts 300.0 nA, which is 44% low — by far the worst error in this essay, and it comes from a circuit nobody would call marginal.

How wide the constant is any good over #

The other way to read the numerator is on its own, without a circuit: over what range of current is 0.7 V within a stated fraction of the true drop?

Half a decade for two per cent. That is the number behind the criticism the constant-drop model usually gets, and taken alone it looks damning: a part running anywhere outside three to ten milliamps has a drop the model is more than two per cent wrong about.

Set it beside the table of current errors above and the two are hard to reconcile. At five volts through a kilohm the current is 4.31 mA — inside the two per cent window, and the current error is 0.17%. At twenty-four volts the current is 23.3 mA, at the very edge of the five per cent window, and the current error is 0.16%: better, not worse.

The reconciliation is the identity. Moving up the axis makes the numerator worse and the denominator better, and the denominator wins because it grows in proportion to the supply while the numerator grows only as its logarithm. A model of a device can get worse while a model of a circuit built from it gets better, and there is no way to see that from a specification of the device alone.

Two models, and which is worse #

It is worth putting the constant-drop model beside the other simple one, because the comparison is not the expected way round.

The ideal-diode model — zero drop when conducting — is the constant-drop model with $V_m = 0$, so the identity gives an error of $V_t/(V_s - V_t)$: 16% at five volts through anything, 3% at twenty-four. It is worse than the constant-drop model everywhere and by a factor of about a hundred at low supplies, which is why nobody uses it for a bias calculation.

The piecewise-linear model — a constant plus an incremental resistance — is better than the constant-drop model near its design point and no better far from it, because far from it the term the resistance corrects is not the term that is wrong. Its numerator is $V_t - V_m - r(i - i_0)$, which is second order near $i_0$ and first order elsewhere, and the range over which it beats the constant is about a decade of current either side.

The one that is not on this scale at all is the exponential itself, which is what Newton solves and what every number in this essay was measured against. It costs about nine iterations from a cold start and two from a warm one, which is why every field on this site that needs a bias point uses it.

Two diodes, and the error that does not cancel #

Almost nothing in practice has one diode in it. A bridge rectifier has two in the path at any instant, a series-pass reference has three or four, and each contributes a drop the model gets wrong by the same amount in the same direction — because they carry the same current.

That is worth saying because the instinct is that errors accumulate randomly. They do not here. Two diodes in series at 5.748 mA are both exact; two at 100 µA are both 105 mV low, so the model is 210 mV low on the pair, and the identity’s numerator has doubled while its denominator is unchanged. A string of n diodes multiplies the current error by n at any current away from the exact one.

The direction of the effect is fixed, which makes it correctable and makes it worth correcting. A bridge feeding a reservoir from a twelve-volt peak has two drops in the path, so the headroom is $12 - 2V_t$ rather than $12 - V_t$, and the current error is $2(V_t - 0.7)/(12 - 2V_t)$. At a hundred milliamps through the diodes the true drop is 0.774 V, so that is $2 \times 0.074/10.45$, which is 1.4% — small, one-directional, and about twice what a single-diode estimate would have given.

The general form is worth writing down because it is the only place the resistance genuinely leaves the answer: for n identical diodes carrying the same current,

$$\frac{i_m - i_t}{i_t} = \frac{n(V_t - V_m)}{V_s - nV_t}$$

which grows faster than n because the denominator shrinks as the numerator grows. Four diodes of 0.8 V in a twelve-volt path, with a hundred millivolts of error each, are 4.5% out where one of them would have been 0.9%.

Where the model earns its place #

Rectifiers and clamps. A bridge rectifier from a twelve-volt transformer is running with sixteen volts of headroom and two drops in the path: the model is right to a per cent and there is nothing to be gained from anything better.

Current-limiting a light-emitting diode. A five-volt rail, a two-volt device and a resistor. The headroom is three volts, the drop’s uncertainty over the batch is a hundred millivolts, and the model’s error is 3% — which is exactly the number a designer needs, and it comes from the spread of the part rather than from the model.

