The noise the cascode device does add

The device that never sees the swing measured what a cascode’s second transistor buys: a stage fourteen times wider in bandwidth, because the lower device’s collector no longer swings and its base-collector capacitance is no longer multiplied by the stage’s gain. It named what the second transistor costs in headroom, and the source that holds to the supply and the drop the lower device did not need measured that cost and took part of it back.

That first essay also left a question about noise, and gave its reason for asking. The usual argument is that the upper device adds no noise at all: it is a common-base stage, driven from the lower device’s collector, which is a high impedance, so any noise current the upper device generates circulates through the upper device itself and never reaches the load. That essay had just found one place where a standard argument about this arrangement had left out a base current. This page asks whether the noise argument leaves out the same current, and it does.

The stage, for reference #

The stage here is the same one: two transistors at a milliampere each, a current gain of 150, the lower one driven from a 1 kΩ source, and 20 pF of base-emitter capacitance and 2 pF of base-collector capacitance in each. The noise is solved on its small-signal netlist, source by source. Every transistor carries two shot-noise currents — the collector’s $2qI_C$, from collector to emitter, and the base’s $2qI_C/\beta$, from base to emitter — and the source resistance carries its Johnson noise. Each is injected alone as a current, the current it drives into a short at the output is read, and the powers add, since the generators are independent.

Nothing is assumed about which generators matter. That is the point of solving it this way: the standard argument is a statement about which terms are zero, and the solve will say whether they are.

Two currents from one device, going two ways #

At low frequency the two noise currents of the upper device arrive at the output in wildly different proportions. Its collector noise contributes 4.88 × 10⁻⁵ of the lower device’s collector noise power — about one part in $\beta^2$, which is as close to nothing as a standard argument could ask. Its base noise contributes 0.667 per cent, one part in β exactly, which is not nothing at all.

The reason is the path each current has to take. A noise current injected from the upper device’s collector to its emitter must leave the emitter node again, and the node has only two ways out: back through the upper device, which presents about $1/g_m$, or down through the lower device, which presents its output resistance of tens of kilohms. It goes back through the upper device. There it is carried by the upper device’s own transconductance to the collector, where it arrives with the opposite sign and cancels the current that was injected — all of it except the share that left through the upper device’s base, which is one part in β of the current, and so one part in $\beta^2$ of the power.

A noise current injected from the upper device’s base to its emitter has a different fate. It too leaves the emitter node through the upper device, and the upper device carries it to the collector. But its entry point is the base, which is signal ground, so there is nothing at the collector for it to cancel against. It arrives at the output whole. The standard argument’s picture — a noise current going round in a loop — is the right picture for the collector’s noise and the wrong one for the base’s.

The two fates can be written in one line each. Call the admittance the rest of the stage presents at the node between the devices $Y$ — at low frequency, just the lower device’s output conductance $1/r_o$. A unit current injected from the upper collector to its emitter reaches the output as $(1/r_\pi + Y)/(g_m + 1/r_\pi + Y)$: the injected current, less what the upper device’s transconductance carries back round the loop. With $Y$ small, that is $1/(\beta + 1)$ in amplitude and about $1/\beta^2$ in power. A unit current injected from the upper base to its emitter reaches the output as $g_m/(g_m + 1/r_\pi + Y)$, which is $\beta/(\beta + 1)$: very nearly all of it. Both expressions are what the solve returns, and both say the same thing the paragraphs above say in words — that the collector’s noise is cancelled by the loop and the base’s is carried by it.

The base’s noise is not an accident of modelling. It is the partition of the emitter current into collector and base: each carrier that arrives at the emitter leaves through one terminal or the other, independently, and that random division is a shot noise on the collector current equal to the base current’s. The two generators that are one current found the same current supplying an input stage’s current noise. Here it supplies an output noise the cascode cannot remove, because the cascode’s whole mechanism is to pass its emitter current to its collector, and the partition happens on the way.

One part in β, at every gain #

Swept against current gain, the base’s share is one over β at every point, and the collector’s falls as one over β squared: two straight lines on logarithmic axes with slopes of −1 and −2. A transistor with a current gain of twenty — a power device, or a small one at high current where its gain has fallen — adds 4.7 per cent of the lower device’s collector noise power through its base.

