The window a field-effect input wins in

The two generators that are one current took an amplifier’s two noise generators, a voltage in series with its input and a current across it, and found that in a bipolar input stage they are one thing. The collector current’s shot noise, referred through the transconductance, is the voltage generator; the base current’s shot noise is the current generator; and since the base current is the collector current over the current gain, their product is fixed at $2kT/\sqrt\beta$ whatever the bias. The best noise figure the stage can reach is therefore fixed too, at $1 + 1/\sqrt\beta$. Bias moves the bottom of the noise-figure bowl along the source-resistance axis and never deepens it.

That essay excluded field-effect inputs, because their two generators are not one current. This page is about the input it excluded, and the consequence is that its bowl can be made as deep as a designer likes, within a window whose edges can be written down.

One current, for reference #

The two lines cross and their product never moves. Every bipolar input, at any bias, into its own optimum source, sits 0.414 dB above a noiseless amplifier at a current gain of a hundred, and only a larger current gain takes it lower: the excess is one over the square root of the gain, which is the fraction of the carriers that leave through the base.

Two currents, and a product that moves #

A field-effect input stage has a voltage generator for the same reason a bipolar one does: its channel carries a current, and the current’s fluctuation, referred through the transconductance, appears as a voltage at the gate. For a long channel it is the channel’s thermal noise, $e_n^2 = 4kT\gamma/g_m$ with $\gamma = 2/3$, and for a device whose drain current follows a square law in its gate voltage, $g_m = \sqrt{2k_p I_D}$. Raising the drain current raises the transconductance, and the voltage noise falls — slowly, as the drain current to the −1/4.

Its current generator has a different origin entirely. No current flows into a field-effect gate by design; what flows is leakage across a reverse-biased junction or through an insulator, typically picoamperes, and that leakage crosses a barrier, so it carries shot noise: $i_n^2 = 2qI_G$. The gate leakage has nothing to do with the drain current. It is set by the gate’s area, its temperature and how it was made.

So the product of the two generators, which fixes the best noise figure through $F_{\min} = 1 + e_n i_n/(2kT)$, is no longer a constant.

Why the product, and not either generator alone, is the thing to watch takes one line. Into a source of resistance $R_s$ the noise factor is $1 + (e_n^2 + i_n^2R_s^2)/(4kTR_s)$: the voltage generator’s share falls as the source grows and the current generator’s rises, and the two are equal at $R_s = e_n/i_n$. There the factor is $1 + 2e_ni_n/(4kT)$. Every input’s deepest point is its two generators multiplied, over twice the thermal energy, and where that point sits is their ratio. A designer choosing a part for a known source wants the ratio near the source and the product small; the bipolar input offers the ratio as a free choice through its bias and the product as a fixed price, and the field-effect input offers both as choices.

The field-effect input’s product sits two to three decades below the bipolar input’s across the whole range drawn, and it slopes. At a microampere of drain current its best noise figure is 0.004 dB; at ten milliamps, 0.0004 dB. Both numbers are a hundred to a thousand times closer to a noiseless amplifier than the bipolar input’s 0.414 dB, because the thing that sets the bipolar floor — the base current, which must flow and must be noisy — has no counterpart. The field-effect input’s current noise comes from a current it does not need, and a better part leaks less. The floor a current sets showed that a current’s shot noise contains no resistance and no device, only the current; the bipolar input is stuck with a base current of a hundredth of its collector current, and the field-effect input’s gate current can be a millionth of that.

The bowl, against a bipolar input biased for every source #

A best noise figure is reached at one source resistance, the ratio $e_n/i_n$, and for a field-effect input with a picoampere of leakage that ratio is megohms. A fair comparison has to ask what each input does into every source, and it has to give the bipolar input its own advantage, which is that its bias can be chosen for the source. With its collector current set to $\sqrt\beta\,V_T/R_s$ for each source resistance $R_s$, a bipolar input with base resistance $r_b$ reaches $1 + r_b/R_s + 1/\sqrt\beta$ — the curve the two generators that are one current drew — and nothing does better into that source.

Below a few kilohms the bipolar input wins, and above five gigohms it wins again. Between them the field-effect input is quieter by up to four tenths of a decibel, and its bowl bottoms out at 0.0007 dB at 4.08 MΩ, which is as close to noiseless as any noise figure in these essays.

