The bias the gain does not set

A bias point is a solution made one argument about one diode: the operating point is not a number read off a characteristic but the root of a transcendental equation that the network and the device pose together, and a damped Newton iteration finds it in thirteen steps from a cold start. The diode was a convenience. It has one parameter worth arguing about — its saturation current — and that parameter enters through a logarithm, so the root barely moves when it does.

A transistor has a parameter that does not enter through a logarithm. Its current gain β multiplies the base current into the collector current, and it is the least controlled number on its data sheet: a general-purpose type is sold in gain groups because one type number, left ungrouped, spans several times over, and temperature, collector current and ageing each move it again. A tenfold range, fifty to five hundred, is the design case, and a bias network is judged by how much of that range arrives at the collector current. That is a question about a root, and it has to be answered by solving for the root at each end.

One resistor, and all of β #

The simplest bias is one resistor from the supply to the base. The base current is then set by that resistor and the supply, nearly independent of the transistor, and the collector current is β times it. The resistor that gives the same current as the network below at β = 100 is 876 kΩ, found by bisection on the solved circuit, with 4.7 kΩ in the collector, 1 kΩ in the emitter and a 12 V supply.

The single resistor’s line is nearly β itself, drawn on logarithmic axes: 0.609 mA at β = 50, 1.155 at 100, 1.646 at 150. At β = 200 it would want 2.1 mA, and 2.1 mA through 4.7 kΩ and 1 kΩ is 11.9 of the twelve volts. The collector falls until the collector junction is forward-biased and the transistor saturates: the collector–emitter voltage is 0.225 V at β = 200 and 0.132 V at 500, and the current stops at 2.08 mA because the resistors, not the transistor, now set it.

That is where the Ebers–Moll model earns its place in the solve. A model with only the forward junction would have reported 2.1 mA at β = 200 and 4.1 at 500, with the collector below the emitter, and nothing in the arithmetic would have objected. The four-junction model makes the saturated answer a root of the same equations, found by the same Newton loop — and every point drawn is checked by rebuilding Kirchhoff’s current law from the device laws at the solution, which closes to a part in 10⁹ at each one.

So a single base resistor is not a bias network in any useful sense. It fixes the base current, and a design whose collector current is β times a fixed number has handed its operating point to the part’s least controlled parameter.

The four-resistor network and the closed form #

The textbook network puts a divider on the base and a resistor in the emitter. Reduced to its Thévenin equivalent — a source $V_{TH}$ behind $R_{TH}$ — the base loop gives the collector current directly:

$$I_C = \frac{\beta\,(V_{TH} - V_{BE})}{R_{TH} + (\beta+1)R_E}$$

and each point on the solid lines is required to match it, to two parts in a thousand, with $V_{BE}$ taken from the solve at that point rather than assumed. The agreement is the check that the netlist is the network the formula describes; the only approximation in the formula is that $V_{BE}$ is a known number, which the solve replaces with the number it actually is — 0.7156 V at β = 50 and 0.7201 V at β = 500 on the 10 kΩ line, moving 4.5 mV because the current it carries moves.

The formula shows at once what the emitter resistor does. With $R_{TH}$ small against $(\beta + 1)R_E$ the collector current is nearly $(V_{TH} - V_{BE})/R_E$ and β has left the expression. With $R_{TH}$ large it is nearly $\beta(V_{TH} - V_{BE})/R_{TH}$, which is the single base resistor again. The divider’s resistance decides which of the two circuits has been built.

The change is measured as a fraction of the current at β = 100, the value the design was made for.

The same root, one device larger #

It is worth seeing how little the method changed between the diode and the transistor, because the claim that an operating point is a solution is only useful if the solution keeps coming from the same place.

The diode’s root is one exponential against one resistor. The transistor’s is four exponentials — two junction currents and two transport currents — against five resistors and two sources, and the loop that solves it is the same loop: each exponential replaced by its tangent at the present guess, the linear network solved, the guess moved by at most 0.15 V, repeated until nothing moves. The divider enters only as its Thévenin equivalent, which the source that is not a source shows is exact for a linear network at any load, so replacing the two divider resistors by one source and one resistor changes nothing in the root.

What the root then gives is more than the current. Every slope a small-signal analysis of this stage will use — the transconductance, the input resistance $\beta r_e$, the base current’s sensitivity — is a derivative at this solved point, and how small is small signal is how far from it those slopes can be trusted. A bias network that moves the point by seventeen per cent across the parts moves every one of those slopes by the same amount, which is the second reason the spread matters: it is the spread of the gain the stage will have, not only of the current it carries.

