The resistor the upper base is held through

The noise the cascode device does add solved a cascode stage’s output noise source by source and found that the usual claim — the upper device adds nothing — is right about its collector noise and wrong about its base noise. The collector’s shot noise circulates inside the upper device and reaches the output at about one part in $\beta^2$, while the base’s arrives whole, one part in β. It also found where the collector’s cancellation ends: at 7.2 MHz, where the admittance the lower half of the stage presents at the node between the devices reaches $g_m/\sqrt{\beta}$.

That essay drew the upper base where every textbook draws it, at signal ground. On a board it is not. It is held at a bias voltage by a network — a divider, a diode string, a follower — and the network has an impedance, and the impedance has noise. The essay’s closing question was what that noise does: whether the upper device’s high-impedance emitter attenuates it, by how much, and whether the frequency at which the attenuation ends is the same one it had just found. This page answers all three, and the third answer is no.

The stage, with a resistance where the ground was #

The stage is the same as before: two transistors at 1 mA with a current gain of 150 and an Early voltage of 80 V, so each has $g_m$ = 38.7 mS and $r_o$ = 80 kΩ; 20 pF at the node between them; 2 pF between each collector and its base; a 1 kΩ source; the output read as the current into a short. The only change is that the upper device’s base is fed through a resistance R to signal ground instead of being tied there. The upper device’s own collector-base capacitance now matters, since it runs from that base to the output, and the output is held at signal ground by the short.

Every source is injected alone and the current it delivers to the output is read, so nothing about which of them matter is assumed. The new source is the resistance’s own thermal noise. The old ones are all still there, and one of them — the upper base’s shot noise — now flows in the resistance too.

At low frequency the resistance is almost silent. Its noise reaches the output at 7.97 × 10⁻⁶ of the lower collector’s noise, and the reason is the one the standard argument would give. The upper device is an emitter follower from its base to the node between the devices: a noise voltage at the base appears, almost unchanged, at that node. What turns the voltage into a current at the output is whatever sinks current from the node, and at low frequency that is only the lower device’s output resistance. A voltage v at the base becomes a current $v/r_o$, and against the lower device’s own noise the ratio of powers is

$$\frac{4kTR/r_o^2}{2qI_C} = \frac{2R}{g_m r_o^2},$$

which is 8.08 × 10⁻⁶ for 1 kΩ — the solve agrees to one and a half per cent. In voltage terms the resistance is divided by $g_m r_o$, 3,094 here, the device’s intrinsic gain. A kilohm at the upper base, referred to the input, is worth about a tenth of a milliohm at the lower one.

The upper base’s own shot noise, now in a resistor #

The same figure has a second curve that the earlier essay also had: the upper base’s shot noise, 0.683 per cent of the lower collector’s, a shade above the 0.667 per cent — one part in β — it was with the base held. The difference is the resistance.

With the base held at ground, the base’s shot current had nowhere to go but out through the emitter, and it arrived at the output whole. With a resistance there it still does that, and it also develops a voltage i·R across the resistance, which the follower carries to the node and $r_o$ turns into a current, in phase with the first. The two add as amplitudes, so the base’s share is one part in β multiplied by $(1 + R/r_o)^2$. At 1 kΩ that is a factor of 1.025; at 100 kΩ, 5.06, and the base’s shot noise is 3.32 per cent of the lower device’s instead of 0.667.

This term grows as $R^2$, and the resistance’s own thermal noise grows as R, so beyond some resistance the base current flowing in the resistor is the louder of the two voltages across it. They are equal where

$$\frac{2qI_C}{\beta}R^2 = 4kTR \quad\Rightarrow\quad R = \frac{2\beta}{g_m} = 2r_\pi,$$

7.76 kΩ here. Above that a bias network’s noise is mostly not its own: it is the upper device’s base current, converted to a voltage by the network’s impedance. The two generators that are one current made the same point about an amplifier’s input, where a base current’s shot noise flowing in the source resistance is what the current-noise term means; the upper base is an input too, to a follower whose output is the node the lower device works into.

At low frequency, then, the bias network is nearly harmless and the textbook picture survives with one small correction. The rest of this page is about how early that stops.

Where the attenuation ends #

The division by $g_m r_o$ lasts only while the node between the devices is a high impedance to everything except the upper device — while the lower half of the stage sinks current from it through $r_o$ and nothing else. The node also carries the 20 pF and, through the lower device’s collector-base capacitance, a Miller-multiplied copy of that capacitance, which the earlier essay found by solving the lower half alone. The follower’s voltage becomes a current in proportion to the whole admittance, not to $1/r_o$.