The window’s two edges are where one of the field-effect input’s generators costs exactly what the bipolar input’s floor does. Write the field-effect input’s noise figure as $1 + R_n/R_s + R_s \cdot 2qI_G/(4kT)$, with $R_n = e_n^2/(4kT)$ its voltage noise expressed as a resistance — 333 Ω for this channel. Near the low edge the leakage term is negligible, and the two inputs are equal where $R_n/R_s = r_b/R_s + 1/\sqrt\beta$:

$$R_{\text{low}} = \sqrt\beta\,(R_n - r_b) = 2.83\ \text{k}\Omega.$$

Near the high edge the voltage terms are negligible, and the two are equal where $R_s\,2qI_G/(4kT) = 1/\sqrt\beta$:

$$R_{\text{high}} = \frac{2V_T}{\sqrt\beta\,I_G} = 5\ \text{G}\Omega.$$

Both edges, found by bisecting the two noise figures without either expression, land on the closed forms. The low edge belongs to the channel and the base resistance; the high edge belongs to the leakage and the thermal voltage; and the bipolar input’s current gain appears in both, as the $1/\sqrt\beta$ that is its floor.

The window’s top is also generous to the bipolar input, and it is worth saying by how much. Its floor at each source assumes a collector current of $\sqrt\beta\,V_T/R_s$, which is 250 µA into a kilohm but 50 pA into five gigohms, and no real transistor holds a current gain of a hundred at fifty picoamperes: the gain collapses at low current as recombination takes a growing share of the base current. So the bipolar curve above a few megohms describes a transistor that does not exist, and in practice the field-effect input wins well past the ceiling drawn. The ceiling is a bound on how far a bipolar input could ever compete, not a place where one does.

The leakage sets the ceiling #

A thousand times more leakage leaves the low edge exactly where it was, at 2.83 kΩ, because the low edge has no leakage in it. It brings the high edge down a thousandfold, from 5 GΩ to 5 MΩ, and it moves the field-effect input’s own best from 4.08 MΩ to 129 kΩ and from 0.0007 to 0.0224 dB — the product has risen by the square root of a thousand. The window is still more than three decades wide, and inside it the field-effect input is still an order of magnitude closer to noiseless than the bipolar input can ever be.

Leakage is the parameter a field-effect input’s temperature moves. A junction’s leakage roughly doubles for every ten kelvin or so, which is how a picoampere at room temperature becomes nanoamperes in a hot enclosure — the same doubling the error a bigger resistor cannot help found turning a photodiode amplifier’s bias current from a tenth of a per cent into several per cent between room temperature and 85 °C, and the same current the current the instrument draws prices as an error rather than a noise. So the window’s top is a temperature-dependent number, and a high-impedance source that was comfortably inside the window on the bench can be outside it in the field. The low edge, set by the channel, barely moves.

Drawn against leakage, the window is a wedge: a flat floor and a ceiling falling as one over the leakage, meeting at 438 nA. Setting the two closed forms equal gives an estimate of the closing leakage, $2V_T/(\beta(R_n - r_b)) = 1.76\ \mu\text{A}$, four times too high. The estimate lets each edge be set by one of the field-effect input’s two terms alone, which is true while the edges are decades apart and false as they approach, where both terms act at once and the field-effect input’s bowl bottom rises into the bipolar floor from both sides. The closed forms describe the window; the solve says when it stops existing.

No field-effect input in ordinary service leaks half a microampere. The practical reading of the wedge is the other way round: across every leakage a real part has, the window is several decades wide, and what changes with leakage is only which high-impedance sources fall outside it.

The channel sets the floor #

The low edge is $\sqrt\beta\,(R_n - r_b)$, and $R_n$ is the channel’s thermal noise, $4kT\gamma/g_m$ divided by $4kT$: simply $\gamma/g_m$. A less-biased channel has a smaller transconductance, a larger noise resistance and a higher floor.

A hundred times less drain current multiplies the voltage noise by $100^{1/4} = 3.16$ and the noise resistance by ten, and the window’s floor moves from 2.83 kΩ to 32.8 kΩ. The ceiling does not move at all — the leakage is the same — and the field-effect input’s own best rises to 0.0022 dB, still two hundred times closer to noiseless than the bipolar floor. So a designer trading power for noise in a field-effect input is trading the bottom edge of the window: a milliampere of drain current wins from a few kilohms, ten microamperes only from a few tens.

That is the design freedom the bipolar input does not have. For the bipolar input, bias moves the optimum and never the floor; for the field-effect input, bias moves the floor and the leakage moves the ceiling, and the two can be chosen separately because they come from two separate currents.

A base resistance larger than the channel’s noise #

The low edge has one more term in it, and it can change the picture completely.

