The refusal, and what it was protecting

The copy, and its two errors builds a current mirror out of two exponential transistors and solves it by Newton’s method on the netlist. Its model declines to be used below two hundred millivolts of collector–emitter voltage, and it says why:

an output at 0.150 V puts the mirror below 0.2 V of collector-emitter voltage, where this model has no base-collector junction and would return a forward-active current for a saturated transistor

The refusal is right and is one of the site’s own habits — a model that declines outside its range rather than returning a plausible number. The list of what that work did not do records it as run in the gate rather than described:

No saturation in the mirror. The model has no base-collector junction, so it refuses below 0.2 V of collector-emitter voltage rather than returning a forward-active current for a saturated transistor. That refusal is run in the gate rather than described.

This essay puts the junction in. Two things come out and only one of them is the expected one.

Both junctions, in the same solver #

A transistor with one junction is what the collection has had until now: an exponential for the base current and an exponential for the collector current, both controlled by the base–emitter voltage. It is exact in forward active operation and has nothing at all to say about a collector that has come down to meet the base.

The transport form of Ebers and Moll is the same model with the second junction restored, and it maps onto this site’s Newton loop without any new machinery — four nonlinear elements instead of two, each controlled by one pair of terminals:

• the base–emitter diode, carrying the forward base current;
• the base–collector diode, which is what conducts in saturation;
• the forward transport current, collector to emitter, an exponential of the base–emitter voltage;
• the reverse transport current, emitter to collector, an exponential of the base–collector voltage.

The one new parameter is the reverse beta: the gain the device has with its collector and emitter swapped. It is one to five for an ordinary planar transistor against a hundred and fifty forwards, because the collector is large and lightly doped and makes a poor emitter. It appears in no small-signal model, on few data sheets, and — as it turns out below — in the saturation boundary.

The refusal was in almost the right place, for the wrong reason #

The comparison that matters is not between the new model and the old essay. It is between the new model and the number the refusal exists to prevent: the forward-active answer, extrapolated into the region where it declines to be used. Both come from the same solver here, with the second junction present or removed, so the difference between them is that junction and nothing else — the Early effect, the base current and the reference are in both.

At two hundred millivolts — the refusal — the extrapolated forward-active answer is 0.0752 per cent low. Seven parts in ten thousand. The model was declining to answer at a point where it was still better than the resistors around it.

The boundary where it is genuinely wrong is lower and is three numbers rather than one:

• 196 mV for a tenth of a per cent,
• 137 mV for one per cent,
• 77 mV for ten.

And it has a closed form, which is one line. With the base–emitter voltage held by the reference, the output current is short by (1 + 1/βᵣ)·exp(−Vcₑ/Vₜ), so the boundary at a stated error is

$$V_{CE} = V_T \ln\!\left(\frac{1 + 1/\beta_R}{\text{error}}\right)$$

a voltage with no current in it — a milliamp mirror and a microamp mirror saturate at the same place — and with the reverse beta in it. Measured against solved, the expression is five per cent low at every level, and the reason is that it holds the base–emitter voltage still while the solve lets it move.

So the refusal’s threshold of 0.2 V was a reasonable guess and is not a computed number. What is computed is that the model it protects survives to 137 mV at one per cent, and that between 137 and 200 millivolts there was an answer available that the model was declining to give.

The failure is not at the output that failed #

The second result is the one worth the essay, and it is not about the saturating transistor at all.

A mirror is usually more than two transistors. One reference sets a base–emitter voltage and several outputs share it, which is what makes a mirror a mirror rather than a current source. So the question that matters when one output saturates is what happens to the others.

The reference current does not move: 0.019 per cent at the worst point measured. That is the reassuring half and it has a one-line reason — a saturated transistor draws more base current, and a base current is a hundred and fiftieth of a collector current, so the extra load on a reference set by a resistor from the supply is nothing.

The base–emitter voltage does move, by 1.75 millivolts. And every output of the mirror is an exponential of that voltage.

