The floor, which bounds from below

The resistor in the same loop

A resistor's noise and a junction's are equal as bare densities at 49.98 millivolts of drop, and a junction in series with the resistor that carries its current is the arrangement every current source is built from. In one loop each noise current has to cross the other element, so the two supply equal shares at 12.50 millivolts, a quarter of the crossing; at the crossing itself the resistor supplies 80 per cent and the loop is 2.553 decibels below both floors, which is further than it gets anywhere else. The same resistor multiplies the stage's input-referred noise by five.

17 min read 6 figures The constants decide Two routes to a number Every model has an edge

Assumes: The floor a current sets · What a resistor in the emitter buys

The floor a current sets put two noise densities on one axis. A resistor’s Johnson current noise, 4kT/R, has no current in it; a current’s shot noise, 2qI, has no resistance in it; and they are equal when the direct voltage across the resistance is 2kT/q — 49.981 millivolts at 290 kelvin, whatever the resistance and whatever the current.

Two floors on one axis, and the 50.0 mV between them. computed by solving, not by drawing. The flat line is the Johnson current noise of 2 kΩ, √(4kT/R), which has no current in it. The rising line is shot noise, √(2qI), which has no resistance in it. They cross at 25 µA — and the direct voltage across the resistance there is 49.981 mV, which is 2kT/q and contains neither quantity. The slider moves the resistance over six decades; the crossing moves with it and the voltage at the crossing does not move at all, to the last bit of a double. — Fig. 1 The crossing this essay starts from, drawn for two kilohms. The resistor’s bare current noise is 2.830 pA/√Hz; a current’s bare shot noise reaches it at 25 µA; and the drop across two kilohms at 25 µA is 49.981 mV, the same voltage the crossing sits at for every resistance.

That essay was careful to say the two lines describe two different objects. A resistor carrying a direct current has no shot noise of its own, so the crossing is a comparison between a resistor and something else carrying the same current — a junction, a vacuum gap, a barrier — rather than a transition inside one component. It is a statement about two floors standing side by side.

In circuits they rarely stand side by side. The commonest way to set a junction’s current is to put a resistor in series with it, and the commonest way to make a quiet current source is to put a resistor in the emitter of a transistor. Both are one loop with a junction and a resistor in it, carrying one current, and the two floors of the crossing are now both inside it. This essay solves that loop and finds that almost every number the side-by-side comparison suggests about it is wrong, while the crossing voltage itself survives with a different meaning.

A divider each noise current has to cross

A junction biased at I is, for small signals, a resistance rᵈ = Vₜ/I with a noise current of 2qI across it. A resistor R is a resistance with 4kT/R across it. Put the two in series and ask what noise current flows in a short circuit connected around the pair — which is what the output of a current source delivers into a low-impedance load.

Each noise current is a Norton source across its own element. The junction’s sees the resistor in series with the short, so the fraction of it that reaches the outside is rᵈ/(rᵈ + R); the rest circulates back through the junction’s own resistance. The resistor’s sees the junction in the same way, and the fraction that escapes is R/(rᵈ + R). With x = IR/Vₜ, which is R/rᵈ, the two powers that reach the outside are

\frac{2qI}{(1+x)^2} \qquad\text{and}\qquad \frac{4kT}{R}\cdot\frac{x^2}{(1+x)^2} = \frac{4qI\,x}{(1+x)^2}

and because they are independent their total is $2qI\,(1+2x)/(1+x)^2$ .

That is algebra, and algebra is worth drawing only with a second route to it. So the loop is also solved as a netlist: the two resistances in series, the loop closed through a zero-volt source standing in for the short, and each noise current injected as a unit source across its own element, the way what a network answers solves any other network. The netlist knows nothing about dividers or about x.

Fig. 2 A junction carrying 25 µA in series with a resistor, against the drop across the resistor. The flat line is the junction’s bare 2qI and the falling one the resistor’s bare 4kT/R; below them are each one’s share of what reaches the outside, and their total. The shares are equal at 12.50 mV (500 Ω); at 49.98 mV (2 kΩ) the resistor supplies 80.0 per cent and the total is 0.5556 of either bare floor. The solved loop matches the closed form at every drop drawn. Drag the bias current and the resistances change while every ratio stays where it is.