If the bipolar input’s base resistance exceeds the field-effect input’s noise resistance, $R_n - r_b$ is negative and there is no low edge: the field-effect input is quieter into every source below the ceiling, down to the lowest drawn. At 400 Ω of base resistance, which a small high-frequency transistor can have, a milliampere-biased field-effect input with 333 Ω of noise resistance wins everywhere from ten ohms to five gigohms.

The base resistance is the ohmic resistance of the transistor’s base region between its terminal and the active junction, and its thermal noise, $4kTr_b$, sits in series with the source exactly as a resistor there would. It does not depend on the bias at all. The field-effect input’s noise resistance plays the same role — a fictitious resistor in series with the gate whose thermal noise is the channel’s — but it does depend on the bias, falling as the drain current rises. So the comparison at low source impedance is between two series resistances, one fixed by the transistor’s geometry and one set by how hard the channel is driven, and a designer who can spend drain current can always push the second below the first.

The common wisdom — a bipolar input for low-impedance sources, a field-effect input for high — is therefore a statement about particular parts rather than about the two kinds of device. It holds when the bipolar input’s base resistance is small against the field-effect input’s channel noise resistance, which is the usual case for a low-noise bipolar transistor biased hard and a modest field-effect input. Choose a large, heavily biased field-effect input and a small bipolar transistor and the ordering at low impedance reverses. The bowl, and the bottom of it draws the noise-figure bowl for one amplifier; here the comparison between two bowls turns on one subtraction.

Where the white-noise picture stops #

Every noise source on this page is white. A field-effect input’s channel noise is not, at low frequency: it has a flicker component whose density rises as one over the frequency, and in many field-effect parts that component dominates below a kilohertz or more. The window above is therefore a statement about the band above the flicker corner. Below it, the channel’s noise resistance grows with falling frequency, the floor of the window rises, and for low-frequency work at low source impedance the bipolar input’s usually lower flicker corner can win back what the white picture says it has lost. The corner where averaging stops working measures what a flicker corner does to a measurement, and it applies here channel for channel.

And the gate leakage is taken as a pure shot noise current. That is right when the leakage crosses a barrier, as in a junction gate, and it is the count under the density that says how far a picoampere’s shot noise can be trusted as a Gaussian in a given bandwidth: at the window’s ceiling, where the leakage term dominates, the gate current’s own statistics are exactly the small-current counts that essay measured.

How the numbers were obtained #

The field-effect input’s voltage noise is $4kT\gamma/g_m$ with $\gamma = 2/3$ and $g_m = \sqrt{2k_p I_D}$ for $k_p$ = 2 mA/V², and its current noise $2qI_G$; its noise factor into a source is $1 + (e_n^2 + i_n^2 R_s^2)/(4kTR_s)$. The bipolar input’s floor is $1 + r_b/R_s + 1/\sqrt\beta$ at the collector current $\sqrt\beta\,V_T/R_s$, the expression the earlier essay checked against a search over collector current. Each window edge is found by bisecting the difference of the two noise factors in the logarithm of the source resistance, bracketed by the field-effect input’s own optimum, and compared with its closed form; the closing leakage is bisected on whether a window exists at all. Everything is at 290 K.

What it leaves out #

Flicker noise, as above. Both kinds of device have it and they have it differently, and a low-frequency comparison needs both corners.

The field-effect input’s gate capacitance, which becomes the dominant current noise at high frequency: the channel’s voltage noise across the input capacitance is a current that rises with frequency, and at high source impedances it closes the window from above long before the leakage does. Where the trouble is at the input measures exactly that mechanism for a photodiode, and it would add a frequency axis to this page’s window.

And correlation between the two generators. Each input’s voltage and current noise are treated as independent, which is right for the field-effect input’s channel and gate and nearly right for the bipolar input’s collector and base. A correlation changes the optimum source from a resistance to an impedance, and the window from a band to a region of a plane.

Still open: the flicker corner, the gate capacitance, and the leakage’s own temperature #

The flicker corner as a second window edge. With flicker noise in both channels, the field-effect input’s noise resistance becomes a function of frequency and so does its window’s floor. Solving the window at each frequency for stated flicker corners would say below which frequency the bipolar input wins at every source, which is the boundary a low-frequency front end is actually chosen on.

The gate capacitance as a current noise. Adding the channel’s voltage noise across the input capacitance gives a current noise rising with frequency; at high source impedance it should bring the window’s ceiling down in proportion to frequency. Where that ceiling crosses the leakage’s would say which of the two limits a high-impedance sensor meets first.

The leakage’s temperature. The ceiling falls as one over the leakage, and the leakage rises with temperature by a law the junction essays already measure. Carrying that law through would turn the window’s ceiling into a temperature at which a given source leaves it, which is the number a high-impedance instrument running warm actually needs.