So with one output at fifty millivolts and 27.3 per cent down, an output sitting comfortably at five volts is 6.54 per cent down — and 6.54 per cent is exactly what a 1.75 millivolt shift in a thermal voltage of 25.85 gives, which is the check that the mechanism is the one named rather than something else the solve happened to produce.

That is a fault whose symptom is nowhere near its cause. A designer measuring the output that is wrong finds a transistor sitting at five volts, in compliance, with a reference that has not moved, reading six per cent low — and every quantity they can measure about that transistor says it is fine.

Why the reference cannot save it #

The instinct is that the reference should hold the base–emitter voltage, because that is what a reference is for. It does not, and the reason is worth stating because it is a property of the arrangement rather than of the parts.

The reference transistor is diode-connected: its collector is its base, and the node they share is fed by a resistor from the supply. That node is a low-impedance point set by an exponential, so its voltage is whatever makes the reference transistor carry the resistor’s current. When a saturated output steals base current, less current is left for the reference transistor, its collector current falls, and its base–emitter voltage falls with it — logarithmically, which is why 1.75 millivolts is the size of it and not fifty.

A mirror with a beta helper — an emitter follower supplying the base node — divides the stolen current by another beta and takes the shift with it. That is what the helper is for and it is usually justified against the ordinary base-current error rather than against this one. The copy, and its two errors measures the ordinary error; this is a third, and it only exists when one output is out of compliance.

What the compliance specification should say #

A data sheet or a design note gives a mirror’s compliance as a single voltage — usually a couple of hundred millivolts, sometimes the saturation voltage plus a margin. The measurement above says three things about that number.

It is not a property of the current. Vₜ·ln of a ratio has no current in it, and the boundary at one per cent is 137 mV whether the mirror is carrying a microamp or a milliamp. That is unusual enough to be worth saying: almost every other boundary in this collection moves with the operating point.

It is a property of the reverse beta, weakly. A reverse beta of one puts the one per cent boundary at 147 mV and a reverse beta of twenty puts it at 121 — a range of twenty-six millivolts from a parameter that varies by a factor of twenty. The logarithm is doing the work, which is why nobody needs the parameter and also why nobody has it.

And it is a property of the error the designer will accept, which is the part a single number cannot carry. Between 77 mV and 196 mV is a factor of two and a half in headroom and a factor of a hundred in accuracy, and a specification that gives one voltage has chosen the accuracy on the designer’s behalf without saying which.

Where the refusal goes now #

The forward-active model still refuses, and it should. What changes is that the refusal is now a choice between two models rather than a boundary of knowledge: below 200 mV the caller is told to use the saturating model, and the saturating model answers everywhere.

That is the shape this collection’s refusals should have wherever a better model exists. Every model has an edge is the essay about the edges themselves, and the answer that is perfect and absurd is about the other kind of refusal — a network that has no answer at all, where there is nothing to fall back to. A refusal of the first kind is a signpost; a refusal of the second is the end of the road, and it is worth knowing which one is being met.

Two solves, one junction apart #

The measurement’s shape is worth a paragraph on its own, because it is what made the two results separable.

The forward-active model and the saturating one are the same function here with two elements removed. Not two implementations, not a model and a formula: one solver, one netlist, one Newton loop, and a filter that drops the base–collector diode and the reverse transport. So the difference between the two answers at a given output voltage is that junction and cannot be anything else — not a different Early voltage, not a different convergence tolerance, not a different reference.

That matters here more than usual because the quantities are small. At two hundred millivolts the difference is seven parts in ten thousand, and the Early effect over the same span of output voltage is nearly five per cent. A comparison made against the high-voltage current — which is the obvious thing to do and was the first thing tried — measures the Early effect with the saturation buried in it, and reports the mirror as one per cent down at a volt, where nothing is saturating at all.

A bias point is a solution, not a choice is where this collection’s habit of solving rather than assuming an operating point is established, and this is a consequence of it: two models that share a solver can be differenced, and two models that share only a subject cannot.

What a measurement of this looks like #

The saturating output’s own droop is easy: sweep the collector voltage of one output with a source measure unit and watch its current. The curve above is what comes back and the boundary is where it leaves the horizontal.