The total never touches either bare line. At a small drop the resistor is nearly a short, the junction’s noise escapes almost whole, and the total is just under 2qI; at a large drop the junction is nearly a short compared with the resistor, the resistor’s noise escapes almost whole, and the total is just under 4kT/R. In between it is below both, and it approaches each of them from underneath.

Equal shares at a quarter of the crossing

The side-by-side comparison says the junction and the resistor are equally noisy at 2kT/q, and it is natural to read that as saying they contribute equally when both are in the loop. They do not. The junction’s share of the total is exactly 1/(1 + 2x), so the shares are equal at x = ½ — a drop of Vₜ/2, 12.50 millivolts, a quarter of the voltage at which the bare floors cross.

Fig. 3 The junction’s share of the loop’s noise power at six drops across the resistor, from the solved loop at 25 µA: 66.7 per cent at 6.25 mV, 50.0 at 12.50 mV, 33.3 at 24.99, 20.0 at 49.98, 11.1 at 99.96 and 4.8 at 249.9. At the drop where the two bare floors cross, the resistor supplies four times what the junction does.

The reason is that the loop is not symmetric in the way the densities are. At the crossing the two Norton currents are equal, but they are shunted by different resistances: the junction’s by its own rᵈ, which at x = 2 is a third of the loop, and the resistor’s by R, which is two thirds of it. Twice as much of the resistor’s noise current escapes as of the junction’s, and power goes as the square, so the resistor supplies four times as much. A smaller resistor moves both things at once: its own noise current rises and the junction’s escapes more easily. They balance at a quarter of the drop, where the resistor’s bare noise power is four times the junction’s and four times as much of the junction’s escapes.

The bench arrangement the floor a current sets describes for building its crossing — two kilohms at 25 µA — is this loop at x = 2. The bare densities are equal there, which is what the crossing says and remains true. The contributions are not: the split is four to one, and the total is five ninths of either bare density in power rather than the sum of two equal ones. The crossing voltage is a statement about densities side by side, and inside a loop it has to be read again.

The crossing survives, as the point of greatest benefit

If equal shares are not at 2kT/q, the crossing voltage might seem to have no meaning inside a loop at all. It has one, and it is a better one.

The loop is below both bare floors everywhere. The useful question for a designer is how far below the lower of them it sits, because that is how much the arrangement buys over simply using whichever element alone would have been quieter.

The loop is 2.55 dB below both floors at 2kT/q, and never further below the lower one. computed by solving, not by drawing. How far a junction-and-resistor loop's noise current sits below the bare junction's 2qI and below the bare resistor's 4kT/R, against the drop across the resistor. Both margins are positive at every drop; the smaller of the two peaks at 49.98 mV, where the two floors are equal, at exactly 10·log(9/5) = 2.553 dB. At 12.50 mV, where the shares are equal, it is 0.512 dB, and at 250 mV 0.616 dB. — Fig. 4 How far the loop’s noise sits below each bare floor, and below the lower of the two, against the drop across the resistor. Both margins are positive at every drop, and the smaller one peaks at 49.98 mV at exactly 10·log(9/5) = 2.553 dB. At 12.50 mV, where the shares are equal, it is 0.512 dB; at 250 mV, 0.616 dB.

The peak is exactly at the crossing, and the reason is short. Below 2kT/q the lower bare floor is the junction’s, and the loop pulls further below it as the drop rises; above 2kT/q the lower floor is the resistor’s, which is itself falling faster than the loop is. The margin rises on one side and falls on the other, so it peaks where the two floors change places, and at that drop x = 2 and the total is (1 + 4)/(1 + 2)² = 5/9 of both. The crossing is not where the two contribute equally; it is where putting them in one loop does the most good.

Two and a half decibels is modest, and it is worth being plain about that. It says the loop is never dramatically quieter than the better of its two parts; what it does is make the choice of resistor forgiving, since at a quarter of the crossing and at five times it the loop is still half a decibel under whichever bare floor is lower. The number that decides a design is elsewhere, and it is the one the next sections are about.

An emitter resistor, at the currents a stage runs at

Twenty-five microamps and kilohms are a bias network’s numbers. A signal stage runs at milliamps with tens of ohms in its emitter, and the loop does not care: every ratio above is a function of x alone, and x is the drop divided by the thermal voltage.