At $1/r_o$ = 12.5 µS the threshold is tiny, and the node’s capacitance passes it early. The admittance has risen by $\sqrt{2}$ at 23.7 kHz, and the resistance’s share of the output has doubled by 23.5 kHz — the two agree, which is the check that the admittance is the whole story. Above that the share rises as the square of the admittance, twenty decibels a decade at first.

That is the answer to the earlier essay’s question, and it is not the frequency the question expected. The upper collector’s noise needed the admittance to reach $g_m/\sqrt{\beta}$, 3.16 mS, before it stopped cancelling, and that happened at 7.21 MHz. The bias network’s noise needs only $1/r_o$, which is 253 times smaller, and so its attenuation begins to fail 307 times lower in frequency. The two effects share a node and a mechanism and nothing else. One is a current trying to circulate and needing a low impedance to leak away through; the other is a voltage trying to become a current and needing only a modest admittance to do it.

It does not follow that the bias network matters at 24 kHz. At that frequency its share is still 1.6 × 10⁻⁵ for 1 kΩ, and it has a long way to rise. What follows is that the attenuation which makes it negligible is not a fixed factor of $g_m r_o$ but a falling one, and the frequency at which it matters depends on the resistance.

How far it rises, and for which resistance #

In noise figure, which is the number a designer compares, nothing happens at audio frequencies: at 1 kHz no resistance from 100 Ω to 100 kΩ costs a hundredth of a decibel. By a megahertz 100 kΩ costs 0.58 dB, most of it the upper base’s shot current in that resistance; 1 kΩ still costs only 0.0013. By 10 MHz, 10 kΩ costs 0.98 dB. And at 100 MHz — a third of the devices’ transit frequency, which is as far as the small-signal model is trusted here — an unbypassed 1 kΩ costs 6.07 dB, which is the stage’s noise figure made several times worse by a component nobody drew.

The frequency at which a given resistance passes the lower device’s own noise moves with it: 96.5 MHz for 100 Ω, 15.2 MHz for 1 kΩ, 2.97 MHz for 10 kΩ. The slider on the figure at the head of the page steps through them, and at 100 kΩ the resistance’s own thermal noise never passes — its base current’s noise, in that resistance, does it instead.

The middle is the worst place to be #

The last row of that list suggests something the cost figure shows plainly: at 100 MHz the cost is not monotonic in the resistance.

Ten ohms costs 0.29 dB. The cost rises to 6.07 dB near a kilohm, and then falls: 3.17 dB at 10 kΩ, 2.28 at 100 kΩ, 2.18 dB at a megohm. A larger resistance is better at this frequency, and the reason is a capacitance the schematic does not show as a bypass. The upper device’s own collector-base capacitance, 2 pF, runs from its base to its collector, and its collector is the output, held at signal ground. At 100 MHz that capacitance is 796 Ω, so above a few kilohms it is the capacitance and not the resistance that sets the base’s impedance, and the resistance is already bypassed by the device itself. What remains is mostly the base’s shot current flowing in that 796 Ω, and it is what holds the large resistances near 2 dB.

It is worth saying plainly what that does not mean. A megohm is not a good bias network: at a megahertz, 100 kΩ already costs 0.58 dB, because at that frequency 2 pF is 80 kΩ and bypasses nothing. The non-monotonic curve is a statement about one frequency, and at every frequency the useful quantity is the same one: the impedance at the upper base, whatever sets it.

What the upper base actually needs #

That turns the question the earlier essay asked — how much bypass does the upper base need — into a question with a number for an answer.

A capacitor across the resistance lowers the base’s impedance wherever the capacitor’s reactance is below the resistance. Ten picofarads, 159 Ω at 100 MHz, leaves 1 kΩ costing 0.60 dB. A hundred picofarads, 15.9 Ω at that frequency, brings 1 kΩ to 0.016 dB, 10 kΩ to 0.010 and 100 kΩ to 0.009 — all of them free, and the same whatever the resistance, because none of them is what the base sees any more.