The sibling’s loss is not easy, and it is the reason the fault survives. It requires measuring one output while a different output is being swept, which means two instruments and a deliberate experiment rather than a characterisation. Nobody does it by accident.

What is done by accident is finding it in a working circuit, and there the symptom is misleading in a specific way: an output current that is low by a few per cent, correlated with something happening elsewhere in the chip, with the reference unchanged and the transistor in question well inside its compliance. The natural conclusion is mismatch — the error that is a distribution is the essay about how large that ought to be — and mismatch does not correlate with what another circuit is doing.

What is not in this model #

No storage time. A saturated transistor has charge in its base that has to be removed before it comes out of saturation, and that is a time rather than a voltage — tens to hundreds of nanoseconds for a small-signal device driven hard. Everything above is a direct-current statement. The switching consequence is the whole reason saturation is avoided in fast circuits, and it is the diode that conducts backwards’s mechanism in a transistor.

No temperature. The boundary is Vₜ times a logarithm, so it is proportional to absolute temperature: 137 mV at 300 K is 106 mV at 233 and 182 at 398. That is a rare case of a boundary in this collection whose temperature dependence is exact and needs no model at all, and it is the opposite direction from the intuition — a mirror has more compliance when it is cold.

And no lateral or substrate transistor. The reverse beta of an ordinary vertical device is one to five; a lateral one on an integrated circuit can be below 0.1, and a substrate transistor’s collector is the substrate, so its saturation puts current somewhere nothing in the schematic accounts for. That is a real failure mode of integrated mirrors and it needs a model with a fourth terminal.

Where else a mirror is asked to go low #

The compliance boundary matters wherever a mirror is the load or the tail of something that swings, which is most places one is used.

A differential pair’s tail sits at a fixed voltage and never approaches it, so the boundary is irrelevant there. A mirror used as an active load is the opposite case: its collector is the output node and swings with the signal, so the boundary is a limit on output swing and the 137 mV above is the honest number for a one per cent load current rather than the 200 the refusal named.

The arrangement that makes it worse is the cascode, and the device that never sees the swing is where its compensation is measured. A cascoded mirror stacks two collector–emitter voltages and its compliance is the sum, which is the price of the output resistance it buys — and the essay that measures the resistance is also the one that owes the price.

What this essay adds to that trade is which number to use for the price. Two hundred millivolts per device, from the refusal, gives four hundred for a cascode; a hundred and thirty-seven at one per cent gives two hundred and seventy-four. On a supply that has come down from fifteen volts to one and a half, that difference is nine per cent of the whole rail.

The habit this belongs to #

A refusal is a claim, and a claim can be checked. This one turned out to be conservative by sixty-three millivolts, which is a small thing, and to be pointing at the wrong output, which is not.

The general form is worth carrying. A model that refuses names where its author stopped believing it, not where it stopped being true, and those are different numbers with no reason to be close. The way to find out is to build the model the refusal was protecting against and take the difference — which costs one extra junction and gives, in this case, a number about a transistor that was never in any trouble.

Two other refusals, and what each was worth #

A refusal placed conservatively costs answers that were available; one placed too late produces plausible wrong ones for a long time first. The collection has an instance of each.

The answer that is perfect and absurd is the late one: a network with a wire written into it as a small resistance returns the right node voltage to fifteen figures, passes both of this site’s verifications with a residual of two parts in $10^{16}$, and reports two hundred thousand amperes. The solver declines it one decade further on, by which time it has been answering for eight decades. That refusal is keyed on a numerical symptom, and a numerical symptom arrives long after the answer stopped meaning anything.

The matrix that is ill, and the answer that is not is the case where refusing would have been wrong: a bridge walked towards balance loses a digit per decade of imbalance on a matrix whose condition number never moves and whose smallest pivot stays four orders above the refusal threshold. Any guard on the matrix would have let it through and any guard tight enough to catch it would have refused a great many good answers.

Which puts this essay’s finding in a useful place between them. A refusal on a modelling question — this element is a stand-in for something the model does not contain — is checkable, because the missing thing can be built and the difference measured. A refusal on a numerical one cannot be checked that way, because there is nothing to build.