What that does to the stage’s noise figure depends on how much of the total the lower device’s collector noise is. Driven from 1 kΩ, most of the noise is the source’s own, and the cascode costs 0.0108 dB at a gain of twenty and 0.0006 dB at a hundred and fifty. From a very small source, where the lower device’s collector noise is most of the total, the cost approaches its ceiling of one part in β of the whole, 0.029 dB at a gain of 150. Either way it is small. The honest statement is “one part in β”, which is smaller than almost anything else in a noise budget, and not “nothing”, which becomes false the moment the frequency rises.

The node the cancellation depends on #

The cancellation of the collector’s noise needed the node between the devices to be a high impedance, so that the injected current had only one way out. At high frequency that node has capacitance on it — the upper device’s base-emitter capacitance, the lower device’s collector capacitances — and a capacitance is a second way out. Once the node’s admittance to ground is comparable to what the upper device offers through its base, the loop is broken and the collector noise stops cancelling.

The figure puts a number on where. The upper device’s collector noise overtakes its base noise at 7.22 MHz. The transfer of the collector noise to the output is $|1/r_\pi + Y|/g_m$, where $Y$ is the admittance of the node excluding the upper device, and the base’s is one; the two are equal where $|1/r_\pi + Y| = g_m/\sqrt\beta$. Solved from the lower half of the stage alone, that admittance reaches $g_m/\sqrt\beta$ at 7.21 MHz, which is the same crossing by a second route.

The 20 pF of node capacitance alone would put the crossing at 25.1 MHz. The difference is the lower device’s own base-collector capacitance, and it is the Miller effect read from the other side. Seen from the lower device’s collector, a current through its 2 pF of $C_\mu$ moves its base, and the lower device’s transconductance answers that movement at its collector, in the direction that draws more current. From the collector the capacitance looks larger by roughly one plus the lower device’s transconductance times the impedance its base sees — a factor of about thirty here. The cascode was built to stop the lower device’s collector moving so that its $C_\mu$ would not be multiplied as seen from its base; nothing stops $C_\mu$ being multiplied as seen from its collector, and that is the node the noise argument depends on.

Cut the node’s own capacitance by a factor of ten, to 2 pF, and the crossing moves only from 7.2 to 11.6 MHz, where the node capacitance alone would have moved it to 251. The lower device dominates its own collector’s admittance. So a designer trying to preserve the cancellation by making the cascode device small, or by laying the node out tightly, recovers very little of it; the capacitance that matters is inside the lower transistor, and its size is set by the gain the lower transistor was put there to provide.

Above the crossing the upper device’s collector noise keeps rising. It passes the lower device’s own collector noise near 316 MHz — around the transistors’ transit frequency, where the node’s admittance is about $g_m$ and the cancellation is gone entirely — and above that the cascode device is the stage’s largest noise source.

A lower current gain #

At a current gain of thirty the low-frequency picture is five times worse in the base’s share, 3.33 per cent, and the collector’s cancellation is only a factor of a thousand rather than twenty thousand. The crossing moves up, to 31.5 MHz, because the base’s share that the collector noise has to overtake is five times larger, so the node’s admittance has to reach $g_m/\sqrt{30}$ rather than $g_m/\sqrt{150}$ before it does. A low-gain device is noisier at low frequency and no worse at high: the two regimes do not trade against each other in the way a single number would suggest.

The noise figure, at the frequency the cascode is bought for #

At 1 kHz the two stages’ noise figures lie on top of each other across four decades of source resistance, never more than 0.0164 dB apart. At 100 MHz they separate, and the cascode is up to 0.55 dB noisier, most where the source resistance is high and the upper device’s collector noise, now uncancelled, is largest against everything else.

That is the uncomfortable conclusion. The cascode is used for bandwidth. The device that never sees the swing measured it taking a stage from 56 kHz to 787 kHz, and the arrangement’s whole reason to exist is the high-frequency end of a response. The standard noise argument is right precisely at the frequencies where nobody needs a cascode, and it fails — through the lower device’s own multiplied capacitance — in the band the cascode was bought to extend.

What a designer can do about it #

The analysis says which knobs move the crossing and which do not. Making the cascode device smaller, or tightening the layout of the node between the devices, reduces the node’s own capacitance and, as the 2 pF figure shows, buys very little, because the node’s admittance is mostly the lower device’s multiplied $C_\mu$. The multiplication is roughly $1 + g_m Z_b$, with $Z_b$ the impedance the lower device’s base sees, so the knob that moves it is the source: a stage driven from a low impedance multiplies its $C_\mu$ less, and keeps the upper device’s collector noise cancelled to a higher frequency than one driven from a high impedance. That is the opposite of what the cascode’s usual justification suggests, which is that it makes the stage indifferent to its source.