And nowhere near a reference. A diode string used as a voltage reference is being used for the value of the drop itself, with no headroom in the answer at all: the identity’s denominator is the quantity being measured. There the model is not crude, it is empty.

The other models quoted as constants #

A constant standing in for a logarithm is exact at one operating point, and this collection meets the same shape three more times. A bias point is a solution, not a choice is the root the constant stands in for. The constant that is a window is the ideality factor, quoted as a number and defined as a derivative. Two millivolts a kelvin, and the wrong sign is the same drop’s temperature coefficient, which is also quoted as a constant and is also a function of the current. Every model has an edge is where all of them are gathered, and How small is small signal is the amplitude boundary the tangent at that one current has.

The gate #

The identity is asserted against the solved operating points — the current error is exactly the drop’s error over the headroom — to $10^{-9}$ over three hundred and eighty-four solves. It is the essay’s spine and it is algebra, so the tolerance is a rounding bound rather than a physical one.

Every curve is required to cross zero at the same current, to a part in $10^4$, which is the claim that the exact point is a property of the device and the constant rather than of the resistor.

And the drop at that current is asserted to be the constant itself, to a microvolt, which closes the loop: the crossing was found by bisecting on the supply and the check reads the drop back out of the device’s own law.

What a constant standing for a curve costs elsewhere #

A model that is exact at one point and wrong away from it is the commonest object in this collection, and three later measurements are what happens when this particular one is used outside its point.

A bias point is a solution, not a choice is the immediate sequel and the mildest case: solved properly the same junction drops 0.692544 V from a five-volt supply through a kilohm and 0.754459 V from forty-eight volts through the same kilohm, because the drop moves about sixty millivolts for every decade of current. Sixty millivolts on seven hundred is under nine per cent, which is why the constant survives in ordinary design at all.

The direct voltage that is a sawtooth is where the same constant is used two and a half decades away from its point and in both directions inside one period. A rectifier’s diode carries two amperes for a twentieth of each cycle and nothing for the rest, so it sits at about 0.8 V during conduction and nowhere at all otherwise — and the loss in the supply is therefore not the rated current times 0.7, which is the arithmetic a parts list is usually sized with.

And the constant that is a window is the sharper version of this essay’s own argument applied to the other number in the same model. An ideality factor is quoted as a number and defined as a derivative, so 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. Asked for the forward voltage at a milliamp, the best window is right to a third of a millivolt and the worst is out by 186, which is a current a hundredth of the truth.

The pair is the useful thing to carry. Seven-tenths of a volt is a constant with one correct current; an ideality factor is a constant with one correct decade; and a design that uses both is using two single-point approximations whose points are not the same point.

Why the resistor decides and the diode does not #

The result that makes this essay worth an essay is the second one: what decides whether the constant-drop model is any good is not the diode at all but how much of the supply the diode is taking. Four different resistors at four different supplies all cross zero error at the same current, because the crossing is a property of the device’s own law — and away from the crossing the error is set by how steeply the circuit’s load line cuts that law.

That is the same structure as several other results here and it is worth naming, because it decides where to look when a model disappoints. A model’s exactness is a property of the model; its usefulness is a property of the circuit it is embedded in. The divider, and the thing it does not know about is the plainest instance — a ratio that is exactly right about an unloaded divider and predicts nothing about a loaded one, with two dividers of identical ratio giving six volts and one volt into the same load — and exact outside and wrong within is the sharpest: six elements reduce to one source and one resistor that no load can distinguish from them, to the last bit of a double, while the reduction is wrong about the heat by a factor of forty-three.

In all three the question “is this model accurate?” has no answer until somebody says what is being asked of it, and in all three there is a circuit for which the answer is exactly. What makes this one worth opening a field with is that the exact case is a single point rather than a range: there is one current at which the constant is right, it is 5.748 milliamperes, and every design that uses the constant is somewhere else on the same curve.