Why it is feedback, and what the ratio is #

The emitter resistor is negative feedback on the collector current, and that description is worth making exact because it turns the table into a single number.

A rise in collector current raises the emitter voltage by $R_E$ times it. The base voltage is held by the divider — as far as the divider can hold it — so the base–emitter voltage falls, and the current falls back. How well the divider holds the base is the whole question: the base current flows through $R_{TH}$, so a stiff divider holds and a soft one follows. The loop gain of that local feedback is the ratio of the resistance the emitter resistor presents at the base, $(\beta + 1)R_E$, to the divider’s own, $R_{TH}$. How much of the amplifier gets through measures the same arithmetic for an amplifier’s gain: a feedback loop passes a fraction $1/(1 + T)$ of any change in the forward element, and β is the forward element here.

So the sensitivity to β is divided by one plus the stiffness ratio. With 10 kΩ against the 101 kΩ that β = 100 reflects, the ratio is about ten — five at β = 50 and fifty at 500, since β is inside it — and the spread falls as the ratio rises. With 1 kΩ the ratio is a hundred at the design point, and the base current’s share of the story is nearly gone.

What a resistor in the emitter buys makes the matching argument for the signal: the same resistor, unbypassed, linearises the stage by the same feedback. Here it is doing the job at direct current, and whether it also does it for the signal is a separate decision — the bypass capacitor that usually sits across it is what makes the two separable.

The floor no divider removes #

A perfectly stiff divider — $R_{TH} = 0$ — fixes the emitter current exactly, at $(V_{TH} - V_{BE})/R_E$. The collector current is then the fraction $\alpha = \beta/(\beta + 1)$ of it, and α still depends on β. From β = 50 to 500 it moves from 0.9804 to 0.9980, which is 1.78 per cent of its value at β = 100.

That is a floor. No choice of divider takes the spread below it, and the stiffest divider drawn — 300 Ω — sits just above it. It is small, and it is the reason a precision current source is not built from one transistor and a divider however stiff: the base current is part of the emitter current and not part of the collector’s, and nothing on the base side can know how large it is. The copy and its two errors meets the identical term in a current mirror, where it is the whole of one of the two errors.

What a stiff divider costs #

If the spread only fell as the divider stiffened, the design would be to make it as stiff as possible. It costs twice.

The divider’s own current. A divider that presents $V_{TH}$ behind $R_{TH}$ from a supply $V_{CC}$ draws $V_{TH}(V_{CC} - V_{TH})/(R_{TH}V_{CC})$ from the supply whether or not the transistor is there — 167 µA at 10 kΩ and 1.67 mA at 1 kΩ, which is more than the collector current being biased. The standing current is in the supply’s budget and, in a battery circuit, in its lifetime.

The stage’s input resistance. The divider is across the input. With the emitter resistor bypassed for the signal, the stage’s input resistance is $R_{TH} \parallel \beta r_e$, and at a milliampere $\beta r_e$ is about 2.2 kΩ at β = 100. A 1 kΩ divider takes the input resistance to under 700 Ω, and whatever drives the stage now sees a load the transistor had nothing to do with — the divider and its load is the same arithmetic from the driving side.

The solid and dashed lines are close to mirror images across the middle of the range, and that is the finding. Well above the floor the spread is proportional to $R_{TH}$ and the standing current to $1/R_{TH}$, so their product is fixed by the emitter resistor, the supply and the range of β, and not by the divider at all. A bias design is a point on a hyperbola, and the choice between a stable current and a cheap divider is a choice of where on it to sit. The two are equal — the spread and the standing current as a fraction of the collector current — at 9.7 kΩ.

The rule of thumb, $R_{TH} = \beta_\mathrm{min}R_E/10$, is 5 kΩ here. It gives a spread of 10.5 per cent across the tenfold range of β and a standing current of 333 µA, 28 per cent of the collector current. That is a reasonable point on the hyperbola, and the figure is what says that it is a point — a factor of ten in either direction is available at the price the other line names.

What changes the hyperbola rather than the point #

Two things move the curve itself, and both are in the circuit rather than in the divider.

