Circuit Response
every model has an edge, and the figure carries it
Series

Operating point — the series

4 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. A diode fed from 5 V through 1.0 kΩ. computed by solving, not by drawing. The operating point is where the exponential meets the load line: 0.692544 V and 4.3075 mA, reached in 13 damped Newton steps from a cold start. The one-line Newton on Vs = v + R·i(v), which touches no matrix, gives 0.692544 V. The "drop" is not a constant: it moves 59.53 mV per decade of current, measured between two solved operating points.

    A bias point is a solution, not a choice

    The phrase "the diode drops 0.7 volts" is a constant standing in for the root of a transcendental equation. Solved properly, from a five-volt supply through a kilohm, it drops 0.692544 V — and from forty-eight volts through the same kilohm it drops 0.754459 V, because the drop moves about sixty millivolts for every decade of current through it.

    part 1 · semiconductors
  2. With Rₜₕ = 10 kΩ, a tenfold range of β moves the collector current by 17.3%. computed by solving, not by drawing. The collector current of a transistor biased from a 2 V Thévenin source through 1, 10 and 100 kΩ, with 4.7 kΩ in the collector, 1 kΩ in the emitter and a 12 V supply, against its current gain from 30 to 1000, each point a Newton solve of the netlist with both junctions in the device. Dotted: one resistor from the supply to the base, 876 kΩ, sized for the same current at β = 100; it is proportional to β until it saturates. With Rₜₕ = 10 kΩ the current moves by 17.3 per cent from β = 50 to 500.

    The bias the gain does not set

    A transistor's current gain varies tenfold across the parts one type number covers, and a bias network is judged by how little of that reaches the collector current. One resistor from the supply to the base passes all of it — 0.61 mA at β = 50, 1.16 at 100, and saturation from 200. The four-resistor network passes 98.5 per cent of a tenfold change with a 100 kΩ divider, 17.3 with 10 kΩ and 3.4 with 1 kΩ, and never less than the 1.78 per cent the ratio β/(β + 1) keeps however stiff the divider. Each tenfold cut in that spread costs a tenfold rise in the divider's standing current, so the design is a point on a hyperbola rather than a rule.

    part 2 · semiconductors
  3. Three bias points below a coupling of 56.6 kΩ, one above — and the middle one is where Newton from nought lands. computed by solving, not by drawing. Two transistors, each with 1 kΩ to a 5 V supply and each biased from the other's collector through a coupling resistor into a 10 kΩ base resistor, solved by Newton on the netlist from six starting guesses and de-duplicated. Below a coupling of 56.6 kΩ there are three operating points: two latched, with one collector at 0.061 V and the other at 4.615 V at 10 kΩ, and a symmetric one at 1.702 V whose Jacobian determinant is negative, so the circuit cannot rest there. Above it the three merge into one symmetric point, and the circuit is an amplifier rather than a memory.

    The circuit with three bias points

    Two transistors, each biased from the other's collector, have three operating points: one collector at 0.061 V and the other at 4.615, the same the other way round, and both at 1.702. All three solve the circuit to a part in 10⁹, and Newton started from nought — where a simulator starts — finds the symmetric one, which is the one the circuit can never sit at: its Jacobian determinant is negative, so any positive capacitance at the nodes makes it run. Weaken the coupling past 56.6 kΩ and the three merge into one, the loop gain around the symmetric point having fallen through one. Which answer a solver returns depends on where it begins, and from 1,681 starting guesses at 10 kΩ, 437 return the one that is not a state.

    part 3 · semiconductors
  4. At θ = 4 K/W the transistor holds at 311.7 K; the second crossing at 333 K runs away. computed by solving, not by drawing. The power a transistor dissipates against its junction temperature (solid), each point an electrical operating point solved at that temperature with the saturation current and thermal voltage it has there, for a transistor biased from a stiff 1 Ω Thévenin source to carry 100 mA from 15 V at 300 K with no emitter resistor; and the heat a sink of 4 K/W removes (dashed). Operating points are the crossings: 311.7 K with a thermal loop gain of 0.64 and 332.5 K with a thermal loop gain of 1.43. Above θ* = 4.43 K/W the two meet and vanish.

    The bias point that heats itself

    A transistor driven from a stiff base voltage to carry 100 mA from 15 V dissipates 1.5 W, and every kelvin that heats it raises the current it carries. On a 4 K/W heat sink the heat balance has two solutions — 311.7 K, which holds, and 332.5 K, which runs away — and above 4.43 K/W it has none. A tenth of an ohm in the emitter moves that edge to 7.69 K/W, and from a quarter of an ohm there is no edge below 520 K at all. The designer's estimate from the loop gain at room temperature puts the edge at 11.04 K/W, two and a half times too high, because the loop gain grows as the part warms.

    part 4 · semiconductors

All series