Fig. 5 The same loop at 1 mA. The junction’s dynamic resistance is 25 Ω; the shares are equal with 12.5 Ω in the emitter and the resistor supplies 80.0 per cent with 50 Ω; the total at 50 Ω is 0.5556 of either bare floor. Every curve has moved up by the square root of forty and not one ratio has moved.

That invariance is what makes the result usable without a calculator. An emitter resistor whose drop is 12.5 millivolts has taken half of the stage’s output noise away from the device; one whose drop is 50 millivolts leaves the device a fifth; one whose drop is 250 millivolts leaves it under five per cent. Those are drops a designer chooses for other reasons — what a resistor in the emitter buys prices them in linearity, and how small is small signal is where the linearity comes from — and it turns out each of them has a noise reading attached.

A current mirror is the circuit a current source usually is, and it has two devices rather than one: a diode-connected reference and the output transistor copying it. Solved as two transconductances with their emitter resistors, the reference fed from an ideal current source, the mirror’s output noise is exactly twice the single loop’s at every degeneration tried, from none to a hundred kilohms at 25 µA. The reference device’s noise reaches the output through the shared base with the same weight as the output device’s own. So every ratio on this page carries into the mirror unchanged, and so does the peak at 2kT/q; only the absolute level doubles. The copy, and its two errors is the mirror’s direct-current account, and this is the noise that sits on top of the copy it makes.

The same resistor, read at the input

Everything so far has been the noise current a stage delivers. An amplifier is judged differently: its noise is referred to its input, because that is where it competes with the signal. The floor a circuit has does all its arithmetic there.

A degenerated stage’s transconductance is gₘ/(1 + x), so referring the output noise current to the input divides it by that squared:

\frac{2qI\,(1+2x)/(1+x)^2}{g_m^2/(1+x)^2} = 2kT\,r_d\,(1 + 2x) = 2kT\,r_d + 4kT\,R

The (1 + x)² cancels, and what is left is the bare device’s input noise — the junction’s own half of the Johnson noise of rᵈ, which the floor a current sets derived — plus the resistor’s thermal noise in full, undivided by anything.

One resistor, one noise, two verdicts: quieter as a current source and noisier as an amplifier. computed by solving, not by drawing. A stage biased at 25 µA with a resistor in its emitter, against the drop across it. The noise current it delivers falls as (1 + 2x)/(1 + x)²; the noise voltage it refers to its input rises as 1 + 2x; and the ratio of the two is (1 + x)², the transconductance the resistor took away. At 49.98 mV the output reading is 0.5556 of the bare stage's and the input reading 5.000 times it; at 250 mV, 0.1735 and 21.01. — Fig. 6 One stage at 25 µA with a resistor in its emitter, read at both ends. The noise current it delivers falls as (1 + 2x)/(1 + x)²; the noise voltage referred to its input rises as 1 + 2x; the ratio between them is (1 + x)², the transconductance the resistor removed. At 49.98 mV the output reading is 0.5556 of the bare stage’s and the input reading 5.000 times it; at 250 mV, 0.1735 and 21.01.

One resistor, one noise, two verdicts. At fifty millivolts of degeneration the stage delivers 0.5556 of the noise current it did without the resistor and refers five times the noise power to its input. Both readings come from the same solved loop, both are exact, and they disagree about whether the resistor made the stage quieter by a factor of nine in power — which is the (1 + x)² = 9 of transconductance that the resistor took and that an amplifier has to buy back with gain somewhere else.

Neither reading is wrong, and which one applies is a property of what the stage is for.

Which reading a circuit is asking for

A current source is read at its output. A bias current, a tail current, an active load, a mirror copying a reference: none of them has an input signal to compete with, and the current they deliver is the product. Its noise is what it delivers, and degeneration reduces it at every resistance. The drop across the resistor is headroom the circuit has to find, so the design question is how much headroom a given reduction costs, and the loop answers it in millivolts — 12.5 for half, 50 for a fifth from the device and 2.55 decibels under the better bare floor.

An amplifier’s input device is read at its input. The signal arrives there and is divided by the same (1 + x) the noise is, so what matters is the noise referred to the input, and there the resistor only ever adds — 4kTR, in full, on top of the device’s 2kT rᵈ. That is the price what a resistor in the emitter buys names in a sentence, now with its size: at 50 millivolts of degeneration the input noise power is five times the device’s.