A larger emitter resistor. The stiffness ratio is $(\beta + 1)R_E/R_{TH}$, so doubling $R_E$ doubles it at the same divider — but $R_E$ takes headroom. At 1.2 mA a kilohm drops 1.2 V of the twelve, and the collector’s swing is what is left between the supply, the collector resistor’s drop and the emitter. A higher $V_{TH}$ with a proportionally larger $R_E$ is the same trade: it also reduces the share of $V_{BE}$ in the numerator, which matters more for temperature than for β.

A narrower β range. Buying the part in a gain group narrows fifty-to-five-hundred to something like two-to-one, and the spread at every divider falls in proportion. That is a procurement decision standing in for a circuit one, and it is common because it costs nothing in the circuit.

The one thing that does not move the hyperbola is the divider, because the divider is the variable along it.

What the other variables do at the same point #

The table holds everything but β fixed, and it is worth saying how large the other effects are beside it, since a designer spends the stiffness once and it has to cover all of them.

$V_{BE}$ with current moves 4.5 mV across the β range on the 10 kΩ line, and against $V_{TH} - V_{BE} \approx 1.28$ V that is a third of a per cent of the current. Small, and already inside the table, because the solve includes it.

$V_{BE}$ with temperature moves by about two millivolts a kelvin, which two millivolts a kelvin, and the wrong sign measures — over a hundred kelvin that is 200 mV, or sixteen per cent of the same 1.28 V. It is the larger effect by far, and a divider does nothing for it: only the size of $V_{TH} - V_{BE}$ does, which is the argument for a larger $R_E$ rather than a stiffer divider.

Resistor tolerance sets the current through $(V_{TH} - V_{BE})/R_E$ directly, so a one-per-cent $R_E$ is a one-per-cent current, independent of the network’s stiffness. The error that is a distribution is how several such contributions combine: as a sum of variances, so the largest dominates.

So the stiffness buys protection against β and nothing else, and at the rule-of-thumb point it has already reduced β’s contribution below temperature’s. Stiffening past that point buys very little that matters.

How the solves were made #

Every point is a Newton solve of the netlist, with the transistor as Ebers–Moll’s four elements — two junction diodes and two transport currents — and a start at a base–emitter voltage of 0.6 V with the collector junction reverse-biased, so that the loop begins on the forward-active side and moves into saturation only if the circuit takes it there.

Each solve is checked two ways. Kirchhoff’s current law is rebuilt from the nonlinear laws at the solution, and closes to better than a part in 10⁹ at every point; and on the four-resistor network the solved current is required to equal the closed form with the solved $V_{BE}$, to two parts in a thousand.

The single base resistor is sized by bisection on the solved circuit, to the four-resistor network’s own current at β = 100 with a 10 kΩ divider, so that the two circuits are compared at the design point rather than at a formula’s idea of it. The single resistor is required to saturate within the range drawn, and it does from β = 200.

The floor is required: the stiffest divider must leave a spread at or above the α floor and within half again of it.

What varying β alone cannot show #

β is the only thing varied. Real gain rises with collector current up to a peak and falls beyond it, and rises with temperature, so a tenfold range at one current is a different range at another. The figures fix the current’s neighbourhood and vary the number.

The Early effect is absent. A collector–emitter voltage that moves with β changes the current slightly through the output conductance; at these voltages it is a fraction of a per cent and it would slightly soften the single resistor’s saturation corner.

And the temperature is fixed, which the next section names as the larger effect and which the operating point meets in its own right when the transistor dissipates enough to warm itself.

Still open: the second root, the self-heating, and the bias that tracks #

A network with more than one answer. Every circuit here has exactly one operating point, and Newton finds it from any reasonable start. A circuit with positive feedback — two transistors each biased from the other’s collector — has several, and which one a solver returns depends on where it starts. That is the case in which the operating point stops being a property of the circuit and becomes a property of its history.

The operating point that sets its own temperature. A transistor dissipating a watt is tens of kelvin above its surroundings, and its $V_{BE}$ has moved by two millivolts for each; the current then depends on a temperature that depends on the current. The emitter resistor is the feedback against that too, and whether it is enough is a solve of the electrical and thermal equations together.

A bias that follows β instead of fighting it. A current mirror sets a transistor’s base from another transistor of the same type carrying a known current, so a high-gain part is biased by a high-gain reference. The spread then depends on how well the two parts’ β match rather than on how large either is, which moves the whole argument from the range of β to its tracking.