A higher current gain in the upper device helps at every frequency. It divides the base’s share by β at low frequency and moves the crossing down, but the base’s share it has to overtake is smaller in the same proportion, so the total the upper device adds is smaller everywhere. A field-effect cascode device has no base current at all, which removes the partition noise entirely; its channel noise then behaves like the bipolar device’s collector noise, cancelled below the crossing and not above it. That is one reason a field-effect device is attractive in the upper position even in an otherwise bipolar circuit.

What the argument should say instead #

The upper device’s noise reaches the output in two parts. Its base’s partition noise arrives whole at every frequency, and is one part in β of the lower device’s collector noise: small, and never zero. Its collector’s noise is cancelled by the lower device’s high output impedance, to one part in $\beta^2$, up to the frequency at which the admittance at the node between the devices reaches $g_m/\sqrt\beta$ — and that admittance is dominated by the lower device’s own base-collector capacitance, multiplied by the lower device’s gain, so the frequency is set by the lower transistor rather than by the cascode device or the layout.

The frequency a device sets for itself measured the Miller multiplication from the base’s side as the thing that limits a common-emitter stage’s bandwidth. The same multiplication, seen from the collector, is what limits the cascode’s noise immunity, and the cascode removes one without the other. The floor a current sets measured shot noise as a property of a current alone; here the relevant current is the base’s, and the arrangement cannot be designed around it except by raising β.

How the numbers were obtained #

The stage is a small-signal netlist: each transistor a transconductance $g_m = I_C/V_T$, an input resistance $r_\pi = \beta/g_m$, an output resistance $r_o = V_A/I_C$ with $V_A$ = 80 V, and the lower device’s 20 pF base-emitter and 2 pF base-collector capacitances, with the named capacitance at the node between the devices. Each noise source is a unit current source injected between the nodes named, the circuit is solved at each frequency by nodal analysis with a zero-volt source holding the output at ground, and the current through it is read; each contribution is that transfer squared times the source’s density, $2qI_C$ for a collector, $2qI_C/\beta$ for a base and $4kT/R_s$ for the source. The noise figure is the total over the source’s own contribution. The node admittance is solved separately on the lower half of the stage alone, by injecting a unit current at the node and reading its voltage, and the crossing frequency is bisected on it.

What it leaves out #

Base resistance. Each transistor’s base has an ohmic resistance with its own Johnson noise, and the upper device’s matters because its base is not a perfect ground: its noise voltage appears at the upper base and is attenuated by the high-impedance emitter node, as a base voltage in a common-base stage is, until the node’s capacitance lowers that impedance too. It would add a third contribution with the same high-frequency behaviour.

Correlation. A transistor’s collector and base shot noises are treated as independent, which is right at low frequency and becomes wrong near the transit frequency, where the two are partly the same carriers observed at different times. The 316 MHz end of the figures is where that matters.

And the load. The output is a short here, which is how a noise current is defined; a real load with its own capacitance changes how the upper device’s collector noise divides between the load and the node, and moves the high-frequency end of every curve.

Still open: the base resistance at the upper base, correlated partition noise, and the cascode in a mirror #

The upper base’s resistance. Driving the upper base through a bias network with a finite impedance, rather than holding it at signal ground, puts that network’s noise at the upper base. At low frequency the high-impedance emitter should attenuate it by about $g_m r_o$; the frequency at which the node’s capacitance ends that attenuation should be the same one found here, and measuring whether it is would say how much bypass the upper base really needs.

The partition noise near the transit frequency. With the collector and base noise correlated by the carriers’ transit time, the base’s share at high frequency is no longer simply one part in β, and the collector’s cancellation, which relies on the two being distinct, changes too. Adding the correlation would say whether the 316 MHz crossing moves.

A cascode mirror’s noise. The source that holds to the supply cascodes a current mirror, where the reference branch’s own noise is mirrored to the output and the upper output device adds its partition noise. Which of the two dominates, and whether the wide-swing arrangement’s separate bias adds a third source, is the same calculation on a larger netlist.