So the design rule is short and has a number in it. The upper base needs an impedance to signal ground of the order of ten ohms at the highest frequency the stage is expected to be quiet at — the cost at 100 MHz of a bare 10 Ω was 0.29 dB, and of 15.9 Ω of capacitance 0.016, the difference being that a capacitor makes no noise of its own and a resistance does. Below a few tens of kilohertz nothing is needed at all, since the device’s own $r_o$ divides the network’s noise by three thousand. In between the requirement rises steadily, and a bias network designed for audio and reused in a stage meant to work at tens of megahertz will pay for it in decibels.

The rule also says something about the networks people actually use. A resistive divider from the supply has a Thévenin resistance of kilohms and sits squarely in the worst part of the curve. A string of diodes carrying a milliampere has about 26 Ω per diode of dynamic resistance, and the junction that is a resistor at zero volts measured a forward-biased junction making half the noise power of a resistor of that value — so two diodes are roughly a 52 Ω resistance making the noise of 26 Ω, well down the low side of the curve and within a factor of a few of the ten-ohm target. They still want the capacitor at the top of the band, but a much smaller one: the requirement is an impedance, and a diode string starts most of the way there.

The reason usually given for bypassing the upper base is bandwidth: the device that never sees the swing measured the cascode’s fourteen-fold bandwidth with its upper base at ground, and that measurement assumed the ground. The noise argument arrives at the same capacitor from a different direction, and it says how large it has to be.

Two routes to one number #

The $2R/(g_m r_o^2)$ share at low frequency is a formula; the solve does not use it. It injects a unit current across the resistance, solves the stage’s small-signal network at that frequency, and reads the current into the output short, then multiplies by 4kT/R. The two agree to one and a half per cent at a kilohm and depart at the largest resistances, where the follower’s own input resistance — $r_\pi + (\beta + 1) r_o$, about 12 MΩ — begins to divide the voltage before it reaches the node. The $(1 + R/r_o)^2$ correction to the base’s shot noise is checked the same way.

The corner is two routes as well. The frequency at which the resistance’s share has doubled is read from the solved output noise; the frequency at which the node admittance reaches $\sqrt{2}/r_o$ is read from the lower half of the stage alone. The first is 23.5 kHz and the second 23.7, and the agreement is what licenses the explanation — the admittance at that node, not anything about the upper device, decides where the attenuation fails.

The stage’s noise figure is the ratio of the total output noise to the part of it due to the source resistance alone, so every cost in decibels is a ratio of two solves of the same stage, one with the resistance and one without, at the same frequency and source.

What was left out #

The transit time. Every number above 30 MHz comes from a hybrid-π model with constant capacitances and frequency-independent current gain. At a third of the transit frequency the model is still describing a device, but the base’s shot noise and the collector’s become correlated as the carriers’ transit time becomes a noticeable fraction of a cycle, and the base’s share is no longer one part in β. The earlier essay named that correlation as open, and it is open here too.

The base’s own spreading resistance. A real transistor has tens or hundreds of ohms of resistance inside its base, between the terminal and the active junction, and no bypass capacitor outside the package can bypass it. At the upper device it is a resistance at the upper base in exactly the sense of this page, and the 10 Ω floor of the design rule is below what many devices carry internally. That makes the internal resistance, rather than the bias network, the likely limit at high frequency.

The lower device’s base network. The lower device’s input sees the source resistance, and the frequency a device sets for itself found the Miller multiplication that sets the node’s admittance through that resistance. A different source would move the 23.5 kHz corner and every frequency above it.

Still open: the spreading resistance, a follower as the bias, and the same node in a mirror #

The base spreading resistance at the upper device. Adding an internal resistance in series with the upper base, inside the bypass, would say at what frequency a well-bypassed cascode is limited by its own transistor, and whether choosing a device with a lower base resistance for the upper position is worth anything below 100 MHz.

A follower as the bias network. An emitter follower holding the upper base presents a low impedance at low frequency and a rising one above its own bandwidth, with its own shot noise behind it. Whether it beats a resistor with a capacitor across it — lower impedance at low frequency, where nothing was needed, and possibly higher at the top, where everything was — is the same calculation with a third transistor. The floor a current sets and the resistor in the same loop have the noise of a junction and a resistor sharing a current, which is what that bias network would be.

The upper base in a mirror. A cascode current mirror holds its upper output base either from its own reference branch or from a separate bias, and in the second case that bias is exactly the network of this page. The source that holds to the supply built both mirrors; their noise, and which of their sources the reference branch copies, is the same kind of solve on a larger network.