An active load in an amplifier is read at the output, and counts against the input. A mirror used as a load puts its noise current straight into the output node, where the signal current is too; its noise is referred to the input through the input device’s transconductance, not its own. There the load’s own degeneration reduces what it contributes, by the output reading above, and costs headroom rather than gain. So the same resistor that should be small in the input device should be large in the load, and the two instructions are one loop read at two ends.

That last case is the most useful thing on the page, and it is also where the loop’s simplicity runs out, because the load’s contribution then depends on the ratio of two different transconductances. The resistor the noise comes from makes the general form of that point for filters: a resistor’s noise is shaped by everything between it and the output, and a budget that assigns each element a noise without its path is assigning the wrong number.

A volt of drop, and what it no longer buys

Past a few hundred millivolts the loop has become its resistor. The junction’s share is 4.8 per cent at 249.9 millivolts, 1.2 per cent at a volt and an eighth of a per cent at ten, and the total sits at 0.868, 0.964 and 0.996 of the resistor’s bare 4kT/R. What is left to buy is what a larger resistor buys on its own, and that is slow: the current noise power of a resistor carrying a fixed current falls as one over the drop across it. The loop is 7.61 decibels below the junction’s 2qI at 250 millivolts, 13.17 at a volt and 23.03 at ten volts.

That is the exchange rate a current source is actually designed against. The first fifty millivolts buy 2.55 decibels against the better bare floor and hand four fifths of the noise to the resistor; going from fifty millivolts to a volt buys another 10.6 decibels against the junction’s floor; going from one volt to ten buys 9.9 more. A bias network on an ordinary supply can afford the second step and rarely the third, so once the device has left the account the noise of a current source is decided by how much headroom its resistor is given, much more than by which transistor sits above it.

Both routes to these numbers agree to the seven figures printed at every drop from a tenth of a millivolt to ten volts: the netlist, which knows nothing about dividers, and the closed form, which knows nothing else.

Base current, output resistance, and a resistor that is only thermal

The loop on this page is a small-signal loop with three things left out, and each would move a number.

The base current. A real transistor’s base carries I/β, and that current has shot noise of its own, 2qI/β, which flows in whatever resistance the base sees. With an ideal base drive it goes nowhere; with a resistive base network it becomes a voltage at the base and is amplified with everything else. At β of a hundred it is a hundredth of the collector’s in power, which is why the loop can leave it out — and why it cannot be left out of an amplifier’s input, where it is the whole of the current noise generator.

The output resistance. A transistor’s finite output resistance is in parallel with the loop, and a real load is not a short. Both change how much of the noise current reaches the load, by the same kind of divider this page already uses. Neither changes the shares between junction and resistor, because both act after the two have combined.

The resistor’s own excess noise. Every resistor here makes 4kTR and nothing else. A real resistor carrying a direct current can add a component that grows with the current and falls with frequency, and it is exactly a degeneration resistor that carries one. That is a property of the part rather than of the loop, and nothing in this collection measures it.

The temperature is 290 kelvin throughout, which fixes the thermal voltage at 24.99 millivolts, and every millivolt figure on the page scales with it.

Still open: the base current, and a stage made of both

The base current is the next question, and it turns the loop into a design. Leaving the base current out was safe for a current source and is not for an amplifier. The base current’s shot noise is an amplifier’s current-noise generator, the collector’s is its voltage-noise generator, and both are the same current divided by β. That fixes the product of the two generators an amplifier’s noise figure is built from, and turns the collector current into the knob that places the optimum source resistance — the argument of the two generators that are one current.

A stage whose input device and load are degenerated differently. The section on active loads states a rule — small in the input device, large in the load — and stops at the point where it needs two transconductances in one budget. Solving an input pair with a mirror load, both degenerated, and asking for the combination that minimises input-referred noise at a fixed headroom would put a number on the rule. The loop here is one of its two halves.

Noise and matching from one resistor. The same emitter resistor that takes the device’s noise out of a mirror’s output also takes the devices’ mismatch out of its copy, and the mismatch that cancels itself measures the second effect on a population of mirrors. Both improve with the drop and both are paid for in the same headroom, so pricing the two together against the voltage they cost would say whether a mirror designed for accuracy is quiet as a side effect or needs a different resistor.